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Learning Objectives
• To describe the electrical properties of a solid using band theory.
To explain the observed properties of metals, a more sophisticated approach is needed than the electron-sea model described in Section 12.5 . The molecular orbital theory we used in Section 6.5 to explain the delocalized π bonding in polyatomic ions and molecules such as NO2, ozone, and 1,3-butadiene can be adapted to accommodate the much higher number of atomic orbitals that interact with one another simultaneously in metals.
Band Theory
In a 1 mol sample of a metal, there can be more than 1024 orbital interactions to consider. In our molecular orbital description of metals, however, we begin by considering a simple one-dimensional example: a linear arrangement of n metal atoms, each containing a single electron in an s orbital. We use this example to describe an approach to metallic bonding called band theoryA theory used to describe the bonding in metals and semiconductors., which assumes that the valence orbitals of the atoms in a solid interact, generating a set of molecular orbitals that extend throughout the solid.
One-Dimensional Systems
If the distance between the metal atoms is short enough for the orbitals to interact, they produce bonding, antibonding, and nonbonding molecular orbitals. The left portion of Figure 12.6.1 shows the pattern of molecular orbitals that results from the interaction of ns orbitals as n increases from 2 to 5.
As we saw in Chapter 6, the lowest-energy orbital is the completely bonding molecular orbital, whereas the highest-energy orbital is the completely antibonding molecular orbital. Molecular orbitals of intermediate energy have fewer nodes than the totally antibonding molecular orbital. The energy separation between adjacent orbitals decreases as the number of interacting orbitals increases. For n = 30, there are still discrete, well-resolved energy levels, but as n increases from 30 to a number close to Avogadro’s number, the spacing between adjacent energy levels becomes almost infinitely small. The result is essentially a continuum of energy levels, as shown on the right in Figure 12.6.1, each of which corresponds to a particular molecular orbital extending throughout the linear array of metal atoms. The levels that are lowest in energy correspond to mostly bonding combinations of atomic orbitals, those highest in energy correspond to mostly antibonding combinations, and those in the middle correspond to essentially nonbonding combinations.
The continuous set of allowed energy levels shown on the right in Figure 12.6.1 is called an energy bandThe continuous set of allowed energy levels generated in band theory when the valence orbitals of the atoms in a solid interact with one another, thus creating a set of molecular orbitals that extend throughout the solid.. The difference in energy between the highest and lowest energy levels is the bandwidthThe difference in energy between the highest and lowest energy levels in an energy band. and is proportional to the strength of the interaction between orbitals on adjacent atoms: the stronger the interaction, the larger the bandwidth. Because the band contains as many energy levels as molecular orbitals, and the number of molecular orbitals is the same as the number of interacting atomic orbitals, the band in Figure 12.6.1 contains n energy levels corresponding to the combining of s orbitals from n metal atoms. Each of the original s orbitals could contain a maximum of two electrons, so the band can accommodate a total of 2n electrons. Recall, however, that each of the metal atoms we started with contained only a single electron in each s orbital, so there are only n electrons to place in the band. Just as with atomic orbitals or molecular orbitals, the electrons occupy the lowest energy levels available. Consequently, only the lower half of the band is filled. This corresponds to filling all of the bonding molecular orbitals in the linear array of metal atoms and results in the strongest possible bonding.
Multidimensional Systems
The previous example was a one-dimensional array of atoms that had only s orbitals. To extrapolate to two- or three-dimensional systems and atoms with electrons in p and d orbitals is straightforward in principle, even though in practice the mathematics becomes more complex, and the resulting molecular orbitals are more difficult to visualize. The resulting energy-level diagrams are essentially the same as the diagram of the one-dimensional example in Figure 12.6.1, with the following exception: they contain as many bands as there are different types of interacting orbitals. Because different atomic orbitals interact differently, each band will have a different bandwidth and will be centered at a different energy, corresponding to the energy of the parent atomic orbital of an isolated atom.
Band Gap
Because the 1s, 2s, and 2p orbitals of a period 3 atom are filled core levels, they do not interact strongly with the corresponding orbitals on adjacent atoms. Hence they form rather narrow bands that are well separated in energy (Figure 12.6.2). These bands are completely filled (both the bonding and antibonding levels are completely populated), so they do not make a net contribution to bonding in the solid. The energy difference between the highest level of one band and the lowest level of the next is the band gapThe difference in energy between the highest level of one energy band and the lowest level of the band above it, which represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals.. It represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals.
Because they extend farther from the nucleus, the valence orbitals of adjacent atoms (3s and 3p in Figure 12.6.2) interact much more strongly with one another than do the filled core levels; as a result, the valence bands have a larger bandwidth. In fact, the bands derived from the 3s and 3p atomic orbitals are wider than the energy gap between them, so the result is overlapping bandsMolecular orbitals derived from two or more different kinds of valence electrons that have similar energies.. These have molecular orbitals derived from two or more valence orbitals with similar energies. As the valence band is filled with one, two, or three electrons per atom for Na, Mg, and Al, respectively, the combined band that arises from the overlap of the 3s and 3p bands is also filling up; it has a total capacity of eight electrons per atom (two electrons for each 3s orbital and six electrons for each set of 3p orbitals). With Na, therefore, which has one valence electron, the combined valence band is one-eighth filled; with Mg (two valence electrons), it is one-fourth filled; and with Al, it is three-eighths filled, as indicated in Figure 12.6.2. The partially filled valence band is absolutely crucial for explaining metallic behavior because it guarantees that there are unoccupied energy levels at an infinitesimally small energy above the highest occupied level.
Band theory can explain virtually all the properties of metals. Metals conduct electricity, for example, because only a very small amount of energy is required to excite an electron from a filled level to an empty one, where it is free to migrate rapidly throughout the crystal in response to an applied electric field. Similarly, metals have high heat capacities (as you no doubt remember from the last time a doctor or a nurse placed a stethoscope on your skin) because the electrons in the valence band can absorb thermal energy by being excited to the low-lying empty energy levels. Finally, metals are lustrous because light of various wavelengths can be absorbed, causing the valence electrons to be excited into any of the empty energy levels above the highest occupied level. When the electrons decay back to low-lying empty levels, they emit light of different wavelengths. Because electrons can be excited from many different filled levels in a metallic solid and can then decay back to any of many empty levels, light of varying wavelengths is absorbed and reemitted, which results in the characteristic shiny appearance that we associate with metals.
Requirements for Metallic Behavior
For a solid to exhibit metallic behavior, it must have a set of delocalized orbitals forming a band of allowed energy levels, and the resulting band must be only partially filled (10%–90%) with electrons. Without a set of delocalized orbitals, there is no pathway by which electrons can move through the solid.
Note the Pattern
Metallic behavior requires a set of delocalized orbitals and a band of allowed energy levels that is partially occupied.
Band theory explains the correlation between the valence electron configuration of a metal and the strength of metallic bonding. The valence electrons of transition metals occupy either their valence ns, (n − 1)d, and np orbitals (with a total capacity of 18 electrons per metal atom) or their ns and (n − 1)d orbitals (a total capacity of 12 electrons per metal atom). These atomic orbitals are close enough in energy that the derived bands overlap, so the valence electrons are not confined to a specific orbital. Metals with 6 to 9 valence electrons (which correspond to groups 6–9) are those most likely to fill the valence bands approximately halfway. Those electrons therefore occupy the highest possible number of bonding levels, while the number of antibonding levels occupied is minimal. Not coincidentally, the elements of these groups exhibit physical properties consistent with the presence of the strongest metallic bonding, such as very high melting points.
Insulators
In contrast to metals, electrical insulatorsA material that conducts electricity poorly because its valence bands are full. are materials that conduct electricity poorly because their valence bands are full. The energy gap between the highest filled levels and the lowest empty levels is so large that the empty levels are inaccessible: thermal energy cannot excite an electron from a filled level to an empty one. The valence-band structure of diamond, for example, is shown in part (a) in Figure 12.6.3. Because diamond has only 4 bonded neighbors rather than the 6 to 12 typical of metals, the carbon 2s and 2p orbitals combine to form two bands in the solid, with the one at lower energy representing bonding molecular orbitals and the one at higher energy representing antibonding molecular orbitals. Each band can accommodate four electrons per atom, so only the lower band is occupied. Because the energy gap between the filled band and the empty band is very large (530 kJ/mol), at normal temperatures thermal energy cannot excite electrons from the filled level into the empty band. Thus there is no pathway by which electrons can move through the solid, so diamond has one of the lowest electrical conductivities known.
Semiconductors
What if the difference in energy between the highest occupied level and the lowest empty level is intermediate between those of electrical conductors and insulators? This is the case for silicon and germanium, which have the same structure as diamond. Because Si–Si and Ge–Ge bonds are substantially weaker than C–C bonds, the energy gap between the filled and empty bands becomes much smaller as we go down group 14 (part (b) and part (c) of Figure 12.6.3"). (For more information on bond strengths, see Section 5.5.) Consequently, thermal energy is able to excite a small number of electrons from the filled valence band of Si and Ge into the empty band above it, which is called the conduction bandThe band of empty molecular orbitals in a semiconductor..
Exciting electrons from the filled valence band to the empty conduction band causes an increase in electrical conductivity for two reasons:
1. The electrons in the previously vacant conduction band are free to migrate through the crystal in response to an applied electric field.
2. Excitation of an electron from the valence band produces a “hole” in the valence band that is equivalent to a positive charge. The hole in the valence band can migrate through the crystal in the direction opposite that of the electron in the conduction band by means of a “bucket brigade” mechanism in which an adjacent electron fills the hole, thus generating a hole where the second electron had been, and so forth.
Consequently, Si is a much better electrical conductor than diamond, and Ge is even better, although both are still much poorer conductors than a typical metal (Figure 12.6.4). Substances such as Si and Ge that have conductivities between those of metals and insulators are called semiconductorsA substance such as Si and Ge that has a conductivity between that of metals and insulators.. Many binary compounds of the main group elements exhibit semiconducting behavior similar to that of Si and Ge. For example, gallium arsenide (GaAs) is isoelectronic with Ge and has the same crystalline structure, with alternating Ga and As atoms; not surprisingly, it is also a semiconductor. The electronic structure of semiconductors is compared with the structures of metals and insulators in Figure 12.6.5.
Figure 12.6.4 A Logarithmic Scale Illustrating the Enormous Range of Electrical Conductivities of Solids
Temperature and Conductivity
Because thermal energy can excite electrons across the band gap in a semiconductor, increasing the temperature increases the number of electrons that have sufficient kinetic energy to be promoted into the conduction band. The electrical conductivity of a semiconductor therefore increases rapidly with increasing temperature, in contrast to the behavior of a purely metallic crystal. In a metal, as an electron travels through the crystal in response to an applied electrical potential, it cannot travel very far before it encounters and collides with a metal nucleus. The more often such encounters occur, the slower the net motion of the electron through the crystal, and the lower the conductivity. As the temperature of the solid increases, the metal atoms in the lattice acquire more and more kinetic energy. Because their positions are fixed in the lattice, however, the increased kinetic energy increases only the extent to which they vibrate about their fixed positions. At higher temperatures, therefore, the metal nuclei collide with the mobile electrons more frequently and with greater energy, thus decreasing the conductivity. This effect is, however, substantially smaller than the increase in conductivity with temperature exhibited by semiconductors. For example, the conductivity of a tungsten wire decreases by a factor of only about two over the temperature range 750–1500 K, whereas the conductivity of silicon increases approximately 100-fold over the same temperature range. These trends are illustrated in Figure 12.6.6.
Note the Pattern
The electrical conductivity of a semiconductor increases with increasing temperature, whereas the electrical conductivity of a metal decreases with increasing temperature.
n- and p-Type Semiconductors
DopingThe process of deliberately introducing small amounts of impurities into commercial semiconductors to tune their electrical properties for specific applications. is a process used to tune the electrical properties of commercial semiconductors by deliberately introducing small amounts of impurities. If an impurity contains more valence electrons than the atoms of the host lattice (e.g., when small amounts of a group 15 atom are introduced into a crystal of a group 14 element), then the doped solid has more electrons available to conduct current than the pure host has. As shown in part (a) in Figure 12.6.7, adding an impurity such as phosphorus to a silicon crystal creates occasional electron-rich sites in the lattice. The electronic energy of these sites lies between those of the filled valence band and the empty conduction band but closer to the conduction band. Because the atoms that were introduced are surrounded by host atoms, and the electrons associated with the impurity are close in energy to the conduction band, those extra electrons are relatively easily excited into the empty conduction band of the host. Such a substance is called an n-type semiconductorA semiconductor that has been doped with an impurity that has more valence electrons than the atoms of the host lattice., with the n indicating that the added charge carriers are negative (they are electrons).
If the impurity atoms contain fewer valence electrons than the atoms of the host (e.g., when small amounts of a group 13 atom are introduced into a crystal of a group 14 element), then the doped solid has fewer electrons than the pure host. Perhaps unexpectedly, this also results in increased conductivity because the impurity atoms generate holes in the valence band. As shown in part (b) in Figure 12.27, adding an impurity such as gallium to a silicon crystal creates isolated electron-deficient sites in the host lattice. The electronic energy of these empty sites also lies between those of the filled valence band and the empty conduction band of the host but much closer to the filled valence band. It is therefore relatively easy to excite electrons from the valence band of the host to the isolated impurity atoms, thus forming holes in the valence band. This kind of substance is called a p-type semiconductorA semiconductor that has been doped with an impurity that has fewer valence electrons than the atoms of the host lattice., with the p standing for positive charge carrier (i.e., a hole). Holes in what was a filled band are just as effective as electrons in an empty band at conducting electricity.
Note the Pattern
n-Type semiconductors are negative charge carriers; the impurity has more valence electrons than the host. p-Type semiconductors are positive charge carriers; the impurity has fewer valence electrons than the host.
The electrical conductivity of a semiconductor is roughly proportional to the number of charge carriers, so doping is a precise way to adjust the conductivity of a semiconductor over a wide range. The entire semiconductor industry is built on methods for preparing samples of Si, Ge, or GaAs doped with precise amounts of desired impurities and assembling silicon chips and other complex devices with junctions between n- and p-type semiconductors in varying numbers and arrangements.
Because silicon does not stand up well to temperatures above approximately 100°C, scientists have been interested in developing semiconductors made from diamonds, a more thermally stable material. A new method has been developed based on vapor deposition, in which a gaseous mixture is heated to a high temperature to produce carbon that then condenses on a diamond kernel. This is the same method now used to create cultured diamonds, which are indistinguishable from natural diamonds. The diamonds are heated to more than 2000°C under high pressure to harden them even further. Doping the diamonds with boron has produced p-type semiconductors, whereas doping them with boron and deuterium achieves n-type behavior. Because of their thermal stability, diamond semiconductors have potential uses as microprocessors in high-voltage applications.
Example 12.6.1
A crystalline solid has the following band structure, with the purple areas representing regions occupied by electrons. The lower band is completely occupied by electrons, and the upper level is about one-third filled with electrons.
1. Predict the electrical properties of this solid.
2. What would happen to the electrical properties if all of the electrons were removed from the upper band? Would you use a chemical oxidant or reductant to effect this change?
3. What would happen to the electrical properties if enough electrons were added to completely fill the upper band? Would you use a chemical oxidant or reductant to effect this change?
Given: band structure
Asked for: variations in electrical properties with conditions
Strategy:
A Based on the occupancy of the lower and upper bands, predict whether the substance will be an electrical conductor. Then predict how its conductivity will change with temperature.
B After all the electrons are removed from the upper band, predict how the band gap would affect the electrical properties of the material. Determine whether you would use a chemical oxidant or reductant to remove electrons from the upper band.
C Predict the effect of a filled upper band on the electrical properties of the solid. Then decide whether you would use an oxidant or a reductant to fill the upper band.
Solution:
1. A The material has a partially filled band, which is critical for metallic behavior. The solid will therefore behave like a metal, with high electrical conductivity that decreases slightly with increasing temperature.
2. B Removing all of the electrons from the partially filled upper band would create a solid with a filled lower band and an empty upper band, separated by an energy gap. If the band gap is large, the material will be an electrical insulator. If the gap is relatively small, the substance will be a semiconductor whose electrical conductivity increases rapidly with increasing temperature. Removing the electrons would require an oxidant because oxidants accept electrons.
3. C Adding enough electrons to completely fill the upper band would produce an electrical insulator. Without another empty band relatively close in energy above the filled band, semiconductor behavior would be impossible. Adding electrons to the solid would require a reductant because reductants are electron donors.
Exercise
A substance has the following band structure, in which the lower band is half-filled with electrons (purple area) and the upper band is empty.
1. Predict the electrical properties of the solid.
2. What would happen to the electrical properties if all of the electrons were removed from the lower band? Would you use a chemical oxidant or reductant to effect this change?
3. What would happen to the electrical properties if enough electrons were added to completely fill the lower band? Would you use a chemical oxidant or reductant to effect this change?
Answer
1. The solid has a partially filled band, so it has the electrical properties of a conductor.
2. Removing all of the electrons from the lower band would produce an electrical insulator with two empty bands. An oxidant is required.
3. Adding enough electrons to completely fill the lower level would result in an electrical insulator if the energy gap between the upper and lower bands is relatively large, or a semiconductor if the band gap is relatively small. A reductant is required.
Summary
Band theory assumes that the valence orbitals of the atoms in a solid interact to generate a set of molecular orbitals that extend throughout the solid; the continuous set of allowed energy levels is an energy band. The difference in energy between the highest and lowest allowed levels within a given band is the bandwidth, and the difference in energy between the highest level of one band and the lowest level of the band above it is the band gap. If the width of adjacent bands is larger than the energy gap between them, overlapping bands result, in which molecular orbitals derived from two or more kinds of valence orbitals have similar energies. Metallic properties depend on a partially occupied band corresponding to a set of molecular orbitals that extend throughout the solid to form a band of energy levels. If a solid has a filled valence band with a relatively low-lying empty band above it (a conduction band), then electrons can be excited by thermal energy from the filled band into the vacant band where they can then migrate through the crystal, resulting in electrical conductivity. Electrical insulators are poor conductors because their valence bands are full. Semiconductors have electrical conductivities intermediate between those of insulators and metals. The electrical conductivity of semiconductors increases rapidly with increasing temperature, whereas the electrical conductivity of metals decreases slowly with increasing temperature. The properties of semiconductors can be modified by doping, or introducing impurities. Adding an element with more valence electrons than the atoms of the host populates the conduction band, resulting in an n-type semiconductor with increased electrical conductivity. Adding an element with fewer valence electrons than the atoms of the host generates holes in the valence band, resulting in a p-type semiconductor that also exhibits increased electrical conductivity.
Key Takeaway
• Bonding in metals and semiconductors can be described using band theory, in which a set of molecular orbitals is generated that extends throughout the solid.
Conceptual Problems
1. Can band theory be applied to metals with two electrons in their valence s orbitals? with no electrons in their valence s orbitals? Why or why not?
2. Given a sample of a metal with 1020 atoms, how does the width of the band arising from p orbital interactions compare with the width of the band arising from s orbital interactions? from d orbital interactions?
3. Diamond has one of the lowest electrical conductivities known. Based on this fact, do you expect diamond to be colored? Why? How do you account for the fact that some diamonds are colored (such as “pink” diamond or “green” diamond)?
4. Why do silver halides, used in the photographic industry, have band gaps typical of semiconducting materials, whereas alkali metal halides have very large band gaps?
5. As the ionic character of a compound increases, does its band gap increase or decrease? Why?
6. Why is silicon, rather than carbon or germanium, used in the semiconductor industry?
7. Carbon is an insulator, and silicon and germanium are semiconductors. Explain the relationship between the valence electron configuration of each element and their band structures. Which will have the higher electrical conductivity at room temperature—silicon or germanium?
8. How does doping affect the electrical conductivity of a semiconductor? Draw the effect of doping on the energy levels of the valence band and the conduction band for both an n-type and a p-type semiconductor.
Answers
1. The low electrical conductivity of diamond implies a very large band gap, corresponding to the energy of a photon of ultraviolet light rather than visible light. Consequently, diamond should be colorless. Pink or green diamonds contain small amounts of highly colored impurities that are responsible for their color.
2. As the ionic character of a compound increases, the band gap will also increase due to a decrease in orbital overlap. Remember that overlap is greatest for orbitals of the same energy, and that the difference in energy between orbitals on adjacent atoms increases as the difference in electronegativity between the atoms increases. Thus, large differences in electronegativity increase the ionic character, decrease the orbital overlap, and increase the band gap.
Numerical Problems
1. Of Ca, N, B, and Ge, which will convert pure silicon into a p-type semiconductor when doping? Explain your reasoning.
2. Of Ga, Si, Br, and P, which will convert pure germanium into an n-type semiconductor when doping? Explain your reasoning.
Contributors
• Anonymous
Thumbnail from Wikimedia | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.06%3A_Metals_and_Semiconductors.txt |
Learning Objectives
• To become familiar with the properties of superconductors.
The phenomenon of superconductivity was discovered by the Danish physicist H. Kamerlingh Onnes (1853–1926; Nobel Prize in Physics, 1913), who found a way to liquefy helium, which boils at 4.2 K and 1 atm pressure. To exploit the very low temperatures made possible by this new cryogenic fluid, he began a systematic study of the properties of metals, especially their electrical properties. Because the electrical resistance of a sample is technically easier to measure than its conductivity, Onnes measured the resistivity of his samples. The resistivity and conductivity of a material are inversely proportional:
$conductivity = \dfrac{1}{resistivity} \tag{12.7.1}$
In 1911, Onnes discovered that at about 4 K, the resistivity of metallic mercury (melting point = 234 K) decreased suddenly to essentially zero, rather than continuing to decrease only slowly with decreasing temperature as expected (Figure 12.7.1). He called this phenomenon superconductivityThe phenomenon in which a solid at low temperatures exhibits zero resistance to the flow of electrical current. because a resistivity of zero means that an electrical current can flow forever. Onnes soon discovered that many other metallic elements exhibit superconductivity at very low temperatures. Each of these superconductorsA solid that at low temperatures exhibits zero resistance to the flow of electrical current. has a characteristic superconducting transition temperature (Tc)The temperature at which the electrical resistance of a material drops to zero. at which its resistivity drops to zero. At temperatures less than their Tc, superconductors also completely expel a magnetic field from their interior (part (a) in Figure 12.7.2 ). This phenomenon is called the Meissner effectThe phenomenon in which a superconductor completely expels a magnetic field from its interior. after one of its discoverers, the German physicist Walther Meissner, who described the phenomenon in 1933. Due to the Meissner effect, a superconductor will actually “float” over a magnet, as shown in part (b) in Figure 12.7.2.
BCS Theory
For many years, the phenomenon of superconductivity could not be satisfactorily explained by the laws of conventional physics. In the early 1950s, however, American physicists John Bardeen, Leon Cooper, and John Schrieffer formulated a theory for superconductivity that earned them the Nobel Prize in Physics in 1972. According to the BCS theoryA theory used to explain the phenomenon of superconductivity. (named for the initials of their last names), electrons are able to travel through a solid with zero resistance because of attractive interactions involving two electrons that are at some distance from each other. As one electron moves through the lattice, the surrounding nuclei are attracted to it. The motion of the nuclei can create a transient (short-lived) hole that pulls the second electron in the same direction as the first. The nuclei then return to their original positions to avoid colliding with the second electron as it approaches. The pairs of electrons, called Cooper pairsPairs of electrons that migrate through a superconducting material as a unit., migrate through the crystal as a unit. The electrons in Cooper pairs change partners frequently, like dancers in a ballet.
According to the BCS theory, as the temperature of the solid increases, the vibrations of the atoms in the lattice increase continuously, until eventually the electrons cannot avoid colliding with them. The collisions result in the loss of superconductivity at higher temperatures.
The phenomenon of superconductivity suggested many exciting technological applications. For example, using superconducting wires in power cables would result in zero power losses, even over distances of hundreds of miles. Additionally, because superconductors expel magnetic fields, a combination of magnetic rails and superconducting wheels (or vice versa) could be used to produce magnetic levitation of, for example, a train over the track, resulting in friction-free transportation.
Unfortunately, for many years the only superconductors known had serious limitations, especially the need for very low temperatures, which required the use of expensive cryogenic fluids such as liquid He. In addition, the superconducting properties of many substances are destroyed by large electrical currents or even moderately large magnetic fields, making them useless for applications in power cables or high-field magnets. The ability of materials such as NbTi, NbSn, Nb3Si, and Nb3Ge to tolerate rather high magnetic fields, however, has led to a number of commercial applications of superconductors, including high-field magnets for nuclear magnetic resonance (NMR) spectrometers and magnetic resonance imaging (MRI) instruments in medicine, which, unlike x-rays, can detect small changes in soft tissues in the body.
High-Temperature Superconductors
Because of these limitations, scientists continued to search for materials that exhibited superconductivity at temperatures greater than 77 K (the temperature of liquid nitrogen, the least expensive cryogenic fluid). In 1986, Johannes G. Bednorz and Karl A. Müller, working for IBM in Zurich, showed that certain mixed-metal oxides containing La, Ba, and Cu exhibited superconductivity above 30 K. These compounds had been prepared by French workers as potential solid catalysts some years earlier, but their electrical properties had never been examined at low temperatures. Although initially the scientific community was extremely skeptical, the compounds were so easy to prepare that the results were confirmed within a few weeks. These high-temperature superconductorsA material that becomes a superconductor at temperatures greater than 30 K. earned Bednorz and Müller the Nobel Prize in Physics in 1987. Subsequent research has produced new compounds with related structures that are superconducting at temperatures as high as 135 K. The best known of these was discovered by Paul Chu and Maw-Kuen Wu Jr. and is called the “Chu–Wu phase” or the 1-2-3 superconductor.
The formula for the 1-2-3 superconductor is YBa2Cu3O7−x, where x is about 0.1 for samples that superconduct at about 95 K. If x ≈ 1.0, giving a formula of YBa2Cu3O6, the material is an electrical insulator. The superconducting phase is thus a nonstoichiometric compound, with a fixed ratio of metal atoms but a variable oxygen content. The overall equation for the synthesis of this material is as follows:
$Y_{2}O_{3}\left ( s \right )+4BaCO_{3}\left ( s \right )+6CuO\left ( s \right )+\dfrac{1}{2}O_{2}\left ( g \right )\overset{\Delta }{\rightarrow}2YBa_{2}Cu_{3}O_{7\left ( s \right )}+4CO_{2}\left ( g \right ) \tag{12.7.2}$
If we assume that the superconducting phase is really stoichiometric YBa2Cu3O7, then the average oxidation states of O, Y, Ba, and Cu are −2, +3, +2, and +7/3 respectively. The simplest way to view the average oxidation state of Cu is to assume that two Cu atoms per formula unit are present as Cu2+ and one is present as the rather unusual Cu3+. In YBa2Cu3O6, the insulating form, the oxidation state of Cu is+5/3 so there are two Cu2+ and one Cu+ per formula unit.
As shown in Figure 12.7.3, the unit cell of the 1-2-3 superconductor is related to the unit cell of the simple perovskite structure (part (b) in Figure 12.7.3). The only difference between the superconducting and insulating forms of the compound is that an O atom has been removed from between the Cu3+ ions, which destroys the chains of Cu atoms and leaves the Cu in the center of the unit cell as Cu+. The chains of Cu atoms are crucial to the formation of the superconducting state.
Table 12.7.1 lists the ideal compositions of some of the known high-temperature superconductors that have been discovered in recent years. Engineers have learned how to process the brittle polycrystalline 1-2-3 and related compounds into wires, tapes, and films that can carry enormous electrical currents. Commercial applications include their use in infrared sensors and in analog signal processing and microwave devices.
Table 12.7.1 The Composition of Various Superconductors
Compound Tc (K)
Ba(Pb1−xBix)O3 13.5
(La2−xSrx)CuO4 35
YBa2Cu3O7−x 95
Bi2(Sr2−xCax)CuO6* 80
Bi2Ca2Sr2Cu3O10* 110
Tl2Ba2Ca2Cu3O10* 125
HgBa2Ca2Cu3O8* 133
K3C60 18
Rb3C60 30
*Nominal compositions only. Oxygen deficiencies or excesses are common in these compounds.
Example 12.7.1
Calculate the average oxidation state of Cu in a sample of YBa2Cu3O7−x with x = 0.5. How do you expect its structure to differ from those shown in Figure 12.7.3 for YBa2Cu3O9 and YBa2Cu3O7?
Given: stoichiometry
Asked for: average oxidation state and structure
Strategy:
A Based on the oxidation states of the other component atoms, calculate the average oxidation state of Cu that would make an electrically neutral compound.
B Compare the stoichiometry of the structures shown in Figure 12.7.3 with the stoichiometry of the given compound to predict how the structures differ.
Solution:
A The net negative charge from oxygen is (7.0 − 0.5) (−2) = −13, and the sum of the charges on the Y and Ba atoms is [1 × (+3)] + [2 × (+2)] = +7. This leaves a net charge of −6 per unit cell, which must be compensated for by the three Cu atoms, for a net charge of <math display="inline" xml:id="av_1.0-ch12_m033"><semantics><mrow><mo>+</mo><mfrac><mn>6</mn><mn>3</mn></mfrac><mo>=</mo><mo>+</mo><mn>2</mn></mrow></semantics>[/itex] per Cu.
B The most likely structure would be one in which every other O atom between the Cu atoms in the Cu chains of YBa2Cu3O7 has been removed.
Exercise
Calculate the average oxidation state of Cu in a sample of HgBa2Ca2Cu3O8. Assume that Hg is present as Hg2+.
Answer: +2
Summary
Superconductors are solids that at low temperatures exhibit zero resistance to the flow of electrical current, a phenomenon known as superconductivity. The temperature at which the electrical resistance of a substance drops to zero is its superconducting transition temperature (Tc). Superconductors also expel a magnetic field from their interior, a phenomenon known as the Meissner effect. Superconductivity can be explained by the BCS theory, which says that electrons are able to travel through a solid with no resistance because they couple to form pairs of electrons (Cooper pairs). High-temperature superconductors have Tc values greater than 30 K.
Key Takeaway
• Superconductivity can be described using the BCS theory, in which Cooper pairs of electrons migrate through the crystal as a unit.
Conceptual Problems
1. Why does the BCS theory predict that superconductivity is not possible at temperatures above approximately 30 K?
2. How does the formation of Cooper pairs lead to superconductivity?
Answer
1. According to BCS theory, the interactions that lead to formation of Cooper pairs of electrons are so weak that they should be disrupted by thermal vibrations of lattice atoms above about 30 K.
Contributors
• Anonymous
Video from Wisconsin MRSEC @ YouTube | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.07%3A_Superconductors.txt |
Learning Objectives
• To understand the differences between synthetic and biological polymers.
Most of the solids discussed so far have been molecules or ions with low molecular masses, ranging from tens to hundreds of atomic mass units. Many of the molecular materials in consumer goods today, however, have very high molecular masses, ranging from thousands to millions of atomic mass units, and are formed from a carefully controlled series of reactions that produce giant molecules called polymersA giant molecule that consists of many basic structural units (monomers) connected in a chain or network by covalent bonds. (from the Greek poly and meros, meaning “many parts”). Polymers are used in corrective eye lenses, plastic containers, clothing and textiles, and medical implant devices, among many other uses. They consist of basic structural units called monomersThe basic structural unit of a polymer., which are repeated many times in each molecule. As shown schematically in Figure 12.8.1 , polymerizationA process by which monomers are connected into chains or networks by covalent bonds. is the process by which monomers are connected into chains or networks by covalent bonds. Polymers can form via a condensation reaction, in which two monomer molecules are joined by a new covalent bond and a small molecule such as water is eliminated, or by an addition reaction, a variant of a condensation reaction in which the components of a species AB are added to adjacent atoms of a multiple bond. (For more information about condensation and addition reactions, see Section 7.5.) Many people confuse the terms plastics and polymers. PlasticThe property of a material that allows it to be molded into almost any shape. is the property of a material that allows it to be molded into almost any shape. Although many plastics are polymers, many polymers are not plastics. In this section, we introduce the reactions that produce naturally occurring and synthetic polymers.
Note the Pattern
Polymers are formed via condensation or addition reactions.
Naturally Occurring Polymers: Peptides and Proteins
Polymers that occur naturally are crucial components of all organisms and form the fabric of our lives. Hair, silk, skin, feathers, muscle, and connective tissue are all primarily composed of proteins, the most familiar kind of naturally occurring, or biological, polymer. The monomers of many biological polymers are the amino acids each called an amino acid residue. The residues are linked together by amide bonds, also called peptide bonds, via a condensation reaction where H2O is eliminated:
In the above equation, R represents an alkyl or aryl group, or hydrogen, depending on the amino acid. We write the structural formula of the product with the free amino group on the left (the N-terminus) and the free carboxylate group on the right (the C-terminus). For example, the structural formula for the product formed from the amino acids glycine and valine (glycyl-valine) is as follows:
The most important difference between synthetic and naturally occurring polymers is that the former usually contain very few different monomers, whereas biological polymers can have as many as 20 different kinds of amino acid residues arranged in many different orders. Chains with less than about 50 amino acid residues are called peptidesBiological polymers with less than about 50 amino acid residues., whereas those with more than about 50 amino acid residues are called proteinsBiological polymers with more than 50 amino acid residues linked together by amide bonds.. Many proteins are enzymesCatalysts that occur naturally in living organisms and that catalyze biological reactions., which are catalysts that increase the rate of a biological reaction.
Note the Pattern
Synthetic polymers usually contain only a few different monomers, whereas biological polymers can have many kinds of monomers, such as amino acids arranged in different orders.
Many small peptides have potent physiological activities. The endorphins, for example, are powerful, naturally occurring painkillers found in the brain. Other important peptides are the hormones vasopressin and oxytocin. Although their structures and amino acid sequences are similar, vasopressin is a blood pressure regulator, whereas oxytocin induces labor in pregnant women and milk production in nursing mothers. Oxytocin was the first biologically active peptide to be prepared in the laboratory by Vincent du Vigneaud (1901–1978), who was awarded the Nobel Prize in Chemistry in 1955.
Synthetic Polymers
Many of the synthetic polymers we use, such as plastics and rubbers, have commercial advantages over naturally occurring polymers because they can be produced inexpensively. Moreover, many synthetic polymers are actually more desirable than their natural counterparts because scientists can select monomer units to tailor the physical properties of the resulting polymer for particular purposes. For example, in many applications, wood has been replaced by plastics that are more durable, lighter, and easier to shape and maintain. Polymers are also increasingly used in engineering applications where weight reduction and corrosion resistance are required. Steel rods used to support concrete structures, for example, are often coated with a polymeric material when the structures are near ocean environments where steel is vulnerable to corrosion (For more information on corrosion, see Section 19.6 .) In fact, the use of polymers in engineering applications is a very active area of research.
Probably the best-known example of a synthetic polymer is nylon (Figure 12.8.2). Its monomers are linked by amide bonds (which are called peptide bonds in biological polymers), so its physical properties are similar to those of some proteins because of their common structural unit—the amide group. Nylon is easily drawn into silky fibersA particle of a synthetic polymer that is more than 100 times longer than it is wide. that are more than a hundred times longer than they are wide and can be woven into fabrics. Nylon fibers are so light and strong that during World War II, all available nylon was commandeered for use in parachutes, ropes, and other military items. With polymer chains that are fully extended and run parallel to the fiber axis, nylon fibers resist stretching, just like naturally occurring silk fibers, although the structures of nylon and silk are otherwise different. Replacing the flexible –CH2– units in nylon by aromatic rings produces a stiffer and stronger polymer, such as the very strong polymer known as Kevlar. Kevlar fibers are so strong and rigid that they are used in lightweight army helmets, bulletproof vests, and even sailboat and canoe hulls, all of which contain multiple layers of Kevlar fabric.
A fiberglass mat (left) and a Kevlar vest (right).
Not all synthetic polymers are linked by amide bonds—for example, polyesters contain monomers that are linked by ester bonds. Polyesters are sold under trade names such as Dacron, Kodel, and Fortrel, which are used in clothing, and Mylar, which is used in magnetic tape, helium-filled balloons, and high-tech sails for sailboats. Although the fibers are flexible, properly prepared Mylar films are almost as strong as steel.
Polymers based on skeletons with only carbon are all synthetic. Most of these are formed from ethylene (CH2=CH2), a two-carbon building block, and its derivatives. The relative lengths of the chains and any branches control the properties of polyethylene. For example, higher numbers of branches produce a softer, more flexible, lower-melting-point polymer called low-density polyethylene (LDPE), whereas high-density polyethylene (HDPE) contains few branches. Substances such as glass that melt at relatively low temperatures can also be formed into fibers, producing fiberglass.
Because most synthetic fibers are neither soluble nor low melting, multistep processes are required to manufacture them and form them into objects. Graphite fibers are formed by heating a precursor polymer at high temperatures to decompose it, a process called pyrolysisA high-temperature decomposition reaction that can be used to form fibers of synthetic polymers.. The usual precursor for graphite is polyacrylonitrile, better known by its trade name—Orlon. A similar approach is used to prepare fibers of silicon carbide using an organosilicon precursor such as polydimethylsilane {[–(CH3)2Si–]n}. A new type of fiber consisting of carbon nanotubes, hollow cylinders of carbon just one atom thick, is lightweight, strong, and impact resistant. Its performance has been compared to that of Kevlar, and it is being considered for use in body armor, flexible solar panels, and bombproof trash bins, among other uses.
Because there are no good polymer precursors for elemental boron or boron nitride, these fibers have to be prepared by time-consuming and costly indirect methods. Even though boron fibers are about eight times stronger than metallic aluminum and 10% lighter, they are significantly more expensive. Consequently, unless an application requires boron’s greater resistance to oxidation, these fibers cannot compete with less costly graphite fibers.
Example 12.8.1
Polyethylene is used in a wide variety of products, including beach balls and the hard plastic bottles used to store solutions in a chemistry laboratory. Which of these products is formed from the more highly branched polyethylene?
Given: type of polymer
Asked for: application
Strategy:
Determine whether the polymer is LDPE, which is used in applications that require flexibility, or HDPE, which is used for its strength and rigidity.
Solution:
A highly branched polymer is less dense and less rigid than a relatively unbranched polymer. Thus hard, strong polyethylene objects such as bottles are made of HDPE with relatively few branches. In contrast, a beach ball must be flexible so it can be inflated. It is therefore made of highly branched LDPE.
Exercise
Which products are manufactured from LDPE and which from HPDE?
1. lawn chair frames
2. rope
3. disposable syringes
4. automobile protective covers
Answer
1. HDPE
2. LDPE
3. HDPE
4. LDPE
Summary
Polymers are giant molecules that consist of long chains of units called monomers connected by covalent bonds. Polymerization is the process of linking monomers together to form a polymer. Plastic is the property of a material that allows it to be molded. Biological polymers formed from amino acid residues are called peptides or proteins, depending on their size. Enzymes are proteins that catalyze a biological reaction. A particle that is more than a hundred times longer than it is wide is a fiber, which can be formed by a high-temperature decomposition reaction called pyrolysis.
Key Takeaway
• Polymers are giant molecules formed from addition or condensation reactions and can be classified as either biological or synthetic polymers.
Conceptual Problems
1. How are amino acids and proteins related to monomers and polymers? Draw the general structure of an amide bond linking two amino acid residues.
2. Although proteins and synthetic polymers (such as nylon) both contain amide bonds, different terms are used to describe the two types of polymer. Compare and contrast the terminology used for the
1. smallest repeating unit.
2. covalent bond connecting the units.
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.08%3A_Polymers.txt |
Learning Objectives
• To become familiar with the properties of some contemporary materials.
In addition to polymers, other materials, such as ceramics, high-strength alloys, and composites, play a major role in almost every aspect of our lives. Until relatively recently, steel was used for any application that required an especially strong and durable material, such as bridges, automobiles, airplanes, golf clubs, and tennis rackets. In the last 15 to 20 years, however, graphite or boron fiber golf clubs and tennis rackets have made wood and steel obsolete for these items. Likewise, a modern jet engine now is largely composed of Ti and Ni by weight rather than steel (Table 12.9.1). The percentage of iron in wings and fuselages is similarly low, which indicates the extent to which other materials have supplanted steel. The Chevrolet Corvette introduced in 1953 was considered unusual because its body was constructed of fiberglass, a composite material, rather than steel; by 1992, Jaguar fabricated an all-aluminum limited-edition vehicle. In fact, the current models of many automobiles have engines that are made mostly of aluminum rather than steel. In this section, we describe some of the chemistry behind three classes of contemporary materials: ceramics, superalloys, and composites.
Table 12.9.1 The Approximate Elemental Composition of a Modern Jet Engine
Element Percentage by Mass
titanium 38
nickel 37
chromium 12
cobalt 6
aluminum 3
niobium 1
tantalum 0.025
Ceramics
A ceramic is any nonmetallic, inorganic solid that is strong enough for use in structural applications. Traditional ceramics, which are based on metal silicates or aluminosilicates, are the materials used to make pottery, china, bricks, and concrete. Modern ceramics contain a much wider range of components and can be classified as either ceramic oxides, which are based on metal oxides such as alumina (Al2O3), zirconia (ZrO2), and beryllia (BeO), or nonoxide ceramics, which are based on metal carbides such as silicon carbide (carborundum, SiC) and tungsten carbide (WC), or nitrides like silicon nitride (Si3N4) and boron nitride (BN).
All modern ceramics are hard, lightweight, and stable at very high temperatures. Unfortunately, however, they are also rather brittle, tending to crack or break under stresses that would cause metals to bend or dent. Thus a major challenge for materials scientists is to take advantage of the desirable properties of ceramics, such as their thermal and oxidative stability, chemical inertness, and toughness, while finding ways to decrease their brittleness to use them in new applications. Few metals can be used in jet engines, for example, because most lose mechanical strength and react with oxygen at the very high operating temperatures inside the engines (approximately 2000°C). In contrast, ceramic oxides such as Al2O3 cannot react with oxygen regardless of the temperature because aluminum is already in its highest possible oxidation state (Al3+). Even nonoxide ceramics such as silicon and boron nitrides and silicon carbide are essentially unreactive in air up to about 1500°C. Producing a high-strength ceramic for service use involves a process called sintering, which fuses the grains into a dense and strong material (Figure 12.9.2).
Ceramics are hard, lightweight, and able to withstand high temperatures, but they are also brittle.
One of the most widely used raw materials for making ceramics is clay. Clay minerals consist of hydrated alumina (Al2O3) and silica (SiO2) that have a broad range of impurities, including barium, calcium, sodium, potassium, and iron. Although the structures of clay minerals are complicated, they all contain layers of metal atoms linked by oxygen atoms. Water molecules fit between the layers to form a thin film of water. When hydrated, clays can be easily molded, but during high-temperature heat treatment, called firing, a dense and strong ceramic is produced.
Because ceramics are so hard, they are easily contaminated by the material used to grind them. In fact, the ceramic often grinds the metal surface of the mill almost as fast as the mill grinds the ceramic! The sol-gel process was developed to address this problem. In this process, a water-soluble precursor species, usually a metal or semimetal alkoxide [M(OR)n] undergoes a hydrolysis reaction to form a cloudy aqueous dispersion called a sol. The sol contains particles of the metal or semimetal hydroxide [M(OH)n], which are typically 1–100 nm in diameter. As the reaction proceeds, molecules of water are eliminated from between the M(OH)n units in a condensation reaction, and the particles fuse together, producing oxide bridges, M–O–M. Eventually, the particles become linked in a three-dimensional network that causes the solution to form a gel, similar to a gelatin dessert. Heating the gel to 200°C–500°C causes more water to be eliminated, thus forming small particles of metal oxide that can be amazingly uniform in size. This chemistry starts with highly pure SiCl4 and proceeds via the following reactions starting with the alkoxide formation
$SiCl_{4}\left ( s \right )+4CH_{3}CH_{2}OH\left ( l \right )+4NH_{3}\left ( g \right ){\rightarrow}SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4NH_{4}Sl\left ( s \right ) \label{12.8.1}$
and then the hydrolysis of the alkoxide
$SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4H_{2}O\left ( l \right ) {\rightarrow} \left ( HO \right )_{3}Si-OH\left ( s \right )+ 4CH_{3}CH_{2}OH\left ( aq \right ) \label{12.8.2}$
ending with the condensation
$\left ( HO_{3} \right )Si-OH\left ( s \right )+nHO-Si\left ( OH \right )_{3}\left ( s \right )\rightarrow \left ( HO_{3} \right )Si\left ( -O-Si\left ( OH \right )_{3} \right )_{n}\left ( s \right )+nH_{2}O\left ( l \right ) \ \label{12.8.3}$
Nature uses the same process to create opal gemstones.
Superalloys
Superalloys are high-strength alloys, often with a complex composition, that are used in systems requiring mechanical strength, high surface stability (minimal flaking or pitting), and resistance to high temperatures. The aerospace industry, for example, requires materials that have high strength-to-weight ratios to improve the fuel efficiency of advanced propulsion systems, and these systems must operate safely at temperatures greater than 1000°C.
Superalloys are used in systems requiring mechanical strength, minimal flaking or pitting, and high-temperature resistance.
Although most superalloys are based on nickel, cobalt, or iron, other metals are used as well. Pure nickel or cobalt is relatively easily oxidized, but adding small amounts of other metals (Al, Co, Cr, Mo, Nb, Ti, and W) results in an alloy that has superior properties. Consequently, most of the internal parts of modern gas turbine jet engines are now made of superalloys based on either nickel (used in blades and disks) or cobalt (used in vanes, combustion chamber liners, and afterburners). The cobalt-based superalloys are not as strong as the nickel-based ones, but they have excellent corrosion resistance at high temperatures.
Other alloys, such as aluminum–lithium and alloys based on titanium, also have applications in the aerospace industry. Because aluminum–lithium alloys are lighter, stiffer, and more resistant to fatigue at high temperatures than aluminum itself, they are used in engine parts and in the metal “skins” that cover wings and bodies. Titanium’s high strength, corrosion resistance, and lightweight properties are equally desirable for applications where minimizing weight is important (as in airplanes). Unfortunately, however, metallic titanium reacts rapidly with air at high temperatures to form TiN and TiO2. The welding of titanium or any similar processes must therefore be carried out in an argon or inert gas atmosphere, which adds significantly to the cost. Initially, titanium and its alloys were primarily used in military applications, but more recently, they have been used as components of the airframes of commercial planes, in ship structures, and in biological implants.
Composite Materials
Composite materials have at least two distinct components: the matrix (which constitutes the bulk of the material) and fibers or granules that are embedded within the matrix and limit the growth of cracks by pinning defects in the bulk material (Figure 12.9.3). The resulting material is stronger, tougher, stiffer, and more resistant to corrosion than either component alone. Composites are thus the nanometer-scale equivalent of reinforced concrete, in which steel rods greatly increase the mechanical strength of the cement matrix, and are extensively used in the aircraft industry, among others. For example, the Boeing 777 is 9% composites by weight, whereas the newly developed Boeing 787 is 50% composites by weight. Not only does the use of composite materials reduce the weight of the aircraft, and therefore its fuel consumption, but it also allows new design concepts because composites can be molded. Moreover, by using composites in the Boeing 787 multiple functions can be integrated into a single system, such as acoustic damping, thermal regulation, and the electrical system.
Three distinct types of composite material are generally recognized, distinguished by the nature of the matrix. These are polymer-matrix composites, metal-matrix composites, and ceramic-matrix composites.
Composites are stronger, tougher, stiffer, and more resistant to corrosion than their components alone.
Fiberglass is a polymer-matrix composite that consists of glass fibers embedded in a polymer, forming tapes that are then arranged in layers impregnated with epoxy. The result is a strong, stiff, lightweight material that is resistant to chemical degradation. It is not strong enough, however, to resist cracking or puncturing on impact. Stronger, stiffer polymer-matrix composites contain fibers of carbon (graphite), boron, or polyamides such as Kevlar. High-tech tennis rackets and golf clubs as well as the skins of modern military aircraft such as the “stealth” F-117A fighters and B-2 bombers are made from both carbon fiber–epoxy and boron fiber–epoxy composites. Compared with metals, these materials are 25%–50% lighter and thus reduce operating costs. Similarly, the space shuttle payload bay doors and panels are made of a carbon fiber–epoxy composite. The structure of the Boeing 787 has been described as essentially one giant macromolecule, where everything is fastened through cross-linked chemical bonds reinforced with carbon fiber.
Metal-matrix composites consist of metals or metal alloys reinforced with fibers. They offer significant advantages for high-temperature applications but pose major manufacturing challenges. For example, obtaining a uniform distribution and alignment of the reinforcing fibers can be difficult, and because organic polymers cannot survive the high temperatures of molten metals, only fibers composed of boron, carbon, or ceramic (such as silicon carbide) can be used. Aluminum alloys reinforced with boron fibers are used in the aerospace industry, where their strength and lightweight properties make up for their relatively high cost. The skins of hypersonic aircraft and structural units in the space shuttle are made of metal-matrix composites.
Ceramic-matrix composites contain ceramic fibers in a ceramic matrix material. A typical example is alumina reinforced with silicon carbide fibers. Combining the two very high-melting-point materials results in a composite that has excellent thermal stability, great strength, and corrosion resistance, while the SiC fibers reduce brittleness and cracking. Consequently, these materials are used in very high-temperature applications, such as the leading edge of wings of hypersonic airplanes and jet engine parts. They are also used in the protective ceramic tiles on the space shuttle, which contain short fibers of pure SiO2 mixed with fibers of an aluminum–boron–silicate ceramic. These tiles are excellent thermal insulators and extremely light (their density is only about 0.2 g/cm3). Although their surface reaches a temperature of about 1250°C during reentry into Earth’s atmosphere, the temperature of the underlying aluminum alloy skin stays below 200°C.
Example $1$:
An engineer is tasked with designing a jet ski hull. What material is most suited to this application? Why?
Given: design objective
Asked for: most suitable material
Strategy:
Determine under what conditions the design will be used. Then decide what type of material is most appropriate.
Solution:
A jet ski hull must be lightweight to maximize speed and fuel efficiency. Because of its use in a marine environment, it must also be resistant to impact and corrosion. A ceramic material provides rigidity but is brittle and therefore tends to break or crack under stress, such as when it impacts waves at high speeds. Superalloys provide strength and stability, but a superalloy is probably too heavy for this application. Depending on the selection of metals, it might not be resistant to corrosion in a marine environment either. Composite materials, however, provide strength, stiffness, and corrosion resistance; they are also lightweight materials. This is not a high-temperature application, so we do not need a metal-matrix composite or a ceramic-matrix composite. The best choice of material is a polymer-matrix composite with Kevlar fibers to increase the strength of the composite on impact.
Exercise $1$
In designing a new generation of space shuttle, National Aeronautics and Space Administration (NASA) engineers are considering thermal-protection devices to protect the skin of the craft. Among the materials being considered are titanium- or nickel-based alloys and silicon-carbide ceramic reinforced with carbon fibers. Why are these materials suitable for this application?
Answer: Ti- or Ni-based alloys have a high strength-to-weight ratio, resist corrosion, and are safe at high temperatures. Reinforced ceramic is lightweight; has high thermal and oxidative stability; and is chemically inert, tough, and impact resistant.
Summary
Ceramics are nonmetallic, inorganic solids that are typically strong; they have high melting points but are brittle. The two major classes of modern ceramics are ceramic oxides and nonoxide ceramics, which are composed of nonmetal carbides or nitrides. The production of ceramics generally involves pressing a powder of the material into the desired shape and sintering at a temperature just below its melting point. The necessary fine powders of ceramic oxides with uniformly sized particles can be produced by the sol-gel process. Superalloys are new metal phases based on cobalt, nickel, or iron that exhibit unusually high temperature stability and resistance to oxidation. Composite materials consist of at least two phases: a matrix that constitutes the bulk of the material and fibers or granules that act as a reinforcement. Polymer-matrix composites have reinforcing fibers embedded in a polymer matrix. Metal-matrix composites have a metal matrix and fibers of boron, graphite, or ceramic. Ceramic-matrix composites use reinforcing fibers, usually also ceramic, to make the matrix phase less brittle.
Key Takeaway
• Materials that have contemporary applications include ceramics, high-strength alloys, and composites, whose properties can be modified as needed.
Conceptual Problems
1. Can a compound based on titanium oxide qualify as a ceramic material? Explain your answer.
2. What features make ceramic materials attractive for use under extreme conditions? What are some potential drawbacks of ceramics?
3. How do composite materials differ from the other classes of materials discussed in this chapter? What advantages do composites have versus other materials?
4. How does the matrix control the properties of a composite material? What is the role of an additive in determining the properties of a composite material?
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.09%3A_Modern_Materials.txt |
We explored the general properties of gases, liquids, and solids. Most of the discussion focused on pure substances containing a single kind of atom, molecule, or cation–anion pair. The substances we encounter in our daily lives, however, are usually mixtures rather than pure substances. Some are heterogeneous mixtures, which consist of at least two phases that are not uniformly dispersed on a microscopic scale; others are homogeneous mixtures, consisting of a single phase in which the components are uniformly distributed. Homogeneous mixtures are also called solutions; they include the air we breathe, the gas we use to cook and heat our homes, the water we drink, the gasoline or diesel fuel that powers engines, and the gold and silver jewelry we wear.
13: Solutions
Learning Objectives
• to understand how enthalpy and entropy changes affect solution formation.
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table 13.1 lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.
Table 13.1: Types of Solutions
Solution Solute Solvent Examples
gas gas gas air, natural gas
liquid gas liquid seltzer water ($CO_2$ gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid $H_2$ in Pd (used for $H_2$ storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)
Forming a Solution
The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:
$Zn(NO_3)_{2(s)} + H_2O(l) \rightarrow Zn^{2+}_{(aq)}+2NO^-_{3(aq)} \tag{13.1}$
Because $Zn(NO_3)_2$ can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:
$Zn_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^-_{(aq)} + H_{2(g)} \tag{13.2}$
Note
Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere.
The Role of Enthalpy in Solution Formation
Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation, or hydration when the solvent is water. Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.
Because enthalpy is a state function, we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by ΔH1, ΔH2, and ΔH3 in Figure 13.1. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \tag{13.3}$
When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent ($\Delta H_1$) and the solute ($\Delta H_2$). Because $ΔH$ is positive for both steps 1 and 2, the solute–solvent interactions ($\Delta H_3$) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic ($\Delta H_{soln} < 0$). When the solute is an ionic solid, ΔH2 corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as $MgO$ (−3791 kJ/mol), are generally insoluble in all solvents.
A positive value for $ΔH_{soln}$ does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high $ΔH_{soln}$ is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive $ΔH_{soln}$ of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure 13.2).
Entropy and Solution Formation
The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detailelsehwere, but for now we can state that entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, is arbitrarily assigned an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.
The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, ΔHsoln should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative $ΔH_{soln}$ but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with $ΔH \ge 0$ are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.
Note
All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.
Table 13.2 summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.
Table 13.2: Relative Changes in Enthalpies for Different Solute–Solvent Combinations*
ΔH1 (separation of solvent molecules) ΔH2 (separation of solute particles) ΔH3 (solute–solvent interactions) ΔHsoln (ΔH1 + ΔH2 +ΔH3) Result of Mixing Solute and Solvent
*ΔH1, ΔH2, and ΔH3 refer to the processes indicated in the thermochemical cycle shown in Figure 13.1.
In all four cases, entropy increases.
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process (ΔHsoln ≈ 0), and the entropic factor due to the increase in disorder is dominant (Figure 13.3). Consequently, all gases dissolve readily in one another in all proportions to form solutions. We will return to a discussion of enthalpy and entropy in Chapter 18, where we treat their relationship quantitatively.
Example 13.1
Considering LiCl, benzoic acid (C6H5CO2H), and naphthalene, which will be most soluble and which will be least soluble in water?
Given: three compounds
Asked for: relative solubilities in water
Strategy: Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure 13.1. Then use Table 13.2 to predict the solubility of each compound in water and arrange them in order of decreasing solubility.
Solution:
The first substance, LiCl, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in Equation 13.3). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect ΔH3 to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect ΔH2 to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak ΔH3 ≈ 0. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than LiCl. We thus predict LiCl to be the most soluble in water and naphthalene to be the least soluble.
Exercise
Exercise 13.1
Considering ammonium chloride, cyclohexane, and ethylene glycol (HOCH2CH2OH), which will be most soluble and which will be least soluble in benzene?
Solution
The most soluble is cyclohexane; the least soluble is ammonium chloride.
Summary
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, ΔHsoln, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (ΔHsoln < 0) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution.
Key Takeaway
The magnitude of the changes in both enthalpy and entropy must be considered when predicting whether a given solute–solvent combination will spontaneously form a solution.
Conceptual Problems
1. Classify each of the following as a heterogeneous mixture or homogeneous mixture. Explain your rationale in each case.
a. aqueous ammonia
b. liquid decongestant
c. vinegar
d. seawater
e. gasoline
f. fog
2. Solutions and heterogeneous mixtures are at the extreme ends of the solubility scale. Name one type of mixture that is intermediate on this scale. How are the properties of the mixture you have chosen different from those of a solution or a heterogeneous mixture?
3. Classify each process as simple dissolution or a chemical reaction.
a. a naphthalene mothball dissolving in benzene
b. a sample of a common drain cleaner that has a mixture of NaOH crystals and Al chunks dissolving in water to give H2 gas and an aqueous solution of Na+, OH, and Al3+ ions
c. an iron ship anchor slowly dissolving in seawater
d. sodium metal dissolving in liquid ammonia
4. Classify each process as simple dissolution or a chemical reaction.
a. a sugar cube dissolving in a cup of hot tea
b. SO3 gas dissolving in water to produce sulfuric acid
c. calcium oxide dissolving in water to produce a basic solution
d. metallic gold dissolving in a small quantity of liquid mercury
5. You notice that a gas is evolved as you are dissolving a solid in a liquid. Will you be able to recover your original solid by evaporation? Why or why not?
6. Why is heat evolved when sodium hydroxide pellets are dissolved in water? Does this process correspond to simple dissolution or a chemical reaction? Justify your answer.
7. Which process(es) is the simple formation of a solution, and which process(es) involves a chemical reaction?
a. mixing an aqueous solution of NaOH with an aqueous solution of HCl
b. bubbling HCl gas through water
c. adding iodine crystals to CCl4
d. adding sodium metal to ethanol to produce sodium ethoxide (C2H5ONa+) and hydrogen gas
8. Using thermochemical arguments, explain why some substances that do not form a solution at room temperature will form a solution when heated. Explain why a solution can form even when ΔHsoln is positive.
9. If you wanted to formulate a new compound that could be used in an instant cold pack, would you select a compound with a positive or negative value of ΔHsoln in water? Justify your answer.
10. Why is entropy the dominant factor in the formation of solutions of two or more gases? Is it possible for two gases to be immiscible? Why or why not?
Answers
1. Homogeneous mixtures: aqueous ammonia, liquid decongestant, vinegar, and gasoline. Heterogeneous mixtures: seawater and fog.
2. All are chemical reactions except dissolving iodine crystals in CCl4. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.01%3A_Factors_Affecting_Solution_Formation.txt |
Learning Objectives
• To understand the relationship between solubility and molecular structure.
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (part (a) in Figure 13.4). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows:
$\text{solute} + \text{solvent} \ce{<=>[\ce{crystallization}][\ce{dissolution}]} \text{solution} \tag{13.4}$
Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles.
• Anonymous
13.03: Units of Concentration
Learning Objectives
• To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
There are several different ways to quantitatively describe the concentration of a solution. For example, molarity was introduced in Chapter 4 as a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions, introduced in Chapter 10, are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example 4 reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.
Example 4
Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Given: mass of substance and mass and density of solution
Asked for: molarity and mole fraction
Strategy:
1. Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
2. Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.
Solution:
A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:
$moles\; CH_3CO_2H=\dfrac{3.78\; \cancel{g}\; CH_3CO_2H}{60.05\; \cancel{g}/mol}=0.0629 \;mol$
$volume=\dfrac{mass}{density}=\dfrac{100.0\; \cancel{g}\; solution}{1.00\; \cancel{g}/mL}=100\; mL$
$molarity\; of\; CH_3CO_2H=\dfrac{moles\; CH_3CO_2H }{\text{liter solution}}=\dfrac{0.0629\; mol\; CH_3CO_2H}{(100\; \cancel{mL})(1\; L/1000\; \cancel{mL})}=0.629\; M \;CH_3CO_2H$
This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than $\frac{1}{2}$ mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.
B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have
$moles\; H_2O=\dfrac{96.2\; \cancel{g}\; H_2O}{18.02\; \cancel{g}/mol}=5.34\; mol\; H_2O$
The mole fraction $X$ of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:
$X_{CH_3CO_2H}=\dfrac{moles\; CH_3CO_2H}{moles \;CH_3CO_2H + moles\; H_2O}=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol}=0.0116=1.16 \times 10^{−2}$
This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.
Exercise 5
A solution of $HCl$ gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of $HCl$ per 100.0 g of solution, and its density is 1.10 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Answer:
1. 6.10 M HCl
2. XHCl = 0.111
The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:
$\text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}$
Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of $H_2O$ under these conditions is very close to 1.0 L, and a 0.50 M solution of $KBr$ in water, for example, has approximately the same concentration as a 0.50 m solution.
Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):
$\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}$
$\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}$
$\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}$
In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.
The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of $H_2SO_4$ per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
Example 5
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
1. What is the molarity of the solution?
2. What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
1. Use the concentration of the solute in parts per million to calculate the molarity.
2. Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
$molarity=\dfrac{moles}{liter solution}=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L}=1.63 \times 10^{-4} M$
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
$mass of benzene=\dfrac{(12.7\; mg\; benzene)(250\; \cancel{mL})}{1000\; mL}=3.18\; mg =3.18 \times 10^{-3}\; g\; benzene$
Exercise 6
The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of $Pb^{2+}$ in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of $Pb^{2+}$ are in an 8 oz glass of water.
Answer: 4.3 × 10−8 M; 2 × 10−6 g
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.
Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as we saw in Chapter 10. As you will learn in Section 13.5, mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.
Table 13.5 summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example 6.
Table 13.5 Different Units for Expressing the Concentrations of Solutions*
Unit Definition Application
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction (X) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
Example 6
Vodka is essentially a solution of pure ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.
1. the mass percentage
2. the mole fraction
3. the molarity
4. the molality
Given: volume percent and density
Asked for: mass percentage, mole fraction, molarity, and molality
Strategy:
1. Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
2. Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
3. Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
4. Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.
Solution:
The key to this problem is to use the density of pure ethanol to determine the mass of ethanol ($CH_3CH_2OH$), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:
$mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH$
If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.
B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:
$\%EtOH=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100)=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) = 34.5\%$
C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:
$moles\; EtOH=(31.6\; \cancel{g\; EtOH}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; EtOH}}\right)=0.686 \;mol\; CH_3CH_2OH$
Similarly, the number of moles of water is
$moles \;H_2O=(60.0\; \cancel{g \; H_2O}) \left(\dfrac{1 \;mol\; H_2O}{18.02\; \cancel{g\; H_2O}}\right)=3.33\; mol\; H_2O$
The mole fraction of ethanol is thus
$X_{EtOH}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171$
D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is
$M_{EtOH}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M$
The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:
$m_{EtOH}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m$
Exercise 7
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
1. mass percentage
2. mole fraction
3. molarity
4. molality
Answer:
1. mass percentage toluene = 24.8%
2. $X_{toluene} = 0.219$
3. 2.35 M toluene
4. 3.59 m toluene
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.02%3A_Solubility_and_Molecular_Structure.txt |
Learning Objectives
• To understand the relationship among temperature, pressure, and solubility.
• The understand that the solubility of a solid may increase or decrease with increasing temperature,
• To understand that the solubility of a gas decreases with an increase in temperature and a decrease in pressure.
Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences.
Effect of Temperature on the Solubility of Solids
Figure $1$ shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and $\ce{CH_3CO_2Na}$) exhibit a dramatic increase in solubility with increasing temperature. Others (such as $\ce{NaCl}$ and $\ce{K_2SO_4}$) exhibit little variation, and still others (such as $\ce{Li_2SO_4}$) become less soluble with increasing temperature.
Notice in particular the curves for $\ce{NH4NO3}$ and $\ce{CaCl2}$. The dissolution of ammonium nitrate in water is endothermic ($ΔH_{soln} = +25.7\; kJ/mol$), whereas the dissolution of calcium chloride is exothermic ($ΔH_{soln} = −68.2 \;kJ/mol$), yet Figure $1$ shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature.
The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate ($\ce{CH_3CO_2Na}$) and 50 g of $\ce{KBr}$, we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure $1$, both compounds dissolve in water at 80°C, and all 50 g of $\ce{KBr}$ remains in solution at 0°C. Only about 36 g of $\ce{CH3CO2Na}$ are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of $\ce{CH_3CO_2Na}$ crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original $\ce{CH_3CO_2Na}$ in essentially pure form in only one step.
Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure $1$ and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was $\ce{KBr}$ in this example) and preferably present in relatively small amounts.
Effect of Temperature on the Solubility of Gases
The solubility of gases in liquids decreases with increasing temperature, as shown in Figure $2$. Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process ($ΔH_{soln} < 0$). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased.
The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure $3$). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved $\ce{CO2}$ from solution. Hard water contains dissolved $\ce{Ca^{2+}}$ and $\ce{HCO3^{-}}$ (bicarbonate) ions. Calcium bicarbonate ($\ce{Ca(HCO3)2}$ is rather soluble in water, but calcium carbonate ($\ce{CaCO3}$) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:
$\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9}$
Heating the solution decreases the solubility of $\ce{CO2}$, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.
Thermal Pollution
In thermal pollution, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of $\ce{O2}$ at higher temperatures (Figure $2$), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water.
A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.
Effect of Pressure on the Solubility of Gases: Henry’s Law
External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure $4$, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.
The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):
$C = kP \label{13.3.1}$
where
• $C$ is the concentration of dissolved gas at equilibrium,
• $P$ is the partial pressure of the gas, and
• $k$ is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature.
Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table $1$.
As the data in Table $1$ demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the Group 18 elements, the Henry’s law constants increase smoothly from $\ce{He}$ to $\ce{Ne}$ to $\ce{Ar}$.
Table $1$: Henry’s Law Constants for Selected Gases in Water at 20°C
Gas Henry’s Law Constant [mol/(L·atm)] × 10−4
$\ce{He}$ 3.9
$\ce{Ne}$ 4.7
$\ce{Ar}$ 15
$\ce{H_2}$ 8.1
$\ce{N_2}$ 7.1
$\ce{O_2}$ 14
$\ce{CO_2}$ 392
Oxygen is Especially Soluble
Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table $1$ shows, $\ce{O2}$ is twice as soluble in water as $\ce{N2}$. Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O2/N2 Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold (J. Chem. Eng. Data 2011, 56, 5036–5044),
Gases that react chemically with water, such as $\ce{HCl}$ and the other hydrogen halides, $\ce{H2S}$, and $\ce{NH3}$, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, $\ce{HCl}$ reacts with water to give $\ce{H^{+}(aq)}$ and $\ce{Cl^{-}(aq)}$, not dissolved $\ce{HCl}$ molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule.
Gases that react with water do not obey Henry’s law.
Henry’s law has important applications. For example, bubbles of $\ce{CO2}$ form as soon as a carbonated beverage is opened because the drink was bottled under $\ce{CO2}$ at a pressure greater than 1 atm. When the bottle is opened, the pressure of $\ce{CO2}$ above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of $\ce{N2}$ to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.
Due to the low Henry’s law constant for $\ce{O2}$ in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the $\ce{O2}$ concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind $\ce{O2}$ reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds $\ce{O2}$ and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four $\ce{O2}$ molecules. Although the concentration of dissolved $\ce{O2}$ in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved $\ce{O2}$ concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for $\ce{O2}$ result in dissolved oxygen concentrations comparable to those in normal blood.
A Video Discussing Henry's Law. Video Link: Henry's Law (The Solubility of Gases in Solvents), YouTube(opens in new window) [youtu.be]
Example $1$
The Henry’s law constant for $\ce{O2}$ in water at 25°C is $1.27 \times 10^{-3} M/atm$, and the mole fraction of $\ce{O2}$ in the atmosphere is 0.21. Calculate the solubility of $\ce{O2}$ in water at 25°C at an atmospheric pressure of 1.00 atm.
Given: Henry’s law constant, mole fraction of $\ce{O2}$, and pressure
Asked for: solubility
Strategy:
1. Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen. (For more information about Dalton’s law of partial pressures)
2. Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.
Solution:
A According to Dalton’s law, the partial pressure of $\ce{O2}$ is proportional to the mole fraction of $\ce{O2}$:
\begin{align*} P_A &= \chi_A P_t \[4pt] &= (0.21)(1.00\; atm) \[4pt] &= 0.21\; atm \end{align*} \nonumber
B From Henry’s law, the concentration of dissolved oxygen under these conditions is
\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber
Exercise $1$
To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved $\ce{CO2}$ in a soft drink:
1. bottled under a pressure of 5.0 atm of $\ce{CO2}$
2. in equilibrium with the normal partial pressure of $\ce{CO_2}$ in the atmosphere (approximately $3 \times 10^{-4} \;atm$). The Henry’s law constant for $\ce{CO2}$ in water at 25°C is $3.4 \times 10^{-2}\; M/atm$.
Answer a
$0.17 M$
Answer b
$1 \times 10^{-5} M$
Summary
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.04%3A_Effects_of_Temperature_and_Pressure_on_Solubility.txt |
Learning Objectives
• To describe the relationship between solute concentration and the physical properties of a solution.
• To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.
Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
Counting concentrations
When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, $\ce{NaCl}$, and $\ce{CaCl_2}$. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both $\ce{NaCl}$ and $\ce{CaCl_2}$ are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of $\ce{NaCl}$ contains 0.01 M $\ce{Na^{+}}$ ions and 0.01 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.02 M. Similarly, the $\ce{CaCl_2}$ solution contains 0.01 M $\ce{Ca^{2+}}$ ions and 0.02 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete.
At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as $\ce{Mg^{2+}}$ or $\ce{Al^{3+}}$), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind.
Vapor Pressure of Solutions and Raoult’s Law
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure $1$, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
Figure $2$ shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.
If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore
$P_A=\chi_AP^0_A \label{13.5.1}$
where $P_A$ is the vapor pressure of component A of the solution (in this case the solvent), $\chi_A$ is the mole fraction of $A$ in solution, and $P^0_A$ is the vapor pressure of pure $A$. Equation $\ref{13.5.1}$ is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then $\chi_A + \chi_B = 1$, and we can substitute $\chi_A = 1 − \chi_B$ to obtain
\begin{align} P_A &=(1−\chi_B)P^0_A \[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align}
Rearranging and defining $ΔP_A=P^0_A−P_A$, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:
\begin{align} P^0_A−P_A &=ΔP_A \[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align}
We can solve vapor pressure problems in either of two ways: by using Equation $\ref{13.5.1}$ to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation $\ref{13.5.3}$ to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example $1$: Anti-Freeze
Ethylene glycol ($\ce{HOCH_2CH_2OH}$), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.
Given: identity of solute, percentage by mass, and vapor pressure of pure solvent
Asked for: vapor pressure of solution
Strategy:
1. Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.
2. Use Raoult’s law to calculate the vapor pressure of the solution.
Solution:
A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:
$moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber$
$moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber$
The mole fraction of water is thus
$\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber$
B From Raoult’s law (Equation $\ref{13.5.1}$), the vapor pressure of the solution is
\begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation $\ref{13.5.3}$ to calculate the resulting decrease in vapor pressure:
$\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber$
$ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber$
$P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber$
The same result is obtained using either method.
Exercise $1$
Seawater is an approximately 3.0% aqueous solution of $\ce{NaCl}$ by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of $\ce{NaCl}$, remembering that 1 mol of $\ce{NaCl}$ produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.
Answer
0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.
Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution ($P_{tot}$) is the sum of the vapor pressures of the components:
$P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4}$
Because $\chi_B = 1 − \chi_A$ for a two-component system,
$P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5}$
Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is
$P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6}$
and the vapor pressure of toluene in the solution is
$P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7}$
Equations $\ref{13.5.6}$ and $\ref{13.5.7}$ are both in the form of the equation for a straight line: $y = mx + b$, where $b = 0$. Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure $3$. Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component.
A solution of two volatile components that behaves like the solution in Figure $3$, which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero ($ΔH_{soln} ≈ 0$), which is one of the defining characteristics of an ideal solution.
Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions.
Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of $\ce{CCl_4}$ and methanol, for example, the nonpolar $\ce{CCl_4}$ molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier $\ce{CCl_4}$ molecules. Consequently, solutions of $\ce{CCl_4}$ and methanol exhibit positive deviations from Raoult’s law.
Example $2$
For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation).
1. cyclohexane and ethanol
2. methanol and acetone
3. n-hexane and isooctane (2,2,4-trimethylpentane)
Given: identity of pure liquids
Asked for: predicted deviation from Raoult’s law (Equation \ref{13.5.1})
Strategy:
Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution.
Solution:
1. Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation).
2. Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation).
3. Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution).
Exercise $2$
For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation):
1. benzene and n-hexane
2. ethylene glycol and $\ce{CCl_4}$
3. acetic acid and n-propanol
Answer a
approximately equal
Answer b
positive deviation (vapor pressure greater than predicted)
Answer c
negative deviation (vapor pressure less than predicted)
A Video Discussing Roult's Law. Video Link: Introduction to the Vapor Pressure of a Solution (Raoult's Law), YouTube(opens in new window) [youtu.be]
A Video Discussing How to find the Vapor Pressure of a Solution. Video Link: Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute), YouTube(opens in new window) [youtu.be]
Boiling Point Elevation
Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure $4$). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.
Figure $4$: Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution.
The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure $5$).
We can define the boiling point elevation ($ΔT_b$) as the difference between the boiling points of the solution and the pure solvent:
$ΔT_b=T_b−T^0_b \label{13.5.8}$
where $T_b$ is the boiling point of the solution and $T^0_b$ is the boiling point of the pure solvent. We can express the relationship between $ΔT_b$ and concentration as follows
$ΔT_b = mK_b \label{13.5.9}$
where m is the concentration of the solute expressed in molality, and $K_b$ is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table $1$ lists characteristic $K_b$ values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
Table $1$: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents
Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m)
acetic acid 117.90 3.22 16.64 3.63
benzene 80.09 2.64 5.49 5.07
d-(+)-camphor 207.4 4.91 178.8 37.8
carbon disulfide 46.2 2.42 −112.1 3.74
carbon tetrachloride 76.8 5.26 −22.62 31.4
chloroform 61.17 3.80 −63.41 4.60
nitrobenzene 210.8 5.24 5.70 6.87
water 100.00 0.51 0.00 1.86
The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.
According to Table $1$, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous $\ce{NaCl}$ solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved ions. Hence a 1.00 m $\ce{NaCl}$ solution will have a boiling point of about 101.02°C.
Example $3$
In Example $1$, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: boiling point
Strategy:
Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation $\ref{13.5.9}$ to calculate the increase in boiling point.
Solution:
From Example $1$, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus
$\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber$
From Equation $\ref{13.5.9}$, the increase in boiling point is therefore
$ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber$
The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation $\ref{13.5.9}$ may not be valid.
Exercise $3$
Assume that a tablespoon (5.00 g) of $\ce{NaCl}$ is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?
Answer
100.04°C, or 100°C to three significant figures. (Recall that 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)
A Video Discussing Boiling Point Elevation and Freezing Point Depression. Video Link: Boiling Point Elevation and Freezing Point Depression, YouTube(opens in new window) [youtu.be] (opens in new window)
Freezing Point Depression
The phase diagram in Figure $4$ shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to $P = 1\,atm$ at a lower temperature than the curve for pure water.
We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression ($ΔT_f$) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
$ΔT_f=T^0_f−T_f \label{13.5.10}$
where $T^0_f$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution.
The order of the terms is reversed compared with Equation $\ref{13.5.8}$ to express the freezing point depression as a positive number. The relationship between $ΔT_f$ and the solute concentration is given by an equation analogous to Equation $\ref{13.5.9}$:
$ΔT_f = mK_f \label{13.5.11}$
where $m$ is the molality of the solution and $K_f$ is the molal freezing point depression constant for the solvent (in units of °C/m).
Like $K_b$, each solvent has a characteristic value of $K_f$ (see Table $1$). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous $\ce{NaCl}$ solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.
The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example $4$: Salting the Roads
In colder regions of the United States, $\ce{NaCl}$ or $\ce{CaCl_2}$ is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of $\ce{NaCl}$ and one of $\ce{CaCl_2}$, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.
Given: solubilities of two compounds
Asked for: concentrations and freezing points
Strategy:
1. Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities.
2. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation $\ref{13.5.11}$ to calculate the freezing point depressions of the solutions.
Solution:
A From Figure above, we can estimate the solubilities of $\ce{NaCl}$ and $\ce{CaCl_2}$ to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are
$m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber$
$m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber$
The lower formula mass of $\ce{NaCl}$ more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than $CaCl_2$.
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of $\ce{NaCl}$ and $CaCl_2$, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for $\ce{NaCl}$ and 3 × 5.4 m = 16 m for $CaCl_2$. The resulting freezing point depressions can be calculated using Equation $\ref{13.5.11}$:
$\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber$
$\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber$
Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that $CaCl_2$ is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, $CaCl_2$ is the salt usually sold for home use, and it is also often used on highways.
Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer
Exercise $4$
Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples $5$ and $5$.
Answer
−13.0°C
Example $5$
Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m $KCl$, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
1. Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.
2. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.
Solution:
A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. $\ce{KCl}$, $\ce{SrCl_2}$, and $\ce{HCl}$ are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: $\ce{KCl}$ and $\ce{HCl}$, 0.2 m; $SrCl_2$, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > $\ce{HCl}$ = $\ce{KCl}$ > $\ce{SrCl_2}$.
Exercise $5$
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m $\ce{NaCl}$, 0.3 m acetic acid, 0.1 m $\ce{CaCl_2}$, and 0.2 m sucrose.
Answer
0.2 m $\ce{NaCl}$ (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m $\ce{CaCl_2}$ < 0.2 m sucrose (highest freezing point)
Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing $K_b$ and $K_f$ values in Table $1$, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of $K_f$ (37.8°C/m), d-(+)-camphor (Table $1$) is often used to determine the molar mass of organic compounds by this method.
Example $6$: Sulfur
A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of $CS_2$ to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved $\ce{S_n}$ molecules (i.e., what is the value of $n$?).
Given: masses of solute and solvent and freezing point
Asked for: molar mass and number of $\ce{S}$ atoms per molecule
Strategy:
1. Use Equation $\ref{13.5.10}$, the measured freezing point of the solution, and the freezing point of $CS_2$ from Table $1$ to calculate the freezing point depression. Then use Equation $\ref{13.5.11}$ and the value of $K_f$ from Table $1$ to calculate the molality of the solution.
2. From the calculated molality, determine the number of moles of solute present.
3. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain $n$, the number of sulfur atoms per mole of dissolved sulfur.
Solution:
A The first step is to calculate the freezing point depression using Equation $\ref{13.5.10}$:
$ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber$
Then Equation $\ref{13.5.11}$ gives
$m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber$
B The total number of moles of solute present in the solution is
$\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber$
C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus
$\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber$
The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of $\ce{S_8}$.
Exercise $6$
One of the byproducts formed during the synthesis of $C_{60}$ is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of $CCl_4$ has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?
Answer
847 g/mol; $\ce{C_{70}}$
A Video Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Video Link: Finding the Molecular Weight of an Unknown using Colligative Properties, YouTube(opens in new window) [youtu.be]
Osmotic Pressure
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure $6$, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure ($\Pi$) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
$\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12}$
where $M$ is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), $R$ is the ideal gas constant, and $T$ is the absolute temperature.
As shown in Example $7$, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example $7$
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% $\ce{NaCl}$ by mass; the solution density is 1.02 g/mL at 25°C.
1. Calculate the osmotic pressure of a 4.0% aqueous $\ce{NaCl}$ solution at 25°C.
2. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of $\ce{NaCl}$ solution; internal osmotic pressure of cell
Asked for: osmotic pressure of $\ce{NaCl}$ solution and concentration of glycerol needed
Strategy:
1. Calculate the molarity of the $\ce{NaCl}$ solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.
2. Use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution.
3. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{13.5.12} to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of $\ce{NaCl}$ per 100 g of solution. Using the formula mass of $\ce{NaCl}$ (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
\begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber
Because 1 mol of $\ce{NaCl}$ produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution:
\begin{align*} \Pi &=MRT \[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\[4pt] &=34 \;atm\end{align*} \nonumber
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure:
\begin{align*} M&=\dfrac{\Pi}{RT}\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the $\ce{NaCl}$ solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise $7$
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer
24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure $7$, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure $8$, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
A Video Discussing Osmotic Pressure. Video Link: Osmotic Pressure, YouTube(opens in new window) [youtu.be] (opens in new window)
Colligative Properties of Electrolyte Solutions
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of $\ce{NaCl}$ in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of $\ce{NaCl}$ and $\ce{KCl}$ are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.
The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor ($i$) and is defined as follows:
$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13}$
Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.
As the solute concentration increases, the van’t Hoff factor decreases.
The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in Table $2$ show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.
Table $2$: van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
$NaCl$ 1.9 2.0
$HCl$ 1.9 2.0
$MgCl_2$ 2.7 3.0
$FeCl_3$ 3.4 4.0
$Ca(NO_3)_2$ 2.5 3.0
$AlCl_3$ 3.2 4.0
$MgSO_4$ 1.4 2.0
Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure $9$). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $Mg^{2+}$, $Al^{3+}$, $SO_4^{2−}$, and $PO_4^{3−}$ have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.
Example $8$: Iron Chloride in Water
A 0.0500 M aqueous solution of $FeCl_3$ has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor $i$ for the solution.
Given: solute concentration, osmotic pressure, and temperature
Asked for: van’t Hoff factor
Strategy:
1. Use Equation $\ref{13.5.12}$ to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
2. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation $\ref{13.5.13}$ to calculate the van’t Hoff factor.
Solution:
A If $\ce{FeCl_3}$ dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be
\begin{align*} \Pi &=MRT \[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber
B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
$4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber$
or after rearranging
$M = 0.170 mol \nonumber$
The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. From Equation $\ref{13.5.13}$, the van’t Hoff factor for the solution is
$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber$
Exercise $8$: Magnesium Chloride in Water
Calculate the van’t Hoff factor for a 0.050 m aqueous solution of $MgCl_2$ that has a measured freezing point of −0.25°C.
Answer
2.7 (versus an ideal value of 3).
A Video Discussing the Colligative Properties in Solutions. Video Link: Colligative Properties in Solutions, YouTube(opens in new window) [youtu.be]
Summary
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation ($ΔT_b$) and freezing point depression ($ΔT_f$) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.
• Henry’s law: $C = kP \nonumber$
• Raoult’s law: $P_A=\chi_AP^0_A \nonumber$
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=\chi_BP^0_A \nonumber$
• vapor pressure of a system containing two volatile components: $P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \nonumber$
• boiling point elevation: $ΔT_b = mK_b \nonumber$
• freezing point depression: $ΔT_f = mK_f \nonumber$
• osmotic pressure: $\Pi=nRTV=MRT \nonumber$
• van ’t Hoff factor: $i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \nonumber$ | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.05%3A_Colligative_Properties_of_Solutions.txt |
Learning Objectives
• To distinguish between true solutions and solutions with aggregate particles.
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspensionA heterogeneous mixture of particles with diameters of about 1 µm that are distributed throughout a second phase and that separate from the dispersing phase on standing. is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloidA heterogeneous mixture of particles with diameters of about 2–500 nm that are distributed throughout a second phase and do not separate from the dispersing phase on standing. is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table 13.6.1.
Table 13.6.1 Properties of Liquid Solutions, Colloids, and Suspensions
Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples
solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water
colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese
suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint
Colloids and Suspensions
Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words solA dispersion of solid particles in a liquid or solid. and gelA semisolid sol in which all of the liquid phase has been absorbed by the solid particles. to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosolsA dispersion of solid or liquid particles in a gas., which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible.
Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effectThe phenomenon of scattering a beam of visible light.,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure 13.6.1
Figure 13.6.1 Tyndall Effect, the Scattering of Light by Colloids The glass on the left has dilute colloidal silver, the one on the right tap water. The red laser pointer beam is scattered by the Tyndall effect in the glass on the left.
Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic.
Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid.
In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine, Figure 9.7.1 5.16). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure 13.6.2). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin.
Figure 13.6.2 Sickle-Cell Anemia The characteristic shape of sickled red blood cells is the result of fibrous aggregation of hemoglobin molecules inside the cell.
Figure 13.6.3 Formation of New Land by the Destabilization of a Colloid Suspension This satellite photograph shows the Mississippi River delta from New Orleans (top) to the Gulf of Mexico (bottom). Where seawater mixes with freshwater from the Mississippi River, colloidal clay particles in the river water precipitate (tan area).
Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure 13.6.3. Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels.
Emulsions
EmulsionsA dispersion of one liquid phase in another liquid with which it is immiscible. are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”:
Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate [CH3(CH2)16CO2Na+], and detergents, such as sodium dodecyl sulfate [CH3(CH2)11OSO3Na+], whose structures are as follows:
When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise.
A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats.
Micelles
Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micellesA spherical or cylindrical aggregate of detergents or soaps in water that minimizes contact between the hydrophobic tails of the detergents or soaps and water., which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (part (a) in Figure 13.6.4).
A large class of biological molecules called phospholipidsA large class of biological, detergent-like molecules that have a hydrophilic head and two hydrophobic tails and that form bilayers. consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayersA two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail., extended sheets consisting of a double layer of molecules. As shown in part (b) in Figure 13.6.4, the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution.
A cell membraneA mixture of phospholipids that form a phospholipid bilayer around the cell. is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cellA collection of molecules, capable of reproducing itself, that is surrounded by a phospholipid bilayer. is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks.
Summary
A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves.
Key Takeaway
• Colloids and suspensions are mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures.
Conceptual Problems
1. How does a colloid differ from a suspension? Which has a greater effect on solvent properties, such as vapor pressure?
2. Is homogenized milk a colloid or a suspension? Is human plasma a colloid or a suspension? Justify your answers.
3. How would you separate the components of an emulsion of fat dispersed in an aqueous solution of sodium chloride?
• Anonymous
13.07: Aggregate Particles in Aqueous Solution
Learning Objectives
• To distinguish between true solutions and solutions with aggregate particles.
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table 13.9.
Table 13.9 Properties of Liquid Solutions, Colloids, and Suspensions
Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples
solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water
colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese
suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.06%3A_Aggregate_Particles.txt |
In Chapter 14, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time.
A smoggy sunset in Shenzhen, China. The reaction of \(O_2\) with \(N_2\) at high temperature in an internal combustion engine produces small amounts of \(NO\), which reacts with atmospheric \(O_2\) to form \(NO_2\), an important component of smog. The reddish-brown color of \(NO_2\) is responsible for the characteristic color of smog, as shown in this true-color photo.
We introduced the concept of equilibrium in Chapter 11, where you learned that a liquid and a vapor are in equilibrium when the number of molecules evaporating from the surface of the liquid per unit time is the same as the number of molecules condensing from the vapor phase. Vapor pressure is an example of a physical equilibrium because only the physical form of the substance changes. Similarly, in Chapter 13, we discussed saturated solutions, another example of a physical equilibrium, in which the rate of dissolution of a solute is the same as the rate at which it crystallizes from solution.
In this chapter, we describe the methods chemists use to quantitatively describe the composition of chemical systems at equilibrium, and we discuss how factors such as temperature and pressure influence the equilibrium composition. As you study these concepts, you will also learn how urban smog forms and how reaction conditions can be altered to produce \(H_2\) rather than the combustion products \(CO_2\) and \(H_2O\) from the methane in natural gas. You will discover how to control the composition of the gases emitted in automobile exhaust and how synthetic polymers such as the polyacrylonitrile used in sweaters and carpets are produced on an industrial scale.
14.02: Factors that Affect Reaction Rates
Learning Objectives
• To understand the factors that affect reaction rates.
Although a balanced chemical equation for a reaction describes the quantitative relationships between the amounts of reactants present and the amounts of products that can be formed, it gives us no information about whether or how fast a given reaction will occur. This information is obtained by studying the chemical kinetics of a reaction, which depend on various factors: reactant concentrations, temperature, physical states and surface areas of reactants, and solvent and catalyst properties if either are present. By studying the kinetics of a reaction, chemists gain insights into how to control reaction conditions to achieve a desired outcome.
Concentration Effects
Two substances cannot possibly react with each other unless their constituent particles (molecules, atoms, or ions) come into contact. If there is no contact, the reaction rate will be zero. Conversely, the more reactant particles that collide per unit time, the more often a reaction between them can occur. Consequently, the reaction rate usually increases as the concentration of the reactants increases. One example of this effect is the reaction of sucrose (table sugar) with sulfuric acid, which is shown in Figure 14.1.1.
Figure 14.1.1 The Effect of Concentration on Reaction Rates Mixing sucrose with dilute sulfuric acid in a beaker (a, right) produces a simple solution. Mixing the same amount of sucrose with concentrated sulfuric acid (a, left) results in a dramatic reaction (b) that eventually produces a column of black porous graphite (c) and an intense smell of burning sugar.
Temperature Effects
You learned in Chapter 10 that increasing the temperature of a system increases the average kinetic energy of its constituent particles. As the average kinetic energy increases, the particles move faster, so they collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature. Conversely, the reaction rate of virtually all reactions decreases with decreasing temperature. For example, refrigeration retards the rate of growth of bacteria in foods by decreasing the reaction rates of biochemical reactions that enable bacteria to reproduce. Figure 14.1.2 shows how temperature affects the light emitted by two chemiluminescent light sticks.
Figure 14.1.2 The Effect of Temperature on Reaction RatesAt high temperature, the reaction that produces light in a chemiluminescent light stick occurs more rapidly, producing more photons of light per unit time. Consequently, the light glows brighter in hot water (left) than in ice water (right).
In systems where more than one reaction is possible, the same reactants can produce different products under different reaction conditions. For example, in the presence of dilute sulfuric acid and at temperatures around 100°C, ethanol is converted to diethyl ether:
$2CH_{3}CH_{2}OH \overset{H_{2}SO_{4}}{\longrightarrow} CH_{3}CH_{2}OCH_{2}CH_{3} + H_{2}O \tag{14.1.1}$
At 180°C, however, a completely different reaction occurs, which produces ethylene as the major product:
$CH_{3}CH_{2}OH \overset{H_{2}SO_{4}}{\longrightarrow} CH_{2}CH_{2} + H_{2}O \tag{14.1.2}$
Phase and Surface Area Effects
When two reactants are in the same fluid phase, their particles collide more frequently than when one or both reactants are solids (or when they are in different fluids that do not mix). If the reactants are uniformly dispersed in a single homogeneous solution, then the number of collisions per unit time depends on concentration and temperature, as we have just seen. If the reaction is heterogeneous, however, the reactants are in two different phases, and collisions between the reactants can occur only at interfaces between phases. The number of collisions between reactants per unit time is substantially reduced relative to the homogeneous case, and, hence, so is the reaction rate. The reaction rate of a heterogeneous reaction depends on the surface area of the more condensed phase.
Automobile engines use surface area effects to increase reaction rates. Gasoline is injected into each cylinder, where it combusts on ignition by a spark from the spark plug. The gasoline is injected in the form of microscopic droplets because in that form it has a much larger surface area and can burn much more rapidly than if it were fed into the cylinder as a stream. Similarly, a pile of finely divided flour burns slowly (or not at all), but spraying finely divided flour into a flame produces a vigorous reaction (Figure 14.1.3 ). Similar phenomena are partially responsible for dust explosions that occasionally destroy grain elevators or coal mines.
< | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.01%3A_Preview_to_Chemical_Equilibria.txt |
Learning Objectives
• To determine the reaction rate.
• To understand the meaning of a rate law.
The factors discussed in Section 14.1 affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate.
Reaction Rates
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.
The progress of a simple reaction (A → B) is shown in Figure 14.2.1 where the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure 14.2.1. The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
$rate=\dfrac{\Delta \left [ B \right ]}{\Delta t}=-\dfrac{\Delta \left [ A \right ]}{\Delta t} \tag{14.2.1}$
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate we would calculate for the reaction A → B using Equation 14.2.1 would be different for each interval. (This is not true for every reaction, as you will see later.) A much greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, which means that the reaction rate is fastest at first. This is consistent with the concentration effects described in Section 14.1because the concentration of A is greatest at the beginning of the reaction.
Note the Pattern
Reaction rates generally decrease with time as reactant concentrations decrease.
Determining the Reaction Rate of Hydrolysis of Aspirin
We can use Equation 14.2.1 to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world. (More than 25,000,000 kg are produced annually worldwide.) Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid.
Figure 14.2.2 Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table 14.2.1 and are shown in the graph in Figure 14.2.3. These data were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).
Table 14.2.1 Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
We can calculate the average reaction rateThe reaction rate calculated for a given time interval from the concentrations of either the reactant or one of the products at the beginning of the interval time
$rate_{\left ( t=0-2.0\;h \right )}=\dfrac{\left [ salicylic\;acid \right ]_{2}-\left [ salicylic\;acid \right ]_{0}}{2.0\;h-0.0\;h}$
$=\dfrac{0.040\times 10^{-3}\;M-0.000\;M}{2.0\;h-0.0\;h}=2\times 10^{-5}\;M/h$
We can also calculate the reaction rate from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:
$rate_{\left ( t=0-2.0\;h \right )}=\dfrac{\left [ aspirin \right ]_{2}-\left [ aspirin \right ]_{0}}{2.0\;h-0.0\;h}$
$rate_{\left ( t=0-2.0\;h \right )}=\dfrac{5.51\times 10^{-3}\;M-5.55\times 10^{-3}\;M}{2.0\;h-0.0\;h}=2.0\times 10^{-5}\;M/h$\;h} \)
If we now calculate the reaction rate during the last interval given in Table 14.2.1 (the interval between 200 h and 300 h after the start of the reaction), we find that the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
$rate_{\left ( t=200-300\;h \right )}=\dfrac{\left [ salicylic\;acid \right ]_{300}-\left [ salicylic\;acid \right ]_{200}}{300\;h-200\;h}$
$=\dfrac{3.73\times 10^{-3}\;M-2.91\times 10^{-3}\;M}{100;h}=8.2\times 10^{-6}\;M/h$
(You should verify from the data in Table 14.2.1 that you get the same rate using the concentrations of aspirin measured at 200 h and 300 h.)
Calculating the Reaction Rate of Fermentation of Sucrose
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Let us look at a reaction in which the coefficients are not all the same: the fermentation of sucrose to ethanol and carbon dioxide, which we encountered in Chapter 8
$\underset{sucrose}{C_{12}H_{22}O_{11}\left ( aq \right )}+H_{2}O\left ( l \right )\rightarrow 4C_{2}H_{5}OH \left ( aq \right ) + CO_{2}\left ( g \right ) \tag{14.2.2}$
The coefficients show us that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, we can find the reaction rate by looking at the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as you learned in Chapter 10, the volume of CO2 gas formed will depend on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
$rate=\dfrac{\Delta \left [ C_{2}H_{5}OH \right ]}{\Delta t}=-\dfrac{4 \Delta \left [ sucrose ] \right )}{\Delta t} \tag{14.2.3}$
The concentration of the reactant—in this case sucrose—decreases with increasing time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation 14.2.3 so that the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with increasing time, so its rate of change is automatically expressed as a positive value.
Often the reaction rate is expressed in terms of the reactant or product that has the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation 14.2.2) corresponds to sucrose, so the reaction rate is generally defined as follows:
$rate=-\dfrac{\Delta \left [ sucrose \right ]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta \left [ C_{2}H_{5}OH \right ]}{\Delta t} \right ) \tag{14.2.4}$
Example 14.2.1
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$2N_{2}O_{5}\left ( g \right )\overset{\Delta }{\rightarrow} 4NO_{2}\left ( g \right )+O_{2}\left ( g \right )$
Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.
Given: balanced chemical equation
Asked for: reaction rate expressions
Strategy:
A Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
B For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.
Solution:
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, we define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B We know from the balanced chemical equation that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that we divide the rate of change of [N2O5] and [NO2] by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, we must divide the rate of production of NO2 by 4. The reaction rate expressions are as follows:
$rate=\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = \dfrac{\Delta \left [ NO_{2} \right ]}{4\Delta t}= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t}$
Exercise
The key step in the industrial production of sulfuric acid is the reaction of SO2 with O2 to produce SO3.
$2SO_{2}\left ( g \right )+ O_{2}\left ( g \right ) \rightarrow 2SO_{3}\left ( g \right )$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$rate=-\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = -\dfrac{\Delta \left [ SO_{2} \right ]}{\Delta t}=\dfrac{\Delta \left [ SO_{3} \right ]}{2\Delta t}$
Example 14.2.2
Using the reaction shown in Example 1, calculate the reaction rate from the following data taken at 56°C:
2N2O5(g) → 4NO2(g) + O2(g)
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175
Given: balanced chemical equation and concentrations at specific times
Asked for: reaction rate
Strategy:
A Using the equations in Example 1, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.
B Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution:
A We are asked to calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example 1, we see that we can evaluate the reaction rate using any of three expressions:
$rate=\dfrac{\Delta \left [ O_{2} \right ]}{\Delta t} = \dfrac{\Delta \left [ NO_{2}\right ]}{4\Delta t}= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t}$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$rate= -\dfrac{\Delta \left [ N_{2}O_{5} \right ]}{2\Delta t} =-\dfrac{\left [ N_{2}O_{5} \right ]_{600}-\left [ N_{2}O_{5} \right ]_{240}}{2\left ( 600\;s-240\;s \right )}$
B Substituting actual values into the expression,
$rate=-\dfrac{0.197\;M-0.0388\;M}{2\left (360 \;s \right )}$
Similarly, we can use NO2 to calculate the reaction rate:
$rate= -\dfrac{\Delta \left [ NO_{2} \right] }{4\Delta t} =-\dfrac{\left [ NO_{2} \right ]_{600}-\left [ NO_{2} \right ]_{240}}{4\left ( 600\;s-240\;s \right )} =\dfrac{0.0699\;M-0.0314\;M}{4\left ( 360 \right )}=2.67\times 10^{-5}\;M/s$
If we allow for experimental error, this is the same rate we obtained using the data for N2O5, as it should be because the reaction rate should be the same no matter which concentration is used. We can also use the data for O2:
$rate= -\dfrac{\Delta \left [ O_{2} \right] }{\Delta t} =-\dfrac{\left [ O_{2} \right ]_{600}-\left [ O_{2} \right ]_{240}}{\left ( 600\;s-240\;s \right )} =\dfrac{0.0175\;M-0.00792\;M}{360\;s}=2.66\times 10^{-5}\;M/s$
Again, this is the same value we obtained from the N2O5 and NO2 data. Thus the reaction rate does not depend on which reactant or product is used to measure it.
Exercise
Using the data in the following table, calculate the reaction rate of SO2(g) with O2(g) to give SO3(g).
2SO2(g) + O2(g) → 2SO3(g)
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148
Answer: 9.0 × 10−6 M/s
Instantaneous Rates of Reaction
So far, we have determined average reaction rates over particular intervals of time. We can also determine the instantaneous rateThe reaction rate of a chemical reaction at any given point in time. of a reaction, which is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate.If you have studied calculus, you may recognize that the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.
Think of the distinction between the instantaneous and average rates of a reaction as being similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although you may travel for a long time at 65 mph on an interstate highway during a long trip, there may be times when you travel only 25 mph in construction zones or 0 mph if you stop for meals or gas. Thus your average speed on the trip may be only 50 mph, whereas your instantaneous speed on the interstate at a given moment may be 65 mph. Whether you are able to stop the car in time to avoid an accident depends on your instantaneous speed, not your average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval normally has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
In chemical kinetics, we generally focus on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.
Rate Laws
In Section 14.1, you learned that reaction rates generally decrease with time because reactant concentrations decrease as reactants are converted to products. You also learned that reaction rates generally increase when reactant concentrations are increased. We now examine the mathematical expressions called rate lawsMathematical expressions that describe the relationships between reactant rates and reactant concentrations in a chemical reaction., which describe the relationships between reactant rates and reactant concentrations. Rate laws are laws as defined in Chapter 1; that is, they are mathematical descriptions of experimentally verifiable data.
Rate laws may be written from either of two different but related perspectives. A differential rate lawA rate law that expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt) expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate lawA rate law that expresses the reaction rate in terms of the initial concentration [R]0 and the measured concentration of one or more reactants ([R]) after a given amount of time (t) describes the reaction rate in terms of the initial concentration ([R]0) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); we will discuss integrated rate laws in Section 14.3. The integrated rate law can be found by using calculus to integrate the differential rate law, although the method of doing so is beyond the scope of this text. Whether you use a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s).
Reaction Orders
For a reaction with the general equation
$aA+bB\rightarrow cC+dD \tag{14.2.5}$
the experimentally determined rate law usually has the following form:
$rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.6}$
The proportionality constant (k) is called the rate constantA proportionality constant whose value is characteristic of the reaction and the reaction conditions and whose numerical value does not change as the reaction progresses under a given set of conditions., and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular value of the rate constant under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions.
Thus the reaction rate depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction orderNumbers that indicate the degree to which the reaction rate depends on the concentration of each reactant., the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation 14.2.6 tells us that Equation 14.2.5 is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n.
Note the Pattern
Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.
Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (We present general forms for integrated rate laws in Section 14.3.) To illustrate how chemists interpret a differential rate law, we turn to the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone. This reaction produces t-butanol according to the following equation:
$\left ( CH_{3} \right )_{3}CBr \left ( soln \right )+H_{2}O\left ( soln \right )\rightarrow \left ( CH_{3} \right )_{3}COH \left( soln \right )+HBr\left ( soln \right ) \tag{14.7.7}$
Combining the rate expression in Equation 14.4 and Equation 14.9 gives us a general expression for the differential rate law:
$rate=-\dfrac{\Delta \left [A \right ]}{\Delta t}= k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.8}$
Inserting the identities of the reactants into Equation 14.2.8 gives the following expression for the differential rate law for the reaction:
$rate=-\dfrac{ \Delta \left [\left (CH_{3} \right )_{3}CBr \right ]}{\Delta t}= k\left [ \left (CH_{3} \right )_{3}CBr \right ]^{m}\left [ H_{2}O \right ]^{n} \tag{14.2.9}$
Experiments done to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of (CH3)3CBr but is independent of the concentration of water. Thus m and n in Equation 14.12 are 1 and 0, respectively, and
$rate= k\left [ \left (CH_{3} \right )_{3}CBr \right ]^{1}\left [ H_{2}O \right ]^{0}=k\left [ \left (CH_{3} \right )_{3}CBr \right ] \tag{14.2.10}$
Because the exponent for the reactant is 1, the reaction is first order in (CH3)3CBr. It is zeroth order in water because the exponent for [H2O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus the overall reaction order is 1 + 0 = 1. What the reaction orders tell us in practical terms is that doubling the concentration of (CH3)3CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH3)3CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when you work with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction.
Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH3Br) is as follows:
$rate=-\dfrac{ \Delta \left [CH_{3} Br \right ]}{\Delta t}= k'\left [ CH_{3} Br \right ] \tag{14.2.11}$
This reaction also has an overall reaction order of 1, but the rate constant in Equation 14.2.11 is approximately 106 times smaller than that for t-butyl bromide. Thus methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level.
Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often change reaction conditions to obtain clues about what is occurring during a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. For methyl bromide, in contrast, the differential rate law becomes rate = k″[CH3Br][OH], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, which again provides clues as to how the reactions differ on a molecular level.
Note the Pattern
Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements.
Example 14.2.3
We present three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled.
1. $2HI\left ( g \right )\overset{Pt}{\rightarrow} H_{2}\left ( g \right )+I_{2}\left ( g \right )$$rate= -\dfrac{1}{2} \dfrac{\Delta \left [HI \right ]}{\Delta t}=k\left [ HI \right ]^{2}$
2. $2N_{2}O\left ( g \right )\overset{\Delta}{\rightarrow} 2N_{2}\left ( g \right )+O_{2}\left ( g \right )$
$rate= \dfrac{1}{2} \dfrac{\Delta \left [ N_{2}O \right ]}{\Delta t}=k$
3. $cyclopropane\left ( g \right )\overset{\Delta}{\rightarrow} propane\left ( g \right )$
$rate= \dfrac{1}{2} \dfrac{\Delta \left [ cyclopropance \right ]}{\Delta t}=k\left [ cyclopropane \right ]$
Given: balanced chemical equations and differential rate laws
Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration
Strategy:
A Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant.
B Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Sum all exponents to obtain the overall reaction order.
C Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate.
Solution:
1. A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]: $\dfrac{M}{s} = kM^{2} \Rightarrow k = \dfrac{M/s}{M^{2}} = \dfrac{1}{M\cdot s} =M^{-1}s^{-1}$
B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall.
C If the concentration of HI is doubled, the reaction rate will increase from k[HI]02 to k(2[HI])02 = 4k[HI]02. The reaction rate will therefore quadruple.
2. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units.
B The rate law tells us that the reaction rate is constant and independent of the N2O concentration. That is, the reaction is zeroth order in N2O and zeroth order overall.
C Because the reaction rate is independent of the N2O concentration, doubling the concentration will have no effect on the reaction rate.
3. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s.
B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.
C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane]0 to 2k[cyclopropane]0. This doubles the reaction rate.
Exercise
Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.
1. $CH_{3}N=CH_{3}N\left ( g \right )\rightarrow C_{2}H_{6}\left ( g \right )+N_{2}\left ( g \right )$
$rate=-\dfrac{\Delta \left [CH_{3}N=CH_{3}N \right]}{\Delta t}=k \left [ CH_{3}N=CH_{3}N \right ]$
2. $2NO_{2}\left ( g \right )+F_{2}\left ( g \right ) \rightarrow 2NO_{2}F\left ( g \right )$
$rate=-\dfrac{\Delta \left [F_{2} \right]}{\Delta t}=-\dfrac{1}{2}\dfrac{\Delta \left [NO_{2} \right]}{\Delta t}=k \left [ NO_{2} \right ]\left [ F_{2} \right ]$
Answer
1. s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
2. M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.
Summary
Reaction rates are reported either as the average rate over a period of time or as the instantaneous rate at a single time.
The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time.
The rate constant (k) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The power to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant.
Key Takeaways
• Reaction rates can be determined over particular time intervals or at a given point in time.
• A rate law describes the relationship between reactant rates and reactant concentrations.
Key Equations
general definition of rate for A → B
Equation 14.2.1: $rate=\dfrac{\Delta \left [ B \right ]}{\Delta t}=-\dfrac{\Delta \left [ A \right ]}{\Delta t}$
general form of rate law when A and B are reactants
Equation 14.2.6: $rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.2.6}$
Conceptual Problems
1. Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction?
2. Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why?
3. Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 1010 L/(mol·s). Would you expect the reactions to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H3O+ + OH → 2H2O and H3O+ + N(CH3)3 → H2O + HN(CH3)3+ in aqueous solution. Which would have the higher rate constant? Why?
4. What information can you get from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation?
5. During the hydrolysis reaction A + H2O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How do you expect this effect to be reflected in the overall reaction order?
Answers
1. Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive.
2. Faster in a less viscous solvent because the rate of diffusion is higher; the H3O+/OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants.
Numerical Problems
1. The reaction rate of a particular reaction in which A and B react to make C is as follows:
$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=\dfrac{1}{2}\dfrac{\Delta \left [ C \right ]}{\Delta t}$
Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C?
2. While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min?
3. Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate.
Time (s) [A] (M)
120 0.158
240 0.089
360 0.062
4. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation.
1. C2H5I → C2H4 + HI: rate = k[C2H5I]
2. SO + O2 → SO2 + O: rate = k[SO][O2]
3. 2CH3 → C2H6: rate = k[CH3]2
4. ClOO → Cl + O2: rate = k
5. Cleavage of C2H6 to produce two CH3· radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with k = 5.46 × 10−4 s−1. How long will it take for the reaction to go to 15% completion? to 50% completion?
6. Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere.
$N_{2}^{+}+O_{2}\overset{k_{1}}{\rightarrow} N_{2}+O_{2}^{+}$
$O_{2}^{+}+O\overset{k_{2}}{\rightarrow} O_{2}+O^{+}$
$O^{+}+N_{2}\overset{k_{3}}{\rightarrow} N+NO^{+}$
Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = k[N2+][O2], which one of the steps is rate limiting?
7. The oxidation of aqueous iodide by arsenic acid to give I3 and arsenous acid proceeds via the following reaction:
$H_{3}AsO_{4}\left ( aq \right )+3I^{-}\left ( aq \right )+2H^{+}\left ( aq \right ) \overset{k_{f}}{\rightleftharpoons} H_{3}AsO_{3}\left ( aq \right )+I_{3}^{-}\left ( aq \right ) +H_{2}O\left ( l \right )$
Write an expression for the initial rate of decrease of [I3], Δ[I3]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: kf/kr = [H3AsO3][I3]/[H3AsO4][I]3[H+]2. Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions?
Answer
1. 298 s; 1270 s
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.03%3A_Reaction_Rates_and_Rate_Laws.txt |
Learning Objectives
• To know how to determine the reaction order from experimental data.
In the examples in this text, the exponents in the rate law are almost always the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reactionA reaction whose rate is independent of concentration. is one whose rate is independent of concentration; its differential rate law is rate = k. We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-k\left [ reactant \right ]^{0}=k(1)=k \tag{14.3.1}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of k, a positive value.
The graph of a zeroth-order reaction. The change in concentration of reactant and product with time produces a straight line.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$\left [ A \right ]=\left [ A_{0} \right ]-kt \tag{14.3.2}$
where [A]0 is the initial concentration of reactant A. (Equation 14.3.2 has the form of the algebraic equation for a straight line, y = mx + b, with y = [A], mx = −kt, and b = [A]0.) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$2N_{2}O \left ( g \right )\overset{Pt}{\rightarrow} 2N_{2} \left ( g \right ) + O_{2} \left ( g \right ) \tag{14.3.3}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate.At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$rate =-\dfrac{1}{2}\left (\dfrac{\Delta \left [N_{2}O \right ]}{\Delta t} \right )=\dfrac{1}{2}\left ( \dfrac{\Delta \left [N_{2} \right ]}{\Delta t} \right )=\dfrac{\Delta \left [O_{2} \right ]}{\Delta t} k\left [N_{2}O \right ]^{0}=k \tag{14.3.4}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure 14.3.2, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
Figure 14.3.2 A Zeroth-Order Reaction This graph shows the concentrations of reactants and products versus time for the zeroth-order catalyzed decomposition of N2O to N2 and O2 on a Pt surface. The change in the concentrations of all species with time is linear.
Note the Pattern
If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzymeA catalyst that occurs naturally in living organisms and catalyzes biological reactions. alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
CH3COH(aq) + NH4+(aq)." href="/@api/deki/files/48697/6d8d2a62963267afdb3d7e7303345615.jpg"> CH3COH(aq) + NH4+(aq)." style="width: 550px; height: 70px;" width="550px" height="70px" src="/@api/deki/files/48697/6d8d2a62963267afdb3d7e7303345615.jpg">
where NAD+ (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in Figure 14.3.3). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure 14.3.3).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reactionA reaction whose rate is directly proportional to the concentration of one reactant., the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right ]^{1}=k\left [ A \right ] \tag{14.3.5}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$rate= \left [ A \right ]= \left [ A_{0} \right ] e^{-kt} \tag{14.3.6}$
where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. (Essential Skills 6 in Section 11.9, discusses natural logarithms.) Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation 14.20 predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation 14.20 and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:
$rate= ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt \tag{14.3.7}$
Because Equation 14.3.7 has the form of the algebraic equation for a straight line, y = mx + b, with y = ln[A] and b = ln[A]0, a plot of ln[A] versus t for a first-order reaction should give a straight line with a slope of −k and an intercept of ln[A]0. Either the differential rate law (Equation 14.3.5) or the integrated rate law (Equation 14.3.7) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin (Figure 14.2.1) and the reaction of t-butyl bromide with water to give t-butanol (Equation 14.7.7). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure 14.11 is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure 14.11 have been studied extensively to find ways of maximizing the concentration of the active species.
Note the Pattern
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table 14.3.1. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table 14.3.1 shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table 14.3.1. For example, substituting the values for Experiment 3 into Equation 14.19,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Table 14.3.1 Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example 14.3.1
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$CH_{3}CH_{2}Cl\left ( g \right )\overset{\Delta }{\rightarrow} HCl\left ( g \right )+C_{2}H_{4}\left ( g \right )$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
A Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
B Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution:
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M) 1.6 × 10−6 s−1 = k
Exercise
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
SO2Cl2(g) → SO2(g) + Cl2(g)
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer: first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in Figure 14.3.5 shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure 14.3.5. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure 14.3.5 for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
$slope = \dfrac{ln\left [ cisplatin \right ]_{1000}- ln \left [ cisplatin \right ]_{100}}{1000\;min-100\;min}$
$-k = \dfrac{ln 0.0022-ln 0.0086}{1000\;min-100\;min}=\dfrac{-6.12-\left ( -4.76 \right )}{900\; min}=-1.51\times 10^{-3} min^{-1}$
$k = 1.51\times 10^{-3} \; min^{-1}$
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure 14.3.5 are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Example 14.3.2
Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.)
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
A Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
B Given a concentration [A], solve the integrated rate law for time t.
Solution:
The exponential form of the integrated rate law for a first-order reaction (Equation 14.3.6) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having calculated the rate constant in Example 4 (k = 1.6 × 10−6 s−1), we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
$\left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=\left [ CH_{3} CH_{2}Cl \right ]_{0}e^{-kt}$
$=0.0200\;M \; exp \left [ \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right ) \right ) ]$
$= 0.0189\;M We could also have used the logarithmic form of the integrated rate law (Equation 14.3.7): \( ln \left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt$
$= ln 0.0200\;M - \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right )$
$= -3.912--0.0576=-3.970$
$\left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=e^{-3.970}\;M$
=0.0189\;M
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for t. Equation 14.3.7 gives the following:
$ln \left [ CH_{3} CH_{2}Cl \right ]_{t}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt$
$kt= ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -ln \left [ CH_{3} CH_{2}Cl \right ]_{t}= ln \; \dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{ \left [ CH_{3} CH_{2}Cl \right ]_{t}}$
$t= \dfrac{1}{k}\; ln\dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{\left [ CH_{3} CH_{2}Cl \right ]_{t}} = \dfrac{1}{1.6\times 10^{-6}\;s^{-1}}\; ln\dfrac{0.0200\;M}{0.0050\;M}$
$t= \dfrac{ln\;4.0}{k}\; =8.7\times 10^{5}\;s=240\;h=2.4\times 10^{2}\;h$
Exercise
In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO2Cl2) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO2Cl2 that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO2Cl2 to decompose?
Answer: 0.0252 M; 29 h
Second-Order Reactions
The simplest kind of second-order reactionA reaction whose rate is proportional to the square of the concentration of the reactant (for a reaction with the general form 2A → products) or is proportional to the product of the concentrations of two reactants (for a reaction with the general form A + B → products). is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$rate=-\dfrac{\Delta \left [ A \right ]}{2 \Delta t}=k\left [ reactant \right ]^{2} \tag{14.3.8}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]} +kt \tag{14.3.9}$
Because Equation 14.3.9 has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Note the Pattern
Second-order reactions generally have the form 2A → products or A + B → products.
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively.The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation 14.3.8) or the integrated rate law (Equation 14.3.9).
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table 14.3.2 . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times 10^{-5}\;M/min }{1.8\times 10^{-5}\;M/mi}=2.8 \;\;and\;\;\dfrac{3.4\times 10^{-3}\;M/min }{2.0\times 10^{-3}\;M/mi}=1.7$
Table 14.3.2 Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
$rate \;\;\alpha\;\;\left [ monomer \right ]^{2}$
This means that the reaction is second order in the monomer. Using Equation 14.3.8 and the data from any row in Table 14.3.2, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
$rate = k\left [ A^{2} \right ]$
$8.0\times 10^{-5}\;M/min = k\left (4.4\times 10^{-3}\;M \right )$
$4.1\;min^{-1} = k$
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in Figure 14.3.6. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in part (b) in Figure 14.3.6. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in Figure 14.4.2 in Section 14.4.
Example 14.3.3
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$2NO_{2}\left ( g \right )\overset{\Delta }{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right )$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
A From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
B Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution:
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
$rate = k\left [ NO_{2} \right ]^{2}$
$5.40\times 10^{-5}\; M/s = k\left (0.010\;M \right )^{2}$
$0.540\; M^{-1}s^{-1} = k$
Exercise
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
2HO2(g) → H2O2(g) + O2(g)
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
Answer: second order in HO2; k = 1.4 × 109 M−1·s−1
Note the Pattern
If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.
Example 14.3.4
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation 14.3.9) and the rate constant calculated in Example 6.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
A Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
B Setting [A] equal to 1/10 of [A]0, use the same equation to solve for t.
Solution:
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation 14.3.9,
$\dfrac{1}{\left [NO_{2} \right ]_{3600}} = \dfrac{1}{\left [NO_{2} \right ]_{0}} +kt=\dfrac{1}{0.056\;M} +\left ( 0.54\;M^{-1}s^{-1} \right )\left (3600\; s \right )$
$=2.0\times 10^{-3} \; M^{-1}$
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation 14.3.9 for t, using the concentrations given.
$\dfrac{\dfrac{1}{\left [NO_{2} \right ]_{3600}} - \dfrac{1}{\left [NO_{2} \right ]_{0}}}{k} =\dfrac{\left ( 1/0.0056\;M \right )-\left ( 1/0.056\;M \right )}{0.54\;M^{-1}s^{-1}}=3.0\times10^{2}\;s=5\;min$
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise
In the exercise in Example 6, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation 14.3.9) and the rate constant calculated in the exercise in Example 6.
Answer: 2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form A + B → products, in which the reaction is first order in A and first order in B. The differential rate law for this reaction is as follows:
$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-\dfrac{\Delta \left [ B \right ]}{\Delta t}=k\left [ A \right ]\left [ B \right ] \tag{14.3.10}$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. We presented one example at the end of Section 14.2, the reaction of CH3Br with OH to produce CH3OH.
Determining the Rate Law of a Reaction
The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanismsThe sequence of events that occur at the molecular level during a reaction. can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction A + B → products, for example, we need to determine k and the exponents m and n in the following equation:
$rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.3.11}$
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments.
Rate data for a hypothetical reaction of the type A + B → products are given in Table 14.3.3. The general rate law for the reaction is given in Equation 14.3.11. We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in Table 14.3.3.
Table 14.3.3 Rate Data for a Hypothetical Reaction of the Form A + B → Products
Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
$\dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{3} \right ]^{m}\left [ B^{3} \right ]^{n}}$
Inserting the appropriate values from Table 14.3.3,
$\dfrac{8.5\times10^{-3}\;\cancel{M/min}}{34.\times10^{-3}\;\cancel{M/min}}= \dfrac{k\left [ 0.50\;\cancel{M} \right ]^{m}\cancel{\left [ 05.0\;M \right ]^{n}}}{k\left [ 1.00\;\cancel{M} \right ]^{m}\cancel{\left [ 0.50\;M \right ]^{n}}}$
Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.
$\dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{5} \right ]^{m}\left [ B^{5} \right ]^{n}}$
Substituting the appropriate values from Table 14.3.3,
$\dfrac{8.5\times10^{-3}\;\cancel{M/min}}{8.5\times10^{-3}\;\cancel{M/min}}= \dfrac{\cancel{k\left [ 0.50\;\cancel{M} \right ]^{m}}\left [ 05.0\;\cancel{M} \right ]^{n}}{\cancel{k\left [ 0.50\;M \right ]^{m}}\left [ 1.00\;\cancel{M} \right ]^{n}}$
Canceling leaves 1.0 = [0.50]n, which gives n = 0; that is, the reaction is zeroth order in B. The experimentally determined rate law is therefore
$rate=k\left [ A \right ]^{2}\left [ B \right ]^{0}$
We can now calculate the rate constant by inserting the data from any row of Table 14.3.3into the experimentally determined rate law and solving for k. Using Experiment 2, we obtain
$19.\times10^-3\; M/min=k\left ( 0.75\;M \right )^{2}$
$3.4\times10^-2\; M^{-1}min^{-1}=k$
You should verify that using data from any other row of Table 14.3.3 gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
Example 14.3.5
Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with O2 to give NO2, which then reacts rapidly with water to give nitrous acid and nitric acid:
These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:
2NO(g) + O2(g) → 2NO2(g)
Determine the rate law for the reaction and calculate the rate constant.
Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: rate law and rate constant
Strategy:
A Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.
B Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.
Solution:
A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:
$rate =k\left [ NO \right ]^{2}\left [ O_{2} \right ]$
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$k=\dfrac{k}{\left [ NO \right ]^{2}\left [ O_{2} \right ]} = \dfrac{7.98\times10{-3}\;M/s}{\left ( 0.0235\;M \right )^{2}\left ( 0.0125\;M \right )}=1.16\times10{3}\;M^{-2}s^{-1}$
The overall reaction order (m + n) is 3, so this is a third-order reaction, a reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise
The peroxydisulfate ion (S2O82−) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$S_{2}O_{8}^{2-}\left ( aq \right )+3I^{-}\left ( aq \right ) \rightarrow 2SO_{4}^{2-}\left ( aq \right )+I_{3}^{-}\left ( aq \right )$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
Experiment [S2O82−]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76
Answer: rate = k[S2O82−][I]; k = 20 M−1·s−1
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
Key Takeaway
• Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data.
Key Equations
zeroth-order reaction
Equation 14.3.1 $rate = -\dfrac{\Delta \left [ A \right ] }{\Delta t} = k$
Equation 14.3.2: $\left [ A \right ]=\left [ A_{0} \right ]-kt$
first-order reaction
Equation 14.3.5: $rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right]$
Equation 14.3.6: $\left [ A \right ]= \left [ A_{0} \right ] e^{-kt}$
Equation 14.3.7: $ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt$
second-order reaction
Equation 14.3.8: $ln \left [ A \right ]= ln \left [ A_{0} \right ] +kt$
Equation 14.3.9: $\dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]}+kt$
Conceptual Problems
1. What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?
2. Predict whether the following reactions are zeroth order and explain your reasoning.
1. a substitution reaction of an alcohol with HCl to form an alkyl halide and water
2. catalytic hydrogenation of an alkene
3. hydrolysis of an alkyl halide to an alcohol
4. enzymatic conversion of nitrate to nitrite in a soil bacterium
3. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form?
4. If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order?
5. The reaction of NO with O2 is found to be second order with respect to NO and first order with respect to O2. What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate?
Numerical Problems
1. Iodide reduces Fe(III) according to the following reaction:
2Fe3+(soln) + 2I(soln) → 2Fe2+(soln) + I2(soln)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?
2. Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:
Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)
1 1.00 2.22 × 10−4
2 0.70 1.64 × 10−4
3 0.50 1.12 × 10−4
4 0.25 0.59 × 10−4
What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?
3. 1-Bromopropane is a colorless liquid that reacts with S2O32− according to the following reaction:
C3H7Br + S2O32− → C3H7S2O3 + Br
The reaction is first order in 1-bromopropane and first order in S2O32−, with a rate constant of 8.05 × 10−4 M−1·s−1. If you began a reaction with 40 mmol/100 mL of C3H7Br and an equivalent concentration of S2O32−, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?
4. The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δt = k[A]2[B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally.
Answers
1. First order in Fe3+; second order in I; third order overall; rate = k[Fe3+][I]2.
2. 1.29 × 10−4 M/s; 3.22 × 10−5 M/s
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.04%3A_Methods_of_Determining_Reaction_Order.txt |
Learning Objectives
• To use graphs to analyze the kinetics of a reaction.
In Section 14.3, you learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.
We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction:
$2NO_{2}\left ( g \right ) \overset{\Delta}{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right ) \tag{14.4.1}$
Experimental data for this reaction at 330°C are listed in Table 14.4.1; they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO2 are plotted versus time in part (a) in Figure 14.4.1. Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in Figure 14.4.1) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in Figure 14.4.1). This plot is a straight line, indicating that the reaction is second order in NO2.
Table 14.4.1 Concentration of NO2 as a Function of Time at 330°C
Time (s) [NO2] (M) ln[NO2] 1/[NO2] (M−1)
0 1.00 × 10−2 −4.605 100
60 6.83 × 10−3 −4.986 146
120 5.18 × 10−3 −5.263 193
180 4.18 × 10−3 −5.477 239
240 3.50 × 10−3 −5.655 286
300 3.01 × 10−3 −5.806 332
360 2.64 × 10−3 −5.937 379
We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure 14.4.2, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in Section 14.3 required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.
Example 14.4.1
Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:
$2N_{2}O_{5}\left ( soln \right ) \rightarrow 4NO_{2}\left ( soln \right )+O_{2}\left ( g \right )$
This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.
Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1)
0 0.0365 −3.310 27.4
600 0.0274 −3.597 36.5
1200 0.0206 −3.882 48.5
1800 0.0157 −4.154 63.7
2400 0.0117 −4.448 85.5
3000 0.00860 −4.756 116
3600 0.00640 −5.051 156
Given: balanced chemical equation, reaction times, and concentrations
Asked for: graph of data, rate law, and rate constant
Strategy:
A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure 14.4.2 to determine the reaction order.
B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.
Solution:
A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t:
The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5].
B The rate law for the reaction is therefore
$rate = k \left [N_{2}O_{5} \right ]$
Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,
$slope= \dfrac{ln\left [N_{2}O_{5} \right ]_{3000}-ln\left [N_{2}O_{5} \right ]_{0}}{3000\;s-0\;s} = \dfrac{\left [-4.756 \right ]-\left [-3.310 \right ]}{3000\;s} =4.820\times 10^{-4}\;s^{-1}$
Thus k = 4.820 × 10−4 s−1.
Exercise
1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction.
Time (s) [C4H6] (M) ln[C4H6] 1/[C4H6] (M−1)
0 1.72 × 10−2 −4.063 58.1
900 1.43 × 10−2 −4.247 69.9
1800 1.23 × 10−2 −4.398 81.3
3600 9.52 × 10−3 −4.654 105
6000 7.30 × 10−3 −4.920 137
Answer
second order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1
Summary
For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
Key Takeaway
• Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.
Conceptual Problems
1. Compare first-order differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law?
1. the magnitude of the rate constant
2. the information needed to determine the order
3. the shape of the graphs
2. In the single-step, second-order reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction A + B → products? Explain.
3. For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why?
Answers
1. For a given reaction under particular conditions, the magnitude of the first-order rate constant does not depend on whether a differential rate law or an integrated rate law is used.
2. The differential rate law requires multiple experiments to determine reactant order; the integrated rate law needs only one experiment.
3. Using the differential rate law, a graph of concentration versus time is a curve with a slope that becomes less negative with time, whereas for the integrated rate law, a graph of ln[reactant] versus time gives a straight line with slope = −k. The integrated rate law allows you to calculate the concentration of a reactant at any time during the reaction; the differential rate law does not.
1. The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent.
Numerical Problems
1. One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zeroth-order reaction:
Relative [A] (M) Relative Rate (M/s)
1 1
2 1
3 1
Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth-, first- and second-order reactions. What does the slope of each line represent?
2. The table below follows the decomposition of N2O5 gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?
Time (s) Pressure (mmHg)
0 348
400 276
1600 156
3200 69
4800 33
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.05%3A_Using_Graphs_to_Determine_Rate_Laws_Rate_Constants_and_Reaction_Orders.txt |
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions.
Half-Lives
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-lifeThe period of time it takes for the concentration of a reactant to decrease to one-half its initial value. of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction (Equation 14.3.6) to produce the following equation:
$ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]} = kt \tag{14.5.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation 14.5.1 gives
$ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]_{0}/2} = ln 2=kt_{1/2} \tag{14.5.2}$
The natural logarithm of 2 (to three decimal places) is 0.693. Substituting this value into the equation, we obtain the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \tag{14.5.3}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure 14.5.1 "The Half-Life of a First-Order Reaction", and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100 \%}{2}=50\%$ $\dfrac{1}{2}\left (100 \% \right )= 50 \%$
2 $\dfrac{50 \%}{2} =25 \%$ $\dfrac{1}{2} \left ( \dfrac{1}{2} \right )\left (100 \% \right )= 25 \%$
3 $\dfrac{25 \%}{2}=12.5 \%$ $\dfrac{1}{2} \left ( \dfrac{1}{2} \right ) \left ( \dfrac{1}{2} \right ) \left (100 \% \right )= 12.5 \%$
n $\dfrac{100 \%}{2^{n}}$ $\left ( \dfrac{1}{2} \right )^{n} \left (100 \% \right )= \left ( \dfrac{1}{2} \right )^{n} \%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
Note the Pattern
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example 14.5.1
The anticancer drug cisplatin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cisplatin has a concentration of 0.053 M, what will be the concentration of cisplatin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
A Use Equation 14.5.3 to calculate the half-life of the reaction.
B Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
C Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution:
A We can calculate the half-life of the reaction using Equation 14.5.3:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5 \times 10^{-3}\;min^{-1}}=4.6 \times 10^{2}\;min$
Thus it takes almost 8 h for half of the cisplatin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cisplatin will be as follows:
$\dfrac{0.053\;M}{2^{5}}=\dfrac{0.053\;M}{32}=0.0017\;M$
After 10 half-lives (77 h), the remaining concentration of cisplatin will be as follows:
$\dfrac{0.053\;M}{2^{10}}=\dfrac{0.053\;M}{1024}=5.2 \times 10^{-5}\;M$
C The percent completion after 5 half-lives will be as follows:
$percent\;completion=\dfrac{\left (0.053\;M-0.0017\;M \right )100}{0.53\;M}=97\%$
The percent completion after 10 half-lives will be as follows:
$percent\;completion=\dfrac{\left (0.053\;M-5.2 \times 10^{-5}\;M \right )100}{0.53\;M}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise
In Example 4 you found that ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C. What is the half-life for the reaction under these conditions? If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer: 4.3 × 105 s = 120 h = 5.0 days; 4.8 × 10−3 M
Radioactive Decay Rates
As you learned in Chapter 1 radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decayThe decrease in the number of a radioisotope’s nuclei per unit time. of the sample, which is also called its activity (A)The decrease in the number of a radioisotope’s nuclei per unit time: A=−ΔN/Δt as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \tag{14.5.4}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A=kN \tag{14.5.5}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation 14.5.4 and Equation 14.5.6 we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \tag{14.5.6}$
Equation 14.5.6 is the same as the equation for the reaction rate of a first-order reaction (Equation 14.3.5), except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation 14.4.7) or the integrated rate law:
$N=N_{0}e^{-kt} \tag{14.5.7}$
$ln \dfrac{N}{N_{0}}=-kt \tag{14.5.8}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.5.1, along with some of their applications.
Table 14.5.1 Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
Note the Pattern
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$^{14}C\rightarrow ^{14}N + \beta ^{-} \tag{14.5.8}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure 14.18). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example 14.5.2
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation 14.5.5 to calculate N0/N. Then substitute the value for the half-life of 14C into Equation 14.5.3 to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation 14.5.7 to obtain the elapsed time.
Solution:
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation 14.5.7) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
$ln \dfrac{N}{N_{0}}=-kt$
$\dfrac{ln \left (N_{0}/N \right )}{k}=t$
A From Equation 14.5.5, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_{0}}{A}= \dfrac{\cancel{k}N_{0}}{\cancel{k}N}=\dfrac{N_{0}}{N}=\dfrac{15.}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation 14.5.3:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\;yr}=1.22 \times 10^{-4} \; yr^{-1}$
B Substituting into the equation for t,
$t=\dfrac{ln\left ( N_{0}/N \right )}{k}=\dfrac{ln\left ( 15./8.0 \right )}{1.22 \times 10^{-4} \; yr^{-1}}=5.2 \times 10^{3} \; yr$
From our calculations, the man died 5200 yr ago.
Exercise
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer: 30,000 yr
Summary
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k.
Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time.
Key Takeaways
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
Key Equations
half-life of first-order reaction
Equation 14.5.3 $t_{1/2}=\dfrac{0.693}{k}$
radioactive decay
Equation 14.5.5: $A=kN$
Conceptual Problems
1. What do chemists mean by the half-life of a reaction?
2. If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare?
Numerical Problems
1. Half-lives for the reaction A + B → C were calculated at three values of [A]0, and [B] was the same in all cases. The data are listed in the following table:
[A]0 (M) t½ (s)
0.50 420
0.75 280
1.0 210
Does this reaction follow first-order kinetics? On what do you base your answer?
2. Ethyl-2-nitrobenzoate (NO2C6H4CO2C2H5) hydrolyzes under basic conditions. A plot of [NO2C6H4CO2C2H5] versus t was used to calculate t½, with the following results:
[NO2C6H4CO2C2H5] (M/cm3) t½ (s)
0.050 240
0.040 300
0.030 400
Is this a first-order reaction? Explain your reasoning.
3. Azomethane (CH3N2CH3) decomposes at 600 K to C2H6 and N2. The decomposition is first order in azomethane. Calculate t½ from the data in the following table:
Time (s) PAzomethane (atm)
0 8.2 × 10−2
2000 3.99 × 10−2
4000 1.94 × 10−2
How long will it take for the decomposition to be 99.9% complete?
4. The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s−1) for this reaction? If you started with a solution that was 7.5 × 10−3 M H2O2, what would be the initial rate of decomposition (M/s)? What would be the concentration of H2O2 after 3.3 h?
Answers
1. No; the reaction is second order in A because the half-life decreases with increasing reactant concentration according to t1/2 = 1/k[A0].
2. t1/2 = 1.92 × 103 s or 1920 s; 19100 s or 5.32 hrs.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.06%3A_Half_Lives_and_Radioactive_Decay_Kinetics.txt |
Learning Objectives
• To determine the individual steps of a simple reaction.
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.
In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
$2C_{8}H_{18}\left ( l \right ) + 25O_{2}\left ( g \right ) \rightarrow 16CO_{2}\left ( g \right ) + 18H_{2}O\left ( g \right ) \tag{14.6.1}$
For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reactionEach of the complex series of reactions that take place in a stepwise fashion to convert reactants to products., involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.
Molecularity and the Rate-Determining Step
To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
$NO_{2}\left ( g \right ) + CO\left ( g \right ) \rightarrow NO\left ( g \right ) + CO_{2} \left ( g \right ) \tag{14.6.2}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO2 with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
$rate = k\left [ NO_{2} \right ]^{2} \tag{14.6.3}$
The fact that the reaction is second order in [NO2] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be rate = k[NO2][CO].
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
$\begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{slow}{\rightarrow} NO_{3} + NO & elementary \;reaction\ step \;2 & \underline{NO_{3} + CO \rightarrow NO_{2} + CO_{2} }& elementary \; reaction\ sum & NO_{2} + CO \rightarrow NO + CO_{2} & overall reaction \end{matrix}$
According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO3 molecule is an intermediateA species in a reaction mechanism that does not appear in the balanced chemical equation for the overall reaction. in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.
Note the Pattern
The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.
Using Molecularity to Describe a Rate Law
The molecularityThe number of molecules that collide during any step in a reaction mechanism. of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)
Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table 14.6.1 ). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = k[A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure 14.6.1 For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = k[A][B].
Table 14.6.1 Common Types of Elementary Reactions and Their Rate Laws
Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third
Identifying the Rate-Determining Step
Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining stepThe slowest step in a reaction mechanism., that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.
Rate-determining step. The phenomenon of a rate-determining step can be compared to a succession of funnels. The smallest-diameter funnel controls the rate at which the bottle is filled, whether it is the first or the last in the series. Pouring liquid into the first funnel faster than it can drain through the smallest results in an overflow.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
$\begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{k_{1}}{\rightarrow} NO_{3}+ NO & rate=k_{1}\left [ NO_{2} \right ]^{2} predicted\ step \;2 & \underline{NO_{3} + CO \overset{k_{2}}{\rightarrow} NO_{2} + CO_{2} } & rate=k_{2}\left [ NO_{3} \right ]\left [ CO_{2} \right ] predicted\ sum & NO_{2} + CO \overset{k}{\rightarrow} NO + CO_{2} & rate=k\left [ NO_{2} \right ]^{2} observed \end{matrix}$
The experimentally determined rate law for the reaction of NO2 with CO is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so k for the overall reaction must equal k1. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.
Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.
Example 14.6.1
In an alternative mechanism for the reaction of NO2 with CO, N2O4 appears as an intermediate.
$\begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{k_{1}}{\rightarrow} N_{2}O_{4} \ step \;2 & \underline{N_{2}O_{4} + CO \overset{k_{2}}{\rightarrow} NO+ NO_{2} + CO_{2} }\ sum & NO_{2} + CO \rightarrow NO + CO_{2} \end{matrix}$
Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?
Given: elementary reactions
Asked for: rate law for each elementary reaction and overall rate law
Strategy:
A Determine the rate law for each elementary reaction in the reaction.
B Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.
Solution:
A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].
B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.
Exercise A
Iodine monochloride (ICl) reacts with H2 as follows:
$2ICl \left ( g \right ) + H_{2}\left ( g \right ) \rightarrow HCl\left ( g \right ) + I_{2} \left ( s \right )$
The experimentally determined rate law is rate = k[ICl][H2]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.)
Answer
$\begin{matrix} step \;1 & ICl+ H_{2} \overset{k_{1}}{\rightarrow} HCl + HI & rate = k_{1}\left [ ICl \right ]\left [ H_{2} \right ] slow\ step \;2 & \underline{HI + ICl \overset{k_{2}}{\rightarrow} HCl + I_{2} }& rate = k_{2}\left [ HI \right ]\left [ ICl \right ] fast\ sum & 2ICl + H_{2} \rightarrow 2HCl + I_{2} \end{matrix}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Exercise B
The reaction between NO and H2 occurs via a three-step process:
$\begin{matrix} step \;1 & NO + NO \overset{k_{1}}{\rightarrow} N_{2}O_{2}+ NO & fast\ step \;2 & N_{2}O_{2} + H_{2} \overset{k_{2}}{\rightarrow} N_{2}O + H_{2}O & slow \ step 3 & N_{2}O + H_{2} \overset{k_{3}}{\rightarrow} N_{2} + H_{2}O & fast \end{matrix}$
Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = k[NO]2[H2]2?
Answer
• rate = k1[NO]2;
• rate = k2[N2O2][H2];
• rate = k3[N2O][H2];
• 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
• step 2
• Yes, because the rate of formation of [N2O2] = k1[NO]2. Substituting k1[NO]2 for [N2O2] in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where k = k1k2.
Chain Reactions
Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are chain reactionsA reaction mechanism in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process., in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere.
Chain reactions are described as having three stages. The first is initiation, a step that produces one or more reactive intermediates. Often these intermediates are radicalsSpecies that have one or more unpaired valence electrons., species that have an unpaired valence electron. In the second stage, propagation, reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, termination, usually by forming stable products.
Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH3Cl), dichloromethane (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4):
$CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl$
$CH_{3}Cl + Cl_{2} \rightarrow CH_{2}Cl_{2} + HCl$
$CH_{2}Cl_{2} + Cl_{2} \rightarrow CHCl_{3} + HCl$
$CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl$
Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH3Cl.
In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·):
$Cl_{2}\rightarrow 2Cl\cdot$
During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH3·, the methyl radical:
$Cl\cdot + CH_{4} \rightarrow CH_{3} \cdot + HCl$
The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·:
$CH_{3} \cdot + Cl_{2}\rightarrow CH_{3}Cl + Cl\cdot$
The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction:
$\begin{matrix} \cancel{Cl\cdot} + CH_{4} \rightarrow \cancel{CH_{3} \cdot} + HCl\ \underline{\cancel{CH_{3} \cdot} + Cl_{2} \rightarrow CH_{3}Cl + \cancel{Cl\cdot} }\ Cl_{2} + CH_{4} \rightarrow CH_{3}Cl + HCl \end{matrix}$
Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways:
$\begin{matrix} CH_{3} \cdot + Cl\cdot \rightarrow CH_{3}Cl\ CH_{3} \cdot + CH_{3} \cdot \rightarrow H_{3}CCH_{3} \ Cl\cdot + Cl\cdot \rightarrow Cl_{2} \end{matrix} Here is the overall chain reaction, with the desired product (CH3Cl) in bold: Initiation: Cl2 → 2Cl· Propagation: Cl· + CH4 → CH3· + HCl CH3· + Cl2CH3Cl + Cl· Termination: CH3· + Cl· → CH3Cl CH3· + CH3· → H3CCH3 Cl· + Cl· → Cl2 The chain reactions responsible for explosions generally have an additional feature: the existence of one or more chain branching steps, in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion. The reaction of H2 and O2, used to propel rockets, is an example of a chain branching reaction: Initiation: H2 + O2 → HO2· + H· Propagation: HO2· + H2 → H2O + OH· OH· + H2 → H2O + H· Termination: H· + O2 → OH· + ·O· ·O· + H2 → OH· + H· Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction. Summary A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. Chain reactions consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often radicals, species that have an unpaired valence electron. Key Takeaway • A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. Conceptual Problems 1. How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation? 2. What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation? 3. When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction? 4. If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step? 5. Give the rate-determining step for each case. 1. Traffic is backed up on a highway because two lanes merge into one. 2. Gas flows from a pressurized cylinder fitted with a gas regulator and then is bubbled through a solution. 3. A document containing text and graphics is downloaded from the Internet. 6. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health? Numerical Problems 1. Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows: where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction? 2. Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest? 3. Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows? \( \begin{matrix} O_{2}NNH_{2} \xrightarrow[k_{-1}]{k_{1}} O_{2}NNH^{-} + H^{+} & fast\ O_{2}NNH^{-} \overset{k_{2}}{\rightarrow} N_{2}O + OH^{-} & slow\ H^{+} + OH^{-} \overset{k_{3}}{\rightarrow} H_{2}O & fast \end{matrix}$
Assume that the rates of the forward and reverse reactions in the first equation are equal.
4. The following reactions are given:
$\begin{matrix} A + B \xrightarrow[k_{-1}]{k_{1}} C + D\ D + E \overset{k_{2}}{\rightarrow} F \end{matrix}$
What is the relationship between the relative magnitudes of k−1 and k2 if these reactions have the rate law Δ[F]/Δt = k[A][B][E]/[C]? How does the magnitude of k1 compare to that of k2? Under what conditions would you expect the rate law to be Δ[F]/Δt = k′[A][B]? Assume that the rates of the forward and reverse reactions in the first equation are equal.
Answers
1. The k2 step is likely to be rate limiting; the rate cannot proceed any faster than the second step.
2. $rate = k_{2} \dfrac{k_{1}\left [ O_{2}NNH_{2} \right ] }{k_{-1}\left [ H^{+} \right ]} = k \dfrac{\left [ O_{2}NNH_{2} \right ] }{\left [ H^{+} \right ]}$
Contributors
• Anonymous
Modified by Joshua B. Halpern
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Learning Objectives
• To understand why and how chemical reactions occur.
In Section 14.6 , you saw that it is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. According to the collision model, a chemical reaction can occur only when the reactant molecules, atoms, or ions collide with more than a certain amount of kinetic energy and in the proper orientation. The collision model explains why, for example, most collisions between molecules do not result in a chemical reaction. Nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 1030 times per second. If every collision produced two molecules of NO, the atmosphere would have been converted to NO and then NO2 a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. The collision model also explains why such chemical reactions occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates.
Activation Energy
In Chapter 10 we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time.
The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy (Ea)The energy barrier or threshold that corresponds to the minimum amount of energy the particles in a reaction must have to react when they colllide.. We will define this concept using the reaction of NO with ozone, which plays an important role in the depletion of ozone in the ozone layer:
$NO\left ( g \right ) + O_{3}\left ( g \right )\rightarrow NO_{2} \left ( g \right ) + O_{2}\left ( g \right ) \tag{14.7.1}$
Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate.
The reaction rate, not the rate constant, will vary with concentration. The rate constant, however, does vary with temperature. Figure 14..7.1 shows a plot of the rate constant of the reaction of NO with O3 at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (Chapter 11) and of conductivity versus temperature (Chapter 12). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier.
Figure 14.7.1 Rate Constant versus Temperature for the Reaction of NO with O3 The nonlinear shape of the curve is caused by a distribution of kinetic energy over a population of molecules. Only a fraction of the particles have enough energy to overcome an energy barrier, but as the temperature is increased, the size of that fraction increases.
In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complexAlso called the transition state, the arrangement of atoms that first forms when molecules are able to overcome the activation energy and react. or the transition stateAlso called the activated complex, the arrangement of atoms that first forms when molecules are able to overcome the activation energy and react. of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily.
Note the Pattern
Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence.
Graphing Energy Changes during a Reaction
We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. Figure 14.7.2 shows a plot for the NO–O3 system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (ΔE) is negative, which means that the reaction releases energy. (In this case, ΔE is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction (Ea is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur.
Part (a) in Figure 14.7.3 illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, part (b) in Figure 14.22 illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and ΔE > 0. Although the energy changes that result from a reaction can be positive, negative, or even zero, in all cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is always positive.
Note the Pattern
For similar reactions under comparable conditions, the one with the smallest Ea will occur most rapidly.
Whereas ΔE is related to the tendency of a reaction to occur spontaneously, Ea gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. (For more information on spontaneous reactions, see Chapter 18.) For two similar reactions under comparable conditions, the reaction with the smallest Ea will occur more rapidly.
Even when the energy of collisions between two reactant species is greater than Ea, however, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For NO and O3 to produce NO2 and O2, a terminal oxygen atom of O3 must collide with the nitrogen atom of NO at an angle that allows O3 to transfer an oxygen atom to NO to produce NO2 (Figure 14.7.4. All other collisions produce no reaction. Because fewer than 1% of all possible orientations of NO and O3 result in a reaction at kinetic energies greater than Ea, most collisions of NO and O3 are unproductive. The fraction of orientations that result in a reaction is called the steric factor (p)The fraction of orientations of particles that result in a chemical reaction., and, in general, its value can range from 0 (no orientations of molecules result in reaction) to 1 (all orientations result in reaction).
The Arrhenius Equation
Figure 14.7.5 shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than Ea; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than Ea. Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier.
For an A + B elementary reaction, all the factors that affect the reaction rate can be summarized in a single series of relationships:
rate = (collision frequency)(steric factor)(fraction of collisions with E > Ea)
where
$rate = k\left [ A \right ]\left [ B \right ] \tag{14.7.2}$
Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, A, called the frequency factorA constant in the Arrhenius equation, it converts concentrations to collisions per second.:
$k = A e^{-E_{a}/RT} \tag{14.7.3}$
The frequency factor is used to convert concentrations to collisions per second.Because the frequency of collisions depends on the temperature, A is actually not constant. Instead, A increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time. Equation 14.39 is known as the Arrhenius equationAn expression that summarizes the collision model of chemical kinetics: k= Aexp(Ea/RT) and summarizes the collision model of chemical kinetics, where T is the absolute temperature (in K) and R is the ideal gas constant [8.314 J/(K·mol)]. Ea indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large Ea increases rapidly with increasing temperature, whereas the reaction rate with a smaller Ea increases much more slowly with increasing temperature.
If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of Equation 14.7.3,
$ln\;k = ln\;A - \dfrac{E_{a}}{RT} = ln\;A + \left [\left (-\dfrac{E_{a}}{R} \right ) \left ( \dfrac{1}{T} \right ) \right ] \tag{14.7.4}$
Equation 14.40 is the equation of a straight line, y = mx + b, where y = ln k and x = 1/T. This means that a plot of ln k versus 1/T is a straight line with a slope of −Ea/R and an intercept of ln A. In fact, we need to measure the reaction rate at only two temperatures to estimate Ea.
Knowing the Ea at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining Ea from reaction rates measured at several temperatures is illustrated in Example 13.
Example 14.7.1
Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping (f) as a function of temperature (T). Use the data in the following table, along with the graph of ln[chirping rate] versus 1/T in Figure 14.7.6 to calculate Ea for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F).
Frequency (f; chirps/min) ln f T (K) 1/T (K)
200 5.30 299 3.34 × 10−3
179 5.19 298 3.36 × 10−3
158 5.06 296 3.38 × 10−3
141 4.95 294 3.40 × 10−3
126 4.84 293 3.41 × 10−3
112 4.72 292 3.42 × 10−3
100 4.61 290 3.45 × 10−3
89 4.49 289 3.46 × 10−3
79 4.37 287 3.48 × 10−3
Given: chirping rate at various temperatures
Asked for: activation energy and chirping rate at specified temperature
Strategy:
A From the plot of ln f versus 1/T in Figure 14.7.6, calculate the slope of the line (−Ea/R) and then solve for the activation energy.
B Express Equation 14.7.4 in terms of k1 and T1 and then in terms of k2 and T2.
C Subtract the two equations; rearrange the result to describe k2/k1 in terms of T2 and T1.
D Using measured data from the table, solve the equation to obtain the ratio k2/k1. Using the value listed in the table for k1, solve for k2.
Solution:
A If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of ln f versus 1/T should give a straight line (Figure 14.7.6 ). Also, the slope of the plot of ln f versus 1/T should be equal to −Ea/R. We can use the two endpoints in Figure 14.7.6 to estimate the slope:
$slope = \dfrac{\Delta \;ln\; f}{\Delta \left ( 1/T \right )} = \dfrac{5.30 - 4.37}{3.34\times 10^{-3} \; K^{-1} - 3.48\times 10^{-3} \; K^{-1}}$
A computer best-fit line through all the points has a slope of −6.67 × 103 K, so our estimate is very close. We now use it to solve for the activation energy:
$E_{a} = \dfrac{0.93}{-0.041\times 10^{-3} \; K^{-1}} = 6.6 \times 10^{3} \; K$
B If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use Equation 14.40 to express the known rate constant (k1) at the first temperature (T1) as follows:
$ln\;k_{1}= ln\; A -\dfrac{E_{a}}{RT_{1}}$
Similarly, we can express the unknown rate constant (k2) at the second temperature (T2) as follows:
$ln\;k_{2}= ln\; A -\dfrac{E_{a}}{RT_{2}}$
C These two equations contain four known quantities (Ea, T1, T2, and k1) and two unknowns (A and k2). We can eliminate A by subtracting the first equation from the second:
$ln\;k_{2} - ln\;k_{1}= \left (ln\; A -\dfrac{E_{a}}{RT_{2}} \right ) - \left (ln\; A -\dfrac{E_{a}}{RT_{1}} \right ) = -\dfrac{E_{a}}{RT_{2}} + \dfrac{E_{a}}{RT_{1}}$
Then
$ln\; \dfrac {k_{2}}{k_{1}}= -\dfrac{E_{a}}{R} \left (\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right )$
D To obtain the best prediction of chirping rate at 308 K (T2), we try to choose for T1 and k1 the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for T1 = 296 K, where f = 158, and using the Ea calculated previously,
$ln\; \dfrac {k_{2}}{k_{1}}= -\dfrac{E_{a}}{R} \left (\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right ) = \dfrac{55\; kJ/mol}{8.314\; J/\left ( K\cdot mol \right )} \left ( \dfrac{1000 \;J}{1 \; kJ} \right ) \left ( \dfrac{1}{296 \; K}-\dfrac{1}{308 \; K} \right ) = 0.87$
Thus k308/k296 = 2.4 and k308 = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute.
Exercise
The equation for the decomposition of NO2 to NO and O2 is second order in NO2:
$2NO_{2}\left ( g \right ) \rightarrow 2NO\left ( g \right ) + O_{2}\left ( g \right )$
Data for the reaction rate as a function of temperature are listed in the following table. Calculate Ea for the reaction and the rate constant at 700 K.
T (K) k (M−1·s−1)
592 522
603 755
627 1700
652 4020
656 5030
Answer: Ea = 114 kJ/mol; k700 = 18,600 M−1·s−1 = 1.86 × 104 M−1·s−1.
What Ea results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C?
Answer: about 51 kJ/mol
Summary
A minimum energy (activation energy, Ea) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: k = Aexp(-Ea/RT). A plot of the natural logarithm of k versus 1/T is a straight line with a slope of −Ea/R.
Key Takeaway
• For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation.
Key Equation
Arrhenius equation
$k = A e^{-E_{a}/RT} \tag{14.7.3}$
Conceptual Problems
1. Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?
2. For any given reaction, what is the relationship between the activation energy and each of the following?
1. electrostatic repulsions
2. bond formation in the activated complex
3. the nature of the activated complex
3. If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction?
4. The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility?
5. Above Ea, molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach Ea? Explain your answer.
6. What is the relationship between A, Ea, and T? How does an increase in A affect the reaction rate?
7. Of two highly exothermic reactions with different values of Ea, which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why?
Numerical Problems
1. What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C?
2. Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate?
T (K) k (M−1·s−1)
720 0.024
740 0.051
760 0.105
800 0.519
3. Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10−5 M/s at 25°C, what would the reaction rate be at the following temperatures?
1. 15°C
2. 30°C
3. 45°C
4. An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the Ea from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature?
5. The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction?
T (K) k (M−1·min−1)
529 1.4
560 3.7
600 25
645 82
6. The reaction rate at 25°C is 1.0 × 10−4 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10−2 M/s. Estimate Ea for this process. If Ea were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10−4 M/s, what would be the reaction rate at 75°C?
Answers
1. The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 21 = 2; 20°C to 70°C, the reaction rate increases by about 25 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2ΔT/10.
1. 1.0 × 10−5 M/s
2. 6.6 × 10−5 M/s
3. 3.5 × 10−4 M/s
2. 100 kJ/mol
Contributors
• Anonymous
Modified by Joshua B. Halpern
Thumbnail from Wikimedia | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.08%3A_The_Collision_Model_of_Chemical_Kinetics.txt |
Learning Objectives
• To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.
Chapter 7 described catalysts A substance that participates in a reaction and causes it to occur more rapidly but that can be recovered unchanged at the end of the reaction and reused. Catalysts may also control which products are formed in a reaction. as substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower Ea, but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst (Figure 14.8.1 ). Nevertheless, because of its lower Ea, the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes.
Note the Pattern
A catalyst affects Ea, not ΔE.
Heterogeneous Catalysis
In heterogeneous catalysisA catalytic reaction in which the catalyst is in a different phase from the reactants., the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.
An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure 14.8.2, the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.
Figure 14.8.2 shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.
Several important examples of industrial heterogeneous catalytic reactions are in Table 14.8.1. Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.
Table 14.8.1 Some Commercially Important Reactions that Employ Heterogeneous Catalysts
Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate CH2=CHCH3+ NH3+3/2O2
CH2=CHCN + 3H2O
CH2=CHCN
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
Homogeneous Catalysis
In homogeneous catalysisA catalytic reaction in which the catalyst is uniformly dispersed throughout the reactant mixture to form a solution., the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds (Table 14.8.2), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis.
Table 14.8.2 Some Commercially Important Reactions that Employ Homogeneous Catalysts
Commercial Process Catalyst Reactants Final Product
Union Carbide [Rh(CO)2I2] CO + CH3OH CH3CO2H
hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H
hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO
adiponitrile process Ni/PR3 complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene
Enzymes
Figure 14.8.3 A Catalytic Defense Mechanism The scalding, foul-smelling spray emitted by this bombardier beetle is produced by the catalytic decomposition of H2O2.
Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrateThe reactant in an enzyme-catalyzed reaction..
Because enzymes can increase reaction rates by enormous factors (up to 1017 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37°C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water (Figure 14.8.3).
Enzyme inhibitorsSubstances that decrease the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of the enzyme, thus slowing or preventing a reaction from occurring. cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research.
Summary
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction.
Key Takeaway
• Catalysts allow a reaction to proceed via a pathway that has a lower activation energy.
Conceptual Problems
1. What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor (A)? What effect does it have on the change in potential energy for the reaction?
2. How is it possible to affect the product distribution of a reaction by using a catalyst?
3. A heterogeneous catalyst works by interacting with a reactant in a process called adsorption. What occurs during this process? Explain how this can lower the activation energy.
4. What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer.
5. Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following.
1. ease of recovery
2. collision frequency
3. temperature sensitivity
4. cost
6. An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred.
7. Consider the following reaction between cerium(IV) and thallium(I) ions:
2Ce4+ + Tl+ → 2Ce3+ + Tl3+
This reaction is slow, but Mn2+ catalyzes it, as shown in the following mechanism:
Ce4+ + Mn2+ → Ce3+ + Mn3+
Ce4+ + Mn3+ → Ce3+ + Mn4+
Mn4+ + Tl+ → Tl3+ + Mn2+
In what way does Mn2+ increase the reaction rate?
8. The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why?
9. Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.)
Answers
1. A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction.
2. In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased.
1. Heterogeneous catalysts are easier to recover.
2. Collision frequency is greater for homogeneous catalysts.
3. Homogeneous catalysts are often more sensitive to temperature.
4. Homogeneous catalysts are often more expensive.
3. The Mn2+ ion donates two electrons to Ce4+, one at a time, and then accepts two electrons from Tl+. Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions.
Numerical Problems
1. At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (ES), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations:
$\begin{matrix} enzyme\;\left ( E \right ) + substrate\;\left ( S \right ) \rightleftharpoons\ enzyme-substrate\;complex\;\left ( ES \right ) \rightleftharpoons\ enzyme\;\left ( E \right ) + product\;\left ( P \right ) \end{matrix}$
This can also be shown as follows:
$E + S \xrightarrow[k_{-1}]{k_{1}} ES \xleftarrow[k_{-2}]{k_{2}} E + P$
Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?
2. A particular reaction was found to proceed via the following mechanism:
A + B → C + D
2C → E
E + A → 3B + F
What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates.
3. A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.
4. The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon.
Answers
1. $\dfrac{\Delta \left ( ES \right )}{\Delta t} =-\left ( k_{2}+k_{-1} \right ) \left [ES \right ] + k_{1} \left [ E \right ]\left [ S \right ] + k_{-2}\left [ E \right ]\left [ P \right ] \approx 0$; zeroth order in substrate.
2. In both cases, the product of pathway A is favored. All of the Z produced in the catalyzed reversible pathway B will eventually be converted to X as X is converted irreversibly to Y by pathway A.
$Z \underset{B}{\rightleftharpoons} X\overset{A}{\rightarrow} Y$
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.09%3A_Catalysis.txt |
Learning Objectives
• To understand what is meant by chemical equilibrium.
Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (N2O4) to nitrogen dioxide (NO2). You may recall from Chapter 14 that NO2 is responsible for the brown color we associate with smog. When a sealed tube containing solid N2O4 (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO2 appears (Figure 15.1.1). The reaction can be followed visually because the product (NO2) is colored, whereas the reactant (N2O4) is colorless:
\[N_2O_4{(g)} colorless \rightleftharpoons 2NO_{2(g)} \; red-brown \tag{15.1.1}\]
The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.”
Figure 15.1.2shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO2 were zero, then it increases as the concentration of N2O4 decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no N2O4 but an initial NO2 concentration twice the initial concentration of N2O4 in part (a) in Figure 15.1.2, in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in Figure 15.1.2. Thus equilibrium can be approached from either direction in a chemical reaction.
Figure 15.1.3 shows the forward and reverse reaction rates for a sample that initially contains pure NO2. Because the initial concentration of N2O4 is zero, the forward reaction rate (dissociation of N2O4) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO2) is initially very high (2.0 × 106 M/s), but it decreases rapidly as the concentration of NO2 decreases. (Recall from Chapter 14 that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO2: rate = kr[NO2]2, where kr is the rate constant for the reverse reaction shown in Equation 15.1.1.) As the concentration of N2O4 increases, the rate of dissociation of N2O4 increases—but more slowly than the dimerization of NO2—because the reaction is only first order in N2O4 (rate = kf[N2O4], where kf is the rate constant for the forward reaction in Equation 15.1.1). Eventually, the forward and reverse reaction rates become identical, kF = kr, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.
The rate of dimerization of NO2 (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N2O4 is zero, the rate of the dissociation reaction (forward reaction) at t = 0 is also zero. As the dimerization reaction proceeds, the N2O4 concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N2O4 and NO2 no longer change.
Note the Pattern
At equilibrium, the forward reaction rate is equal to the reverse reaction rate.
Example 15.1.1
The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation <math display="inline" xml:id="av_1.0-15_m004"><semantics><mrow><mn>2</mn><mtext>A</mtext><mo>⇌</mo><mtext>B,</mtext></mrow></semantics>[/itex] where the blue circles are A and the purple ovals are B. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?
Given: three reaction systems
Asked for: relative time to reach chemical equilibrium
Strategy:
Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.
Solution:
In systems 1 and 3, the concentration of A decreases from t0 through t2 but is the same at both t2 and t3. Thus systems 1 and 3 are at equilibrium by t3. In system 2, the concentrations of A and B are still changing between t2 and t3, so system 2 may not yet have reached equilibrium by t3. Thus system 2 took the longest to reach chemical equilibrium.
Exercise
In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?
Answer: system 2
Summary
Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.
Key Takeaway
• At equilibrium, the forward and reverse reactions of a system proceed at equal rates.
Conceptual Problems
1. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the concentrations or amounts of the reactants and the products?
2. Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.
3. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level?
4. Which of these systems exists in a state of chemical equilibrium?
1. oxygen and hemoglobin in the human circulatory system
2. iodine crystals in an open beaker
3. the combustion of wood
4. the amount of 14C in a decomposing organism
Answer
1. Both forward and reverse reactions occur but at the same rate. Na+ and Cl ions continuously leave the surface of an NaCl crystal to enter solution, while at the same time Na+ and Cl ions in solution precipitate on the surface of the crystal.
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.01%3A_The_Concept_of_Chemical_Equilibrium.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To write an equilibrium constant expression for any reaction.
Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in Equation 15.2.1, the decomposition of N2O4 to NO2. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
$\text{forward rate} = k_f[N_2O_4] \tag{15.2.1}$
and
$\text{reverse rate} = k_r[NO_2]^2 \tag{15.2.2}$
At equilibrium, the forward rate equals the reverse rate:
$k_f[N_2O_4] = k_r[NO_2]^2 \tag{15.2.3}$
so
$\dfrac{k_f}{k_r}=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.2.4}$
The ratio of the rate constants gives us a new constant, the equilibrium constant (K)The ratio of the rate constants for the forward reaction and the reverse reaction; that is, K =kf/kr It is also the equilibrium constant calculated from solution concentrations: K =[C]c[D]d/[A]a[B]b for the general reaction aA + bB ⇌cC +dD in which each component is in solution. which is defined as follows:
$K=\dfrac{k_f}{k_r} \tag{15.2.5}$
Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.
Note the Pattern
The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
Table 15.2.1 lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation 15.1.1. At equilibrium the magnitude of the quantity [NO2]2/[N2O4] is essentially the same for all five experiments. In fact, no matter what the initial concentrations of NO2 and N2O4 are, at equilibrium the quantity [NO2]2/[N2O4] will always be 6.53 ± 0.03 × 10−3 at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.
Table 15.2.1 Initial and Equilibrium Concentrations for $NO_2:N_2O_4$ Mixtures at 25°C
Initial Concentrations Concentrations at Equilibrium
Experiment [N2O4] (M) [NO2] (M) [N2O4] (M) [NO2] (M) K = [NO2]2/[N2O4]
1 0.0500 0.0000 0.0417 0.0165 6.54 × 10−3
2 0.0000 0.1000 0.0417 0.0165 6.54 × 10−3
3 0.0750 0.0000 0.0647 0.0206 6.56 × 10−3
4 0.0000 0.0750 0.0304 0.0141 6.54 × 10−3
5 0.0250 0.0750 0.0532 0.0186 6.50 × 10−3
Developing an Equilibrium Constant Expression
In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form
$aA+bB \rightleftharpoons cC+dD \tag{15.2.6}$
where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass actionFor the general balanced chemical equation aA+bB = cC+ dD the equilibrium constant expression is K =[C]c[D]d/[A]a[B]b and can be stated as follows:
$K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{15.2.7}$
where K is the equilibrium constant for the reaction. Equation 15.7 is called the equilibrium equationFor the general balanced chemical equation aA+bB = cC+ dD the equilibrium constant expression is K =[C]c[D]d/[A]a[B]b, and the right side of Equation 15.2.7 is called the equilibrium constant expression. The relationship shown in Equation 15.2.7 is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.
The equilibrium constant can vary over a wide range of values. The values of K shown in Table 15.2.2 , for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than 103 indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between H2 and Cl2 to produce HCl, which has an equilibrium constant of 1.6 × 1033 at 300 K. Because H2 is a good reductant and Cl2 is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of K less than 10−3 indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.
Table 15.2.2 Equilibrium Constants for Selected Reactions*
Reaction Temperature (K) Equilibrium Constant (K)
$S_{(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}$ 300 4.4 × 1053
$2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H2O_{(g)}$ 500 2.4 × 1047
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}$ 300 1.6 × 1033
$H_{2(g)}+Br_{2(g)} \rightleftharpoons 2HBr_{(g)}$ 300 4.1 × 1018
$2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ 300 4.2 × 1013
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}$ 300 2.7 × 108
$H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)}$ 100 1.92
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$ 300 2.9 × 10−1
$I_{2(g)} \rightleftharpoons 2I_{(g)}$ 800 4.6 × 10−7
$Br_{2(g)} \rightleftharpoons 2Br_{(g)}$ 1000 4.0 × 10−7
$Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$ 1000 1.8 × 10−9
$F_{2(g)} \rightleftharpoons 2F_{(g)}$ 500 7.4 × 10−13
*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures.
You will also notice in Table 15.2.2 that equilibrium constants have no units, even though Equation 15.2.7 suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation 15.2.8, the units of concentration cancel, which makes K unitless as well:
$\dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \tag{15.2.8}$
Many reactions have equilibrium constants between 1000 and 0.001 (103K ≥ 10−3), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD:
$H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \tag{15.10}$
The equilibrium constant expression for this reaction is
$K= \dfrac{[HD]^2}{[H_2][D_2]}$
with K varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of H2, D2, and HD contains significant concentrations of both product and reactants.
Figure 15.2.1summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants ⇌ products Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equation 15.2.8 and Equation 15.2.7), when kf >> kr, K is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when kf << kr, K is a very small number, and the reaction produces almost no products as written. Systems for which kfkr have significant concentrations of both reactants and products at equilibrium.
Note the Pattern
A large value of the equilibrium constant K means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.
Example 15.2.1
Write the equilibrium constant expression for each reaction.
1. $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
2. $CO_{(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$
3. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}$
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
Refer to Equation 15.2.7. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.
Solution:
1. The only product is ammonia, which has a coefficient of 2. For the reactants, N2 has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows: $\dfrac{[NH_3]^2}{[N_2][H_2]^3} \notag$
2. The only product is carbon dioxide, which has a coefficient of 1. The reactants are CO, with a coefficient of 1, and O2, with a coefficient of 1/2 Thus the equilibrium constant expression is as follows: $\dfrac{[CO_2]}{[CO][O_2]^{1/2}} \notag$
3. This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for O2. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2: $\dfrac{[CO]^2[O_2]}{[CO_2]^2} \notag$
Exercise
Write the equilibrium constant expression for each reaction.
1. $N_2O_{(g)} \rightleftharpoons N_{2(g)}+\frac{1}{2}O_{2(g)}$
2. $2C_8H_{18(g)}+25O_{2(g)} \rightleftharpoons 16CO_{2(g)}+18H_2O_{(g)}$
3. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$
Answer
1. $K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}$
2. $K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}$
3. $K=\dfrac{[HI]^2}{[H_2][I_2]}$
Example 15.2.2
Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54$
2. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{−18}$
3. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97$
4. $2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}$
Given: systems and values of K
Asked for: composition of systems at equilibrium
Strategy:
Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.
Solution:
1. Only system 4 has K >> 103, so at equilibrium it will consist of essentially only products.
2. System 2 has K << 10−3, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
3. Both systems 1 and 3 have equilibrium constants in the range 103 ≥ K ≥ 10−3, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag$
Values of the equilibrium constant at various temperatures were reported as K25°C = 3.3 × 108, K177°C = 2.6 × 103, and K327°C = 4.1.
1. At which temperature would you expect to find the highest proportion of H2 and N2 in the equilibrium mixture?
2. Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?
Answer
1. 327°C, where K is smallest
2. 25°C
Variations in the Form of the Equilibrium Constant Expression
Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation 15.7 in reverse, we obtain the following:
$cC+dD \rightleftharpoons aA+bB \tag{15.2.10}$
The corresponding equilibrium constant $K′$ is as follows:
$K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \tag{15.2.11}$
This expression is the inverse of the expression for the original equilibrium constant, so K′ = 1/K. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction N2O4 ⇌2NO2 is as follows:
$K=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.2.12}$
but for the opposite reaction, 2NO2N2O4 the equilibrium constant K′ is given by the inverse expression:
$K'=\dfrac{[N_2O_4]}{[NO_2]^2} \tag{15.2.13}$
Consider another example, the formation of water: 2H2(g) + O2(g) 2H2O(g) Because H2 is a good reductant and O2 is a good oxidant, this reaction has a very large equilibrium constant (K = 2.4 × 1047 at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form O2 and H2, is very small: K′ = 1/K = 1/(2.4 × 1047) = 4.2 × 10−48. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into H2 and O2.
Note the Pattern
The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.
Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction
$2NO_2 \rightleftharpoons N_2O_4$
as
$NO_2 \rightleftharpoons \frac{1}{2}N_2O_4$
with the equilibrium constant K″ is as follows:
$K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \tag{15.2.14}$
The values for K′ (Equation 15.2.13) and K″ are related as follows:
$K′′=(K')^{1/2}=\sqrt{K'} \tag{15.2.15}$
In general, if all the coefficients in a balanced chemical equation are subsequently multiplied by n, then the new equilibrium constant is the original equilibrium constant raised to the nth power.
Example 15.2.3
At 745 K, K is 0.118 for the following reaction:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag$
What is the equilibrium constant for each related reaction at 745 K?
1. $2NH_{3(g)} \rightleftharpoons N2(g)+3H_{2(g)}$
2. $\frac{1}{2}N_{2(g)}+\frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$
Given: balanced equilibrium equation, K at a given temperature, and equations of related reactions
Asked for: values of K for related reactions
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate K for each reaction.
Solution:
The equilibrium constant expression for the given reaction of N2(g) with H2(g) to produce NH3(g) at 745 K is as follows:
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118 \notag$
1. This reaction is the reverse of the one given, so its equilibrium constant expression is as follows: $K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47 \notag$
2. In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows: $K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344 \notag$
Exercise
At 527°C, the equilibrium constant for the reaction
$2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)} \notag$
is 7.9 × 104. Calculate the equilibrium constant for the following reaction at the same temperature:
$SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g) \notag$
Answer: 3.6 × 10−3
Equilibrium Constant Expressions for Systems that Contain Gases
For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol KpAn equilibrium constant expressed as the ratio of the partial pressures of the products and reactants, each raised to its coefficient in the chemical equation. is used to denote equilibrium constants calculated from partial pressures. For the general reaction aA + bB ⇌ cC + dD in which all the components are gases, we can write the equilibrium constant expression as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
$K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \tag{15.2.16}$
Thus Kp for the decomposition of N2O4 (Equation 15.1.1) is as follows:
$K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \tag{15.2.17}$
Like K, Kp is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity.The “effective pressure” is called the fugacity, just as activity is the effective concentration.
Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of K and Kp are usually different. They are, however, related by the ideal gas constant (R) and the temperature (T):
$K_p = K(RT)^{Δn} \tag{15.2.18}$
where K is the equilibrium constant expressed in units of concentration and Δn is the difference between the numbers of moles of gaseous products and gaseous reactants (npnr). The temperature is expressed as the absolute temperature in kelvins. According to Equation 15.19, Kp = K only if the moles of gaseous products and gaseous reactants are the same (i.e., Δn = 0). For the decomposition of N2O4, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so Δn = 1. Thus, for this reaction, Kp = K(RT)1 = KRT.
Example 15.2.4
The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag$
What is Kp for this reaction at the same temperature?
Given: equilibrium equation, equilibrium constant, and temperature
Asked for: K p
Strategy:
Use the coefficients in the balanced chemical equation to calculate Δn. Then use Equation 15.19 to calculate K from Kp.
Solution:
This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so Δn = (2 − 4) = −2. We know K, and T = 745 K. Thus, from Equation 15.2.15, we have the following:
$K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K)][745\; K]\}^2}=3.16 \times 10^{−5} \notag$
Because Kp is a unitless quantity, the answer is Kp = 3.16 × 10−5.
Exercise
Calculate Kp for the reaction
$2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)} \notag$
at 527°C, if K = 7.9 × 104 at this temperature.
Answer: Kp = 1.2 × 103
Homogeneous and Heterogeneous Equilibriums
When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibriumAn equilibrium in which the reactants and products of an equilibrium reaction form a single phase, whether gas or liquid.. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibriumAn equilibrium in which the reactants of an equilibrium reaction, the products, or both are in more than one phase., such as the reaction of a gas with a solid or liquid.
Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are assigned an activity of 1. (Recall from Chapter 11, for example, that the density of water, and thus its volume, changes by only a few percentage points between 0°C and 100°C.)
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$CO_{2(g)}+C_{(s)} \rightleftharpoons 2CO_{(g)} \tag{15.2.19}$
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{[CO]^2}{[CO_2][C]} \tag{15.2.20}$
Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:
$[C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \tag{15.2.21}$
We can rearrange Equation 15.2.17 so that the constant terms are on one side:
$K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \tag{15.2.22}$
Incorporating the constant value of [C] into the equilibrium equation for the reaction in Equation 15.2.16,
$K'=\dfrac{[CO]^2}{[CO_2]} \tag{15.2.23}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \tag{15.2.24}$
Incorporating all the constant values into K′ or Kp allows us to focus on the substances whose concentrations change during the reaction.
Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of CO and CO2, the system described in Equation 15.2.16 will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure 15.2.2, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.
Example 15.2.5
Write each expression for K, incorporating all constants, and Kp for the following equilibrium reactions.
1. $PCl_{3(l)}+Cl_{2(g)} \rightleftharpoons PCl_{5(s)}$
2. $Fe_3O_{4(s)}+4H_{2(g)} \rightleftharpoons 3Fe_{(s)}+4H_2O_{(g)}$
Given: balanced equilibrium equations
Asked for: expressions for K and Kp
Strategy:
Find K by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express Kp as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation.
Solution:
1. This reaction contains a pure solid (PCl5) and a pure liquid (PCl3). Their concentrations do not appear in the equilibrium constant expression because they do not change significantly. So $K=\dfrac{1}{[Cl_2]} \notag$
and
$K_p=\dfrac{1}{P_{Cl_2}} \notag$
2. This reaction contains two pure solids (Fe3O4 and Fe), which do not appear in the equilibrium constant expressions. The two gases do, however, appear in the expressions: $K=\dfrac{[H_2O]^4}{[H_2]^4} \notag$
Exercise
Write the expressions for K and Kp for the following reactions.
1. $CaCO_{3(s)} \rightleftharpoons CaO_{(s)}+CO_{2(g)}$
2. $\underset{glucose}{C_6H_{12}O_{6(s)}} + 6O_{2(g)} \rightleftharpoons 6CO_{2(g)}+6H_2O_{(g)}$
Answer
1. K = [CO2]; and Kp = P(CO2)
2. $K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction.
Note the Pattern
The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium.
Equilibrium Constant Expressions for the Sums of Reactions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of N2 with O2 to give NO2. As we stated in Section 15.1, this reaction is an important source of the NO2 that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), N2 reacts with O2 at the high temperatures inside an internal combustion engine to give NO. The released NO then reacts with additional O2 to give NO2 (2). The equilibrium constant for each reaction at 100°C is also given.
1. $N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\;\; K_1=2.0 \times 10^{−25}$
2. $2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9$
3. Summing reactions (1) and (2) gives the overall reaction of N2 with O2:
4. $N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=?$
The equilibrium constant expressions for the reactions are as follows:
$K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2} \notag$v
What is the relationship between K1, K2, and K3, all at 100°C? The expression for K1 has [NO]2 in the numerator, the expression for K2 has [NO]2 in the denominator, and [NO]2 does not appear in the expression for K3. Multiplying K1 by K2 and canceling the [NO]2 terms,
$K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3 \notag$
Thus the product of the equilibrium constant expressions for K1 and K2 is the same as the equilibrium constant expression for K3:
$K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15} \notag$
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, ΔH for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
Note the Pattern
To determine K for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example 15.2.6
The following reactions occur at 1200°C:
1. $CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{−2}$
2. $CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4$
Calculate the equilibrium constant for the following reaction at the same temperature.
3. $CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?$
Given: two balanced equilibrium equations, values of K, and an equilibrium equation for the overall reaction
Asked for: equilibrium constant for the overall reaction
Strategy:
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of K for that equation. Calculate K for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
$CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)} \notag$
$\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)} \notag$
$CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)} \notag$
The values for K1 and K2 are given, so it is straightforward to calculate K3:
$K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3 \notag$
Exercise
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
1. $\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}$
2. $SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}$
3. $\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?$
Answer: K3 = 1.1 × 1066
Summary
The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same K. For a system at equilibrium, the law of mass action relates K to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, K and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (Kp) is related to K by the ideal gas constant (R), the temperature (T), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.
Key Takeaways
• The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
• For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
Key Equations
Definition of equilibrium constant in terms of forward and reverse rate constants
Equation 15.2.5: $K=\dfrac{k_f}{k_r}$
Equilibrium constant expression (law of mass action)
Equation 15.2.7: <$K=\dfrac{\left [ C \right ]^c \left [ D \right ]^d}{\left [ A \right ]^a \left [ B \right ]^b}$
Equilibrium constant expression for reactions involving gases using partial pressures
Equation 15.2.1617: $K_p=\dfrac{P_C^c P_D^d}{P_A^aP_B^b}$
Relationship between K p and K
Equation 15.2.18: $K_p=K\left ( RT \right )^{\Delta n}$
Conceptual Problems
1. For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?
2. Which of the following equilibriums are homogeneous and which are heterogeneous?
1. $2HF_{(g)} \rightleftharpoons H_{2(g)}+F_{2(g)}$
2. $C_{(s)} + 2H_{2(g)} \rightleftharpoons CH_{4(g)}$
3. $\ce{H_2C=CH_{2(g)}} + H_{2(g)} \rightleftharpoons C_2H_{6(g)}$
4. $2Hg_{(l)} + O_{2(g)} \rightleftharpoons 2HgO_{(s)}$
3. Classify each equilibrium system as either homogeneous or heterogeneous.
1. $NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)}$
2. $C_{(s)} + O_{2(g)} \rightleftharpoons CO_{2(g)}$
3. $2Mg_{(s)} + O_{2(g)} \rightleftharpoons 2MgO_{(s)}$
4. $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)}+Cl^−_{(aq)}$
4. If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?
5. Industrial production of NO by the reaction <math xml:id="av_1.0-15_m114" display="inline"><semantics><mrow><msub><mtext>N</mtext><mn>2</mn></msub><mtext>(g)</mtext><mo>+</mo><msub><mtext>O</mtext><mn>2</mn></msub><mtext>(g)</mtext><mo>⇌</mo><mtext>2NO(g)</mtext></mrow></semantics>[/itex] is carried out at elevated temperatures to drive the reaction toward the formation of product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?
6. How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?
7. What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?
8. Write the equilibrium constant expressions for K and Kp for each reaction.
1. $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)}+H_{2(g)}$
2. $PCl_{3(g)}+Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$
3. $2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
9. Write the equilibrium constant expressions for K and Kp as appropriate for each reaction.
1. $2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}$
2. $\frac{1}{2}H_{2(g)}+12I_{2(g)} \rightleftharpoons HI_{(g)}$
3. $cis-stilbene_{(soln)} \rightleftharpoons trans-stilbene_{(soln)}$
10. Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?
11. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.
1. $2S_{(s)}+3O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
2. $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$
3. $2ZnS_{(s)}+3O_{2(g)} \rightleftharpoons 2ZnO_{(s)}+2SO_{2(g)}$
12. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.
1. $2HgO_{(s)} \rightleftharpoons 2Hg_{(l)}+O_{2(g)}$
2. $H_{2(g)}+I_{2(s)} \rightleftharpoons 2HI_{(g)}$
3. $NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)}$
13. At room temperature, the equilibrium constant for the reaction <math xml:id="av_1.0-15_m134" display="inline"><semantics><mrow><mn>2</mn><mtext>A(g)</mtext><mo>⇌</mo><mtext>B(g)</mtext></mrow></semantics>[/itex] is 1. What does this indicate about the concentrations of A and B at equilibrium? Would you expect K and Kp to vary significantly from each other? If so, how would their difference be affected by temperature?
14. For a certain series of reactions, if [OH][HCO3]/[CO32−] = K1 and [OH][H2CO3]/[HCO3] = K2, what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.
15. In the equation for an enzymatic reaction, ES represents the complex formed between the substrate S and the enzyme protein E. In the final step of the following oxidation reaction, the product P dissociates from the ESO2 complex, which regenerates the active enzyme:
$E + S \rightleftharpoons ES\;\;\; K_1$
$ES + O_2 \rightleftharpoons ESO_2 \;\;\; K_2$
$ESO_2 \rightleftharpoons E+P\;\;\;\; K_3$
Give the overall reaction equation and show that K = K1 × K2 × K3.
Answers
1. The equilibrium constant for the reaction written in reverse: K′ = 1/K.
2. Each system is heterogeneous.
3. Rapid cooling “quenches” the reaction mixture and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.
4. $K=\dfrac{k_f}{k_r}\;\;\; K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]) 1. \(K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_p=\dfrac{(P_{N_2O})^2}{(P_{NO})^2(P_{O_2}})$
2. $K=\dfrac{[HI]}{[H_2]^{1/2}[I_2]^{1/2}}\;\;\; K_p=\dfrac{P_{HI}}{(P_{H_2})^{1/2}(P_{I_2})}$
3. $K=\dfrac{[\text{trans-stilbene}]}{[\text{cis-stilbene}]}$
1. $K=\dfrac{[SO_3]^2}{[O_2]^3}\;\;\; K_p=\dfrac{(P_{SO_3})^2}{(P_{O_2})^3}$
2. $K=\dfrac{[CO]^2}{[CO_2]}\;\;\; K_p=\dfrac{(P_{CO})^2}{P_{CO_2}}$
3. $K=\dfrac{[SO_2]^2}{[O_2]^3}\;\;\; K_p=\dfrac{(P_{SO_2})^2}{(P_{O_2})^3}$
5. At equilibrium,
$[A]=\sqrt{B}$ and $Δn =−1$ so $K_p=K(RT)^{Δn}=\dfrac{K}{RT}$
the difference increases as T increases.
Numerical Problems
1. Explain what each of the following values for K tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: K = 0.892; K = 3.25 × 108; K = 5.26 × 10−11. Are products or reactants favored at equilibrium?
2. Write the equilibrium constant expression for each reaction. Are these equilibrium constant expressions equivalent? Explain.
1. $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
2. $\frac{1}{2}N_2O_{4(g)} \rightleftharpoons NO_{2(g)}$
3. Write the equilibrium constant expression for each reaction.
1. $\frac{1}{2}N_{2(g)}+\frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$
2. $\frac{1}{3}N_{2(g)}+H_{2(g)} \rightleftharpoons \frac{2}{3}NH_{3(g)}$
How are these two expressions mathematically related to the equilibrium constant expression for
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} ? \notag$
4. Write an equilibrium constant expression for each reaction.
1. $C_{(s)} + 2H_2O_{(g)}⇌CO_{2(g)}+2H_{2(g)}$
2. $SbCl_{3(g)}+Cl_{2(g)}⇌SbCl_{5(g)}$
3. $2O_{3(g)}⇌3O_{2(g)}$
5. Give an equilibrium constant expression for each reaction.
1. $2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$
2. $\frac{1}{2}H_{2(g)}+\frac{1}{2}I_{2(g)}⇌HI_{(g)}$
3. $CaCO_{3(s)} + 2HOCl_{(aq)} \rightleftharpoons Ca^{2+}_{(aq)} + 2OCl^−_{(aq)} + H_2O_{(l)} + CO_{2(g)}$
6. Calculate K and Kp for each reaction.
1. $2NOBr_{(g)}⇌2NO_{(g)}+Br_{(g)}$ at 727°C, the equilibrium concentration of NO is 1.29 M, Br2 is 10.52 M, and NOBr is 0.423 M.
2. $C_{(s)} + CO_{2(g)}⇌2CO_{(g)}$: at 1200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm CO2 and 76.8 atm CO, and the vessel contains 3.55 g of carbon.
7. Calculate K and Kp for each reaction.
1. $N_2O4_{(g)}⇌2NO_{2(g)}$: at the equilibrium temperature of −40°C, a 0.150 M sample of N2O4 undergoes a decomposition of 0.456%.
2. $CO_{(g)}+2H_{2(g)}⇌CH_3OH_{(g)}$: an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of 6.71 × 102 atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7193 g of methanol.
8. Determine K and Kp (where applicable) for each reaction.
1. $2H_2S_{(g)}⇌2H_{2(g)}+S_{2(g)}$: at 1065°C, an equilibrium mixture consists of $1.00 \times 10^{−3}$ M $H_2$, $1.20 \times 10^{−3}$ M $S_2$, and $3.32 \times 10^{−3}$ M $H_2S$.
2. $Ba(OH)_{2(s)}⇌2OH^−_{(aq)}+Ba^{2+}_{(aq)}$: at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.
9. Determine K and Kp for each reaction.
1. $2NOCl_{(g)}⇌2NO_{(g)}+Cl_{2(g)}$: at 500 K, a 24.3 mM sample of $NOCl$ has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of $NOCl$.
2. $Cl_{2(g)}+PCl_{3(g)}⇌PCl_{5(g)}$: at 250°C, a 500 mL reaction vessel contains 16.9 g of $Cl_2$ gas, 0.500 g of $PCl_3$, and 10.2 g of $PCl_5$ at equilibrium.
10. The equilibrium constant expression for a reaction is [CO2]2/[SO2]2[O2]. What is the balanced chemical equation for the overall reaction if one of the reactants is Na2CO3(s)?
11. The equilibrium constant expression for a reaction is [NO][H2O]3/2/[NH3][O2]5/4. What is the balanced chemical equation for the overall reaction?
12. Given K = kf/kr, what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?
13. The value of the equilibrium constant for
$2H_{2(g)}+S_{2(g)}⇌2H_2S_{(g)} \notag$
is 1.08 × 107 at 700°C. What is the value of the equilibrium constant for the following related reactions?
1. $H_{2(g)}+12S_{2(g)}⇌H_2S_{(g)}$
2. $4H_{2(g)}+2S_{2(g)}⇌4H_2S_{(g)}$
3. $H_2S_{(g)}⇌H_{2(g)}+12S_{2(g)}$
Answers
1. K = 0.892: the concentrations of the products and the reactants are approximately equal at equilibrium so neither is favored; K = 3.25 × 108: the ratio of the concentration of the products to the reactants at equilibrium is very large so the formation of products is favored; K = 5.26 × 10−11: the ratio of the concentration of the products to the reactants at equilibrium is very small so the formation of products is not favored.
1. $K'=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}$
2. $K''=\dfrac{[NH_3]^{2/3}}{[N_2]^{1/3}[H_2]}\;\;\; K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\;\;\; K′ = K^{1/2},\text{ and} K″ = K^{1/3}$
1. $K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}$
2. $K=\dfrac{[HI]}{[H_2]^{1/2}[I_2]^{1/2}}$
3. $K=\dfrac{[Ca^{2+}][OCl^−]^2[P_{CO_2}]}{[HOCl]^2}$
1. $K = 1.25 \times 10^{−5}\;\;\; K_p = 2.39 \times 10^{−4}$
2. $K = 9.43\;\;\; K_p = 5.60 \times 10^{−3}$
1. $K=\dfrac{[Cl_2][NO]^2}{[NOCl]^2}=4.59 \times 10^{−4}\;\;\; K_p=1.88 \times 10^{−2}$
2. $K=\dfrac{[PCl_5]}{[PCl_3][Cl_2]}=28.3\;\;\; K_p=0.658$
2. $NH_3 + \frac{5}{4}O_2⇌NO +\frac{3}{2}H_2O$, which can also be written as follows: $4NH_{3(g)} + 5O_{2(g)}⇌4NO_{(g)} + 6H_2O_{(g)}$
1. 3.29 × 103
2. 1.17 × 1014
3. 3.04 × 10−4
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.02%3A_The_Equilibrium_Constant.txt |
Learning Objectives
• To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of CaCO3(s) to CaO(s) and CO2(g) is K = [CO2]. At 800°C, the concentration of CO2 in equilibrium with solid CaCO3 and CaO is 2.5 × 10−3 M. Thus K at 800°C is 2.5 × 10−3. (Remember that equilibrium constants are unitless.)
A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:
$n-butane \left ( g \right ) \rightleftharpoons isobutane \left ( g \right ) \tag{15.3.1}$
and the equilibrium constant K = [isobutane]/[n-butane]. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,
$K= \dfrac{isobutane}{n-butane}=\dfrac{0.041\;\cancel{M}}{0.016 \; \cancel{M}} = 2.6 \tag{15.3.2}$
Thus the equilibrium constant for the reaction as written is 2.6.
Example 15.3.1
The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:
$2SO_{2}\left ( g \right ) + O_{2}\left ( g \right ) \rightleftharpoons 2SO_{3}\left ( g \right )$
A mixture of SO2 and O2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10−2 M SO3, 3.5 × 10−3 M O2, and 3.0 × 10−3 M SO2. Calculate K and Kp at this temperature.
Given: balanced equilibrium equation and composition of equilibrium mixture
Asked for: equilibrium constant
Strategy:
Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K.
Solution:
Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,
$K=\dfrac{\left [ SO_{3} \right ]^{2}}{\left [ SO_{2} \right ]^{2}\left [ O_{2} \right ]}=\dfrac{\left ( 5.0\times 10^{-2} \right )^{2}}{\left ( 3.0\times 10^{-3} \right )^{2}\left ( 3.5\times 10^{-3} \right )}=7.9\times 10^{4}$
To solve for Kp, we use Equation 15.2.17, where Δn = 2 − 3 = −1:
$K_{p}= K\left ( RT \right )^{\Delta n}$
$=7.9\times 10^{4}\left [ \left (0.082606\; L\cdot atm/mol\cdot \cancel{K} \right ) \left ( 800 \; \cancel{K} \right )\right ]$
$=1.2\times 10^{3}$
Exercise
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
$H_{2}\left ( g \right ) + I_{2}\left ( g \right ) \rightleftharpoons 2HI\left ( g \right )$
A mixture of H2 and I2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10−2 M HI, 6.47 × 10−3 M H2, and 5.94 × 10−4 M I2. Calculate K and Kp for this reaction.
Answer: K = 48.8; Kp = 48.8
Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this.
Example 15.3.2
A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl2. Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl2 is as follows:
$2NOCl \left ( g \right ) \rightleftharpoons 2NO\left ( g \right ) + Cl_{2}\left ( g \right )$
Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium
Asked for: K
Strategy:
A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).
B Calculate all possible initial concentrations from the data given and insert them in the table.
C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.
D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.
Solution:
A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:
$K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}$
To obtain the concentrations of NOCl, NO, and Cl2 at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.
2NOCl(g) ⇌ 2NO(g) + Cl2
[NOCl] [NO] [Cl2]
initial
change
final
B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl]i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl2 are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl2 in a 2.00 L container, so [Cl2]f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:
2NOCl(g) ⇌ 2NO(g) + Cl2
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change
final 0.028
C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl2, the substance for which initial and final concentrations are known:
Δ[Cl2] = [0.028 M (final) − 0.00 M (initial)] = +0.028 M
According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl2, so the change in the NO concentration is as follows:
$\Delta \left [ NO \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=0.056\; M$
Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl2 produced, so the change in the NOCl concentration is as follows:
$\Delta \left [ NOCl \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{-2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=-0.056\; M$
We insert these values into our table:
2NOCl(g) ⇌ 2NO(g) + Cl2
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.028
D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:
[NO]f = 0.000 M + 0.056 M = 0.056 M [NOCl]f = 0.500 M + (−0.056 M) = 0.444 M
We can now complete the table:
2NOCl(g) ⇌ 2NO(g) + Cl2
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.444 0.056 0.028
We can now calculate the equilibrium constant for the reaction:
$K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}=\dfrac{\left ( 0.056 \right )^{2}\left ( 0.028 \right )}{0.444}^{2}=4.5\times 10^{-4}$
Exercise
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH3) by reacting 0.1248 M H2 and 0.0416 M N2 at about 500°C. At equilibrium, the mixture contained 0.00272 M NH3. What is K for the reaction N2 + 3 H2 ⇌ 2NH3at this temperature? What is Kp?
Answer: K = 0.105; Kp = 2.61 × 10−5
The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.
Calculating Equilibrium Concentrations from the Equilibrium Constant
To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation 15.26), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9.
nbutane(g) ⇌isobutane(g)
[n-Butane] [Isobutane]
initial
change
final
The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as +x, then the change in the concentration of n-butane is Δ[n-butane] = −x. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.
nbutane(g) ⇌isobutane(g)
[n-Butane] [Isobutane]
initial 1.00 0
change x +x
final (1.00 − x) (0 + x) = x
Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,
$K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{x}{1.00-x}=2.6$
Rearranging and solving for x,
$x = 2.6\left ( 1.00-x \right )=2.6-2.6x$
$x + 2.6x =2.6$
$x = 0.72$
We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n-butane and isobutane listed in the table:
[n-butane]f = (1.00 − x) M = (1.00 − 0.72) M = 0.28 M [isobutane]f = (0.00 + x) M = (0.00 + 0.72) M = 0.72 M
We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same K that we used in the calculation:
$K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{0.72 \; \cancel{M}}{0.28 \; \cancel{M}}=2.6$
This is the same K we were given, so we can be confident of our results.
Example 10 illustrates a common type of equilibrium problem that you are likely to encounter.
Example 15.3.3
The water–gas shift reaction is important in several chemical processes, such as the production of H2 for fuel cells. This reaction can be written as follows:
$H_{2}\left ( g \right ) + CO_{2}\left ( g \right ) \rightleftharpoons H_{2}O\left ( g \right ) + CO\left ( g \right )$
K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H2 and 0.0150 M CO2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?
Given: balanced equilibrium equation, K, and initial concentrations
Asked for: final concentrations
Strategy:
A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x. From the values in the table, calculate the final concentrations.
B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x.
C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K.
Solution:
A The initial concentrations of the reactants are [H2]i = [CO2]i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H2O as x, then Δ[H2O] = +x. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x. For example, 1 mol of CO is produced for every 1 mol of H2O, so the change in the CO concentration can be expressed as Δ[CO] = +x. Similarly, for every 1 mol of H2O produced, 1 mol each of H2 and CO2 are consumed, so the change in the concentration of the reactants is Δ[H2] = Δ[CO2] = −x. We enter the values in the following table and calculate the final concentrations.
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
[H2] [CO2] [H2O] [CO]
initial 0.0150 0.0150 0 0
change x x +x +x
final (0.0150 − x) (0.0150 − x) x x
B We can now use the equilibrium equation and the given K to solve for x:
$K=\dfrac{\left [ H_{2}O] \right ] \left [ CO \right ]}{\left [ H_{2} \right ]\left [ CO_{2} \right ]}=\dfrac{\left (x \right )\left ( x \right ) }{\left ( 0.0150-x \right )\left ( 0.0150-x \right )}=\dfrac{x^{2}}{\left ( 0.0150-x \right )^{2}}=0.160 \notag$
We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,
$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \notag$
(The quadratic formula is presented in Essential Skills 7 in Section 15.7 .) Taking the square root of the middle and right terms,
$\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \notag$
$x =(0.326)(0.0150)−0.326x \notag$
$1.326x=0.00489 \notag$
$x =0.00369=3.69 \times 10^{−3} \notag$
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M$
• $[H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M$
• $[CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M$
We can check our work by inserting the calculated values back into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107 \notag$
To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.
Exercise
Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)} \notag$
K = 54 at 425°C. If 0.172 M H2 and I2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?
Answer: [HI]f = 0.270 M; [H2]f = [I2]f = 0.037 M
In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11.
Example 15.3.4
In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO2 and 0.570 M H2 is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?
Given: balanced equilibrium equation, concentrations of reactants, and K
Asked for: composition of reaction mixture at equilibrium
Strategy:
A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations (x) and the final concentrations.
B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x.
C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K.
Solution:
A [CO2]i = 0.632 M and [H2]i = 0.570 M. Again, x is defined as the change in the concentration of H2O: Δ[H2O] = +x. Because 1 mol of CO is produced for every 1 mol of H2O, the change in the concentration of CO is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of H2 and CO2 are consumed for every 1 mol of H2O produced, Δ[H2] = Δ[CO2] = −x. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}$
[H2] [CO2] [H2O] [CO]
initial 0.570 0.632 0 0
change x x +x +x
final (0.570 − x) (0.632 − x) x x
B We can now use the equilibrium equation and the known K value to solve for x:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \notag$
In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:
$x^2 = 0.106(0.360 − 1.20x + x^2) \notag$
Collecting terms on one side of the equation,
$0.894x^2 + 0.127x − 0.0382 = 0 \notag$
This equation can be solved using the quadratic formula:
$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \notag$
$x =0.148 \text{ and } −0.290 \notag$
Only the answer with the positive value has any physical significance, so Δ[H2O] = Δ[CO] = +0.148 M, and Δ[H2] = Δ[CO2] = −0.148 M.
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M$
• $[H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M$
• $[CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M$
We can check our work by substituting these values into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \notag$
Because K is essentially the same as the value given in the problem, our calculations are confirmed.
Exercise
The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H2 and 0.0450 M I2 is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?
Answer: [HI]f = 0.0882 M; [H2]f = 0.156 M; [I2]f = 9.2 × 10−4 M
In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small (K ≤ 10−3) or very large (K ≥ 103), which means that the change in the concentration (defined as x) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12.
Example 15.3.5
Atmospheric nitrogen and oxygen react to form nitric oxide:
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)} \notag$
Kp = 2.0 × 10−31 at 25°C. What is the partial pressure of NO in equilibrium with N2 and O2 in the atmosphere (at 1 atm, P{N2} = 0.78 atm and P{O2} = 0.21 atm
Given: balanced equilibrium equation and values of Kp, P{O2} and P{N2}
Asked for: partial pressure of NO
Strategy:
A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.
B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration (x).
C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K.
Solution:
A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O2 is 0.21 atm and that of N2 is 0.78 atm. If we define the change in the partial pressure of NO as 2x, then the change in the partial pressure of O2 and of N2 is −x because 1 mol each of N2 and of O2 is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}$
P{N2} P{N2} P{NO}
initial P 0.78 0.21 0
change in P x x +2x
final P (0.78 − x) (0.21 − x) 2x
B Substituting these values into the equation for the equilibrium constant,
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \notag$
In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation,
$\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31} \notag$
$\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \notag$
$x^2=\dfrac{0.33 \times 10^{−31}}{4} \notag$
$x^=9.1 \times 10^{−17} \notag$
C Substituting this value of x into our expressions for the final partial pressures of the substances,
• $P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm$
• $P_{N_2}=(0.78−x) \;atm=0.78 \;atm$
• $P_{O_2}=(0.21−x) \;atm=0.21\; atm$
From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10−16 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10−3 > K > 103, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N2 and O2 to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 times 10^{−31} \notag$
The final Kp agrees with the value given at the beginning of this example.
Exercise
Under certain conditions, oxygen will react to form ozone, as shown in the following equation:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag$
Kp = 2.5 × 10−59 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere P(O2) =0.21 atm?
Answer: 4.8 × 10−31 atm
Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (K ≥ 103). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example 13.
Example 15.3.6
The chemical equation for the reaction of hydrogen with ethylene (C2H4) to give ethane (C2H6) is as follows:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag$
K = 9.6 × 1018 at 25°C. If a mixture of 0.200 M H2 and 0.155 M C2H4 is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?
Given: balanced chemical equation, K, and initial concentrations of reactants
Asked for: equilibrium concentrations
Strategy:
A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.
B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).
C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.
Solution:
A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}$
[H2] [C2H4] [C2H6]
initial 0.200 0.155 0
assuming 100% reaction 0.045 0 0.155
change +x +x x
final (0.045 + x) (0 + x) (0.155 − x)
B Substituting values into the equilibrium constant expression,
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \notag$
Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows:
$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \notag$
$x=3.6 \times 10^{−19} \notag$
C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:
• $[C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M$
• $[C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M$
• $[H_2]_f = (0.045 + x) \;M = 0.045 \;M$
We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \notag$
This K value agrees with our initial value at the beginning of the example.
Exercise
Hydrogen reacts with chlorine gas to form hydrogen chloride:
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \notag$
Kp = 4.0 × 1031 at 47°C. If a mixture of 0.257 M H2 and 0.392 M Cl2 is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?
Answer:
• $[H_2]_f = 4.8 \times 10^{−32}\; M$
• $[Cl_2]_f = 0.135\; M$
• $[HCl]_f = 0.514\; M$
Summary
When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.
Key Takeaway
• Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.
Conceptual Problems
1. Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.
2. Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A + 2B ⇌ C for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?
Numerical Problems
Please be sure you are familiar with the topics discussed in Essential Skills 7 (Section 15.7 ) before proceeding to the Numerical Problems.
1. In the equilibrium reaction A + B ⇌C, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction A ⇌B + C?
2. The following table shows the reported values of the equilibrium P{O2} at three temperatures for the reaction Ag2O(s) ⇌ 2 Ag(s) + 1/2 O2(g) for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?
T (°C) P(O2) mm Hg
150 182
184 143
191 126
3. Given the equilibrium system N2O4(g) ⇌ 2 NO2(g), what happens to Kp if the initial pressure of N2O4 is doubled? If Kp is 1.7 × 10−1 at 2300°C, and the system initially contains 100% N2O4 at a pressure of 2.6 × 102 atm, what is the equilibrium pressure of each component?
4. At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H2(g) + I2(g) ⇌2HI(g) At equilibrium, [H2] = 0.047 M and [HI] = 0.345 M. What are K and Kp for this reaction?
5. Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO(g) + 2H2(g) ⇌ CH3OH(g) and Kp = 1.3 × 10−4. If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?
6. Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reactionA(s ⇌ 2 B(g) + C(g), what is Kp?
7. The decomposition of ammonium carbamate to NH3 and CO2 at 40°C is written as NH4CO2NH2(s) ⇌ 2NH3(g) + CO2 If the partial pressure of NH3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO2? What is the total gas pressure of the system? What is Kp?
8. At 375 K, Kp for the reaction SO2Cl2(g) ⇌ SO2(g) + Cl2g) is 2.4, with pressures expressed in atmospheres. At 303 K, Kp is 2.9 × 10−2.
1. What is K for the reaction at each temperature?
2. If a sample at 375 K has 0.100 M Cl2 and 0.200 M SO2 at equilibrium, what is the concentration of SO2Cl2?
3. If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
9. For the gas-phase reaction aA ⇌ bB, show that Kp = K(RT)Δn assuming ideal gas behavior.
10. For the gas-phase reaction I2 ⇌2I, show that the total pressure is related to the equilibrium pressure by the following equation:
$P_T=\sqrt{K_pP_{I_2}} + P_{I_2} \notag$
11. Experimental data on the system Br2(l) ⇌ Br2(aq) are given in the following table. Graph [Br2] versus moles of Br2(l) present; then write the equilibrium constant expression and determine K.
Grams Br2 in 100 mL Water [Br2] (M)
1.0 0.0626
2.5 0.156
3.0 0.188
4.0 0.219
4.5 0.219
12. Data accumulated for the reaction n-butane(g) ⇌ isobutane(g) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?
Moles n-butane Moles Isobutane
0.5 1.25
1.0 2.5
1.50 3.75
13. Solid ammonium carbamate (NH4CO2NH2) dissociates completely to ammonia and carbon dioxide when it vaporizes:
$NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)} \notag$
At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is Kp? If the concentration of CO2 is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the NH3 concentration is necessary for the system to restore equilibrium?
14. The equilibrium constant for the reaction COCl2(g) ⇌ CO(g) + Cl2(g) is Kp = 2.2 × 10−10 at 100°C. If the initial concentration of COCl2 is 3.05 × 10−3 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?
15. Aqueous dilution of IO4 results in the following reaction:
$IO^−_{4(aq)}+2H_2O_{(l)} \rightleftharpoons H_4IO^−_{6(aq)} \notag$
and K = 3.5 × 10−2. If you begin with 50 mL of a 0.896 M solution of IO4 that is diluted to 250 mL with water, how many moles of H4IO6 are formed at equilibrium?
16. Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation:
$I_{2(g)}+Br_{2(g)} \rightleftharpoons 2IBr_{(g)} \notag$
Kp = 1.2 × 102. If you begin the reaction with 7.4 g of I2 vapor and 6.3 g of Br2 vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?
17. For the reaction
$C_{(s)} + 12N_{2(g)}+\frac{5}{2}H_{2(g)} \rightleftharpoons CH3NH2(g) \notag$
K = 1.8 × 10−6. If you begin the reaction with 1.0 mol of N2, 2.0 mol of H2, and sufficient C(s) in a 2.00 L container, what are the concentrations of N2 and CH3NH2 at equilibrium? What happens to K if the concentration of H2 is doubled?
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.03%3A_Solving_Equilibrium_Problems.txt |
Learning Objectives
• To predict in which direction a reaction will proceed.
In Section 15.3, we saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination.
The Reaction Quotient
To determine whether a system has reached equilibrium, chemists use a quantity called the reaction quotient (Q)A quantity derived from a set of values measured at any time during the reaction of any mixture of reactants and products, regardless of whether the system is at equilibrium: Q = [C]c[D]d/[A]a[B]b for the general balanced chemical equation aA + bB ⇌ cC + dD. The expression for the reaction quotient has precisely the same form as the equilibrium constant expression, except that Q may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction:
$aA+bB \rightarrow cC+dD \notag$
the reaction quotient is defined as follows:
$Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{15.4.1}$
The reaction quotient Qp analogous to Kp, can be written for any reaction that involves gases by using the partial pressures of the components.
To understand how information is obtained using a reaction quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide,
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)} \notag$
for which K = 4.65 × 10−3 at 298 K. We can write Q for this reaction as follows:
$Q=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.4.2}$
The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of Q were calculated for each. Each experiment begins with different proportions of product and reactant:
Experiment [NO2] (M) [N2O4] (M) Q = [NO2]2/[N2O4]
1 0 0.0400 $\dfrac{0^2}{0.0400}=0$
2 0.0600 0 $\dfrac{(0.0600)^2}{0}=\text{undefined}$
3 0.0200 0.0600 $\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{−3}$
As these calculations demonstrate, Q can have any numerical value between 0 and infinity (undefined); that is, Q can be greater than, less than, or equal to K.
Comparing the magnitudes of Q and K enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes Q approach K. If Q = K, for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed. If Q < K, then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants. Conversely, if Q > K, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products. These points are illustrated graphically in Figure 15.4.1
Note the Pattern
If Q < K, the reaction will proceed to the right as written. If Q > K, the reaction will proceed to the left as written. If Q = K, then the system is at equilibrium.
Example 15.4.1
At elevated temperatures, methane (CH4) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction:
$CH_{4(g)}+H_2O_{(g)} \rightleftharpoons CO_{(g)}+3H_{2(g)} \notag$
K = 2.4 × 10−4 at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If 1.2 × 10−2 mol of CH4, 8.0 × 10−3 mol of H2O, 1.6 × 10−2 mol of CO, and 6.0 × 10−3 mol of H2 are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce CO and H2 or to the left to form CH4 and H2O?
Given: balanced chemical equation, K, amounts of reactants and products, and volume
Asked for: direction of reaction
Strategy:
A Calculate the molar concentrations of the reactants and the products.
B Use Equation 15.4.1 to determine Q. Compare Q and K to determine in which direction the reaction will proceed.
Solution:
A We must first find the initial concentrations of the substances present. For example, we have 1.2 × 10−2 mol of CH4 in a 2.0 L container, so
$[CH_4]=\dfrac{1.2\times 10^{−2} mol}{2.0\; L}=6.0 \times 10^{−3} M \notag$
We can calculate the other concentrations in a similar way:
• $[H_2O] = 4.0 \times 10^{−3} M$,
• $[CO] = 8.0 \times 10^{−3} M$, and
• $[H_2] = 3.0 \times 10^{−3} M$.
B We now compute Q and compare it with K:
$Q=\dfrac{[CO][H_2]^3}{[CH_4][H_2O}=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})}=9.0 \times 10^{−6} \notag$
Because K = 2.4 × 10−4, we see that Q < K. Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming H2 and CO at the expense of H2O and CH4.
Exercise
In the water–gas shift reaction introduced in Example 10, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen:
$CO_{(g)}+H_2O_{(g)} \rightleftharpoons CO_{2(g)}+H_{2(g)} \notag$
K = 0.64 at 900 K. If 0.010 mol of both CO and H2O, 0.0080 mol of CO2, and 0.012 mol of H2 are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written?
Answer: Q = 0.96 (Q > K), so the reaction will proceed to the left, and CO and H2O will form.
Predicting the Direction of a Reaction with a Graph
By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which Q = K. Such a graph allows us to predict what will happen to a reaction when conditions change so that Q no longer equals K, such as when a reactant concentration or a product concentration is increased or decreased.
Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:
$PbCO_{3(s)} \rightleftharpoons PbO_{(s)}+CO_{2(g)} \tag{15.4.3}$
Because PbCO3 and PbO are solids, the equilibrium constant is simply K = [CO2]. At a given temperature, therefore, any system that contains solid PbCO3 and solid PbO will have exactly the same concentration of CO2 at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure 15.4.2, which shows a plot of [CO2] versus the amount of PbCO3 added. Initially, the added PbCO3 decomposes completely to CO2 because the amount of PbCO3 is not sufficient to give a CO2 concentration equal to K. Thus the left portion of the graph represents a system that is not at equilibrium because it contains only CO2(g) and PbO(s). In contrast, when just enough PbCO3 has been added to give [CO2] = K, the system has reached equilibrium, and adding more PbCO3 has no effect on the CO2 concentration: the graph is a horizontal line. Thus any CO2 concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough PbCO3 and PbO are present. For example, the point labeled A in Figure 15.4.2 lies above the horizontal line, so it corresponds to a [CO2] that is greater than the equilibrium concentration of CO2 (Q > K). To reach equilibrium, the system must decrease [CO2], which it can do only by reacting CO2 with solid PbO to form solid PbCO3. Thus the reaction in Equation 15.4.3 will proceed to the left as written, until [CO2] = K. Conversely, the point labeled B in Figure 15.4.2 lies below the horizontal line, so it corresponds to a [CO2] that is less than the equilibrium concentration of CO2 (Q < K). To reach equilibrium, the system must increase [CO2], which it can do only by decomposing solid PbCO3 to form CO2 and solid PbO. The reaction in Equation 15.4.2 will therefore proceed to the right as written, until [CO2] = K.
In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:
$CdO_{(s)}+H_{2(g)} \rightleftharpoons Cd_{(s)}+H_2O_{(g)} \tag{15.4.4}$
and the equilibrium constant K is [H2O]/[H2]. If [H2O] is doubled at equilibrium, then [H2] must also be doubled for the system to remain at equilibrium. A plot of [H2O] versus [H2] at equilibrium is a straight line with a slope of K (Figure 15.4.3). Again, only those pairs of concentrations of H2O and H2 that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation 15.4.4 will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in Figure 15.4.3 lies below the line, indicating that the [H2O]/[H2] ratio is less than the ratio of an equilibrium mixture (Q < K). Thus the reaction in Equation 15.4.3 will proceed to the right as written, consuming H2 and producing H2O, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in Figure 15.4.3 lies above the line, indicating that the [H2O]/[H2] ratio is greater than the ratio of an equilibrium mixture (Q > K). Thus the reaction in Equation 15.4.3 will proceed to the left as written, consuming H2O and producing H2, which causes the concentration ratio to move down and to the right toward the equilibrium line.
In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:
$NH_4I_{(s)} \rightleftharpoons NH_{3(g)}+HI_{(g)} \tag{15.4.5}$
For this system, K is equal to the product of the concentrations of the two products: [NH3][HI]. If we double the concentration of NH3, the concentration of HI must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure 15.4.4. As a result, for a given concentration of either HI or NH3, only a single equilibrium composition that contains equal concentrations of both NH3 and HI is possible, for which [NH3] = [HI] = K1/2. Any point that lies below and to the left of the equilibrium curve (such as point A in Figure 15.4.4) corresponds to Q < K, and the reaction in Equation 15.4.4 will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure 15.4.3 ) corresponds to Q > K, and the reaction in Equation 15.4.4 will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium.
Le Chatelier’s Principle
When a system at equilibrium is perturbed in some way, the effects of the perturbation can be predicted qualitatively using Le Chatelier’s principleIf a stress is applied to a system at equilibrium, the composition of the system will change to relieve the applied stress. (named after the French chemist Henri Louis Le Chatelier, 1850–1936).The name is pronounced “Luh SHOT-lee-ay.” This principle can be stated as follows: if a stress is applied to a system at equilibrium, the composition of the system will change to counteract the applied stress. Stress occurs when any change in a system affects the magnitude of Q or K. In Equation 15.4.4, for example, increasing [NH3] produces a stress on the system that requires a decrease in [HI] for the system to return to equilibrium. As a further example, consider esters, which are one of the products of an equilibrium reaction between a carboxylic acid and an alcohol. (For more information on this type of reaction, see Section 7.5 .) Esters are responsible for the scents we associate with fruits (such as oranges and bananas), and they are also used as scents in perfumes. Applying a stress to the reaction of a carboxylic acid and an alcohol will change the composition of the system, leading to an increase or a decrease in the amount of ester produced. In Section 15.5 and Section 15.6 , we explore how chemists control reactions conditions to affect equilibrium concentrations.
Note the Pattern
In all reactions, if a stress is applied to a system at equilibrium, the composition of the system will change to counteract the applied stress (Le Chatelier’s principle).
Example 15.4.2
Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium.
1. 2HgO(s) ⇌ 2Hg(l) + O2(g): the amount of HgO is doubled.
2. NH4HS(s) ⇌ NH3(g) + H2S(g): the concentration of H2S is tripled.
3. n-butane(g) ⇌ isobutane(g): the concentration of isobutane is halved.
Given: equilibrium systems and changes
Asked for: equilibrium constant expressions and effects of changes
Strategy:
Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made.
Solution:
1. Because HgO(s) and Hg(l) are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, K = [O2]. The equilibrium concentration of O2 is a constant and does not depend on the amount of HgO present. Hence adding more HgO will not affect the equilibrium concentration of O2, so no compensatory change is necessary.
2. NH4HS does not appear in the equilibrium constant expression because it is a solid. Thus K = [NH3][H2S], which means that the concentrations of the products are inversely proportional. If adding H2S triples the H2S concentration, for example, then the NH3 concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals K.
3. For this reaction, K = [isobutane]/[n-butane], so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium.
Exercise
Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium?
1. HBr(g) + NaH(s) ⇌ NaBr(s) + H2(g): the concentration of HBr is decreased by a factor of 3.
2. 6Li(s) + N2(g) ⇌ 2Li3N(s): the amount of Li is tripled.
3. SO2(g) + Cl2(g) ⇌ SO2Cl2(l): the concentration of Cl2 is doubled.
Answer
1. K = [H2]/[HBr]; [H2] must decrease by about a factor of 3.
2. K = 1/[N2]; solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary.
3. K = 1/[SO2][Cl2]; [SO2] must decrease by about half.
Summary
The reaction quotient (Q or Qp) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, Q = K. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. Le Chatelier’s principle states that if a stress is applied to a system at equilibrium, the composition of the system will adjust to counteract the stress.
Key Takeaway
• The reaction quotient (Q) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction.
Key Equation
Reaction quotient
Equation 15.4.1: $Q =\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
Conceptual Problems
1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.4.4 and Figure 15.4.4 as your guides, sketch the shape of each graph using appropriate labels.
1. $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$
2. $2MgO_{(s)} \rightleftharpoons 2Mg_{(s)}+O_{2(g)}$
3. $2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
4. $2PbS_{(s)}+3O_{2(g)} \rightleftharpoons 2PbO_{(s)}+2SO_{2(g)}$
2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?
1. 2NaHCO3(s) ⇌ Na2CO3(s) + CO2(g) + H2O(g): [CO2] is doubled.
2. N2F4(g) ⇌ 2NF2(g): [NF] is decreased by a factor of 2.
3. H2(g) + I2(g) ⇌ 2HI(g): [I2] is doubled.
3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?
1. CS2(g) + 4H2(g) ⇌ CH4(g) + 2H2S(g): [CS2] is doubled.
2. PCl5(g) ⇌ PCl3(g) + Cl2(g): [Cl2] is decreased by a factor of 2.
3. 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g): [NO] is doubled.
Answer
1. $K=\dfrac{[CH_4][H_2S]^2}{[CS_2][H_2]^4}$; doubling $[CS_2]$ would require decreasing $[H_2]$ by a factor of 2√4≅ 1.189.
2. $K=\dfrac{[PCl_3]}{[Cl_2][PCl_5]}$; if $[Cl_2]$ is halved, $[PCl_5]$ must also be halved.
3. $K=\dfrac{[NO]^4[H_2O]^6}{[NH_3][O_2]^5}$; if $[NO]$ is doubled, $[H_2O]$ is multiplied by 22/3≅1.587.
Numerical Problems
1. The data in the following table were collected at 450°C for the reaction
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Equilibrium Partial Pressure (atm)
P (atm) NH3 N2 H2
30 (equilibrium) 1.740 6.588 21.58
100 15.20 19.17 65.13
600 321.6 56.74 220.8
The reaction equilibrates at a pressure of 30 atm. The pressure on the system is first increased to 100 atm and then to 600 atm. Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?
2. For the reaction 2A ⇌ B + 3 C, K at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?
Experiment A B C
1 2.50 M 2.50 M 2.50 M
2 1.30 atm 1.75 atm 14.15 atm
3 12.61 mol 18.72 mol 6.51 mol
3. The following two reactions are carried out at 823 K:
$CoO_{(s)}+H_{2(g)} \rightleftharpoons Co_{(s)}+H_2O_{(g)}$ with $K=67$
$CoO_{(s)}+CO_{(g)} \rightleftharpoons Co_{(s)}+CO_{2(g)}$ with $K=490$
1. Write the equilibrium expression for each reaction.
2. Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of H2 or CO plus 0.500 mol CoO.
3. Using the information provided, calculate Kp for the following reaction: $H_{2(g)}+CO_{2(g)} \rightleftharpoons CO_{(g)}+H_2O_{(g)}$
4. Describe the shape of the graphs of [reactants] versus [products] as the amount of CoO changes.
4. Hydrogen iodide (HI) is synthesized via H2(g) + I2(g) ⇌ 2HI(g), for which Kp = 54.5 at 425°C. Given a 2.0 L vessel containing 1.12 × 10−2 mol of H2 and 1.8 × 10−3 mol of I2 at equilibrium, what is the concentration of HI? Excess hydrogen is added to the vessel so that the vessel now contains 3.64 × 10−1 mol of H2. Calculate Q and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?
Answers
1. Not at equilibrium; in both cases, the sum of the equilibrium partial pressures is less than the total pressure, so the reaction will proceed to the right to decrease the pressure.
1. $K=[H_2O][H_2]$; K=[CO2][CO]
2. $P_{H_2O} = 21.0\; atm$; $P_{H_2} = 0.27\; atm$; $P_{CO_2} = 21.3\; atm$; $P_{CO} = 0.07\; atm$
3. $K_p = 0.14$
4. The amount of CoO has no effect on the shape of a graph of products versus reactants as long as some solid CoO is present.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.04%3A_Non-equilibrium_Conditions.txt |
Learning Objectives
• To predict the effects of stresses on a system at equilibrium.
Chemists use various strategies to increase the yield of the desired products of reactions. When synthesizing an ester, for example, how can a chemist control the reaction conditions to obtain the maximum amount of the desired product? Only three types of stresses can change the composition of an equilibrium mixture: (1) a change in the concentrations (or partial pressures) of the components by adding or removing reactants or products, (2) a change in the total pressure or volume, and (3) a change in the temperature of the system. In this section, we explore how changes in reaction conditions can affect the equilibrium composition of a system. We will explore each of these possibilities in turn.
Changes in Concentration
If we add a small volume of carbon tetrachloride (CCl4) solvent to a flask containing crystals of iodine, we obtain a saturated solution of I2 in CCl4, along with undissolved crystals:
$I_{2(s)} \rightleftharpoons I_{2(soln)}\tag{15.5.1}$
The system reaches equilibrium, with K = [I2]. If we add more CCl4, thereby diluting the solution, Q is now less than K. Le Chatelier’s principle tells us that the system will react to relieve the stress—but how? Adding solvent stressed the system by decreasing the concentration of dissolved I2. Hence more crystals will dissolve, thereby increasing the concentration of dissolved I2 until the system again reaches equilibrium if enough solid I2 is available (Figure 15.5.1). By adding solvent, we drove the reaction shown in Equation 15.5.1 to the right as written.
We encounter a more complex system in the reaction of hydrogen and nitrogen to form ammonia:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{15.5.2}$
The Kp for this reaction is 2.14 × 10−2 at about 540 K. Under one set of equilibrium conditions, the partial pressure of ammonia is P(NH3) = 0.454 atm, that of hydrogen is P(H2) = 2.319 atm, and that of nitrogen is P(N2) = 0.773 atm. If an additional 1 atm of hydrogen is added to the reactor to give P(H2) = 3.319 atm, how will the system respond? Because the stress is an increase in P(H2) the system must respond in some way that decreases the partial pressure of hydrogen to counteract the stress. The reaction will therefore proceed to the right as written, consuming H2 and N2 and forming additional NH3. Initially, the partial pressures of H2 and N2 will decrease, and the partial pressure of NH3 will increase until the system eventually reaches a new equilibrium composition, which will have a net increase in P(H2)
We can confirm that this is indeed what will happen by evaluating Qp under the new conditions and comparing its value with Kp. The equations used to evaluate Kp and Qp have the same form: substituting the values after adding hydrogen into the expression for Qp results in the following:
$Q_p=\dfrac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}=\dfrac{(0.454)^2}{(0.773)(2.319+1.00)^3}=7.29 \times 10^{-3}$
Thus Qp < Kp, which tells us that the ratio of products to reactants is less than at equilibrium. To reach equilibrium, the reaction must proceed to the right as written: the partial pressures of the products will increase, and the partial pressures of the reactants will decrease. Qp will thereby increase until it equals Kp, and the system will once again be at equilibrium. Changes in the partial pressures of the various substances in the reaction mixture (Equation 15.5.2) as a function of time are shown in Figure 15.5.2.
We can force a reaction to go essentially to completion, regardless of the magnitude of K, by continually removing one of the products from the reaction mixture. Consider, for example, the methanation reaction, in which hydrogen reacts with carbon monoxide to form methane and water:
$CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \tag{15.5.3}$
This reaction is used for the industrial production of methane, whereas the reverse reaction is used for the production of H2 (Example 14). The expression for Q has the following form:
$Q=\dfrac{[CH_4][H_2O]}{[CO][H_2]^3} \tag{15.5.4}$
Regardless of the magnitude of K, if either H2O or CH4 can be removed from the reaction mixture so that [H2O] or [CH4] is approximately zero, then Q ≈ 0. In other words, when product is removed, the system is stressed (Q << K), and more product will form to counter the stress. Because water (bp = 100°C) is much less volatile than methane, hydrogen, or carbon monoxide (all of which have boiling points below −100°C), passing the gaseous reaction mixture through a cold coil will cause the water vapor to condense to a liquid that can be drawn off. Continuing to remove water from the system forces the reaction to the right as the system attempts to equilibrate, thus enriching the reaction mixture in methane. This technique, referred to as driving a reaction to completion, can be used to force a reaction to completion even if K is relatively small. For example, esters are usually synthesized by removing water. The products of the condensation reaction are shown here. In Chapter 19, we will describe the thermodynamic basis for the change in the equilibrium position caused by changes in the concentrations of reaction components.
Example 15.5.1
For each equilibrium system, predict the effect of the indicated stress on the specified quantity.
1. 2SO2(g) + O2(g) ⇌ 2SO3(g):
(1) the effect of removing O2 on P(SO2)
(2) the effect of removing O2 on P(SO3)
2. CaCO3(s) ⇌ CaO(s) + CO2(g):
(1) the effect of removing CO2 on the amount of CaCO3;
(2) the effect of adding CaCO3 on P(CO2)
Given: balanced chemical equations and changes
Asked for: effects of indicated stresses
Strategy:
Use Q and K to predict the effect of the stress on each reaction.
Solution:
1. (1) Removing O2 will decrease P(O2)thereby decreasing the denominator in the reaction quotient and making Qp > Kp. The reaction will proceed to the left as written, increasing the partial pressures of SO2 and O2 until Qp once again equals Kp.
(2) Removing O2 will decrease P(O2) and thus increase Qp, so the reaction will proceed to the left. The partial pressure of SO3 will decrease.
2. Kp and Qp are both equal to P(CO2).
(1) Removing CO2 from the system causes more CaCO3 to react to produce CO2, which increases P(CO2) to the partial pressure required by Kp. (2) Adding (or removing) solid CaCO3 has no effect on P(CO2) because it does not appear in the expression for Kp (or Qp).
Exercise
For each equilibrium system, predict the effect that the indicated stress will have on the specified quantity.
1. $H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}$:
1. the effect of adding $CO$ on $[H_2]$;
2. the effect of adding $CO_2$ on $[H_2]$
2. $CuO_{(s)}+CO_{(g)} \rightleftharpoons Cu_{(s)}+CO_{2(g)}$:
1. the effect of adding $CO$ on the amount of $Cu$;
2. the effect of adding $CO_2$ on $[CO]$
Answer
1. (1) [H2] increases; (2) [H2] decreases.
2. (1) the amount of Cu increases; (2) [CO] increases.
Changes in Total Pressure or Volume
Because liquids are relatively incompressible, changing the pressure above a liquid solution has little effect on the concentrations of dissolved substances. Consequently, changes in external pressure have very little effect on equilibrium systems that contain only solids or liquids. In contrast, because gases are highly compressible, their concentrations vary dramatically with pressure. From the ideal gas law, PV = nRT, described in Chapter 10, the concentration (C) of a gas is related to its pressure as follows:
$C=\dfrac{n}{V}=\dfrac{P}{RT} \tag{15.5.5}$
Hence the concentration of any gaseous reactant or product is directly proportional to the applied pressure (P) and inversely proportional to the total volume (V). Consequently, the equilibrium compositions of systems that contain gaseous substances are quite sensitive to changes in pressure, volume, and temperature.
These principles can be illustrated using the reversible dissociation of gaseous N2O4 to gaseous NO2 (Equation 15.2.1).
$N_2O_4 \rightleftharpoons 2 NO_2 \tag{15.1.1}$
The syringe shown in Figure 15.5.3 initially contains an equilibrium mixture of colorless N2O4 and red-brown NO2. Decreasing the volume by 50% causes the mixture to become darker because all concentrations have doubled. Decreasing the volume also constitutes a stress, however, as we can see by examining the effect of a change in volume on Q. At equilibrium,
$Q = K = \dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.5.6a}$
If the volume is decreased by half, the concentrations of the substances in the mixture are doubled, so the new reaction quotient is as follows:
$Q=\dfrac{(2[NO_2]_i)^2}{2[N_2O_4]_i}=\dfrac{4([NO_2]_i)^2}{2[N_2O_4]_i}=2K \tag{15.5.6b}$
Because Q is now greater than K, the system is no longer at equilibrium. The stress can be relieved if the reaction proceeds to the left, consuming 2 mol of NO2 for every 1 mol of N2O4 produced. This will decrease the concentration of NO2 and increase the concentration of N2O4, causing Q to decrease until it once again equals K. Thus, as shown in part (c) in Figure 15.5.3, the intensity of the brown color due to NO2 decreases with time following the change in volume.
Figure 15.5.3 The Effect of Changing the Volume (and Thus the Pressure) of an Equilibrium Mixture of N2O4 and NO2 at Constant Temperature(a) The syringe with a total volume of 15 mL contains an equilibrium mixture of N2O4 and NO2; the red-brown color is proportional to the NO2 concentration. (b) If the volume is rapidly decreased by a factor of 2 to 7.5 mL, the initial effect is to double the concentrations of all species present, including NO2. Hence the color becomes more intense. (c) With time, the system adjusts its composition in response to the stress as predicted by Le Chatelier’s principle, forming colorless N2O4 at the expense of red-brown NO2, which decreases the intensity of the color of the mixture.
Note the Pattern
Increasing the pressure of a system (or decreasing the volume) favors the side of the reaction that has fewer gaseous molecules and vice versa.
In general, if a balanced chemical equation contains different numbers of gaseous reactant and product molecules, the equilibrium will be sensitive to changes in volume or pressure. Increasing the pressure on a system (or decreasing the volume) will favor the side of the reaction that has fewer gaseous molecules and vice versa.
Example 15.5.2
For each equilibrium system, write the reaction quotient for the system if the pressure is decreased by a factor of 2 (i.e., if the volume is doubled) at constant temperature and then predict the direction of the reaction.
1. $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
2. $C_2H_{2(g)}+C_2H_{6(g)} \rightleftharpoons 2C_2H_{4(g)}$
3. $2 NO_{2 (g)} \rightleftharpoons 2NO_{(g)}+O_{2(g)}$
Given: balanced chemical equations
Asked for: direction of reaction if pressure is halved
Strategy:
Use Le Chatelier’s principle to predict the effect of the stress.
Solution:
1. Two moles of gaseous products are formed from 4 mol of gaseous reactants. Decreasing the pressure will cause the reaction to shift to the left because that side contains the larger number of moles of gas. Thus the pressure increases, counteracting the stress. K for this reaction is [NH3]2/[N2][H2]3. When the pressure is decreased by a factor of 2, the concentrations are halved, which means that the new reaction quotient is as follows:
$Q=\dfrac{[1/2\;NH_3]^2}{[1/2\;N_2][1/2H_2]^3}=\dfrac{1/4\;[NH_3]^2}{1/16\;[N_2][H_2]^3}=4K \notag$
2. Two moles of gaseous products form from 2 mol of gaseous reactants. Decreasing the pressure will have no effect on the equilibrium composition because both sides of the balanced chemical equation have the same number of moles of gas. Here K = [C2H4]2/[C2H2][C2H6]. The new reaction quotient is as follows:
$Q=\dfrac{[C_2H_4]^2}{[C_2H_2][C_2H_6]}=\dfrac{[1/2C_2H_4]^2}{[1/2C_2H_2][1/2C_2H_6]}=\dfrac{1/4[C_2H_4]^2}{1/4[C_2H_2][C_2H_6]}=K \notag$
3. Three moles of gaseous products are formed from 2 mol of gaseous reactants. Decreasing the pressure will favor the side that contains more moles of gas, so the reaction will shift toward the products to increase the pressure. For this reaction K = [NO]2[O2]/[NO2]2. Under the new reaction conditions the reaction quotient is as follows:
$Q=\dfrac{[1/2NO]^2[1/2O_2]}{[1/2 NO_2]^2}=\dfrac{1/8[NO]^2[O_2]}{1/4[NO_2]^2}=1/2K \notag$
Exercise
For each equilibrium system, write a new reaction quotient for the system if the pressure is increased by a factor of 2 (i.e., if the volume is halved) at constant temperature and then predict the direction in which the reaction will shift.
1. $H_2O_{(g)}+CO_{(g)} \rightleftharpoons H_{2(g)}+CO_{2(g)}$
2. $H_{2(g)}+C_2H_{4(g)} \rightleftharpoons C_2H_{6(g)}$
3. $2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
Answer
1. Q = K; no effect
2. Q = 1/2 K; to the right
3. Q = 1/2 K; to the right
Changes in Temperature
In all the cases we have considered so far, the magnitude of the equilibrium constant, K or Kp, was constant. Changes in temperature can, however, change the value of the equilibrium constant without immediately affecting the reaction quotient (QK). In this case, the system is no longer at equilibrium; the composition of the system will change until Q equals K at the new temperature.
To predict how an equilibrium system will respond to a change in temperature, we must know something about the enthalpy change of the reaction (ΔHrxn). As you learned in Chapter 9 , heat is released to the surroundings in an exothermic reaction (ΔHrxn < 0), and heat is absorbed from the surroundings in an endothermic reaction (ΔHrxn > 0). We can express these changes in the following way:
• Exothermic (ΔH<0): $\text{reactants} \rightleftharpoons \text{products} + \text{heat} \tag{15.5.7}$
• Endothermic (ΔH>0): $\text{reactants} + \text{heat} \rightleftharpoons \text{products} \tag{15.5.8}$
Thus heat can be thought of as a product in an exothermic reaction and as a reactant in an endothermic reaction. Increasing the temperature of a system corresponds to adding heat. Le Chatelier’s principle predicts that an exothermic reaction will shift to the left (toward the reactants) if the temperature of the system is increased (heat is added). Conversely, an endothermic reaction will shift to the right (toward the products) if the temperature of the system is increased. If a reaction is thermochemically neutral (ΔHrxn = 0), then a change in temperature will not affect the equilibrium composition.
We can examine the effects of temperature on the dissociation of N2O4 to NO2, for which ΔH = +58 kJ/mol. This reaction can be written as follows:
$58\; kJ+N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)} \tag{15.5.9}$
Increasing the temperature (adding heat to the system) is a stress that will drive the reaction to the right, as illustrated in Figure 15.5.4. Thus increasing the temperature increases the ratio of NO2 to N2O4 at equilibrium, which increases K.
The effect of increasing the temperature on a system at equilibrium can be summarized as follows: increasing the temperature increases the magnitude of the equilibrium constant for an endothermic reaction, decreases the equilibrium constant for an exothermic reaction, and has no effect on the equilibrium constant for a thermally neutral reaction. Table 15.5.1 shows the temperature dependence of the equilibrium constants for the synthesis of ammonia from hydrogen and nitrogen, which is an exothermic reaction with ΔH° = −91.8 kJ/mol. The values of both K and Kp decrease dramatically with increasing temperature, as predicted for an exothermic reaction.
Table 15.5.1 Temperature Dependence of K and Kp for N2(g) + 3 H2 (g) ⇌ 2 NH3(g)
Temperature (K) K K p
298 3.3 × 108 5.6 × 105
400 3.9 × 104 3.6 × 101
450 2.6 × 103 1.9
500 1.7 × 102 1.0 × 10−1
550 2.6 × 101 1.3 × 10−2
600 4.1 1.7 × 10−3
Note the Pattern
Increasing the temperature causes endothermic reactions to favor products and exothermic reactions to favor reactants.
Example 15.5.3
For each equilibrium reaction, predict the effect of decreasing the temperature:
1. $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\;\;\; ΔH_{rxn}=−91.8\; kJ/mol$
2. $CaCO_{3(s)} \rightleftharpoons CaO_{(s)}+CO_{2(g)}\;\;\; ΔH_{rxn}=178\; kJ/mol$
Given: balanced chemical equations and values of ΔHrxn
Asked for: effects of decreasing temperature
Strategy:
Use Le Chatelier’s principle to predict the effect of decreasing the temperature on each reaction.
Solution:
1. The formation of NH3 is exothermic, so we can view heat as one of the products: $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}+91.8\; kJ \notag$
2. If the temperature of the mixture is decreased, heat (one of the products) is being removed from the system, which causes the equilibrium to shift to the right. Hence the formation of ammonia is favored at lower temperatures.
3. The decomposition of calcium carbonate is endothermic, so heat can be viewed as one of the reactants: $CaCO_{3(s)}+178 \;kJ \rightleftharpoons CaO_{(s)}+CO_{2(g)} \notag$
If the temperature of the mixture is decreased, heat (one of the reactants) is being removed from the system, which causes the equilibrium to shift to the left. Hence the thermal decomposition of calcium carbonate is less favored at lower temperatures.
Exercise
For each equilibrium system, predict the effect of increasing the temperature on the reaction mixture:
1. $2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $ΔH_{rxn}=−198\; kJ/mol$
2. $N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}$ with $ΔH_{rxn}=+181\; kJ/mol$
Answer
1. Reaction shifts to the left.
2. Reaction shifts to the right.
Summary
Three types of stresses can alter the composition of an equilibrium system: adding or removing reactants or products, changing the total pressure or volume, and changing the temperature of the system. A reaction with an unfavorable equilibrium constant can be driven to completion by continually removing one of the products of the reaction. Equilibriums that contain different numbers of gaseous reactant and product molecules are sensitive to changes in volume or pressure; higher pressures favor the side with fewer gaseous molecules. Removing heat from an exothermic reaction favors the formation of products, whereas removing heat from an endothermic reaction favors the formation of reactants.
Key Takeaway
• Equilibriums are affected by changes in concentration, total pressure or volume, and temperature.
Conceptual Problems
1. If an equilibrium reaction is endothermic in the forward direction, what is the expected change in the concentration of each component of the system if the temperature of the reaction is increased? If the temperature is decreased?
2. Write the equilibrium equation for the following system:
\[4NH_{3(g)} +5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}\
Would you expect the equilibrium to shift toward the products or reactants with an increase in pressure? Why?
3. The reaction rate approximately doubles for every 10°C rise in temperature. What happens to K?
4. The formation of A2B2(g) via the equilibrium reaction 2AB(g) ⇌ A2B2(g) is exothermic. What happens to the ratio kf/kr if the temperature is increased? If both temperature and pressure are increased?
5. In each system, predict the effect that the indicated change will have on the specified quantity at equilibrium:
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$:
H2 is removed; what is the effect on P(I2)?
2. $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$
If Br2 is removed; what is the effect on PNOBr?
3. $2NaHCO_{3(s)} \rightleftharpoons Na_2CO_{3(g)} + CO_{2(g)}+H_2O_{(g)}$
If CO2 is removed; what is the effect on P(NaHCO3)?
6. What effect will the indicated change have on the specified quantity at equilibrium?
1. $NH_4Cl_{(s)} \rightleftharpoons NH_{3(g)}+HCl_{(g)}$
If NH4Cl is increased; what is the effect on PHCl?
2. $2H2O_{(g)} \rightleftharpoons 2H_{2(g)}+O_{2(g)}$
If O2 is added; what is the effect on P(H2)?<
3. $PCl_{3(g)}+Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$
If Cl2 is removed; what is the effect on P(PCl5)?
Numerical Problems
1. For each equilibrium reaction, describe how Q and K change when the pressure is increased, the temperature is increased, the volume of the system is increased, and the concentration(s) of the reactant(s) is increased.
1. A(g) ⇌ B(g) Δ H = −20.6 kJ/mol
2. 2A(g) ⇌ B(g) Δ H = −0.3 kJ/mol
3. A(g) + B(g) ⇌ 2C(g) Δ H = 46 kJ/mol
2. For each equilibrium reaction, describe how Q and K change when the pressure is decreased, the temperature is increased, the volume of the system is decreased, and the concentration(s) of the reactant(s) is increased.
1. 2A(g) ⇌ B(g) Δ H = −80 kJ/mol
2. A(g) ⇌ 2B(g) Δ H = 0.3 kJ/mol
3. 2A(g) ⇌ 2B(g) + C(g) Δ H = 46 kJ/mol
3. Le Chatelier’s principle states that a system will change its composition to counteract stress. For the system CO(g) + Cl2(g) ⇌ COCl2(g), write the equilibrium constant expression Kp. What changes in the values of Q and K would you anticipate when (a) the volume is doubled, (b) the pressure is increased by a factor of 2, and (c) COCl2 is removed from the system?
4. For the equilibrium system 3O2(g) ⇌ 2 O3(g), ΔH° = 284 kJ, write the equilibrium constant expression Kp. What happens to the values of Q and K if the reaction temperature is increased? What happens to these values if both the temperature and pressure are increased?
5. Carbon and oxygen react to form CO2 gas via C(s) + O2(g) ⇌ CO2(g), for which K = 1.2 × 1069. Would you expect K to increase or decrease if the volume of the system were tripled? Why?
6. The reaction COCl2(g) ⇌ CO(g) + Cl2 (g) has K = 2.2 × 10−10 at 100°C. Starting with an initial P( COCl2) of 1.0 atm, you determine the following values of PCO at three successive time intervals: 6.32 × 10−6 atm, 1.78 × 10−6 atm, and 1.02 × 10−5 atm. Based on these data, in which direction will the reaction proceed after each measurement? If chlorine gas is added to the system, what will be the effect on Q?
7. The following table lists experimentally determined partial pressures at three temperatures for the reaction Br2(g) ⇌ 2 Br(g)
T (K) 1123 1173 1273
P(Br2) (atm) 3.000 0.3333 6.755 × 10−2
P(Br) (atm) 3.477 × 10−2 2.159 × 10−2 2.191 × 10−2
Is this an endothermic or an exothermic reaction? Explain your reasoning.
8. The dissociation of water vapor proceeds according to the following reaction:H2O(g) ⇌ 1/2 O2(g) + H2 (g). At 1300 K, there is 0.0027% dissociation, whereas at 2155 K, the dissociation is 1.18%. Calculate K and Kp. Is this an endothermic reaction or an exothermic reaction? How do the magnitudes of the two equilibriums compare? Would increasing the pressure improve the yield of H2 gas at either temperature? (Hint: assume that the system initially contains 1.00 mol of H2O in a 1.00 L container.)
9. When 1.33 mol of CO2 and 1.33 mol of H2 are mixed in a 0.750 L container and heated to 395°C, they react according to the following equation: CO2(g) + H2 (g) ⇌ CO(g) + H2O(g) If K = 0.802, what are the equilibrium concentrations of each component of the equilibrium mixture? What happens to K if H2O is removed during the course of the reaction?
10. The equilibrium reaction H2(g) + Br2(g) ⇌ 2 HBr(g) has Kp = 2.2 × 109 at 298 K. If you begin with 2.0 mol of Br2 and 2.0 mol of H2 in a 5.0 L container, what is the partial pressure of HBr at equilibrium? What is the partial pressure of H2 at equilibrium? If H2 is removed from the system, what is the effect on the partial pressure of Br2?
11. Iron(II) oxide reacts with carbon monoxide according to the following equation: FeO(s) + CO(g) ⇌ Fe(s) + CO2(g) At 800°C, K = 0.34; at 1000°C, K = 0.40.
1. A 20.0 L container is charged with 800.0 g of CO2, 1436 g of FeO, and 1120 g of iron. What are the equilibrium concentrations of all components of the mixture at each temperature?
2. What are the partial pressures of the gases at each temperature?
3. If CO were removed, what would be the effect on P(CO2) at each temperature?
12. The equilibrium constant K for the reaction C(s) + CO2 g) >⇌ 2 CO(g) is 1.9 at 1000 K and 0.133 at 298 K.
1. If excess C is allowed to react with 25.0 g of CO2 in a 3.00 L flask, how many grams of CO are produced at each temperature?
2. What are the partial pressures of each gas at 298 K? at 1000 K?
3. Would you expect K to increase or decrease if the pressure were increased at constant temperature and volume?
13. Data for the oxidation of methane, CH4(g) + 2 O2(g) ⇌ CO2(g) + 2 H2O(g), in a closed 5.0 L vessel are listed in the following table. Fill in the blanks and determine the missing values of Q and K (indicated by ?) as the reaction is driven to completion.
CH4 O2 CO2 H2O Q K
initial (moles) 0.45 0.90 0 0 ?
at equilibrium 1.29
add 0.50 mol of methane 0.95 ?
new equilibrium ?
remove water 0 ?
new equilibrium 1.29
Answers
1. Kp = P(COCl2)/(P(CO) P(Cl2)) P
None of the changes would affect K; (a) Q doubles; (b) Q is halved; Q decreases.
2. K would not change; it does not depend on volume.
3. [CO] = [H2O] = 0.839 M, [CO2] = [H2] = 0.930 M; no effect on K
1. At 800°C, [CO] = 0.678 M, [CO2] = 0.231 M; at 1000°C, [CO] = 0.645 M, [CO2] = 0.264 M
2. At 800°C, PCO = 59.7 atm, P(CO2) = 20.3 atm; at 1000°C, P(CO) = 67.4 atm, P(CO2) = 27.6 atm.
3. Removing CO would cause the reaction to shift to the right, causing P(CO2) to decrease.
4. CH4 O2 CO2 H2O Q K
initial (moles) 0.45 0.90 0 0 0 1.29
at equilibrium 0.215 0.43 0.235 0.47 K 1.29
add 0.50 mol of methane 0.715 0.43 0.235 0.47 0.39 1.29
new equilibrium 0.665 0.33 0.285 0.57 K 1.29
remove water 0.665 0.33 0.285 0 0 1.29
new equilibrium 0.57 0.14 0.38 0.19 K 1.29
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.05%3A_Factors_That_Affect_Equilibrium.txt |
Learning Objectives
• To understand different ways to control the products of a reaction.
Whether in the synthetic laboratory or in industrial settings, one of the primary goals of modern chemistry is to control the identity and quantity of the products of chemical reactions. For example, a process aimed at synthesizing ammonia is designed to maximize the amount of ammonia produced using a given amount of energy. Alternatively, other processes may be designed to minimize the creation of undesired products, such as pollutants emitted from an internal combustion engine. To achieve these goals, chemists must consider the competing effects of the reaction conditions that they can control.
One way to obtain a high yield of a desired compound is to make the reaction rate of the desired reaction much faster than the reaction rates of any other possible reactions that might occur in the system. Altering reaction conditions to control reaction rates, thereby obtaining a single product or set of products, is called kinetic controlThe altering of reaction conditions to control reaction rates, thereby obtaining a single desired product or set of products. A second approach, called thermodynamic control. The altering of reaction conditions so that a single desired product or set of products is present in significant quantities at equilibrium., consists of adjusting conditions so that at equilibrium only the desired products are present in significant quantities.
An example of thermodynamic control is the Haber-Bosch process. Karl Bosch (1874–1940) was a German chemical engineer who was responsible for designing the process that took advantage of Fritz Haber’s discoveries regarding the N2 + H2/NH3 equilibrium to make ammonia synthesis via this route cost-effective. He received the Nobel Prize in Chemistry in 1931 for his work. The industrial process is called either the Haber process or the Haber-Bosch process. used to synthesize ammonia via the following reaction:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{15.6.1a}$
with
$ΔH_{rxn}=−91.8\; kJ/mol \tag{15.6.1b}$
Because the reaction converts 4 mol of gaseous reactants to only 2 mol of gaseous product, Le Chatelier’s principle predicts that the formation of NH3 will be favored when the pressure is increased. The reaction is exothermic, however (ΔHrxn = −91.8 kJ/mol), so the equilibrium constant decreases with increasing temperature, which causes an equilibrium mixture to contain only relatively small amounts of ammonia at high temperatures (Figure 15.6.1). Taken together, these considerations suggest that the maximum yield of NH3 will be obtained if the reaction is carried out at as low a temperature and as high a pressure as possible. Unfortunately, at temperatures less than approximately 300°C, where the equilibrium yield of ammonia would be relatively high, the reaction is too slow to be of any commercial use. The industrial process therefore uses a mixed oxide (Fe2O3/K2O) catalyst that enables the reaction to proceed at a significant rate at temperatures of 400°C–530°C, where the formation of ammonia is less unfavorable than at higher temperatures.
Because of the low value of the equilibrium constant at high temperatures (e.g., K = 0.039 at 800 K), there is no way to produce an equilibrium mixture that contains large proportions of ammonia at high temperatures. We can, however, control the temperature and the pressure while using a catalyst to convert a fraction of the N2 and H2 in the reaction mixture to NH3, as is done in the Haber-Bosch process. This process also makes use of the fact that the product—ammonia—is less volatile than the reactants. Because NH3 is a liquid at room temperature at pressures greater than 10 atm, cooling the reaction mixture causes NH3 to condense from the vapor as liquid ammonia, which is easily separated from unreacted N2 and H2. The unreacted gases are recycled until complete conversion of hydrogen and nitrogen to ammonia is eventually achieved. Figure 15.6.2 is a simplified layout of a Haber-Bosch process plant.
The Sohio acrylonitrile process, in which propene and ammonia react with oxygen to form acrylonitrile, is an example of a kinetically controlled reaction:
$\ce{CH_2=CHCH_{3(g)} + NH_{3(g)} + 3/2 O_{2(g)} <=> CH_2=CHC≡N_{(g)} + 3H_2O_{(g)}} \tag{15.6.2}$
Like most oxidation reactions of organic compounds, this reaction is highly exothermic (ΔH° = −519 kJ/mol) and has a very large equilibrium constant (K = 1.2 × 1094). Nonetheless, the reaction shown in Equation 15.43 is not the reaction a chemist would expect to occur when propene or ammonia is heated in the presence of oxygen. Competing combustion reactions that produce CO2 and N2 from the reactants, such as those shown in Equation 15.6.3 and Equation 15.6.4, are even more exothermic and have even larger equilibrium constants, thereby reducing the yield of the desired product, acrylonitrile:
$\ce{CH_2=CHCH_{3(g)} + 9/2 O_{2(g)} <=> 3CO_{2(g)} + 3H_2O_{(g)}} \tag{15.6.3a}$
with
$ΔH^°=−1926.1 \;kJ/mol \tag{15.6.3b}$
and
$K=4.5 \times 10^{338} \tag{15.6.3c}$
$\ce{2NH_{3(g)} + 3 O_{2(g)} <=> N_{2(g)} + 6H_2O_(g)} \tag{15.6.4a}$
with
$ΔH°=−1359.2 \;kJ/mol, \tag{15.6.4c}$
and
$K=4.4 \times 10^{234} \tag{15.6.4c}$
In fact, the formation of acrylonitrile in Equation 15.6.2. is accompanied by the release of approximately 760 kJ/mol of heat due to partial combustion of propene during the reaction.
The Sohio process uses a catalyst that selectively accelerates the rate of formation of acrylonitrile without significantly affecting the reaction rates of competing combustion reactions. Consequently, acrylonitrile is formed more rapidly than CO2 and N2 under the optimized reaction conditions (approximately 1.5 atm and 450°C). The reaction mixture is rapidly cooled to prevent further oxidation or combustion of acrylonitrile, which is then washed out of the vapor with a liquid water spray. Thus controlling the kinetics of the reaction causes the desired product to be formed under conditions where equilibrium is not established. In industry, this reaction is carried out on an enormous scale. Acrylonitrile is the building block of the polymer called polyacrylonitrile, found in all the products referred to collectively as acrylics, whose wide range of uses includes the synthesis of fibers woven into clothing and carpets.
Note the Pattern
Controlling the amount of product formed requires that both thermodynamic and kinetic factors be considered.
Example 15.6.1
Recall that methanation is the reaction of hydrogen with carbon monoxide to form methane and water:
$CO_{(g)} +3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)}$
This reaction is the reverse of the steam reforming of methane described in Example 14. The reaction is exothermic (ΔH° = −206 kJ/mol), with an equilibrium constant at room temperature of Kp = 7.6 × 1024. Unfortunately, however, CO and H2 do not react at an appreciable rate at room temperature. What conditions would you select to maximize the amount of methane formed per unit time by this reaction?
Given: balanced chemical equation and values of ΔH° and K
Asked for: conditions to maximize yield of product
Strategy:
Consider the effect of changes in temperature and pressure and the addition of an effective catalyst on the reaction rate and equilibrium of the reaction. Determine which combination of reaction conditions will result in the maximum production of methane.
Solution:
The products are highly favored at equilibrium, but the rate at which equilibrium is reached is too slow to be useful. You learned in Chapter 14 that the reaction rate can often be increased dramatically by increasing the temperature of the reactants. Unfortunately, however, because the reaction is quite exothermic, an increase in temperature will shift the equilibrium to the left, causing more reactants to form and relieving the stress on the system by absorbing the added heat. If we increase the temperature too much, the equilibrium will no longer favor methane formation. (In fact, the equilibrium constant for this reaction is very temperature sensitive, decreasing to only 1.9 × 10−3 at 1000°C.) To increase the reaction rate, we can try to find a catalyst that will operate at lower temperatures where equilibrium favors the formation of products. Higher pressures will also favor the formation of products because 4 mol of gaseous reactant are converted to only 2 mol of gaseous product. Very high pressures should not be needed, however, because the equilibrium constant favors the formation of products. Thus optimal conditions for the reaction include carrying it out at temperatures greater than room temperature (but not too high), adding a catalyst, and using pressures greater than atmospheric pressure.
Industrially, catalytic methanation is typically carried out at pressures of 1–100 atm and temperatures of 250°C–450°C in the presence of a nickel catalyst. (At 425°C°C, Kp is 3.7 × 103, so the formation of products is still favored.) The synthesis of methane can also be favored by removing either H2O or CH4 from the reaction mixture by condensation as they form.
Exercise
As you learned in Example 10, the water–gas shift reaction is as follows:
$H_{2(g)} + CO_{2(g)} \rightleftharpoons H_2O_{(g)} + CO_{(g)}$
Kp = 0.106 and ΔH = 41.2 kJ/mol at 700 K. What reaction conditions would you use to maximize the yield of carbon monoxide?
Answer: high temperatures to increase the reaction rate and favor product formation, a catalyst to increase the reaction rate, and atmospheric pressure because the equilibrium will not be greatly affected by pressure
Summary
Changing conditions to affect the reaction rates to obtain a single product is called kinetic control of the system. In contrast, thermodynamic control is adjusting the conditions to ensure that only the desired product or products are present in significant concentrations at equilibrium.
Key Takeaway
• Both kinetic and thermodynamic factors can be used to control reaction products.
Conceptual Problems
1. A reaction mixture will produce either product A or B depending on the reaction pathway. In the absence of a catalyst, product A is formed; in the presence of a catalyst, product B is formed. What conclusions can you draw about the forward and reverse rates of the reaction that produces A versus the reaction that produces B in (a) the absence of a catalyst and (b) the presence of a catalyst?
2. Describe how you would design an experiment to determine the equilibrium constant for the synthesis of ammonia:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag$
The forward reaction is exothermic (ΔH° = −91.8 kJ). What effect would an increase in temperature have on the equilibrium constant?
3. What effect does a catalyst have on each of the following?
1. the equilibrium position of a reaction
2. the rate at which equilibrium is reached
3. the equilibrium constant?
4. How can the ratio Q/K be used to determine in which direction a reaction will proceed to reach equilibrium?
5. Industrial reactions are frequently run under conditions in which competing reactions can occur. Explain how a catalyst can be used to achieve reaction selectivity. Does the ratio Q/K for the selected reaction change in the presence of a catalyst?
Numerical Problems
1. The oxidation of acetylene via 2C2H2 (g) + 5O2(g) ⇌ 4 CO2(g) + 2 H2O(l) has ΔH° = −2600 kJ. What strategy would you use with regard to temperature, volume, and pressure to maximize the yield of product?
2. The oxidation of carbon monoxide via CO(g) + 1/2 O2(g) ⇌ CO2(g) has ΔH° = −283 kJ. If you were interested in maximizing the yield of CO2, what general conditions would you select with regard to temperature, pressure, and volume?
3. You are interested in maximizing the product yield of the system 2SO2(g) + O2 (g) 2 SO3 (g) K = 280 and ΔH° = −158 kJ. What general conditions would you select with regard to temperature, pressure, and volume? If SO2 has an initial concentration of 0.200 M and the amount of O2 is stoichiometric, what amount of SO3 is produced at equilibrium?
Answer
1. Use low temperature and high pressure (small volume).
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.06%3A_Controlling_the_Products_of_Reactions.txt |
Learning Objectives
• The quadratic formula
Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter.
The Quadratic Formula
Mathematical expressions that involve a sum of powers in one or more variables (e.g., x) multiplied by coefficients (such as a) are called polynomials. Polynomials of a single variable have the general form
$a_nx^n + ... + a_2x^2 + a_1x + a_0 \tag{15.7.1}$
The highest power to which the variable in a polynomial is raised is called its order. Thus the polynomial shown here is of the nth order. For example, if n were 3, the polynomial would be third order.
A quadratic equation is a second-order polynomial equation in a single variable x:
$ax^2 + bx + c = 0 \tag{15.7.2}$
According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called roots—that can be found using a method called completing the square. In this method, we solve for x by first adding −c to both sides of the quadratic equation and then divide both sides by a:
$x^2+\dfrac{bx}{a} =−\dfrac{c}{a} \tag{15.7.3}$
We can convert the left side of this equation to a perfect square by adding b2/4a2, which is equal to (b/2a)2:
Left side: $x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=(x+\dfrac{b}{2a})^2 \tag{15.7.4}$
Having added a value to the left side, we must now add that same value, b2 ⁄ 4a2, to the right side:
$(x+\dfrac{b}{2a})^2=−\dfrac{c}{a}+\dfrac{b^2}{4a^2} \tag{15.7.5}$
The common denominator on the right side is 4a2. Rearranging the right side, we obtain the following:
$(x+\dfrac{b}{2a})^2=\dfrac{b^2−4ac}{4a^2} \tag{15.7.6}$
Taking the square root of both sides and solving for x,
$x+\dfrac{b}{2a}= \dfrac{\pm \sqrt{b^2−4ac}}{2a} \tag{15.7.7}$
$x= \dfrac{−b \pm \sqrt{b^2−4ac}}{2a} \tag{15.7.8}$
This equation, known as the quadratic formula, has two roots:
$x= \dfrac{−b + \sqrt{b^2−4ac}}{2a} \tag{15.7.9}$
and
$x= \dfrac{−b - \sqrt{b^2−4ac}}{2a} \tag{15.7.10}$
Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients (a, b, c) into the quadratic formula.
When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Skill Builder ES1 gives you practice using the quadratic formula.
Skill Builder ES1
Use the quadratic formula to solve for x in each equation. Report your answers to three significant figures.
1. x2 + 8x − 5 = 0
2. 2x2 − 6x + 3 = 0
3. 3x2 − 5x − 4 = 6
4. 2x(−x + 2) + 1 = 0
5. 3x(2x + 1) − 4 = 5
Solution:
1. $9x=−8+82−4(1)(−5)√2(1)=0.583$ and $x=−8−82−4(1)(−5)√2(1)=−8.58$
2. $x=−(−6)+(−62)−4(2)(3)√2(2)=2.37$ and $x=−(−6)−(−62)−4(2)(3)√2(2)=0.634$
3. $x=−(−5)+(−52)−4(3)(−10)√2(3)=2.84$ and $x=−(−5)−(−52)−4(3)(−10)√2(3)=−1.17$
4. $x=−4+42−4(−2)(1)√2(−2)=−0.225$ and $x=−4−42−4(−2)(1)√2((−2))=2.22$
5. $x=−1+12−4(2)(−3)√2(2)=1.00$ and $x=−1−12−4(2)(−3)√2(2)=1.50$
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.07%3A_Essential_Skills_7.txt |
Learning Objectives
• To understand the autoionization reaction of liquid water.
• To know the relationship among pH, pOH, and pKw.
As you learned in Chapter 8 and Chapter 4, acids and bases can be defined in several different ways (Table 16.1.1 ). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce H+ ions (protons), and an Arrhenius base is a substance that dissociates in water to produce OH (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton, as we demonstrated in Chapter 8 . Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, discussed in Chapter 8, focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.
Table 16.1.1 Definitions of Acids and Bases
Acids Bases
Arrhenius H+ donor OH donor
Brønsted–Lowry H+ donor H+ acceptor
Lewis electron-pair acceptor electron-pair donor
Because this chapter deals with acid–base equilibriums in aqueous solution, our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base.
In Chapter 8, we also introduced the acid–base properties of water, its autoionization reaction, and the definition of pH. The purpose of this section is to review those concepts and describe them using the concepts of chemical equilibrium developed in Chapter 15 .
Acid–Base Properties of Water
The structure of the water molecule, with its polar O–H bonds and two lone pairs of electrons on the oxygen atom, was described in Chapter 8 and Chapter 4, and the structure of liquid water was discussed in Chapter 13 . Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions (Cl) and protons (H+). As you learned in Chapter 8, the proton, in turn, reacts with a water molecule to form the hydronium ion (H3O+):
$\underset{aicd}{HCl_{(aq)}} + \underset{base}{H_2O_{(l)}} \rightarrow \underset{acid}{H_3O^+_{(aq)}} + \underset{base}{Cl^-_{(aq)}} \tag{16.1.1}$
In this reaction, HCl is the acid, and water acts as a base by accepting an H+ ion. The reaction in Equation 16.1.1 is often written in a simpler form by removing H2O from each side:
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \tag{16.1.2}$
In Equation 16.1.2, the hydronium ion is represented by H+, although free H+ ions do not exist in liquid water.
Water can also act as an acid, as shown in Equation 16.1.3. In this equilibrium reaction, H2O donates a proton to NH3, which acts as a base:
$\underset{aicd}{H_2O_{(aq)}} + \underset{base}{NH_{3(aq)}} \rightleftharpoons \underset{acid}{NH^+_{4 (aq)}} + \underset{base}{OH^-_{(aq)}} \tag{16.1.3}$
Thus water is amphiproticSubstances that can behave as either an acid or a base in a chemical reaction, depending on the nature of the other reactant(s)., meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that Equation 16.3 is an equilibrium reaction as indicated by the double arrow.
The Ion-Product Constant of Liquid Water
Because water is amphiprotic, one water molecule can react with another to form an OH ion and an H3O+ ion in an autoionization process:
$2H_2O(l) \rightleftharpoons H_3O^+_{(aq)}+OH^−_{(aq)} \tag{16.1.4}$
The equilibrium constant K for this reaction can be written as follows:
$K=\dfrac{[H_3O^+(aq)][OH^−(aq)]}{[H_2O(l)]^2} \tag{16.1.5}$
When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25°C, the concentrations of the hydronium ion and the hydroxide ion are equal: [H3O+] = [OH] = 1.003 × 10−7 M. Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb. We can calculate [H2O] at 25°C from the density of water at this temperature (0.997 g/mL):
$[H_2O(l)]=mol/L=(0.997\; \cancel{g}/mL)\left(\dfrac{1 \;mol}{18.02\; \cancel{g}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=55.3\; M \tag{16.1.6}$
With so few water molecules dissociated, the equilibrium of the autoionization reaction (Equation 16.4) lies far to the left. Consequently, [H2O] is essentially unchanged by the autoionization reaction and can be treated as a constant. Incorporating this constant into the equilibrium expression allows us to rearrange Equation 16.1.5 to define a new equilibrium constant, the ion-product constant of liquid water (Kw)
$K=\dfrac{K_w}{[H_2O(l)]^2} \tag{16.1.7a}$
with
$K_w = [H_3O^+(aq)][OH^−(aq)]=[H_3O^+(aq)][OH^−(aq)] \tag{16.1.7b}$
Substituting the values for [H3O+] and [OH] at 25°C into this expression,
$K_w=(1.003 \times10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \tag{16.1.8}$
Thus, to three significant figures, Kw = 1.01 × 10−14 M. Like any other equilibrium constant, Kw varies with temperature, ranging from 1.15 × 10−15 at 0°C to 4.99 × 10−13 at 100°C.
In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If [H3O+] > [OH], however, the solution is acidic, whereas if [H3O+] < [OH], the solution is basic. For an aqueous solution, the H3O+ concentration is a quantitative measure of acidity: the higher the H3O+ concentration, the more acidic the solution. Conversely, the higher the OH concentration, the more basic the solution. In most situations that you will encounter, the H3O+ and OH concentrations from the dissociation of water are so small (1.003 × 10−7 M) that they can be ignored in calculating the H3O+ or OH concentrations of solutions of acids and bases, but this is not always the case.
The Relationship among pH, pOH, and pKw
The pH scale is a concise way of describing the H3O+ concentration and hence the acidity or basicity of a solution. Recall from Chapter 8 that pH and the H+ (H3O+) concentration are related as follows:
$pH=−log_{10}[H^{+}(aq)] \tag{16.1.9}$
$[H^{+}(aq)]=10^{−pH} \tag{16.1.10}$
Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. (Refer to Essential Skills 3 in Section 8.11 , if you need to refresh your memory about how to use logarithms.) Recall also that the pH of a neutral solution is 7.00 ([H3O+] = 1.0 × 10−7 M), whereas acidic solutions have pH < 7.00 (corresponding to [H3O+] > 1.0 × 10−7) and basic solutions have pH > 7.00 (corresponding to [H3O+] < 1.0 × 10−7).
Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and [OH] are related as follows:
$pOH=−log_{10}[OH^{−}(aq)] \tag{16.1.11}$
$[OH^{−}(aq)]=10^{−pOH} \tag{16.1.12}$
The constant Kw can also be expressed using this notation, where pKw = −log Kw.
Because a neutral solution has [OH] = 1.0 × 10−7, the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25°C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25°C by taking the negative logarithm of both sides of Equation 16.7:
$−logKw=−log([H_{3}O^{+}(aq)][OH{−}(aq)])=(−log[H_{3}O^{+}(aq)])+(−log[OH^{−}(aq)])=pH+pOH \tag{16.1.13}$
Thus at any temperature, pH + pOH = pKw, so at 25°C, where Kw = 1.0 × 10−14, pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the pKw at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in Figure 16.1.1 over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales.
Note the Pattern
For any neutral solution, pH + pOH = 14.00 (at 25°C) and and pH=12pKw.
Example 16.1.1
The Kw for water at 100°C is 4.99 × 10−13. Calculate pKw for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100°C. Report pH and pOH values to two decimal places.
Given: K w
Asked for: pKw, pH, and pOH
Strategy:
A Calculate pKw by taking the negative logarithm of Kw.
B For a neutral aqueous solution, [H3O+] = [OH]. Use this relationship and Equation 16.7 to calculate [H3O+] and [OH]. Then determine the pH and the pOH for the solution.
Solution:
A Because pKw is the negative logarithm of Kw, we can write
$pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302 \notag$
The answer is reasonable: Kw is between 10−13 and 10−12, so pKw must be between 12 and 13.
B Equation 16.1.7 shows that Kw = [H3O+][OH]. Because [H3O+] = [OH] in a neutral solution, we can let x = [H3O+] = [OH]:
$K_w =[H_3O^+][OH^−]=(x)(x)=x^2 \notag$
$x=\sqrt{K_w} =\sqrt{4.99 \times 10^{−13}} =7.06 \times 10^{−7}\; M \notag$
Because x is equal to both [H3O+] and [OH],
pH = pOH = −log(7.06 × 10−7) = 6.15 (to two decimal places)
We could obtain the same answer more easily (without using logarithms) by using the pKw. In this case, we know that pKw = 12.302, and from Equation 16.13, we know that pKw = pH + pOH. Because pH = pOH in a neutral solution, we can use Equation 16.13 directly, setting pH = pOH = y. Solving to two decimal places we obtain the following:
$pK_w pH + pOH = y + y = 2y \notag$
$y=\dfrac{pK_w}{2}=\dfrac{12.302}{2}=6.15=pH=pOH \notag$
Exercise
Humans maintain an internal temperature of about 37°C. At this temperature, Kw = 3.55 × 10−14. Calculate pKw and the pH and the pOH of a neutral solution at 37°C. Report pH and pOH values to two decimal places.
Answer: pKw = 13.45 pH = pOH = 6.73
Summary
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion (H3O+). The autoionization of liquid water produces OH and H3O+ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as Kw = [H3O+][OH]. At 25°C, Kw is 1.01 × 10−14; hence pH + pOH = pKw = 14.00.
Key Takeaway
• For any neutral solution, pH + pOH = 14.00 (at 25°C) and pH = 1/2 pKw.
Key Equations
Ion-product constant of liquid water
Equation 16.1.7: Kw = [H3O+][OH]
Definition of pH
Equation 16.1.9: pH = −log10[H+]
Equation 16.1.10: [H+] = 10−pH
Definition of pOH
Equation 16.1.11: pOH = −log10[OH+]
Equation 16.1.12: [OH] = 10−pOH
Relationship among pH, pOH, and p K w
Equation 16.1.13: pKw= pH + pOH
Conceptual Problems
1. What is the relationship between the value of the equilibrium constant for the autoionization of liquid water and the tabulated value of the ion-product constant of liquid water (Kw)?
2. The density of liquid water decreases as the temperature increases from 25°C to 50°C. Will this effect cause Kw to increase or decrease? Why?
3. Show that water is amphiprotic by writing balanced chemical equations for the reactions of water with HNO3 and NH3. In which reaction does water act as the acid? In which does it act as the base?
4. Write a chemical equation for each of the following.
1. Nitric acid is added to water.
2. Potassium hydroxide is added to water.
3. Calcium hydroxide is added to water.
4. Sulfuric acid is added to water.
5. Show that K for the sum of the following reactions is equal to Kw.
1. $HMnO_{4(aq)} \rightleftharpoons H^+_{(aq)} + MnO^−_{4(aq)} \notag$
Answers
1. $K_{auto} = \dfrac{[H_3O^+][OH^−]}{[H_2O]^2} \notag$
$K_w = [H_3O^+][OH^−] = K_{auto}[H_2O]^2 \notag$
2. water is the base: $H_2O_{(l)} + HNO_{3(g)} \rightarrow H_3O^+_{(aq)} + NO^−_{3(aq)} \notag$
water is the acid: $H_2O_{(l)} + NH_{3(g)} \rightarrow OH^−_{(aq)} + NH^−_{4(aq)} \notag$
Numerical Problems
1. The autoionization of sulfuric acid can be described by the following chemical equation:
$H_2SO_{4(l)}+H_2SO_{4(l)} \rightleftharpoons H_3SO^+_{4(soln)}+H_SO^−_{4(soln)} \notag$
At 25°C, K = 3 × 10−4. Write an equilibrium constant expression for K(H2SO4) that is analogous to Kw. The density of H2SO4 is 1.8 g/cm3 at 25°C. What is the concentration of H3SO4+? What fraction of H2SO4 is ionized?
2. An aqueous solution of a substance is found to have [H3O]+ = 2.48 × 10−8 M. Is the solution acidic, neutral, or basic?
3. The pH of a solution is 5.63. What is its pOH? What is the [OH]? Is the solution acidic or basic?
4. State whether each solution is acidic, neutral, or basic.
1. [H3O+] = 8.6 × 10−3 M
2. [H3O+] = 3.7 × 10−9 M
3. [H3O+] = 2.1 × 10−7 M
4. [H3O+] = 1.4 × 10−6 M
5. Calculate the pH and the pOH of each solution.
1. 0.15 M HBr
2. 0.03 M KOH
3. 2.3 × 10−3 M HNO3
4. 9.78 × 10−2 M NaOH
5. 0.00017 M HCl
6. 5.78 M HI
6. Calculate the pH and the pOH of each solution.
1. 25.0 mL of 2.3 × 10−2 M HCl, diluted to 100 mL
2. 5.0 mL of 1.87 M NaOH, diluted to 125 mL
3. 5.0 mL of 5.98 M HCl added to 100 mL of water
4. 25.0 mL of 3.7 M HNO3 added to 250 mL of water
5. 35.0 mL of 0.046 M HI added to 500 mL of water
6. 15.0 mL of 0.0087 M KOH added to 250 mL of water.
7. The pH of stomach acid is approximately 1.5. What is the [H+]?
8. Given the pH values in parentheses, what is the [H+] of each solution?
1. household bleach (11.4)
2. milk (6.5)
3. orange juice (3.5)
4. seawater (8.5)
5. tomato juice (4.2)
9. A reaction requires the addition of 250.0 mL of a solution with a pH of 3.50. What mass of HCl (in milligrams) must be dissolved in 250 mL of water to produce a solution with this pH?
10. If you require 333 mL of a pH 12.50 solution, how would you prepare it using a 0.500 M sodium hydroxide stock solution?
Answers
1. $K_{H_2SO_4}=[H_3SO_4^+][HSO_4^−]=K[H_2SO_4]_2 \notag$
$[H_3SO_4^+] = 0.3 M \notag$
[H3SO4+] = 0.3 M; the fraction ionized is 0.02.
2. pOH = 8.37; [OH] = 4.3 × 10−9 M; acidic
1. pH = 0.82; pOH = 13.18
2. pH = 12.5; pOH = 1.5
3. pH = 2.64; pOH = 11.36
4. pH = 12.990; pOH = 1.010
5. pH = 3.77; pOH = 10.23
6. pH = −0.762; pOH = 14.762
3. 2.9 mg HCl
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.01%3A_The_Autoionization_of_Water.txt |
Learning Objectives
• To understand the concept of conjugate acid–base pairs.
• To know the relationship between acid or base strength and the magnitude of Ka, Kb, pKa, and pKb.
• To understand the leveling effect.
We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibriums in terms of the Brønsted–Lowry model and then proceed to a quantitative description in Section 16.4 .
Conjugate Acid–Base Pairs
In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pair. For example, in the reaction of HCl with water (Equation 16.1.1), HCl, the parent acid, donates a proton to a water molecule, the parent base, thereby forming Cl. Thus HCl and Cl constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl ion in solution acts as a base to accept a proton from H3O+, forming H2O and HCl. Thus H3O+ and H2O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain two conjugate acid–base pairs, which in this case are HCl/Cl and H3O+/H2O.
Note the Pattern
All acid–base reactions contain two conjugate acid–base pairs.
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, H3O+ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH3CO2H/CH3CO2) and the parent base and its conjugate acid (H3O+/H2O).
In the reaction of ammonia with water to give ammonium ions and hydroxide ions (Equation 16.1.3), ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH4+/NH3 and H2O/OH. Some common conjugate acid–base pairs are shown in Figure 16.2.1.
Acid–Base Equilibrium Constants: Ka, Kb, pKa, and pKb
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A is its conjugate base, is as follows:
$HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \tag{16.2.1}$
The equilibrium constant for this dissociation is as follows:
$K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \tag{16.2.2}$
As we noted earlier, the concentration of water is essentially constant for all reactions in aqueous solution, so [H2O] in Equation 16.2.2 can be incorporated into a new quantity, the acid ionization constant (Ka)An equilibrium constant for the ionization (dissociation) of a weak acid (HA) with water, HA(aq) + H2O(l)H3O(aq)+ + A(aq), in which the concentration of water is treated as a constant: Ka = [H3O(aq)+][A(aq)]/[HA], also called the acid dissociation constant:
$K_a=K[H_2O]=\dfrac{[H_3O^+][A^−]}{[HA]} \tag{16.2.3}$
Thus the numerical values of K and Ka differ by the concentration of water (55.3 M). Again, for simplicity, H3O+ can be written as H+ in Equation 16.2.3. Keep in mind, though, that free H+ does not exist in aqueous solutions and that a proton is transferred to H2O in all acid ionization reactions to form H3O+. The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium.Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of H+ or OH, thus making them unitless. The values of Ka for a number of common acids are given in Table 16.2.1.
Table 16.2.1 Values of Ka, pKa, Kb, and pKb for Selected Acids (HA) and Their Conjugate Bases (A)
Acid HA K a pKa A K b pKb
hydroiodic acid HI 2 × 109 −9.3 I 5.5 × 10−24 23.26
sulfuric acid (1)* H2SO4 1 × 102 −2.0 HSO4 1 × 10−16 16.0
nitric acid HNO3 2.3 × 101 −1.37 NO3 4.3 × 10−16 15.37
hydronium ion H3O+ 1.0 0.00 H2O 1.0 × 10−14 14.00
sulfuric acid (2)* HSO4 1.0 × 10−2 1.99 SO42− 9.8 × 10−13 12.01
hydrofluoric acid HF 6.3 × 10−4 3.20 F 1.6 × 10−11 10.80
nitrous acid HNO2 5.6 × 10−4 3.25 NO2 1.8 × 10−11 10.75
formic acid HCO2H 1.78 × 10−4 3.750 HCO2 5.6 × 10−11 10.25
benzoic acid C6H5CO2H 6.3 × 10−5 4.20 C6H5CO2 1.6 × 10−10 9.80
acetic acid CH3CO2H 1.7 × 10−5 4.76 CH3CO2 5.8 × 10−10 9.24
pyridinium ion C5H5NH+ 5.9 × 10−6 5.23 C5H5N 1.7 × 10−9 8.77
hypochlorous acid HOCl 4.0 × 10−8 7.40 OCl 2.5 × 10−7 6.60
hydrocyanic acid HCN 6.2 × 10−10 9.21 CN 1.6 × 10−5 4.79
ammonium ion NH4+ 5.6 × 10−10 9.25 NH3 1.8 × 10−5 4.75
water H2O 1.0 × 10−14 14.00 OH 1.00 0.00
acetylene C2H2 1 × 10−26 26.0 HC2 1 × 1012 −12.0
ammonia NH3 1 × 10−35 35.0 NH2 1 × 1021 −21.0
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:
$B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \tag{16.2.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb)An equilibrium constant for the reaction of a weak base (B) with water, B(aq) + H2O(l)BH+(aq) + OH(aq), in which the concentration of water is treated as a constant: Kb = [BH+][OH-]/[B], also called the base dissociation constant:
$K_b=K[H_2O]=[BH^+][OH^−][B] \tag{16.2.5}$
Once again, the concentration of water is constant, so it does not appear in the equilibrium constant expression; instead, it is included in the Kb. The larger the Kb, the stronger the base and the higher the OH concentration at equilibrium. The values of Kb for a number of common weak bases are given in Table 16.2.2
Table 16.2.2 Values of Kb, pKb, Ka, and pKa for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
Base B K b pKb BH+ K a pKa
hydroxide ion OH 1.0 0.00* H2O 1.0 × 10−14 14.00
phosphate ion PO43− 2.1 × 10−2 1.68 HPO42− 4.8 × 10−13 12.32
dimethylamine (CH3)2NH 5.4 × 10−4 3.27 (CH3)2NH2+ 1.9 × 10−11 10.73
methylamine CH3NH2 4.6 × 10−4 3.34 CH3NH3+ 2.2 × 10−11 10.66
trimethylamine (CH3)3N 6.3 × 10−5 4.20 (CH3)3NH+ 1.6 × 10−10 9.80
ammonia NH3 1.8 × 10−5 4.75 NH4+ 5.6 × 10−10 9.25
pyridine C5H5N 1.7 × 10−9 8.77 C5H5NH+ 5.9 × 10−6 5.23
aniline C6H5NH2 7.4 × 10−10 9.13 C6H5NH3+ 1.3 × 10−5 4.87
water H2O 1.0 × 10−14 14.00 H3O+ 1.0* 0.00
*As in Table 16.2.
There is a simple relationship between the magnitude of Ka for an acid and Kb for its conjugate base. Consider, for example, the ionization of hydrocyanic acid (HCN) in water to produce an acidic solution, and the reaction of CN with water to produce a basic solution:
$HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \tag{16.2.6}$
$CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \tag{16.2.7}$
The equilibrium constant expression for the ionization of HCN is as follows:
$K_a=\dfrac{[H^+][CN^−]}{[HCN]} \tag{16.2.8}$
The corresponding expression for the reaction of cyanide with water is as follows:
$K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \tag{16.2.9}$
If we add Equation 16.2.8 and Equation 16.2.9, we obtain the following (recall from Chapter 15 that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions):
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]} \notag$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]} \notag$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+][OH^−] \notag$
In this case, the sum of the reactions described by Ka and Kb is the equation for the autoionization of water, and the product of the two equilibrium constants is Kw:
$K_aK_b = K_w \tag{16.2.10}$
Thus if we know either Ka for an acid or Kb for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.
Just as with pH, pOH, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining pKa as follows:
$pKa = −\log_{10}K_a \tag{16.2.11}$
$K_a=10^{−pK_a} \tag{16.2.12}$
and $pK_b$ as
$pK_b = −\log_{10}K_b \tag{16.2.13}$
$K_b=10^{−pK_b} \tag{16.2.14}$
Similarly, Equation 16.2.10, which expresses the relationship between Ka and Kb, can be written in logarithmic form as follows:
$pK_a + pK_b = pK_w \tag{16.2.15}$
At 25°C, this becomes
$pK_a + pK_b = 14.00 \tag{16.2.16}$
The values of pKa and pKb are given for several common acids and bases in Table 16.2.1 and Table 16.2.2 , respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of pKa correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (HNO2), with a pKa of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a pKa of 9.21. Conversely, smaller values of pKb correspond to larger base ionization constants and hence stronger bases.
The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.2.1 . The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of pKa. This order corresponds to decreasing strength of the conjugate base or increasing values of pKb. At the bottom left of Figure 16.2 are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
Note the Pattern
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:
$\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \notag$
In an acid–base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce H3O+ and Cl; only negligible amounts of HCl molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \tag{16.2.17}$
In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of H3O+ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:
$\ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \notag$
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:
$H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \notag$
Note the Pattern
All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.
Example 16.2.1
1. Calculate Kb and pKb of the butyrate ion (CH3CH2CH2CO2). The pKa of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
2. Calculate Ka and pKa of the dimethylammonium ion [(CH3)2NH2+]. The base ionization constant Kb of dimethylamine [(CH3)2NH] is 5.4 × 10−4 at 25°C.
Given: pKa and Kb
Asked for: corresponding Kb and pKb, Ka and pKa
Strategy:
The constants Ka and Kb are related as shown in Equation 16.2.10. The pKa and pKb for an acid and its conjugate base are related as shown in Equation 16.2.15 and Equation 16.2.16. Use the relationships pK = −log K and K = 10−pK (Equation 16.24 and Equation 16.26) to convert between Ka and pKa or Kb and pKb.
Solution:
1. We are given the pKa for butyric acid and asked to calculate the Kb and the pKb for its conjugate base, the butyrate ion. Because the pKa value cited is for a temperature of 25°C, we can use Equation 16.2.16: pKa + pKb = pKw = 14.00. Substituting the pKa and solving for the pKb,
$4.83+pK_b=14.00 \notag$
$pK_b=14.00−4.83=9.17 \notag$
Because pKb = −log Kb, Kb is 10−9.17 = 6.8 × 10−10.
2. In this case, we are given Kb for a base (dimethylamine) and asked to calculate Ka and pKa for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is Kb rather than pKb, we can use Equation 16.23: KaKb = Kw. Substituting the values of Kb and Kw at 25°C and solving for Ka,
$K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14} \notag$
$K_a=1.9 \times 10^{−11} \notag$
Because pKa = −log Ka, we have pKa = −log(1.9 × 10−11) = 10.72. We could also have converted Kb to pKb to obtain the same answer:
$pK_b=−\log(5.4 \times 10^{−4})=3.27 \notag$
$pKa+pK_b=14.00 \notag$
$pK_a=10.73 \notag$
$K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11} \notag$
If we are given any one of these four quantities for an acid or a base (Ka, pKa, Kb, or pKb), we can calculate the other three.
Exercise
Lactic acid [CH3CH(OH)CO2H] is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its pKa is 3.86 at 25°C. Calculate Ka for lactic acid and pKb and Kb for the lactate ion.
Answer: Ka = 1.4 × 10−4 for lactic acid; pKb = 10.14 and Kb = 7.2 × 10−11 for the lactate ion
Solutions of Strong Acids and Bases: The Leveling Effect
You will notice in Table 16.2.1 that acids like H2SO4 and HNO3 lie above the hydronium ion, meaning that they have pKa values less than zero and are stronger acids than the H3O+ ion.Recall from Chapter 8 that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as HONO2. Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving HNO3 instead. In fact, all six of the common strong acids that we first encountered in Chapter 8 have pKa values less than zero, which means that they have a greater tendency to lose a proton than does the H3O+ ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibriums for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the H3O+ ion and the conjugate base of the acid.
Although Ka for HI is about 108 greater than Ka for HNO3, the reaction of either HI or HNO3 with water gives an essentially stoichiometric solution of H3O+ and I or NO3. In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M H3O+, regardless of the identity of the strong acid. This phenomenon is called the leveling effectThe phenomenon that makes H3O+ the strongest acid that can exist in water. Any species that is a stronger acid than H3O+ is leveled to the strength of H3O+ in aqueous solution.: any species that is a stronger acid than the conjugate acid of water (H3O+) is leveled to the strength of H3O+ in aqueous solution because H3O+ is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths.
One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than H2O. Measurements of the conductivity of 0.1 M solutions of both HI and HNO3 in acetic acid show that HI is completely dissociated, but HNO3 is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than HNO3. The relative order of acid strengths and approximate Ka and pKa values for the strong acids at the top of Table 16.2.1 were determined using measurements like this and different nonaqueous solvents.
Note the Pattern
In aqueous solutions, [H3O+] is the strongest acid and OH is the strongest base that can exist in equilibrium with H2O.
The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH is leveled to the strength of OH because OH is the strongest base that can exist in equilibrium with water. Salts such as K2O, NaOCH3 (sodium methoxide), and NaNH2 (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table 16.3, are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of OH and the corresponding cation:
$K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \tag{16.2.18}$
$NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \tag{16.2.19}$
$NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \tag{16.2.20}$
Other examples that you may encounter are potassium hydride (KH) and organometallic compounds such as methyl lithium (CH3Li).
Polyprotic Acids and Bases
As you learned in Chapter 8 , polyprotic acids such as H2SO4, H3PO4, and H2CO3 contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the pKa increases. Consider H2SO4, for example:
$HSO^−_{4 (aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \notag$
The equilibrium in the first reaction lies far to the right, consistent with H2SO4 being a strong acid. In contrast, in the second reaction, appreciable quantities of both HSO4 and SO42− are present at equilibrium.
Note the Pattern
For a polyprotic acid, acid strength decreases and the pKa increases with the sequential loss of each proton.
The hydrogen sulfate ion (HSO4) is both the conjugate base of H2SO4 and the conjugate acid of SO42−. Just like water, HSO4 can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion (SO42−) is a polyprotic base that is capable of accepting two protons in a stepwise manner:
$SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^- \notag$
$HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \notag$
Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by pKa + pKb = pKw. Consider, for example, the HSO4/ SO42− conjugate acid–base pair. From Table 16.2.1, we see that the pKa of HSO4 is 1.99. Hence the pKb of SO42− is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas OH is a strong base, so the equilibrium shown above lies to the left. The HSO4 ion is also a very weak base [pKa of H2SO4 = 2.0, pKb of HSO4 = 14 − (−2.0) = 16], which is consistent with what we expect for the conjugate base of a strong acid. Thus the equilibrium also lies almost completely to the left. Once again, equilibrium favors the formation of the weaker acid–base pair.
Example 16.2.2
Predict whether the equilibrium for each reaction lies to the left or the right as written.
• $NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}$
• $CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)} \rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}$
Given: balanced chemical equation
Asked for: equilibrium position
Strategy:
Identify the conjugate acid–base pairs in each reaction. Then refer to Table 16.2.1 Table 16.2.2, and Figure 16.2.1 to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.
Solution:
1. The conjugate acid–base pairs are NH4+/NH3 and HPO42−/PO43−. According to Table 16.2 and Table 16.3, NH4+ is a stronger acid (pKa = 9.25) than HPO42− (pKa = 12.32), and PO43− is a stronger base (pKb = 1.68) than NH3 (pKb = 4.75). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \notag$
2. The conjugate acid–base pairs are CH3CH2CO2H/CH3CH2CO2 and HCN/CN. According to Table 16.2.1, HCN is a weak acid (pKa = 9.21) and CN is a moderately weak base (pKb = 4.79). Propionic acid (CH3CH2CO2H) is not listed in Table 16.2.1, however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid (−CH2CH3 versus −CH3), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the pKa of propionic acid to be similar in magnitude to the pKa of acetic acid. (In fact, the pKa of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than HCN. Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \notag$
Exercise
Predict whether the equilibrium for each reaction lies to the left or the right as written.
1. $H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}$
2. $HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}$
Answer
1. left
2. left
Acid–Base Properties of Solutions of Salts
We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. In Chapter 8, you learned that a neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as Na+, replaces the proton on the acid. An example is the reaction of CH3CO2H, a weak acid, with NaOH, a strong base:
$\underset{acid}{CH_3CO_2H_{(l)}} +\underset{base}{NaOH_{(s)}} \overset{H_2O}{\longrightarrow} \underset{salt}{H_2OCH_3CO_2Na_{(aq)} }+\underset{water}{H_2O_{(l)}} \tag{16.2.22}$
Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.
When a salt such as NaCl dissolves in water, it produces Na+(aq) and Cl(aq) ions. Using a Lewis approach, the Na+ ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The Cl ion is the conjugate base of the strong acid HCl, so it has essentially no basic character. Consequently, dissolving NaCl in water has no effect on the pH of a solution, and the solution remains neutral.
Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations (K+ and Na+) have essentially no acidic character, but the anions (CN and CH3CO2) are weak bases that can react with water because they are the conjugate bases of the weak acids HCN and acetic acid, respectively.
$CN^-_{(aq)} + H_2O_{(l)} \ce{ <<=>} HCN_{(aq)} + OH^-_{(aq)} \tag{16.2.23}$
$CH_3CO^2_{2(aq)} + H_2O_{(l)} \ce{<<=>} CH_3CO_2H_{(aq)} + OH^-_{(aq)} \tag{16.2.24}$
Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both HCN and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the pH of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table 16.2.2 and Figure 16.2.1, we can see that CN is a stronger base (pKb = 4.79) than acetate (pKb = 9.24), which is consistent with KCN producing a more basic solution than sodium acetate at the same concentration.
In contrast, the conjugate acid of a weak base should be a weak acid (Figure 16.2.1). For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with HCl. As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows:
$NH^+_{4(aq)} + H_2O_{(l)} \ce{ <<=>} HH_{3(aq)} + H_3O^+_{(aq)} \tag{16.2.25}$
$C_5H_5NH^+_{(aq)} + H_2O_{(l)} \ce{<<=>} C_5H_5NH_{(aq)} + H_3O^+_{(aq)} \tag{16.2.26}$
Figure 16.2.1 shows that H3O+ is a stronger acid than either NH4+ or C5H5NH+, and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The H3O+ concentration produced by the reactions is great enough, however, to decrease the pH of the solution significantly: the pH of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. This is consistent with the information shown in Figure 16.2.1, indicating that the pyridinium ion is more acidic than the ammonium ion.
What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Eq 16.2.25, the ammonium ion will lower the pH, while according to Eq 16.2.24, the acetate ion will raise the pH. This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a pH < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a pH > 7.00.
Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce H3O+. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (part (a) in Figure 16.2.2), as discussed in Chapter 8 . A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.
Second, the positive charge on the Al3+ ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the O–H bonds, as shown in part (b) in Figure 16.2.2. With less electron density between the O atoms and the H atoms, the O–H bonds are weaker than in a free H2O molecule, making it easier to lose a H+ ion.
The magnitude of this effect depends on the following two factors (Figure 16.2.3):
1. The charge on the metal ion. A divalent ion (M2+) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion (M+) of the same radius.
2. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.
Thus aqueous solutions of small, highly charged metal ions, such as Al3+ and Fe3+, are acidic:
$[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \tag{16.2.27}$
The [Al(H2O)6]3+ ion has a pKa of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as Li+ and Mg2+ or Ca2+ and Y3+, have different sizes and charges but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well.
Note the Pattern
Solutions of small, highly charged metal ions in water are acidic.
Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactionsA chemical reaction in which a salt reacts with water to yield an acidic or a basic solution.. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions.
Note the Pattern
A hydrolysis reaction is an acid–base reaction.
Example 16.2.3
Predict whether aqueous solutions of these compounds are acidic, basic, or neutral.
1. KNO3
2. CrBr3·6H2O
3. Na2SO4
Given: compound
Asked for: acidity or basicity of aqueous solution
Strategy:
A Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the pH of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution.
B If the anion is the conjugate base of a strong acid, it will not affect the pH of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic.
Solution:
1. A The K+ cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid.
B The NO3 anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce H+ or OH, and the solution will be neutral.
2. A The Cr3+ ion is a relatively highly charged metal cation that should behave similarly to the Al3+ ion and form the [Cr(H2O)6]3+ complex, which will behave as a weak acid:
$Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_({aq)} \notag$
B The Br anion is a very weak base (it is the conjugate base of the strong acid HBr), so it does not affect the pH of the solution. Hence the solution will be acidic.
3. A The Na+ ion, like the K+, is a very weak acid, so it should not affect the acidity of the solution.
B In contrast, SO42− is the conjugate base of HSO4, which is a weak acid. Hence the SO42− ion will react with water as shown in Figure 16.6 to give a slightly basic solution.
Exercise
Predict whether aqueous solutions of the following are acidic, basic, or neutral.
1. KI
2. Mg(ClO4)2
3. NaHS
Answer
1. neutral
2. acidic
3. basic (due to the reaction of HS with water to form H2S and OH)
Summary
Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, KaKb = Kw. Smaller values of pKa correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of pKb correspond to larger base ionization constants and hence stronger bases. At 25°C, pKa + pKb = 14.00. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than H3O+ and no base stronger than OH can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base. A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A), the conjugate acid of a weak base as the cation (BH+), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
Key Takeaways
• Acid–base reactions always contain two conjugate acid–base pairs.
• Each acid and each base has an associated ionization constant that corresponds to its acid or base strength.
Key Equations
Acid ionization constant
Equation 16.2.13: $K_a=K[H_2O]=\dfrac{[H_3O^+][A^−]}{[HA]}$
Base ionization constant
Equation 16.2.15: $K_b=K[H_2O]=[BH^+][OH^−][B]$
Relationship between K a and K b of a conjugate acid–base pair
Equation 16.2.20: $K_aK_b = K_w$
Definition of p K a
Equation 16.2.21: $pKa = −\log_{10}K_a$
Equation 16.2.22: $K_a=10^{−pK_a}$
Definition of p K b
Equation 16.2.23: $pK_b = −\log_{10}K_b$
Equation 16.2.25: $K_b=10^{−pK_b}$
Relationship between p K a and p K b of a conjugate acid–base pair
Equation 16.2.25: $pK_a + pK_b = pK_w$
Equation 16.29: $pK_a + pK_b = 14.00 \; \text{at 25°C}$
Conceptual Problems
1. Identify the conjugate acid–base pairs in each equilibrium.
1. $HSO^−_{4(aq)}+H_2O_{(l)} \rightleftharpoons SO^{2−}_{4(aq)}+H_3O^+_{(aq)}$
2. $C_3H_7NO_{2(aq)}+H_3O^+_{(aq)} \rightleftharpoons C_3H_8NO^+_{2(aq)}+H_2O_{(l)}$
3. $CH_3CO_2H_{(aq)}+NH_{3(aq)} \rightleftharpoons CH_3CO^−_{2(aq)}+NH^+_{4(aq)}$
4. $SbF_{5(aq)}+2HF_{(aq)} \rightleftharpoons H_2F^+_{(aq)}+SbF^−_{6(aq)}$
2. Identify the conjugate acid–base pairs in each equilibrium.
1. $HF(aq)+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+F^−_{(aq)}$
2. $CH_3CH_2NH_{2(aq)}+H_2O_{(l)} \rightleftharpoons CH_3CH_2NH^+_{3(aq)}+OH^−_{(aq)}$
3. $C_3H_7NO_{2(aq)}+OH^−_{(aq)} \rightleftharpoons C_3H_6NO^−_{2(aq)}+H_2O_{(l)}$
4. $CH_3CO_2H_{(aq)}+2HF_{(aq)} \rightleftharpoons CH_3C(OH)_2+(aq)+HF^−_{2(aq)}$
3. Salts such as NaH contain the hydride ion (H). When sodium hydride is added to water, it produces hydrogen gas in a highly vigorous reaction. Write a balanced chemical equation for this reaction and identify the conjugate acid–base pairs.
4. Write the expression for Ka for each reaction.
1. $HCO^−_{3(aq)}+H_2O_{(l)} \rightleftharpoons CO^{2−}_{3(aq)}+H_3O^+_{(aq)}$
2. $formic\; acid_{(aq)}+H_2O_{(l)} \rightleftharpoons formate_{(aq)}+H_3O^+_{(aq)}$
3. $H_3PO_{4(aq)}+H_2O_{(l)} \rightleftharpoons H_2PO^−_{4(aq)}+H_3O^+_{(aq)}$
5. Write an expression for the ionization constant Kb for each reaction.
1. $OCH^−_{3(aq)}+H_2O_{(l)} \rightleftharpoons HOCH_{3(aq)}+OH−(aq)$
2. $NH^−_{2(aq)}+H_2O_{(l)} \rightleftharpoons NH_{3(aq)}+OH^−_{(aq)}$
3. $S^{2−}_{(aq)}+H_2O_{(l)} \rightleftharpoons HS^−_{(aq)}+OH^−_{(aq)}$
6. Predict whether each equilibrium lies primarily to the left or to the right.
1. $HBr_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+Br^−_{(aq)}$
2. $NaH_{(soln)}+NH_{3(l)} \rightleftharpoons H2(soln)+NaNH_{2(soln)}$
3. $OCH^−_{3(aq)}+NH_{3(aq)} \rightleftharpoons CH3OH(aq)+NH^−_{2(aq)}$
4. $NH_{3(aq)}+HCl_{(aq)} \rightleftharpoons NH^+_{4(aq)}+Cl^−_{(aq)}$
7. Species that are strong bases in water, such as CH3, NH2, and S2−, are leveled to the strength of OH, the conjugate base of H2O. Because their relative base strengths are indistinguishable in water, suggest a method for identifying which is the strongest base. How would you distinguish between the strength of the acids HIO3, H2SO4, and HClO4?
8. Is it accurate to say that a 2.0 M solution of H2SO4, which contains two acidic protons per molecule, is 4.0 M in H+? Explain your answer.
9. The alkalinity of soil is defined by the following equation: alkalinity = [HCO3] + 2[CO32−] + [OH] − [H+]. The source of both HCO3 and CO32− is H2CO3. Explain why the basicity of soil is defined in this way.
10. Why are aqueous solutions of salts such as CaCl2 neutral? Why is an aqueous solution of NaNH2 basic?
11. Predict whether aqueous solutions of the following are acidic, basic, or neutral.
1. Li3N
2. NaH
3. KBr
4. C2H5NH3+Cl
12. When each compound is added to water, would you expect the pH of the solution to increase, decrease, or remain the same?
1. LiCH3
2. MgCl2
3. K2O
4. (CH3)2NH2+Br
13. Which complex ion would you expect to be more acidic—Pb(H2O)42+ or Sn(H2O)42+? Why?
14. Would you expect Sn(H2O)42+ or Sn(H2O)64+ to be more acidic? Why?
15. Is it possible to arrange the hydrides LiH, RbH, KH, CsH, and NaH in order of increasing base strength in aqueous solution? Why or why not?
Answer
1. $\overset{\text{acid 1}}{HSO^−_{4(aq)}} + \underset{\text{base 2}}{H_2O_{(l)}} \rightleftharpoons \overset{\text{base 1}}{SO^{2−}_{4(aq)}} + \underset{\text{acid 2}}{H_3O^+_{(aq)}}$
2. $\underset{\text{base 2}}{C_3H_7NO_{2(aq)}} + \overset{\text{acid 1}}{H_3O^+_{(aq)}} \rightleftharpoons \underset{\text{acid 2}}{C_3H_8NO^+_{2(aq)}} + \overset{\text{base 1}}{H_2O_{(l)}}$
3. $\overset{\text{acid 1}}{HOAc_{(aq)}} + \underset{\text{base 2}}{NH_{3(aq)}} \rightleftharpoons \overset{\text{base 1}}{CH_3CO^−_{2(aq)}} + \underset{\text{acid 2}}{NH^+_{4(aq)}}$
4. $\overset{\text{acid 1}}{SbF_{5(aq)}} + \underset{\text{base 2}}{2HF_{(aq)}} \rightleftharpoons \underset{\text{acid 2}}{H_2F^+_{(aq)}} + \overset{\text{base 1}}{SbF_6^−(aq)}$
Numerical Problems
1. Arrange these acids in order of increasing strength.
• acid A: pKa = 1.52
• acid B: pKa = 6.93
• acid C: pKa = 3.86
Given solutions with the same initial concentration of each acid, which would have the highest percent ionization?
2. Arrange these bases in order of increasing strength:
• base A: pKb = 13.10
• base B: pKb = 8.74
• base C: pKb = 11.87
Given solutions with the same initial concentration of each base, which would have the highest percent ionization?
3. Calculate the Ka and the pKa of the conjugate acid of a base with each pKb value.
1. 3.80
2. 7.90
3. 13.70
4. 1.40
5. −2.50
4. Benzoic acid is a food preservative with a pKa of 4.20. Determine the Kb and the pKb for the benzoate ion.
5. Determine Ka and pKa of boric acid [B(OH)3], solutions of which are occasionally used as an eyewash; the pKb of its conjugate base is 4.80.
Answers
1. acid B < acid C < acid A (strongest)
1. Ka = 6.3 × 10−11; pKa = 10.20
2. Ka = 7.9 × 10−7; pKa = 6.10
3. Ka = 0.50; pKa = 0.30
4. Ka = 2.5 × 10−13; pKa = 12.60
5. Ka = 3.2 × 10−17; pKa = 16.50
2. Ka = 6.3 × 10−10 pKa = 9.20
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.02%3A_A_Qualitative_Description_of__Acid-Base_Equilibria.txt |
Learning Objectives
• To understand how molecular structure affects the strength of an acid or base.
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the A–H or B–H+ bond, the less likely the bond is to break to form H+ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
Relative Acid Strength HF < HCl < HBr < HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 is as follows, with pKa values in parentheses: H2O (14.00 = pKw) < H2S (7.05) < H2Se (3.89) < H2Te (2.6).
Stability of the Conjugate Base
Whether we write an acid–base reaction as \(AH \rightleftharpoons A^−+H^+\) or as \(BH^+ \rightleftharpoons B+H^+\) the conjugate base (A or B) contains one more lone pair of electrons than the parent acid (AH or BH+). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of H+ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses: CH4 (~50) << NH3 (~36) < H2O (14.00) < HF (3.20).
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of CH4 is CH3, and the conjugate base of HF is F. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the F ion than carbon can stabilize the negative charge in the CH3 ion. Consequently, HF has a greater tendency to dissociate to form H+ and F than does methane to form H+ and CH3, making HF a much stronger acid than CH4.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula HE, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form E and H+. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Note the Pattern
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.
Inductive Effects
Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
\[ HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \tag{16.3.1}\]
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5
As the electronegativity of X increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as H+.
The acidity of oxoacids, with the general formula HOXOn (n = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom X. As shown in Figure 16.3.1, the Ka values of the oxoacids of chlorine increase by a factor of about 104 to 106 with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
Note the Pattern
Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure 16.3.1 show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure 16.3.1 and Figure 16.3.2, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from HClO to HClO4 (also written as HOClO3), while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as H+ ions, thereby increasing the strength of the acid.
At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure 16.3.2 , the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms. The electrostatic potential plots in Figure 16.3.2 demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in ClO4, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion (ClO4), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (OCl), the negative charge is largely localized on a single oxygen atom (Figure 16.3.2). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
Note the Pattern
Electron delocalization in the conjugate base increases acid strength.
Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, H3PO4 is a weak acid, H2SO4 is a strong acid, and HClO4 is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to Cl, which causes electrons to be drawn from oxygen to the central atom, weakening the O–H bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table 16.3.1 shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid (H2CO3) were a discrete molecule with the structure (HO)2C=O, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (H3PO4), for which pKa1 = 2.16. Instead, the tabulated value of pKa1 for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see in Section 16.6, however, H2CO3 is only a minor component of the aqueous solutions of CO2 that are referred to as carbonic acid. Similarly, if phosphorous acid (H3PO3) actually had the structure (HO)3P, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as HOCl (pKa = 7.40). In fact, the pKa1 for phosphorous acid is 1.30, and the structure of phosphorous acid is (HO)2P(=O)H with one H atom directly bonded to P and one P=O bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as H3PO4. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table 16.3.1 Values of pKa for Selected Polyprotic Acids and Bases
Polyprotic Acids Formula p K a1 p K a2 p K a3
carbonic acid* “H2CO3 6.35 10.33
citric acid HO2CCH2C(OH)(CO2H)CH2CO2H 3.13 4.76 6.40
malonic acid HO2CCH2CO2H 2.85 5.70
oxalic acid HO2CCO2H 1.25 3.81
phosphoric acid H3PO4 2.16 7.21 12.32
phosphorous acid H3PO3 1.3 6.70
succinic acid HO2CCH2CH2CO2H 4.21 5.64
sulfuric acid H2SO4 −2.0 1.99
sulfurous acid* “H2SO3 1.85 7.21
Polyprotic Bases Formula p K b1 p K b2
ethylenediamine H2N(CH2)2NH2 4.08 7.14
piperazine HN(CH2CH2)2NH 4.27 8.67
propylenediamine H2N(CH2)3NH2 3.45 5.12
*H2CO3 and H2SO3 are at best minor components of aqueous solutions of CO2(g) and SO2(g), respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
\[pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \notag \]
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the –CH3 group of acetic acid by a –CF3 group results in about a 10,000-fold increase in acidity!
Example 16.3.1
Arrange the compounds of each series in order of increasing acid or base strength.
1. sulfuric acid [H2SO4, or (HO)2SO2], fluorosulfonic acid (FSO3H, or FSO2OH), and sulfurous acid [H2SO3, or (HO)2SO]
2. ammonia (NH3), trifluoramine (NF3), and hydroxylamine (NH2OH)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or base strengths
Strategy:
Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.
Solution:
1. Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, FSO3H is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
\[pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \notag \]
2. The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an H+ ion. Thus NF3 is predicted to be a much weaker base than NH3. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in NH3 by OH will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured pKb values: \[pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \notag \]
Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has not been measured.
Exercise
Arrange the compounds of each series in order of
1. decreasing acid strength: H3PO4, CH3PO3H2, and HClO3.
2. increasing base strength: CH3S, OH, and CF3S.
Answer
1. HClO3 > CH3PO3H2 > H3PO4
2. CF3S < CH3S < OH
Summary
The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an H+ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of H+, making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an O–H bond and allow hydrogen to be more easily lost as H+ ions.
Key Takeaway
• Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound.
Conceptual Problems
1. Section 16.3 presented several factors that affect the relative strengths of acids and bases. For each pair, identify the most important factor in determining which is the stronger acid or base in aqueous solution.
1. CH3CCl2CH2CO2H versus CH3CH2CH2CO2H
2. CH3CO2H versus CH3CH2OH
3. HClO versus HBrO
4. CH3C(=O)NH2 versus CH3CH2NH2
5. H3AsO4 versus H3AsO3
2. The stability of the conjugate base is an important factor in determining the strength of an acid. Which would you expect to be the stronger acid in aqueous solution—C6H5NH3+ or NH4+? Justify your reasoning.
3. Explain why H2Se is a weaker acid than HBr.
4. Arrange the following in order of decreasing acid strength in aqueous solution: H3PO4, CH3PO3H2, and HClO3.
5. Arrange the following in order of increasing base strength in aqueous solution: CH3S, OH, and CF3S.
6. Arrange the following in order of increasing acid strength in aqueous solution: HClO2, HNO2, and HNO3.
7. Do you expect H2SO3 or H2SeO3 to be the stronger acid? Why?
8. Give a plausible explanation for why CF3OH is a stronger acid than CH3OH in aqueous solution. Do you expect CHCl2CH2OH to be a stronger or a weaker acid than CH3OH? Why?
9. Do you expect Cl2NH or NH3 to be the stronger base in aqueous solution? Why?
Answers
1. CF3S < CH3S < OH (strongest base)
2. NH3; Cl atoms withdraw electron density from N in Cl2NH.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.03%3A_Molecular_Structure_and_Acid-Base_Strength.txt |
Learning Objectives
• To use Ka and Kb values to calculate the percent ionization and the pH of a solution of an acid or a base.
This section presents a quantitative approach to analyzing acid–base equilibriums. You will learn how to determine the values of Ka and Kb, how to use Ka or Kb to calculate the percent ionization and the pH of an aqueous solution of an acid or a base, and how to calculate the equilibrium constant for the reaction of an acid with a base from the Ka and Kb of the reactants.
Determining Ka and Kb
The ionization constants Ka and Kb are equilibrium constants that are calculated from experimentally measured concentrations, just like the equilibrium constants discussed in Chapter 15. Before proceeding further, it is important to understand exactly what is meant when we describe the concentration of an aqueous solution of a weak acid or a weak base. Suppose, for example, we have a bottle labeled 1.0 M acetic acid or 1.0 M ammonia. As you learned in Chapter 8, such a solution is usually prepared by dissolving 1.0 mol of acetic acid or ammonia in water and adding enough water to give a final volume of exactly 1.0 L. If, however, we were to list the actual concentrations of all the species present in either solution, we would find that none of the values is exactly 1.0 M because a weak acid such as acetic acid or a weak base such as ammonia always reacts with water to some extent. The extent of the reaction depends on the Ka or the Kb, the concentration of the acid or the base, and the temperature. Consequently, only the total concentration of both the ionized and unionized species is equal to 1.0 M.
The analytical concentration (C) is defined as the total concentration of all forms of an acid or a base that are present in solution, regardless of their state of protonation. Thus a “1.0 M” solution of acetic acid has an analytical concentration of 1.0 M, which is the sum of the actual concentrations of unionized acetic acid (CH3CO2H) and the ionized form (CH3CO2):
$C_{CH_3CO_2H}=[CH_3CO_2H] + [CH_3CO_2^−] \tag{16.4.1}$
As we shall see shortly, if we know the analytical concentration and the Ka, we can calculate the actual values of [CH3CO2H] and [CH3CO2].
The equilibrium equations for the reaction of acetic acid and ammonia with water are as follows:
$K_a=\dfrac{[H^+][CH_3CO_2^−]}{[CH_3CO_2H]} \tag{16.4.2}$
$K_b=\dfrac{[NH_4^+][OH^−]}{[NH_3]} \tag{16.4.3}$
where Ka and Kb are the ionization constants for acetic acid and ammonia, respectively. In addition to the analytical concentration of the acid (or the base), we must have a way to measure the concentration of at least one of the species in the equilibrium constant expression to determine the Ka (or the Kb). There are two common ways to obtain the concentrations: (1) measure the electrical conductivity of the solution, which is related to the total concentration of ions present, and (2) measure the pH of the solution, which gives [H+] or [OH].
Example 6 and Example 7 illustrate the procedure for determining Ka for a weak acid and Kb for a weak base. In both cases, we will follow the procedure developed in Chapter 15: the analytical concentration of the acid or the base is the initial concentration, and the stoichiometry of the reaction with water determines the change in concentrations. The final concentrations of all species are calculated from the initial concentrations and the changes in the concentrations. Inserting the final concentrations into the equilibrium constant expression enables us to calculate the Ka or the Kb.
Example 16.4.1
Electrical conductivity measurements indicate that 0.42% of the acetic acid molecules in a 1.00 M solution are ionized at 25°C. Calculate Ka and pKa for acetic acid at this temperature.
Given: analytical concentration and percent ionization
Asked for: Ka and pKa
Strategy:
A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.
B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.
C Substitute the final concentrations into the equilibrium constant expression and calculate the Ka. Take the negative logarithm of Ka to obtain the pKa.
Solution:
A The balanced equilibrium equation for the dissociation of acetic acid is as follows:
$CH_3CO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+CH3CO^−_{2(aq)}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][CH_3CO_2^−]}{[CH_3CO_2H]}$
B To calculate the Ka, we need to know the equilibrium concentrations of CH3CO2H, CH3CO2, and H+. The most direct way to do this is to construct a table that lists the initial concentrations and the changes in concentrations that occur during the reaction to give the final concentrations, using the procedure introduced in Chapter 15. The initial concentration of unionized acetic acid ([CH3CO2H]i) is the analytical concentration, 1.00 M, and the initial acetate concentration ([CH3CO2]i) is zero. The initial concentration of H+ is not zero, however; [H+]i is 1.00 × 10−7 M due to the autoionization of water. The measured percent ionization tells us that 0.42% of the acetic acid molecules are ionized at equilibrium. Consequently, the change in the concentration of acetic acid is Δ[CH3CO2H] = −(4.2 × 10−3)(1.00 M) = −0.0042 M. Conversely, the change in the acetate concentration is Δ[CH3CO2] = +0.0042 M because every 1 mol of acetic acid that ionizes gives 1 mol of acetate. Because one proton is produced for each acetate ion formed, Δ[H+] = +0.0042 M as well. These results are summarized in the following table.
$CH_3CO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+CH_3CO^−_{2(aq)}$
[CH3CO2H] [H+] [CH3CO2]
initial 1.00 1.00 × 10−7 0
change −0.0042 +0.0042 +0.0042
final
The final concentrations of all species are therefore as follows:
• $[CH_3CO_2H]_f=[CH_3CO_2H]_i+Δ[CH_3CO_2H]=1.00\; M+(−0.0042\; M)=1.00 \;M$
• $[CH_3CO_2^−]_f=[CH_3CO_2^−]_i+Δ[CH_3CO_2^−]=0\; M+(+0.0042\; M)=0.0042\; M$
• $[H^+]_f=[H^+]_i+Δ[H^+]=1.00 \times 10^{−7} M + (+0.0042\; M)= 0.0042\; M$
C We can now calculate Ka by inserting the final concentrations into the equilibrium constant expression:
$K_a=\dfrac{[H^+][CH_3CO_2^−]}{[CH_3CO_2H]}=\dfrac{(0.0042)(0.0042)}{1.00}=1.8 \times 10^{−5}$
The pKa is the negative logarithm of Ka: pKa = −log Ka = −log(1.8 × 10−5) = 4.74.
Exercise
Picric acid is the common name for 2,4,6-trinitrophenol, a derivative of phenol (C6H5OH) in which three H atoms are replaced by nitro (–NO2) groups. The presence of the nitro groups removes electron density from the phenyl ring, making picric acid a much stronger acid than phenol (pKa = 9.99). The nitro groups also make picric acid potentially explosive, as you might expect based on its chemical similarity to 2,4,6-trinitrotoluene, better known as TNT. A 0.20 M solution of picric acid is 73% ionized at 25°C. Calculate Ka and pKa for picric acid.
Answer: Ka = 0.39; pKa = 0.41
Example 16.4.2
A 1.0 M aqueous solution of ammonia has a pH of 11.63 at 25°C. Calculate Kb and pKb for ammonia.
Given: analytical concentration and pH
Asked for: Kb and pKb
Strategy:
A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.
B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.
C Substitute the final concentrations into the equilibrium constant expression and calculate the Kb. Take the negative logarithm of Kb to obtain the pKb.
Solution:
A The balanced equilibrium equation for the reaction of ammonia with water is as follows:
$NH_{3(aq)}+H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)}+OH^−_{(aq)} \notag$
and the equilibrium constant expression is as follows:
$K_b=\dfrac{[NH_4^+][OH^−]}{[NH_3]} \notag$
Remember that water does not appear in the equilibrium constant expression for Kb.
B To calculate Kb, we need to know the equilibrium concentrations of NH3, NH4+, and OH. The initial concentration of NH3 is the analytical concentration, 1.0 M, and the initial concentrations of NH4+ and OH are 0 M and 1.00 × 10−7 M, respectively. In this case, we are given the pH of the solution, which allows us to calculate the final concentration of one species (OH) directly, rather than the change in concentration. Recall that pKw = pH + pOH = 14.00 at 25°C. Thus pOH = 14.00 − pH = 14.00 − 11.63 = 2.37, and [OH]f = 10−2.37 = 4.3 × 10−3 M. Our data thus far are listed in the following table.
$NH_{3(aq)} \rightleftharpoons NH^+_{4(aq)}+OH^−_{(aq)}$
[NH3] [NH4+] [OH]
initial 1.0 0 1.00 × 10−7
change
final 4.3 × 10−3
The final [OH] is much greater than the initial [H+], so the change in [OH] is as follows:
Δ[OH] = (4.3 × 10−3 M) − (1.00 × 10−7 M) ≈ 4.3 × 10−3 M
The stoichiometry of the reaction tells us that 1 mol of NH3 is converted to NH4+ for each 1 mol of OH formed, so
Δ[NH4+] = +4.3 × 10−3 M and Δ[NH3] = −4.3 ×10−3 M
We can now insert these values for the changes in concentrations into the table, which enables us to complete the table.
$H_2O_{(l)}+NH3_{(aq)} \rightleftharpoons NH^+_{4(aq)}+OH^−_{(aq)}$
[NH3] [NH4+] [OH]
initial 1.0 0 1.00 × 10−7
change −4.3 × 10−3 +4.3 × 10−3 +4.3 × 10−3
final 1.0 4.3 × 10−3 4.3 × 10−3
C Inserting the final concentrations into the equilibrium constant expression gives Kb:
$K_b=\dfrac{[NH_4^+][OH^−]}{[NH_3]}=\dfrac{(4.3 \times 10^{−3})^2}{1.0}=1.8 \times 10^{−5} \notag$
and pKb = −log Kb = 4.74.
The Kb and the pKb for ammonia are almost exactly the same as the Ka and the pKa for acetic acid at 25°C. In other words, ammonia is almost exactly as strong a base as acetic acid is an acid. Consequently, the extent of the ionization reaction in an aqueous solution of ammonia at a given concentration is the same as in an aqueous solution of acetic acid at the same concentration.
Exercise
The pH of a 0.050 M solution of pyridine (C6H5N) is 8.96 at 25°C. Calculate Kb and pKb for pyridine.
Answer: Kb = 1.7 × 10−9; pKb = 8.77
Calculating Percent Ionization from Ka or Kb
When carrying out a laboratory analysis, chemists frequently need to know the concentrations of all species in solution. Because the reactivity of a weak acid or a weak base is usually very different from the reactivity of its conjugate base or acid, we often need to know the percent ionization of a solution of an acid or a base to understand a chemical reaction. The percent ionization is defined as follows:
$\text{percent ionization οf acid} =\dfrac{[H^+]}{C_{HA}}×100 \tag{16.4.4}$
$\text{percent ionization οf base}=\dfrac{[OH^−]}{C_B}×100 \tag{16.4.5}$
One way to determine the concentrations of species in solutions of weak acids and bases is a variation of the tabular method we used previously to determine Ka and Kb values. As a demonstration, we will calculate the concentrations of all species and the percent ionization in a 0.150 M solution of formic acid at 25°C. The data in Table E1 show that formic acid (Ka = 1.8 × 10−4 at 25°C) is a slightly stronger acid than acetic acid. The equilibrium equation for the ionization of formic acid in water is as follows:
$HCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+HCO^−_{2(aq)} \tag{16.4.6}$
and the equilibrium constant expression for this reaction is as follows:
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} \tag{16.4.7}$
We set the initial concentration of HCO2H equal to 0.150 M, and that of HCO2 is 0 M. The initial concentration of H+ is 1.00 × 10−7 M due to the autoionization of water. Because the equilibrium constant for the ionization reaction is small, the equilibrium will lie to the left, favoring the unionized form of the acid. Hence we can define x as the amount of formic acid that dissociates.
If the change in [HCO2H] is −x, then the change in [H+] and [HCO2] is +x. The final concentration of each species is the sum of its initial concentration and the change in concentration, as summarized in the following table.
$HCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+HCO^−_{2(aq)}$
[HCO2H] [H+] [HCO2]
initial 0.150 1.00 × 10−7 0
change x +x +x
final (0.150 − x) (1.00 × 10−7 + x) x
We can calculate x by substituting the final concentrations from the table into the equilibrium constant expression:
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(1.00 \times 10^{−7}+x) x}{0.150−x}$
Because the ionization constant Ka is small, x is likely to be small compared with the initial concentration of formic acid: (0.150 − x) M ≈ 0.150 M. Moreover, [H+] due to the autoionization of water (1.00 × 10−7 M) is likely to be negligible compared with [H+] due to the dissociation of formic acid: (1.00 × 10−7 + x) M ≈ x M. Inserting these values into the equilibrium constant expression and solving for x,
$K_a=\dfrac{x^2}{0.150} =1.8 \times 10^{−4}$
$x=5.2 \times 10^{−3}$
We can now calculate the concentrations of the species present in a 0.150 M formic acid solution by inserting this value of x into the expressions in the last line of the table:
$[HCO_2H]=(0.150−x)\; M=0.145 \;M$
$[HCO_2]=x=5.2 \times 10^{−3}\; M$
$[H^+]=(1.00×10−7+x) M=5.2 \times 10^{−3} M$
Thus the pH of the solution is –log(5.2 × 10−3) = 2.28. We can also use these concentrations to calculate the fraction of the original acid that is ionized. In this case, the percent ionization is the ratio of [H+] (or [HCO2]) to the analytical concentration, multiplied by 100 to give a percentage:
$\text{percent ionization}=\dfrac{[H^+]}{C_{HA}}×100=\dfrac{5.2 \times 10^{−3}\; M}{0.150} \times 100=3.5\%$
Always check to make sure that any simplifying assumption was valid. As a general rule of thumb, approximations such as those used here are valid only if the quantity being neglected is no more than about 5% of the quantity to which it is being added or from which it is being subtracted. If the quantity that was neglected is much greater than about 5%, then the approximation is probably not valid, and you should go back and solve the problem using the quadratic formula. In the previous demonstration, both simplifying assumptions were justified: the percent ionization is only 3.5%, which is well below the approximately 5% limit, and the 1.00 × 10−7 M [H+] due to the autoionization of water is much, much less than the 5.2 × 10−3 M [H+] due to the ionization of formic acid.
As a general rule, the [H+] contribution due to the autoionization of water can be ignored as long as the product of the acid or the base ionization constant and the analytical concentration of the acid or the base is at least 10 times greater than the [H+] or [OH] from the autoionization of water—that is, if
$K_aC_{HA} \ge 10(1.00 \times 10^{−7}) = 1.0 \times 10^{−6} \tag{16.4.8}$
or
$K_bC_B \ge 10(1.00 \times 10^{−7}) = 1.0 \times 10^{−6} \tag{16.4.9}$
By substituting the appropriate values for the formic acid solution into Equation 16.45, we see that the simplifying assumption is valid in this case:
$K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} > 1.0 \times 10^{−6} \tag{16.4.10}$
Doing this simple calculation before solving this type of problem saves time and allows you to write simplified expressions for the final concentrations of the species present. In practice, it is necessary to include the [H+] contribution due to the autoionization of water only for extremely dilute solutions of very weak acids or bases.
Example 8 illustrates how the procedure outlined previously can be used to calculate the pH of a solution of a weak base.
Example 16.4.3
Calculate the pH and percent ionization of a 0.225 M solution of ethylamine (CH3CH2NH2), which is used in the synthesis of some dyes and medicines. The pKb of ethylamine is 3.19 at 20°C.
Given: concentration and pKb
Asked for: pH and percent ionization
Strategy:
A Write the balanced equilibrium equation for the reaction and the equilibrium constant expression. Calculate Kb from pKb.
B Use Equation 16.4.8 to see whether you can ignore [H+] due to the autoionization of water. Then use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [OH]. Use Equation 16.4.5 to calculate the percent ionization.
C Use the relationship Kw = [OH][H+] to obtain [H+]. Then calculate the pH of the solution.
Solution:
A We begin by writing the balanced equilibrium equation for the reaction:
$CH_3CH_2NH_{2(aq)}+H_2O_{(l)} \rightleftharpoons CH_3CH_2NH^+_{3(aq)}+OH^−_{(aq)} \notag$
The corresponding equilibrium constant expression is as follows:
$K_b=\dfrac{[CH_3CH_2NH_3^+][OH^−]}{[CH_3CH_2NH_2}] \notag$
From the $pK_b$, we have $K_b = 10−3.19 = 6.5 \times 10^{−4}$.
B To calculate the pH, we need to determine the H+ concentration. Unfortunately, H+ does not appear in either the chemical equation or the equilibrium constant expression. However, [H+] and [OH] in an aqueous solution are related by Kw = [H+][OH]. Hence if we can determine [OH], we can calculate [H+] and then the pH. The initial concentration of CH3CH2NH2 is 0.225 M, and the initial [OH] is 1.00 × 10−7 M. Because ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to let x equal the amount of CH3CH2NH2 that reacts with water. The change in [CH3CH2NH2] is therefore −x, and the change in both [CH3CH2NH3+] and [OH] is +x. To see whether the autoionization of water can safely be ignored, we substitute Kb and CB into Equation 16.4.10:
KbCB = (6.5 × 10−4)(0.225) = 1.5 × 10−4 > 1.0 × 10−6
Thus the simplifying assumption is valid, and we will not include [OH] due to the autoionization of water in our calculations.
$H_2O_{(l)}+CH_3CH_2NH_{2(aq)} \rightleftharpoons CH_3CH_2NH^+_{3(aq)}+OH^−_{(aq)}$
[CH3CH2NH2] [CH3CH2NH3+] [OH]
initial 0.225 0 1.00 × 10−7
change x +x +x
final (0.225 − x) x x
Substituting the quantities from the last line of the table into the equilibrium constant expression,
$K_b=\dfrac{[CH_3CH_2NH_3^+][OH^−]}{[CH_3CH_2NH_2]}=\dfrac{(x)(x)}{0.225−x}=6.5\times 10^{−4} \notag$
As before, we assume the amount of CH3CH2NH2 that ionizes is small compared with the initial concentration, so [CH3CH2NH2]f = 0.225 − x ≈ 0.225. With this assumption, we can simplify the equilibrium equation and solve for x:
$K_b=\dfrac{x^2}{0.225} =6.5 \times 10^{−4} \notag$
$x=0.012=[CH_3CH_2NH_3^+]_f=[OH^−]_f \notag$
The percent ionization is therefore
$\text{percent ionization}=\dfrac{[OH^−]}{C_B} \times 100=\dfrac{0.012\; M}{0.225\; M} \times 100=5.4\%$
which is at the upper limit of the approximately 5% range that can be ignored. The final hydroxide concentration is thus 0.012 M.
C We can now determine the [H+] using the expression for Kw:
$K_w=[OH^−][H^+] \notag$
$1.01 \times 10^{−14} =(0.012 \;M)[H^+] \notag$
$8.4 \times 10^{−13}\; M=[H^+] \notag$
The pH of the solution is −log(8.4 × 10−13) = 12.08. Alternatively, we could have calculated pOH as −log(0.012) = 1.92 and determined the pH as follows:
$pH + pOH =pKw=14.00 \notag$
$pH=14.00−1.92=12.08 \notag$
The two methods are equivalent.
Exercise
Aromatic amines, in which the nitrogen atom is bonded directly to a phenyl ring (−C6H5) tend to be much weaker bases than simple alkylamines. For example, aniline (C6H5NH2) has a pKb of 9.13 at 25°C. What is the pH of a 0.050 M solution of aniline?
Answer: 8.78
The previous examples illustrate a key difference between solutions of strong acids and bases and solutions of weak acids and bases. Because strong acids and bases ionize essentially completely in water, the percent ionization is always approximately 100%, regardless of the concentration. In contrast, the percent ionization in solutions of weak acids and bases is small and depends on the analytical concentration of the weak acid or base. As illustrated for benzoic acid in Figure 16.4.1, the percent ionization of a weak acid or a weak base actually increases as its analytical concentration decreases. The percent ionization also increases as the magnitude of Ka and Kb increases.
Unlike the Ka or the Kb, the percent ionization is not a constant for weak acids and bases but depends on both the Ka or the Kb and the analytical concentration. Consequently, the procedure in Example 8 must be used to calculate the percent ionization and pH for solutions of weak acids and bases. Example 9 and its corresponding exercise demonstrate that the combination of a dilute solution and a relatively large Ka or Kb can give a percent ionization much greater than 5%, making it necessary to use the quadratic equation to determine the concentrations of species in solution.
Note the Pattern
The percent ionization in a solution of a weak acid or a weak base increases as the analytical concentration decreases and as the Ka or the Kb increases.
Example 16.4.4
Benzoic acid (C6H5CO2H) is used in the food industry as a preservative and medically as an antifungal agent. Its pKa at 25°C is 4.20, making it a somewhat stronger acid than acetic acid. Calculate the percentage of benzoic acid molecules that are ionized in each solution.
1. a 0.0500 M solution
2. a 0.00500 M solution
Given: concentrations and pKa
Asked for: percent ionization
Strategy:
A Write both the balanced equilibrium equation for the ionization reaction and the equilibrium equation (Equation 16.2.2). Use Equation 16.2.13 to calculate the Ka from the pKa.
B For both the concentrated solution and the dilute solution, use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [C6H5CO2]f for each solution.
C Use the values of [C6H5CO2]f and Equation 16.4.4 to calculate the percent ionization.
Solution:
A If we abbreviate benzoic acid as PhCO2H where Ph = –C6H5, the balanced equilibrium equation for the ionization reaction and the equilibrium equation can be written as follows:
$PhCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+PhCO^−_{2(aq)} \notag$
$K_a=\dfrac{[H^+][PhCO_2^−]}{[PhCO_2H]} \notag$
From the pKa, we have Ka = 10−4.20 = 6.3 × 10−5.
1. B For the more concentrated solution, we set up our table of initial concentrations, changes in concentrations, and final concentrations:
$PhCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+PhCO^−_{2(aq)}$
[PhCO2H] [H+] [PhCO2]
initial 0.0500 1.00 × 10−7 0
change x +x +x
final (0.0500 − x) (1.00 × 10−7 + x) x
Inserting the expressions for the final concentrations into the equilibrium equation and making our usual assumptions, that [PhCO2] and [H+] are negligible due to the autoionization of water,
$K_a=\dfrac{[H^+][PhCO_2^−]}{[PhCO_2H]}=\dfrac{(x)(x)}{0.0500−x}=\dfrac{x^2}{0.0500}=6.3 \times 10^{−5} \notag$
$1.8 \times 10^{-3}=x \notag$
This value is less than 5% of 0.0500, so our simplifying assumption is justified, and [PhCO2] at equilibrium is 1.8 × 10−3 M. We reach the same conclusion using CHA: KaCHA = (6.3 × 10−5)(0.0500) = 3.2 × 10−6 > 1.0 × 10−6.
C The percent ionized is the ratio of the concentration of PhCO2 to the analytical concentration, multiplied by 100:
$\text{percent ionized}=\dfrac{[PhCO_2^−]}{C_{PhCO_2H}} \times 100=\dfrac{1.8 \times 10^{−30}}{0.0500} \times 100=3.6\% \notag$
Because only 3.6% of the benzoic acid molecules are ionized in a 0.0500 M solution, our simplifying assumptions are confirmed.
2. B For the more dilute solution, we proceed in exactly the same manner. Our table of concentrations is therefore as follows:
$PhCO_2H_{(aq)} \rightleftharpoons H^+_{(aq)}+PhCO^−_{2(aq)}$
[PhCO2H] [H+] [PhCO2]
initial 0.00500 1.00 × 10−7 0
change x +x +x
final (0.00500 − x) (1.00 × 10−7 + x) x
Inserting the expressions for the final concentrations into the equilibrium equation and making our usual simplifying assumptions,
$K_a=\dfrac{[H^+][PhCO_2^−]}{[PhCO_2H]}=\dfrac{(x)(x)}{0.00500−x}=\dfrac{x^2}{0.00500}=6.3 \times 10^{−5} \notag$
$5.6 \times 10^{−4}=x \notag$
Unfortunately, this number is greater than 10% of 0.00500, so our assumption that the fraction of benzoic acid that is ionized in this solution could be neglected and that (0.00500 − x) ≈ x is not valid. Furthermore, we see that KaCHA = (6.3 × 10−5)(0.00500) = 3.2 × 10−7 < 1.0 × 10−6. Thus the relevant equation is as follows:
$\dfrac{x^2}{0.00500−x}=6.3 \times 10^{−5} \notag$
which must be solved using the quadratic formula. Multiplying out the quantities,
x2 = (6.3 × 10−5)(0.00500 − x) = (3.2 × 10−7) − (6.3 × 10−5)x
Rearranging the equation to fit the standard quadratic equation format,
x2 + (6.3 × 10−5)x − (3.2 × 10−7) = 0
This equation can be solved by using the quadratic formula:
$x=\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a} \notag$
$x=\dfrac{−(6.3 \times10^{−5}) \pm \sqrt{(6.3 \times 10^{−5})^2−4(1)(−3.2 \times 10^{−7})}}{2(1)} \notag$
$x=\dfrac{−(6.3 \times 10^{−5}) \pm (1.1 \times 10^{−3})}{2} = 5.3 \times 10^{−4} \notag$
or
$x= −5.9\times 10^{−4} \notag$
Because a negative x value corresponds to a negative [PhCO2], which is not physically meaningful, we use the positive solution: x = 5.3 × 10−4. Thus [PhCO2] = 5.3 × 10−4 M.
C The percent ionized is therefore
$\text{percent ionized}=\dfrac{[PhCO_2^−]C_{PhCO_2H}} \times 100=\dfrac{5.3 \times 10^{−4}}{0.00500} \times 100=11\% \notag$
In the more dilute solution (C = 0.00500 M), 11% of the benzoic acid molecules are ionized versus only 3.6% in the more concentrated solution (C = 0.0500 M). Decreasing the analytical concentration by a factor of 10 results in an approximately threefold increase in the percentage of benzoic acid molecules that are ionized.
Exercise
Lactic acid (CH3CH(OH)CO2H) is a weak acid with a pKa of 3.86 at 25°C. What percentage of the lactic acid is ionized in each solution?
1. a 0.10 M solution
2. a 0.0020 M solution
Answer
1. 3.7%
2. 23%
Determining Keq from Ka and Kb
In Section 16.2 , you learned how to use Ka and Kb values to qualitatively predict whether reactants or products are favored in an acid–base reaction. Tabulated values of Ka (or pKa) and Kb (or pKb), plus the Kw, enable us to quantitatively determine the direction and extent of reaction for a weak acid and a weak base by calculating K for the reaction. To illustrate how to do this, we begin by writing the dissociation equilibriums for a weak acid and a weak base and then summing them:
$acid:\;\;HA \rightleftharpoons H^++A^− \;\;\; K_a \tag{16.4.11a}$
$\;base:\;\;B + H_2O \rightleftharpoons HB^++OH^− \;\;\; K_b\tag{16.4.11b}$
$sum:\;\;HA + B+ H_2O \rightleftharpoons H^++A^−+HB^++OH^− \;\;\; K_{sum}=K_aK_b \tag{16.4.11c}$
The overall reaction has H2O on the left and H+ and OH on the right, which means it involves the autoionization of water ($H_2O \rightleftharpoons H^++OH^−$) in addition to the acid–base equilibrium in which we are interested. We can obtain an equation that includes only the acid–base equilibrium by simply adding the equation for the reverse of the autoionization of water ($H^++OH^− \rightleftharpoons H_2O$), for which K = 1/Kw, to the overall equilibrium in Equation 16.4.11 and canceling:
$HA + B+ \cancel{H_2O} \rightleftharpoons \cancel{H^+} + A^−+HB^++\cancel{OH^−} \;\;\; K_{sum}=K_aK_b \tag{16.4.12a}$
$\cancel{ H^+} + \cancel{OH^−} \rightleftharpoons \cancel{H_2O} \;\;\; 1/K_w \tag{16.4.12b}$
$HA + B \rightleftharpoons A^- + HB^+ \;\;\; K=(K_aK_b)/K_w \tag{16.4.12c}$
Thus the equilibrium constant for the reaction of a weak acid with a weak base is the product of the ionization constants of the acid and the base divided by Kw. Example 10 illustrates how to calculate the equilibrium constant for the reaction of a weak acid with a weak base.
Example 16.4.5
Fish tend to spoil rapidly, even when refrigerated. The cause of the resulting “fishy” odor is a mixture of amines, particularly methylamine (CH3NH2), a volatile weak base (pKb = 3.34). Fish is often served with a wedge of lemon because lemon juice contains citric acid, a triprotic acid with pKa values of 3.13, 4.76, and 6.40 that can neutralize amines. Calculate the equilibrium constant for the reaction of excess citric acid with methylamine, assuming that only the first dissociation constant of citric acid is important.
Given: pKb for base and pKa for acid
Asked for: K
Strategy:
A Write the balanced equilibrium equation and the equilibrium constant expression for the reaction.
B Convert pKa and pKb to Ka and Kb and then use Equation 16.4.12 to calculate K.
Solution:
A If we abbreviate citric acid as H3citrate, the equilibrium equation for its reaction with methylamine is as follows:
$CH_3NH_{2(aq)}+H_3citrate_{(aq)} \rightleftharpoons CH_3NH^+_{3(aq)}+H_2citrate^−_{(aq)} \notag$
The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{[CH_3NH_3^+][H_2citrate^−]}{[CH_3NH_2][H_3citrate]} \notag$
B Equation 16.49 is K = (KaKb)/Kw. Converting pKa and pKb to Ka and Kb gives Ka = 10−3.13 = 7.4 × 10−4 for citric acid and Kb = 10−3.34 = 4.6 × 10−4 for methylamine. Substituting these values into the equilibrium equation,
$K=\dfrac{K_aK_b}{K_w}=\dfrac{(7.4 \times 10^{−4})(4.6 \times 10^{−4})}{1.01 \times 10^{−14}}=3.4 \times 10^7 \notag$
The value of pK can also be calculated directly by taking the negative logarithm of both sides of Equation 16.49, which gives
$pK = pK_a + pK_b − pK_w = 3.13 + 3.34 − 14.00 = −7.53 \notag \] Thus K = 10−(−7.53) = 3.4 × 107, in agreement with the earlier value. In either case, the K values show that the reaction of citric acid with the volatile, foul-smelling methylamine lies very far to the right, favoring the formation of a much less volatile salt with no odor. This is one reason a little lemon juice helps make less-than-fresh fish more appetizing. Exercise Dilute aqueous ammonia solution, often used as a cleaning agent, is also effective as a deodorizing agent. To see why, calculate the equilibrium constant for the reaction of aqueous ammonia with butyric acid (CH3CH2CH2CO2H), a particularly foul-smelling substance associated with the odor of rancid butter and smelly socks. The pKb of ammonia is 4.75, and the pKa of butyric acid is 4.83. Answer: 2.6 × 104 Summary If the concentration of one or more of the species in a solution of an acid or a base is determined experimentally, Ka and Kb can be calculated, and Ka, pKa, Kb, and pKb can be used to quantitatively describe the composition of solutions of acids and bases. The concentrations of all species present in solution can be determined, as can the pH of the solution and the percentage of the acid or base that is ionized. The equilibrium constant for the reaction of a weak acid with a weak base can be calculated from Ka (or pKa), Kb (or pKb), and Kw. Key Takeaway • For a solution of a weak acid or a weak base, the percent ionization increases as the Ka or the Kb increases and as the analytical concentration decreases. Key Equations Percent ionization of acid Equation 16.4.4: \(=\dfrac{[H^+]}{C_{HA}}×100$
Percent ionization of base
Equation 16.4.5: $=\dfrac{[OH^−]}{C_B}×100$
Equilibrium constant for reaction of a weak acid with a weak base
Equation 16.4.12: $K=\dfrac{K_aK_b}{K_w}$
Conceptual Problems
1. Explain why the analytical concentration (C) of H2SO4 is equal to [H2SO4] + [HSO4] + [SO42−].
2. Write an expression for the analytical concentration (C) of H3PO4 in terms of the concentrations of the species actually present in solution.
3. For relatively dilute solutions of a weak acid such as acetic acid (CH3CO2H), the concentration of undissociated acetic acid in solution is often assumed to be the same as the analytical concentration. Explain why this is a valid practice.
4. How does dilution affect the percent ionization of a weak acid or a weak base?
5. What is the relationship between the Ka of a weak acid and its percent ionization? Does a compound with a large pKa value have a higher or a lower percent ionization than a compound with a small pKa value (assuming the same analytical concentration in both cases)? Explain.
6. For a dilute solution of a weak acid (HA), show that the pH of the solution can be approximated using the following equation (where CHA is the analytical concentration of the weak acid):
$pH=−\log \sqrt{K_a C_{HA}}$
Under what conditions is this approximation valid?
Numerical Problems
1. The pKa of NH3 is estimated to be 35. Its conjugate base, the amide ion (NH2), can be isolated as an alkali metal salt, such as sodium amide (NaNH2). Calculate the pH of a solution prepared by adding 0.100 mol of sodium amide to 1.00 L of water. Does the pH differ appreciably from the pH of a NaOH solution of the same concentration? Why or why not?
2. Phenol is a topical anesthetic that has been used in throat lozenges to relieve sore throat pain. Describe in detail how you would prepare a 2.00 M solution of phenol (C6H5OH) in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of phenol with water? Use the information in Table E1 and Table E2 to calculate the pH of the phenol solution.
3. Describe in detail how you would prepare a 1.50 M solution of methylamine in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of methylamine with water? Use the information in Table E1 and Table E2 to calculate the pH of the solution.
4. A 0.200 M solution of diethylamine, a substance used in insecticides and fungicides, is only 3.9% ionized at 25°C. Write an equation showing the equilibrium reaction and then calculate the pKb of diethylamine. What is the pKa of its conjugate acid, the diethylammonium ion? What is the equilibrium constant expression for the reaction of diethylammonium chloride with water?
5. A 1.00 M solution of fluoroacetic acid (FCH2CO2H) is 5% dissociated in water. What is the equilibrium constant expression for the dissociation reaction? Calculate the concentration of each species in solution and then calculate the pKa of FCH2CO2H.
6. The pKa of 3-chlorobutanoic acid (CH3CHClCH2CO2H) is 4.05. What percentage is dissociated in a 1.0 M solution? Do you expect the pKa of butanoic acid to be greater than or less than the pKa of 3-chlorobutanoic acid? Why?
7. The pKa of the ethylammonium ion (C2H5NH3+) is 10.64. What percentage of ethylamine is ionized in a 1.00 M solution of ethylamine?
8. The pKa of Cl3CCO2H is 0.64. What is the pH of a 0.580 M solution? What percentage of the Cl3CCO2H is dissociated?
9. The pH of a 0.150 M solution of aniline hydrochloride (C6H5NH3+Cl) is 2.70. What is the pKb of the conjugate base, aniline (C6H5NH2)? Do you expect the pKb of (CH3)2CHNH2 to be greater than or less than the pKb of C6H5NH2? Why?
10. What is the pH of a 0.620 M solution of CH3NH3+Br if the pKb of CH3NH2 is 10.62?
11. The pKb of 4-hydroxypyridine is 10.80 at 25°C. What is the pH of a 0.0250 M solution?
12. The pKa values of formic acid and the methylammonium ion are 3.75 and 10.62, respectively. Calculate K for the following reaction:
$HCO^−_{2(aq)} + CH_3NH^+_{3(aq)} \rightleftharpoons HCO_2H_{(aq)}+CH_3NH_{2(aq)}$
13. The pKa values of butanoic acid and the ammonium ion are 4.82 and 9.24, respectively. Calculate K for the following reaction:
$CH_3CH_2CH_2CO^−_{2(aq)}+NH^+_{4(aq)} \rightleftharpoons CH_3CH_2CH_2CO_2H_{(aq)}+NH_{3(aq)}$
14. Use the information in Tables E1 and E2 to calculate the pH of a 0.0968 M solution of calcium formate.
15. Calculate the pH of a 0.24 M solution of sodium lactate. The pKa of lactic acid is 3.86.
16. Use the information in Tables E1 and E2 to determine the pH of a solution prepared by dissolving 750.0 mg of methylammonium chloride (CH3NH3+Cl) in enough water to make 150.0 mL of solution.
17. Use the information in Tables E1 and E2 to determine the pH of a solution prepared by dissolving 855 mg of sodium nitrite (NaNO2) in enough water to make 100.0 mL of solution.
Answers
1. pKb = 9.43; (CH3)2CHNH2 will be a stronger base and have a lower pKb; aniline is a weaker base because the lone pair on the nitrogen atom can be delocalized on the aromatic ring.
2. 3.8 × 10−5
3. 8.18
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.04%3A_Quantitative_Aspects_of_Acid-Base_Equilibria.txt |
Learning Objectives
• To calculate the pH at any point in an acid–base titration.
In Chapter 8, you learned that in an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curveA plot of the pH of the solution being titrated versus the amount of acid or base (of known concentration) added.. The shape of the curve provides important information about what is occurring in solution during the titration.
Titrations of Strong Acids and Bases
Part (a) of Figure 16.5.1 shows a plot of the pH as 0.20 M HCl is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as HCl is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of HCl (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M NaOH is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of NaOH as shown in part (b) in Figure 16.5.1 . As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.
Suppose that we now add 0.20 M NaOH to 50.0 mL of a 0.10 M solution of HCl. Because HCl is a strong acid that is completely ionized in water, the initial [H+] is 0.10 M, and the initial pH is 1.00. Adding NaOH decreases the concentration of H+ because of the neutralization reaction: OH+ H+ ⇌ H2O (in part (a) in Figure 16.5.2). Thus the pH of the solution increases gradually. Near the equivalence pointThe point in a titration where a stoichiometric amount of the titrant has been added., however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the H+ ions originally present have been consumed. (For more information on titrations and the equivalence point, see Section 8.9 .) For the titration of a monoprotic strong acid (HCl) with a monobasic strong base (NaOH), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
$moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \tag{16.5.1}$
If 0.20 M NaOH is added to 50.0 mL of a 0.10 M solution of HCl, we solve for Vb:
$V_b(0.20 Me)=0.025 L=25 mL$
At the equivalence point (when 25.0 mL of NaOH solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more NaOH produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M NaOH.
As shown in part (b) in Figure 16.5.2, the titration of 50.0 mL of a 0.10 M solution of NaOH with 0.20 M HCl produces a titration curve that is nearly the mirror image of the titration curve in part (a) in Figure 16.5.2. The pH is initially 13.00, and it slowly decreases as HCl is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M HCl.
The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
Note the Pattern
The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.
Example 16.5.1
Calculate the pH of the solution after 24.90 mL of 0.200 M NaOH has been added to 50.00 mL of 0.100 M HCl.
Given: volumes and concentrations of strong base and acid
Asked for: pH
Strategy:
A Calculate the number of millimoles of H+ and OH to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.
B Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.
Solution:
A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of H+ in 50.00 mL of 0.100 M HCl can be calculated as follows:
$50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \notag$
The number of millimoles of NaOH added is as follows:
$24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \notag$
Thus H+ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of OH to the HCl solution. Because only 4.98 mmol of OH has been added, the amount of excess H+ is 5.00 mmol − 4.98 mmol = 0.02 mmol of H+.
B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of H+ is as follows:
$\left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{=4} \; M \notag$
The pH is −log[H+] = −log(3 × 10−4) = 3.5, which is significantly less than the pH of 7.00 for a neutral solution.
Exercise
Calculate the pH of a solution prepared by adding 40.00 mL of 0.237 M HCl to 75.00 mL of a 0.133 M solution of NaOH.
Answer: 11.6
Titrations of Weak Acids and Bases
In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding Ka or Kb. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned in Section 16.4, [H+] of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its pKa and its concentration. Because only a fraction of a weak acid dissociates, [H+] is less than [HA]. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Part (a) in Figure 16.5.3 shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M NaOH superimposed on the curve for the titration of 0.100 M HCl shown in part (a) in Figure 16.5.2 . Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the HCl solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the HCl titration; the magnitude of the pH change at the equivalence point depends on the pKa of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess NaOH present, regardless of whether the acid is weak or strong.
Note the Pattern
The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the Ka or Kb.
The titration curve in part (a) in Figure 16.5.3 was created by calculating the starting pH of the acetic acid solution before any NaOH is added and then calculating the pH of the solution after adding increasing volumes of NaOH. The procedure is illustrated in the following subsection and Example 12 for three points on the titration curve, using the pKa of acetic acid (4.76 at 25°C; Ka = 1.7 × 10−5).
Calculating the pH of a Solution of a Weak Acid or a Weak Base
As explained Section 16.4, if we know Ka or Kb and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a table of initial concentrations, changes in concentrations, and final concentrations. In this situation, the initial concentration of acetic acid is 0.100 M. If we define x as [H+] due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:
$CH_3CO_2H_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3CO_2^−$
[CH3CO2H] [H+] [CH3CO2]
initial 0.100 1.00 × 10−7 0
change x +x +x
final (0.100 − x) x x
In this and all subsequent examples, we will ignore [H+] and [OH] due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.4.8 and Equation 16.4.9 to check that this assumption is justified.
Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),
$K_a=\dfrac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]}=\dfrac{(x)(x)}{0.100 - x} \approx \dfrac{x^2}{0.100}=1.74 \times 10^{-5} \notag$
Solving this equation gives $x = [H^+] = 1.32 \times 10^{-3}\; M$. Thus the pH of a 0.100 M solution of acetic acid is as follows:
$pH = −\log(1.32 \times 10^{-3}) = 2.879 \notag$
Calculating the pH during the Titration of a Weak Acid or a Weak Base
Now consider what happens when we add 5.00 mL of 0.200 M NaOH to 50.00 mL of 0.100 M CH3CO2H (part (a) in Figure 16.5.4). Because the neutralization reaction proceeds to completion, all of the OH ions added will react with the acetic acid to generate acetate ion and water:
$CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \tag{16.5.2}$
All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation (Equation 16.2.2) to determine [H+] of the resulting solution.
Step 1: To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of CH3CO2H in the original solution and the amount of OH in the NaOH solution that was added. The acetic acid solution contained
$50.00 \; \cancel{mL} (0.100 \;mmol (CH_3CO_2H)/\cancel{mL} )=5.00\; mmol (CH_3CO_2H) \notag$
The NaOH solution contained
5.00 mL=1.00 mmol (NaOH)
Comparing the amounts shows that CH3CO2H is in excess. Because OH reacts with CH3CO2H in a 1:1 stoichiometry, the amount of excess CH3CO2H is as follows:
5.00 mmol CH3CO2H − 1.00 mmol OH = 4.00 mmol CH3CO2H
Each 1 mmol of OH reacts to produce 1 mmol of acetate ion, so the final amount of CH3CO2 is 1.00 mmol.
The stoichiometry of the reaction is summarized in the following table, which shows the numbers of moles of the various species, not their concentrations.
$CH_3CO_2H(aq)+OH^{-}(aq)\rightarrow CH_3CO_2^{-}(aq)+H_2O(l)$
[CH3CO2H] [OH] [CH3CO2]
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol
This table gives the initial amount of acetate and the final amount of OH ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of CH3CO2 in equilibrium is insignificant compared to the amount of OH added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of OH, but the amount of OH due to the autoionization of water is insignificant compared to the amount of OH added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.
Step 2: To calculate [H+] at equilibrium following the addition of NaOH, we must first calculate [CH3CO2H] and [CH3CO2] using the number of millimoles of each and the total volume of the solution at this point in the titration:
$final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \notag$
$\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \notag$
$\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \notag$
Knowing the concentrations of acetic acid and acetate ion at equilibrium and Ka for acetic acid (1.74 × 10−5), we can use Equation 16.2.2 to calculate [H+] at equilibrium:
$K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \notag$
$\left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \notag$
Calculating −log[H+] gives pH = −log(6.95 × 10−5) = 4.158.
Comparing the titration curves for HCl and acetic acid in part (a) in Figure 16.5.3, we see that adding the same amount (5.00 mL) of 0.200 M NaOH to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for HCl (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example 16.5.2, we calculate another point for constructing the titration curve of acetic acid.
Example 16.5.2
What is the pH of the solution after 25.00 mL of 0.200 M NaOH is added to 50.00 mL of 0.100 M acetic acid?
Given: volume and molarity of base and acid
Asked for: pH
Strategy:
1. Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of OH and CH3CO2H. Determine which species, if either, is present in excess.
2. Tabulate the results showing initial numbers, changes, and final numbers of millimoles.
3. If excess acetate is present after the reaction with OH, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.
4. Calculate Kb using the relationship Kw = KaKb (Equation 16.2.10). Calculate [OH] and use this to calculate the pH of the solution.
Solution:
A Ignoring the spectator ion (Na+), the equation for this reaction is as follows:
$CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \notag$
The initial numbers of millimoles of OH and CH3CO2H are as follows:
$25.00 \;\cancel{mL}\left ( \dfrac{0.200 \; mmol\; OH^{-}}{\cancel{mL}} \right )= 5.00 \; mmol \; OH^{-} \notag$
$50.00 \;\cancel{mL}\left ( \dfrac{0.100 \; mmol\; CH_{3}CO_{2}H}{\cancel{mL}} \right )= 5.00 \; mmol \; CH_{3}CO_{2}H \notag$
The number of millimoles of OH equals the number of millimoles of CH3CO2H, so neither species is present in excess.
B Because the number of millimoles of OH added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.
$CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l)$
[CH3CO2H] [OH] [CH3CO2]
initial 5.00 mmol 5.00 mmol 0 mmol
final 0 mmol 0 mmol 5.00 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:
$[CH_3CO_2]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \notag$
The equilibrium reaction of acetate with water is as follows:
$CH_3CO^-_{2(aq)}+H_2O_{(l)} \rightleftharpoons CH_3CO_2H_{(aq)}+OH^-_{(aq)} \notag$
The equilibrium constant for this reaction is Kb = Kw/Ka, where Ka is the acid ionization constant of acetic acid. We therefore define x as [OH] produced by the reaction of acetate with water. Here is the completed table of concentrations:
$H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)}$
[CH3CO2] [CH3CO2H] [OH]
initial 0.0667 0 1.00 × 10−7
change x +x +x
final (0.0667 − x) x x
D Substituting the expressions for the final values from this table into Equation 16.2.5,
$K_{b}= \dfrac{\left [CH_3CO_2H \right ]\left [ OH^{-} \right ]}{\left [CH_3CO_2^{-} \right ]}\approx =\dfrac{x\cdot x}{0.0667-x} = \dfrac{x^{2}}{0.0667} \notag$
We can obtain Kb by rearranging Equation 16.2.10 and substituting the known values:
$K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10}=\dfrac{x^{2}}{0.0667} \notag$
which we can solve to get x = 6.22 × 10−6. Thus [OH] = 6.22 × 10−6 M, and the pH of the final solution is 8.794 (part (a) in Figure 16.5.4). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce OH.
Exercise
Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M HCl solution to 125.0 mL of a 0.150 M solution of ammonia. The pKb of ammonia is 4.75 at 25°C.
Answer: 9.23
As shown in part (b) in Figure 16.5.3, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.
The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure 16.5.4 illustrates the shape of titration curves as a function of the pKa or the pKb. As the acid or the base being titrated becomes weaker (its pKa or pKb becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.
One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in part (a) in Figure 16.5.5 and part (b) in Figure 16.5.5 for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A]. Recall from Equation 16.2.2 that the ionization constant for a weak acid is as follows:
$K_{a}=\dfrac{\left [ H_{3}O^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]} \notag$
If [HA] = [A], this reduces to Ka = [H3O+]. Taking the negative logarithm of both sides,
$log \;K_{a}= -log \left [ H_{3}O^{+} \right ] \notag$
From the definitions of pKa and pH, we see that this is identical to
$pK_{a}= pH \tag{16.5.3} \notag$
Thus the pH at the midpoint of the titration of a weak acid is equal to the pKa of the weak acid, as indicated in part (a) in Figure 16.5.4 for the weakest acid where we see that the midpoint for pKa = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the pKa (or the pKb) of a weak acid (or a weak base).
Note the Pattern
The pH at the midpoint of the titration of a weak acid is equal to the pKa of the weak acid.
Titrations of Polyprotic Acids or Bases
When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the pKa values are separated by at least three pKa units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid H3PO4 with NaOH is illustrated in Figure 16.5.5 and shows two well-defined steps: the first midpoint corresponds to pKa1, and the second midpoint corresponds to pKa2. Because HPO42− is such a weak acid, pKa3 has such a high value that the third step cannot be resolved using 0.100 M NaOH as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure 16.5.5 The initial pH is high, but as acid is added, the pH decreases in steps if the successive pKb values are well separated. Table16.3.1 lists the ionization constants and pKa values for some common polyprotic acids and bases.
Example 16.5.3
Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M NaOH solution to 100.0 mL of a 0.0510 M solution of oxalic acid (HO2CCO2H), a diprotic acid (abbreviated as H2ox). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (O2CCO2, abbreviated ox2−).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as Ca2+ are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.
Given: volume and concentration of acid and base
Asked for: pH
Strategy:
A Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.
B Calculate the concentrations of all the species in the final solution. Use Equation 16.2.3 to determine [H+] and convert this value to pH.
Solution:
A Table 16.3.1 gives the pKa values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:
$100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \notag$
$55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \notag$
The strongest acid (H2ox) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of OH to react with Hox, forming ox2− and H2O. The reactions can be written as follows:
$\underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \notag$
$\underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \notag$
In tabular form,
H2ox OH Hox ox2−
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol
B The equilibrium between the weak acid (Hox) and its conjugate base (ox2−) in the final solution is determined by the magnitude of the second ionization constant, Ka2 = 10−3.81 = 1.6 × 10−4. To calculate the pH of the solution, we need to know [H+], which is determined using exactly the same method as in the acetic acid titration in Example 12:
final volume of solution = 100.0 mL + 55.0 mL = 155.0 mL
Thus the concentrations of Hox and ox2− are as follows:
$\left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \notag$
$\left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \notag$
We can now calculate [H+] at equilibrium using the following equation:
$K_{a2}= =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \notag$
Rearranging this equation and substituting the values for the concentrations of Hox and ox2−,
$\left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \notag$
So
$pH = -log\left [ H^{+} \right ]= -log\left ( 3.7 \times 10^{-4} \right )= 3.43 \notag$
This answer makes chemical sense because the pH is between the first and second pKa values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than pKa1), but we added only enough to titrate less than half of the second, less acidic proton, with pKa2. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to pKa2.
Exercise
Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (pKb1 = 4.27, pKb2 = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M HCl (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.
Answer: 4.9 (Video Solution)
Indicators
In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.
We can describe the chemistry of indicators by the following general equation:
$HIn\left ( aq \right ) \rightleftharpoons H^{+}\left ( aq \right ) + In^{-}\left ( aq \right ) \notag$
where the protonated form is designated by HIn and the conjugate base by In. The ionization constant for the deprotonation of indicator HIn is as follows:
$K_{In} =\dfrac{\left [ H^{+} \right ]\left [ In^{-} \right ]}{HIn} \tag{16.5.4}$
The pKin (its pKa) determines the pH at which the indicator changes color.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH (Figure 16.22). In all cases, though, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a pKin that is within one pH unit of the expected pH at the equivalence point of the titration.
Figure 16.5.6 Naturally Occurring pH Indicators in Red Cabbage Juice Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue. (From 99 Chemical on YouTube)
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure 16.5.7 shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In increases, the color of the solution slowly changes from the characteristic color of HIn to that of In. As we will see in Section 16.6, the [In]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure 16.5.8. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. The pH ranges over which two common indicators (methyl red, pKin = 5.0, and phenolphthalein, pKin = 9.5) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of NaOH has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of NaOH has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of NaOH will therefore cause the methyl red indicator to change color, resulting in a huge error.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure 16.25).
Summary
The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the pKa of the weak acid or the pKb of the weak base. Thus titration methods can be used to determine both the concentration and the pKa (or the pKb) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself.
Key Takeaway
• Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the pKa, and the pKb of the system.
Conceptual Problems
1. Why is the portion of the titration curve that lies below the equivalence point of a solution of a weak acid displaced upward relative to the titration curve of a strong acid? How are the slopes of the curves different at the equivalence point? Why?
2. Predict whether each solution will be neutral, basic, or acidic at the equivalence point of each titration.
1. An aqueous solution of NaOH is titrated with 0.100 M HCl.
2. An aqueous solution of ethylamine (CH3CH2NH2) is titrated with 0.150 M HNO3
3. An aqueous solution of aniline hydrochloride (C6H5NH3+Cl) is titrated with 0.050 M KOH.
3. The pKa values of phenol red, bromophenol blue, and phenolphthalein are 7.4, 4.1, and 9.5, respectively. Which indicator is best suited for each acid–base titration?
1. titrating a solution of Ba(OH)2 with 0.100 M HCl
2. titrating a solution of trimethylamine (Me3N) with 0.150 M HNO3
3. titrating a solution of aniline hydrochloride (C6H5NH3+Cl) with 0.050 M KOH
4. For the titration of any strong acid with any strong base, the pH at the equivalence point is 7.0. Why is this not usually the case in titrations of weak acids or weak bases?
5. Why are the titration curves for a strong acid with a strong base and a weak acid with a strong base identical in shape above the equivalence points but not below?
6. Describe what is occurring on a molecular level during the titration of a weak acid, such as acetic acid, with a strong base, such as NaOH, at the following points along the titration curve. Which of these points corresponds to pH = pKa?
1. at the beginning of the titration
2. at the midpoint of the titration
3. at the equivalence point
4. when excess titrant has been added
7. On a molecular level, describe what is happening during the titration of a weak base, such as ammonia, with a strong acid, such as HCl, at the following points along the titration curve. Which of these points corresponds to pOH = pKb?
1. at the beginning of the titration
2. at the midpoint of the titration
3. at the equivalence point
4. when excess titrant has been added
8. For the titration of a weak acid with a strong base, use the Ka expression to show that pH = pKa at the midpoint of the titration.
9. Chemical indicators can be used to monitor pH rapidly and inexpensively. Nevertheless, electronic methods are generally preferred. Why?
10. Why does adding ammonium chloride to a solution of ammonia in water decrease the pH of the solution?
11. Given the equilibrium system CH3CO2H(aq) ⇌ CH3CO2(aq) + H+(aq), explain what happens to the position of the equilibrium and the pH in each case.
1. Dilute HCl is added.
2. Dilute NaOH is added.
3. Solid sodium acetate is added.
12. Given the equilibrium system CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH(aq), explain what happens to the position of the equilibrium and the pH in each case.
1. Dilute HCl is added.
2. Dilute NaOH is added.
3. Solid CH3NH3+Cl is added.
Answer
1. shifts to left; pH decreases
2. shifts to right; pH increases
3. shifts to left; pH increases
Numerical Problems
1. Calculate the pH of each solution.
1. A volume of 25.0 mL of 6.09 M HCl is added to 100.0 mL of distilled water
2. A volume of 5.0 mL of 2.55 M NaOH is added to 75.0 mL of distilled water.
2. What is the pH of a solution prepared by mixing 50.0 mL of 0.225 M HCl with 100.0 mL of a 0.184 M solution of NaOH?
3. What volume of 0.50 M HCl is needed to completely neutralize 25.00 mL of 0.86 M NaOH?
4. Calculate the final pH when each pair of solutions is mixed.
1. 100 mL of 0.105 M HCl and 100 mL of 0.115 M sodium acetate
2. 50 mL of 0.10 M HCl and 100 mL of 0.15 M sodium acetate
3. 100 mL of 0.109 M acetic acid and 100 mL of 0.118 M NaOH
4. 100 mL of 0.998 M acetic acid and 50.0 mL of 0.110 M NaOH
5. Calculate the final pH when each pair of solutions is mixed.
1. 100 mL of 0.983 M HCl and 100 mL of 0.102 M sodium fluoride
2. 50.0 mL of 0.115 M HCl and 100 mL of 0.109 M sodium fluoride
3. 100 mL of 0.106 M hydrofluoric acid and 50.0 mL of 0.996 M NaOH
4. 100 mL of 0.107 M sodium acetate and 50.0 mL of 0.987 M acetic acid
6. Calcium carbonate is a major contributor to the “hardness” of water. The amount of CaCO3 in a water sample can be determined by titrating the sample with an acid, such as HCl, which produces water and CO2. Write a balanced chemical equation for this reaction. Generate a plot of solution pH versus volume of 0.100 M HCl added for the titration of a solution of 250 mg of CaCO3 in 200.0 mL of water with 0.100 M HCl; assume that the HCl solution is added in 5.00 mL increments. What volume of HCl corresponds to the equivalence point?
7. For a titration of 50.0 mL of 0.288 M NaOH, you would like to prepare a 0.200 M HCl solution. The only HCl solution available to you, however, is 12.0 M.
1. How would you prepare 500 mL of a 0.200 M HCl solution?
2. Approximately what volume of your 0.200 M HCl solution is needed to neutralize the NaOH solution?
3. After completing the titration, you find that your “0.200 M” HCl solution is actually 0.187 M. What was the exact volume of titrant used in the neutralization?
8. While titrating 50.0 mL of a 0.582 M solution of HCl with a solution labeled “0.500 M KOH,” you overshoot the endpoint. To correct the problem, you add 10.00 mL of the HCl solution to your flask and then carefully continue the titration. The total volume of titrant needed for neutralization is 71.9 mL.
1. What is the actual molarity of your KOH solution?
2. What volume of titrant was needed to neutralize 50.0 mL of the acid?
9. Complete the following table and generate a titration curve showing the pH versus volume of added base for the titration of 50.0 mL of 0.288 M HCl with 0.321 M NaOH. Clearly indicate the equivalence point.
Base Added (mL) 10.0 30.0 40.0 45.0 50.0 55.0 65.0 75.0
pH
10. The following data were obtained while titrating 25.0 mL of 0.156 M NaOH with a solution labeled “0.202 M HCl.” Plot the pH versus volume of titrant added. Then determine the equivalence point from your graph and calculate the exact concentration of your HCl solution.
Volume of HCl (mL) 5 10 15 20 25 30 35
pH 11.46 11.29 10.98 4.40 2.99 2.70 2.52
11. Fill in the data for the titration of 50.0 mL of 0.241 M formic acid with 0.0982 M KOH. The pKa of formic acid is 3.75. What is the pH of the solution at the equivalence point?
Volume of Base Added (mL) 0 5 10 15 20 25
pH
12. Glycine hydrochloride, which contains the fully protonated form of the amino acid glycine, has the following structure:
It is a strong electrolyte that completely dissociates in water. Titration with base gives two equivalence points: the first corresponds to the deprotonation of the carboxylic acid group and the second to loss of the proton from the ammonium group. The corresponding equilibrium equations are as follows:
$^{+}NH_{3}-CH_{2}-CO_{2}H\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a1}=2.3 \notag$
$^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right )+ H^{+}$
$^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a2}=9.6 \notag$
$NH_{2}-CH_{2}-COO\left ( aq \right )+ H^{+} \notag$
1. Given 50.0 mL of solution that is 0.430 M glycine hydrochloride, how many milliliters of 0.150 M KOH are needed to fully deprotonate the carboxylic acid group?
2. How many additional milliliters of KOH are needed to deprotonate the ammonium group?
3. What is the pH of the solution at each equivalence point?
4. How many milliliters of titrant are needed to obtain a solution in which glycine has no net electrical charge? The pH at which a molecule such as glycine has no net charge is its isoelectric point. What is the isoelectric point of glycine?
13. What is the pH of a solution prepared by adding 38.2 mL of 0.197 M HCl to 150.0 mL of 0.242 M pyridine? The pKb of pyridine is 8.77.
14. What is the pH of a solution prepared by adding 40.3 mL of 0.289 M NaOH to 150.0 mL of 0.564 M succinic acid (HO2CCH2CH2CO2H)? (For succinic acid, pKa1 = 4.21 and pKa2 = 5.64).
15. Calculate the pH of a 0.15 M solution of malonic acid (HO2CCH2CO2H), whose pKa values are as follows: pKa1 = 2.85 and pKa2 = 5.70.
Answers
1. 43 mL
2. Video Solution
3. Video Solution
4. Video Solution
1. dilute 8.33 mL of 12.0 M HCl to 500.0 mL
2. 72 mL
3. 77.0 mL
5. pH at equivalence point = 8.28
Volume of Base Added (mL) 0 5 10 15 20 25
pH 2.19 2.38 2.70 2.89 3.04 3.15
6. 1.85
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.05%3A_Acid-Base_Titrations.txt |
Learning Objectives
• To understand how adding a common ion affects the position of an acid–base equilibrium.
• To know how to use the Henderson-Hasselbalch equation to calculate the pH of a buffer.
BuffersSolutions that maintain a relatively constant pH when an acid or a base is added. are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (HA) and its conjugate base (A) or a weak base (B) and its conjugate acid (BH+), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of H+). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution.
A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. Recall that the dissociation reaction of acetic acid is as follows:
$CH_3CO_2H_{(aq)} \leftrightharpoons CH_3CO^−_{(aq)}+H^+_{(aq)} \tag{16.6.1}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][CH_3CO_2]}{[CH_3CO_2H]} \tag{16.6.2}$
Sodium acetate (CH3CO2Na) is a strong electrolyte that ionizes completely in aqueous solution to produce Na+ and CH3CO2 ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation 16.6.1 will shift to the left, consuming some of the added CH3CO2 and some of the H+ ions originally present in solution:
Because Na+ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which [H+] is less than the initial value. Because [H+] has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as HCl to the system, [H+] increases. Once again the equilibrium is temporarily disturbed, but the excess H+ ions react with the conjugate base (CH3CO2), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [CH3CO2] than before. In both cases, only the equilibrium composition has changed; the ionization constant Ka for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case CH3CO2, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is called the common ion effectThe shift in equilibrium that results when a strong electrolyte containing one ion in common with a reaction system that is at equilibrium is added to the system..
Note the Pattern
Adding a common ion to a system at equilibrium affects the equilibrium composition but not the ionization constant.
Example 16.6.1
In Section 16.4, we calculated that a 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given: solution concentration and pH, pKa, and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for: pH and percent ionization of formic acid
Strategy:
A Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
B Substitute the expressions for the final concentrations into the expression for Ka. Calculate [H+] and the pH of the solution.
C Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution:
1. A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The Na+ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium: $HCO_2H_{(aq)} \leftrightharpoons HCO^−_{2(aq)}+H^+_{(aq)} \notag$
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
$HCO_2H_{(aq)} \leftrightharpoons HCO^−_{2(aq)}+H^+_{(aq)}$
[HCO2H] [H+] [HCO2]
initial 0.150 1.00 × 10−7 0.100
change x +x +x
final (0.150 − x) x (0.100 + x)
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
$K_{a} = \dfrac{[H^{+}][HCO_2^{-}]}{[HCO_2H]}=\dfrac{\left ( x \right )\left (0.100+x \right )}{0.150+x}\approx \dfrac{x(0.100)}{0.150}=10^{-3.75}= 1.8 \times 10^{-4} \notag$
Rearranging and solving for x,
$x=(1.80 \times 10^{-4}) \times \dfrac{ 0.150\; M}{ 0.200\; M}=1.35 \times 10^{-4}=[HCO_2^{-}] \notag$
The value of x is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, KaCHA = (1.8 × 10−4)(0.150) = 2.7 × 10−5, which is greater than 1.0 × 10−6, so again, our assumption is justified. The final pH is −log(2.7 × 10−4) = 3.57, compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of H+ ions, driving the equilibrium to the left.
2. C Because HCl is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
$HCO_2H_{(aq)} \leftrightharpoons H^+_{(aq)}+HCO^−_{2(aq)}$
[HCO2H] [H+] [HCO2]
initial 0.150 0.200 0
change x +x +x
final (0.150 − x) (0.200 + x) x
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final [HCO2]. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
$Ka=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \notag$
Rearranging and solving for x,
$x=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M}=1.35 \times 10^{−4}=[HCO_2^−] \notag$
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
$\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100%=0.0900%$
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding H+ ions drives the dissociation equlibrium to the left.
Exercise
As you learned in Example 8, a 0.225 M solution of ethylamine (CH3CH2NH2, pKb = 3.19) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
1. the pH of the solution if enough solid ethylamine hydrochloride (EtNH3Cl) is added to make the solution 0.100 M in EtNH3+
2. the percentage of ethylamine that is ionized if enough solid NaOH is added to the original solution to give a final concentration of 0.050 M NaOH
Answer
1. 11.16
2. 1.3%
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (B) and its conjugate acid (BH+). The general equation for the ionization of a weak base is as follows:
$B_{(aq)}+H_2O_{(l)} \leftrightharpoons BH^+_{(aq)}+OH^−_{(aq)} \tag{16.6.3}$
If the equilibrium constant for the reaction as written in Equation 16.6.3 is small, for example Kb = 10−5, then the equilibrium constant for the reverse reaction is very large: K = 1/Kb = 105. Adding a strong base such as OH to the solution therefore causes the equilibrium in Equation 16.6.3 to shift to the left, consuming the added OH. As a result, the OH ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the OH ion concentration, the reaction will proceed to the left to counteract the stress.
If the pKb of the base is 5.0, the pKa of its conjugate acid is pKa = pKw − pKb = 14.0 – 5.0 = 9.0. Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
$BH^+_{(aq)}+H_2O_{(l)} \leftrightharpoons B_{(aq)}+H_3O^+_{(aq)} \tag{16.6.4}$
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation 16.6.4 shifts to the left. As a result, the H+ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb H+ and OH ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacityThe amount of strong acid or strong base that a buffer solution can absorb before the pH changes dramatically., the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on K), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure 16.6.1, when NaOH is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the pKa or pKb of the weak acid or weak base. The procedure is analogous to that used in Example 14 to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is HA ⇌H+ + A for which the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][A^-]}{[HA]} \tag{16.6.5}$
This equation can be rearranged as follows:
$[H^+]=K_a\dfrac{[HA]}{[A^−]} \tag{16.6.6}$
Taking the logarithm of both sides and multiplying both sides by −1,
$−\log[H^+]=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right)=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \tag{16.6.7}$
Replacing the negative logarithms in Equation 16.60,
$pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \tag{16.6.8}$
or, more generally,
$pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \tag{16.6.9}$
Equation 16.6.8 and Equation 16.6.9 are both forms of the Henderson-Hasselbalch equationA rearranged version of the equilibrium constant expression that provides a direct way to calculate the pH of a buffer solution: pH = pKa + log([base]/[acid])., named after the two early-20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch equation may be limited to solutions whose concentrations are at least 100 times greater than their Ka values.
There are three special cases where the Henderson-Hasselbalch equation is easily interpreted without the need for calculations:
1. [base] = [acid]. Under these conditions, [base]/[acid] = 1 in Equation 16.6.9. Because log 1 = 0, pH = pKa, regardless of the actual concentrations of the acid and base. Recall from Section 16.5 that this corresponds to the midpoint in the titration of a weak acid or a weak base.
2. [base]/[acid] = 10. In Equation 16.6.9, because log 10 = 1, pH = pKa + 1.
3. [base]/[acid] = 100. In Equation 16.6.9, because log 100 = 2, pH = pKa + 2.
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = pKa − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
Note the Pattern
If [base] = [acid] for a buffer, then pH = pKa. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example 16.6.2
What is the pH of a solution that contains
1. 0.135 M HCO2H and 0.215 M HCO2Na? (The pKa of formic acid is 3.75.)
2. 0.0135 M HCO2H and 0.0215 M HCO2Na?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The pKb of pyridine is 8.77.)
Given: concentration of acid, conjugate base, and pKa; concentration of base, conjugate acid, and pKb
Asked for: pH
Strategy:
Substitute values into either form of the Henderson-Hasselbalch equation (Equation 16.61 or Equation 16.62) to calculate the pH.
Solution:
1. According to the Henderson-Hasselbalch equation, the pH of a solution that contains both a weak acid and its conjugate base is pH = pKa + log([A]/[HA]). Inserting the given values into the equation, $pH=3.75+\log\left(\dfrac{0.215}{0.135}\right)=3.75+\log 1.593=3.95 \notag$
This result makes sense because the [A]/[HA] ratio is between 1 and 10, so the pH of the buffer must be between the pKa (3.75) and pKa + 1, or 4.75.
2. This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch equation, $pH=3.75+\log\left(\dfrac{0.0215}{0.0135}\right)=3.75+\log 1.593=3.95 \notag$
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
3. In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (HPy+). We will therefore use Equation 16.62, the more general form of the Henderson-Hasselbalch equation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and [acid] = [HPy+] = 0.234 M. We also are given pKb = 8.77 for pyridine, but we need pKa for the pyridinium ion. Recall from Equation 16.2.10 that the pKb of a weak base and the pKa of its conjugate acid are related: pKa + pKb = pKw. Thus pKa for the pyridinium ion is pKw − pKb = 14.00 − 8.77 = 5.23. Substituting this pKa value into the Henderson-Hasselbalch equation, $pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right)=5.23+\log\left(\dfrac{0.119}{0.234}\right)=5.23 −0.294=4.94 \notag$
Once again, this result makes sense: the [B]/[BH+] ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the pKa (5.23) and pKa − 1, or 4.23.
Exercise
What is the pH of a solution that contains
1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The pKa of benzoic acid is 4.20, and the pKb of trimethylamine is also 4.20.
Answer
1. 4.08
2. 9.68
The Henderson-Hasselbalch equation can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example 16.
Example 16.6.3
The buffer solution in Example 15 contained 0.135 M HCO2H and 0.215 M HCO2Na and had a pH of 3.95.
1. What is the final pH if 5.00 mL of 1.00 M HCl are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M NaOH are added?
Given: composition and pH of buffer; concentration and volume of added acid or base
Asked for: final pH
Strategy:
A Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example 14. Then calculate the amount of acid or base added.
B Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch equation (Equation 16.6.9) to obtain the pH.
Solution:
The added HCl (a strong acid) or NaOH (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
1. A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer: $100\; \cancel{mL}\left ( \dfrac{0.135 \; mmol\; HCO_2H}{\cancel{mL}} \right )=13.5 \; mmol\; HCO_2H \notag$
$100\; \cancel{mL}\left ( \dfrac{0.215 \; mmol\; HCO_2^{-}}{\cancel{mL}} \right )=21.5 \; mmol\; HCO_2^{-} \notag$
The millimoles of H+ in 5.00 mL of 1.00 M HCl is as follows:
$5.00\; \cancel{mL}\left ( \dfrac{1.00 \; mmol\; H^{+}}{\cancel{mL}} \right )=5.00 \; mmol\; H^{+} \notag$
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
$HCO^{2−}_{(aq)} + H^+_{(aq)} \rightarrow HCO_2H_{(aq)} \notag$
HCO2(aq) + H+(aq) → HCO2H(aq)
[HCO2] [H+] [HCO2H]
initial 21.5 mmol 5.00 mmol 13.5 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 16.5 mmol ∼0 mmol 18.5 mmol
The final amount of H+ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final [H+] and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example 14 or the Henderson–Hasselbach equation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch equation requires the concentrations of HCO2 and HCO2H, which can be calculated using the number of millimoles (n) of each and the total volume (VT). Substituting these values into the Henderson-Hasselbalch equation,
$pH=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right)=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right)=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \notag$
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
$pH=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right)=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right)=3.75 −0.050=3.70 \notag$
Once again, this result makes sense on two levels. First, the addition of HCl has decreased the pH from 3.95, as expected. Second, the ratio of HCO2 to HCO2H is slightly less than 1, so the pH should be between the pKa and pKa − 1.
1. A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of HCO2H and 21.5 mmol of HCO2. The number of millimoles of OH in 5.00 mL of 1.00 M NaOH is as follows: $5.00\; \cancel{mL}\left ( \dfrac{1.00 \; mmol\; H^{+}}{\cancel{mL}} \right )=5.00 \; mmol\; H^{+} \notag$
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
HCO2H(aq) + OH(aq) → HCO2(aq) + H2O(l)
[HCO2H] [OH] [HCO2]
initial 13.5 mmol 5.00 mmol 21.5 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 8.5 mmol ∼0 mmol 26.5 mmol
The final amount of OH in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both HCO2 and HCO2H into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
$pH=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right)=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right)=3.75+0.494=4.24 \notag$
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the pKa and pKa + 1, as expected for a solution with a HCO2/HCO2H ratio between 1 and 10.
Exercise
The buffer solution from Example 15 contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
1. What is the final pH if 12.0 mL of 1.5 M NaOH are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M HCl are added?
Answer
1. 5.30
2. 4.42
Note the Pattern
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch equation, not their concentrations.
Note the Pattern
The most effective buffers contain equal concentrations of an acid and its conjugate base.
The results obtained in Example 16 and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of HCl or NaOH solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to 1.1 × 10−4 M HCl). In this case, adding 5.00 mL of 1.00 M HCl would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M NaOH would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure 16.6.2 for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of CH3CO2 to CH3CO2H from 1:1 reduces the buffer capacity of the solution.
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure 16.6.3. As indicated by the labels, the region around pKa corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the pKa to approximately a pH value of 1 unit greater than the pKa, which is why buffer solutions usually have a pH that is within ±1 pH units of the pKa of the acid component of the buffer.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to Ka. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to Kb. However, we can calculate either Ka or Kb from the other because they are related by Kw.
Blood: A Most Important Buffer
Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO2/HCO3 system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the CO2/HCO3 buffer system is usually written as follows:
$H_2CO_{3(aq)} \leftrightharpoons H^+_{(aq)}+HCO^−_{3(aq)} \tag{16.6.10}$
with Ka = 4.5 × 10−7 and pKa = 6.35 at 25°C. In fact, Equation 16.6.10 is a grossly oversimplified version of the CO2/HCO3 system because a solution of CO2 in water contains only rather small amounts of H2CO3. Thus Equation 16.6.10 does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C. As shown in Equation 16.6.11, CO2 is in equilibrium with H2CO3, but the equilibrium lies far to the left, with an H2CO3/CO2 ratio less than 0.01 under most conditions:
$CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H_2CO_{3(aq)} \tag{16.6.11}$
with K′ = 4.0 × 10−3 at 37°C. The true pKa of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a Ka of 2.0 × 10−4, which makes it a much stronger acid than Equation 16.6.10 suggests. Adding Equation 16.6.10 and Equation 16.6.11 and canceling H2CO3 from both sides give the following overall equation for the reaction of CO2 with water to give a proton and the bicarbonate ion:
$CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H_2CO_{3(aq)}\tag{16.6.12a}$
with $K'=4.0 \times 10^{−3} (37°C)$
$H_2CO_{3(aq)} \leftrightharpoons H^+_{(aq)} + HCO^−_{3(aq)}\tag{16.6.12b}$
with $K_a=2.0 \times 10^{−4} (37°C)$
$CO_{2(aq)}+H_2O_{(l)} \leftrightharpoons H^+_{(aq)}+HCO^−_{3(aq)} \tag{16.6.12c}$
with $K=8.0 \times 10^{−7} (37°C)$
The K value for the reaction in Equation 16.6.12 is the product of the true ionization constant for carbonic acid (Ka) and the equilibrium constant (K) for the reaction of CO2(aq) with water to give carbonic acid. The equilibrium equation for the reaction of CO2 with water to give bicarbonate and a proton is therefore
$K=\dfrac{[H^+][HCO_3^−]}{[CO_2]}=8.0 \times 10^{−7} \tag{16.6.13}$
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law, [CO2] = kP(CO2) where k is the Henry’s law constant for CO2, which is 3.0 × 10−5 M/mmHg at 37°C. (For more information about Henry’s law, see Section 13.4 .) Substituting this expression for [CO2] in Equation 16.6.13,
$K=\dfrac{[H^+][HCO_3^−]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{CO_2})} \tag{16.6.14}$
where P(CO2) is in mmHg. Taking the negative logarithm of both sides and rearranging,
$pH=6.10+\log \left( \dfrac{ [HCO_3^−]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{CO_2}) } \right) \tag{16.6.15}$
Thus the pH of the solution depends on both the CO2 pressure over the solution and [HCO3]. Figure 16.6.4 plots the relationship between pH and [HCO3] under physiological conditions for several different values of P(CO2) with normal pH and [HCO3] values indicated by the dashed lines.
According to Equation 16.6.12, adding a strong acid to the CO2/HCO3 system causes [HCO3] to decrease as HCO3 is converted to CO2. Excess CO2 is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in P(CO2). Because the change in [HCO3]/P(CO2) is small, Equation 16.6.15 predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the OH reacts with CO2 to form [HCO3], but CO2 is replenished by the body, again limiting the change in both [HCO3]/P(CO2) and pH. The CO2/HCO3 buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower P(CO2) at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and [HCO3]. The increase in pH and decrease in [HCO3] in response to the decrease in P(CO2) are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (HA) and its conjugate weak base (A). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the Ka or Kb), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch equation, which is valid for solutions whose concentrations are at least 100 times greater than their Ka values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO2/HCO3 system, which dominates the buffering action of blood plasma.
Key Equations
Henderson-Hasselbalch equation
Equation 16.6.8: $pH=pK_a +\log \left(\dfrac{[A^−]}{[HA]} \right) \notag$
Equation 16.6.9: $pH=pK_a +\log \left(\dfrac{[base]}{[acid]} \right) \notag$
Key Takeaway
• The common ion effect allows solutions to act as buffers, whose pH can be calculated using the Henderson-Hasselbalch equation.
Conceptual Problems
1. Explain why buffers are crucial for the proper functioning of biological systems.
2. What is the role of a buffer in chemistry and biology? Is it correct to say that buffers prevent a change in [H3O+]? Explain your reasoning.
3. Explain why the most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base.
4. Which region of the titration curve of a weak acid or a weak base corresponds to the region of the smallest change in pH per amount of added strong acid or strong base?
5. If you were given a solution of sodium acetate, describe two ways you could convert the solution to a buffer.
6. Why are buffers usually used only within approximately one pH unit of the pKa or pKb of the parent weak acid or base?
7. The titration curve for a monoprotic acid can be divided into four regions: the starting point, the region around the midpoint of the titration, the equivalence point, and the region after the equivalence point. For which region would you use each approach to describe the behavior of the solution?
1. a buffer
2. a solution of a salt of a weak base
3. a solution of a weak acid
4. diluting a strong base
8. Which of the following will produce a buffer solution? Explain your reasoning in each case.
1. mixing 100 mL of 0.1 M HCl and 100 mL of 0.1 M sodium fluoride
2. mixing 50 mL of 0.1 M HCl and 100 mL of 0.1 M sodium fluoride
3. mixing 100 mL of 0.1 M hydrofluoric acid and 100 mL of 0.1 M HCl
4. mixing 100 mL of 0.1 M hydrofluoric acid and 50 mL of 0.1 M NaOH
5. mixing 100 mL of 0.1 M sodium fluoride and 50 mL of 0.1 M NaOH.
9. Which of the following will produce a buffer solution? Explain your reasoning in each case.
1. mixing 100 mL of 0.1 M HCl and 100 mL of 0.1 M sodium acetate
2. mixing 50 mL of 0.1 M HCl and 100 mL of 0.1 M sodium acetate
3. mixing 100 mL of 0.1 M acetic acid and 100 mL of 0.1 M NaOH
4. mixing 100 mL of 0.1 M acetic acid and 50 mL of 0.1 M NaOH
5. mixing 100 mL of 0.1 M sodium acetate and 50 mL of 0.1 M acetic acid
10. Use the definition of Kb for a weak base to derive the following expression, which is analogous to the Henderson-Hasselbalch equation but for a weak base (B) rather than a weak acid (HA):
$pOH=pK_b−\log\left(\dfrac{[base]}{[acid]}\right)$
11. Why do biological systems use overlapping buffer systems to maintain a constant pH?
12. The CO2/HCO3 buffer system of blood has an effective pKa of approximately 6.1, yet the normal pH of blood is 7.4. Why is CO2/HCO3 an effective buffer when the pKa is more than 1 unit below the pH of blood? What happens to the pH of blood when the CO2 pressure increases? when the O2 pressure increases?
13. Carbon dioxide produced during respiration is converted to carbonic acid (H2CO3). The pKa1 of carbonic acid is 6.35, and its pKa2 is 10.33. Write the equations corresponding to each pK value and predict the equilibrium position for each reaction.
Answer
1. Not a buffer; the HCl completely neutralizes the sodium acetate to give acetic acid and NaCl(aq).
2. Buffer; the HCl neutralizes only half of the sodium acetate to give a solution containing equal amounts of acetic acid and sodium acetate.
3. Not a buffer; the NaOH completely neutralizes the acetic acid to give sodium acetate.
4. Buffer; the NaOH neutralizes only half of the acetic acid to give a solution containing equal amounts of acetic acid and sodium acetate.
5. Buffer; the solution will contain a 2:1 ratio of sodium acetate and acetic acid.
Numerical Problems
1. Benzenesulfonic acid (pKa = 0.70) is synthesized by treating benzene with concentrated sulfuric acid. Calculate the following:
1. the pH of a 0.286 M solution of benzenesulfonic acid
2. the pH after adding enough sodium benzenesulfonate to give a final benzenesulfonate concentration of 0.100 M
2. Phenol has a pKa of 9.99. Calculate the following:
1. the pH of a 0.195 M solution
2. the percent increase in the concentration of phenol after adding enough solid sodium phenoxide (the sodium salt of the conjugate base) to give a total phenoxide concentration of 0.100 M
3. Salicylic acid is used in the synthesis of acetylsalicylic acid, or aspirin. One gram dissolves in 460 mL of water to create a saturated solution with a pH of 2.40.
1. What is the Ka of salicylic acid?
2. What is the final pH of a saturated solution that is also 0.238 M in sodium salicylate?
3. What is the final pH if 10.00 mL of 0.100 M HCl are added to 150.0 mL of the buffered solution?
4. What is the final pH if 10.00 mL of 0.100 M NaOH are added to 150.0 mL of the buffered solution?
4. An intermediate used in the synthesis of perfumes is valeric acid, also called pentanoic acid. The pKa of pentanoic acid is 4.84 at 25°C.
1. What is the pH of a 0.259 M solution of pentanoic acid?
2. Sodium pentanoate is added to make a buffered solution. What is the pH of the solution if it is 0.210 M in sodium pentanoate?
3. What is the final pH if 8.00 mL of 0.100 M HCl are added to 75.0 mL of the buffered solution?
4. What is the final pH if 8.00 mL of 0.100 M NaOH are added to 75.0 mL of the buffered solution?
Answer
1. 1.35 × 10−3
2. 4.03
3. 3.88
4. 4.30
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.06%3A_Buffers.txt |
Although Chapter 16 focused exclusively on acid–base equilibriums in aqueous solutions, equilibrium concepts can also be applied to many other kinds of reactions that occur in aqueous solution. In this chapter, we describe the equilibriums involved in the solubility of ionic compounds and the formation of complex ions.
Solubility equilibriums involving ionic compounds are important in fields as diverse as medicine, biology, geology, and industrial chemistry. Carefully controlled precipitation reactions of calcium salts, for example, are used by many organisms to produce structural materials, such as bone and the shells that surround mollusks and bird eggs. In contrast, uncontrolled precipitation reactions of calcium salts are partially or wholly responsible for the formation of scale in coffee makers and boilers, “bathtub rings,” and kidney stones, which can be excruciatingly painful. The principles discussed in this chapter will enable you to understand how these apparently diverse phenomena are related. Solubility equilibriums are also responsible for the formation of caves and their striking features, such as stalactites and stalagmites, through a long process involving the repeated dissolution and precipitation of calcium carbonate. In addition to all of these phenomena, by the end of this chapter you will understand why barium sulfate is ideally suited for x-ray imaging of the digestive tract, and why soluble complexes of gadolinium can be used for imaging soft tissue and blood vessels using magnetic resonance imaging (MRI), even though most simple salts of both metals are toxic to humans.
Scanning electron micrograph of kettle scale. Hard water is a solution that consists largely of calcium and magnesium carbonate in CO2-rich water. When the water is heated, CO2 gas is released, and the carbonate salts precipitate from solution and produce a solid called scale.
Contributors
• Anonymous
Modified by Joshua B. Halpern
17.02: Determining the Solubility of Ionic Compounds
Learning Objectives
• To calculate the solubility of an ionic compound from its Ksp.
We begin our discussion of solubility and complexation equilibriums—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibriums, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
The Solubility Product
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:As you will discover in Section 17.4 and in more advanced chemistry courses, basic anions, such as S2−, PO43−, and CO32−, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \tag{17.1.1}$
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp)The equilibrium constant expression for the dissolution of a sparingly soluble salt that includes the concentration of a pure solid, which is a constant. of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. (For more information on the equilibrium constant expression, see Section 15.2.) The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \tag{17.1.2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \tag{17.1.2b}$
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43− ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table 17.1.1; they show that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equation 17.1, changes in pH can affect the solubility of a compound, as you will discover in Section 17.4.
Note the Pattern
As with K, the concentration of a pure solid does not appear explicitly in Ksp.
Table 17.1.1 Solubility Products for Selected Ionic Substances at 25°C
Solid Color K sp Solid Color K sp
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39
*These contain the Hg22+ ion.
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of m
ass of solute per 100 mL of solvent, Ksp, like K, is defined in terms of the molar concentrations of the component ions.
A kidney stone. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O2CCO2)·H2O and Ca3(PO4)2.
Example 17.1.1
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: K sp
Strategy:
A Write the balanced dissolution equilibrium and the corresponding solubility product expression.
B Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution:
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2−) are as follows:
$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2−] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2− ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \notag$
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise
One crystalline form of calcium carbonate (CaCO3) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer: 4.5 × 10−9
Note the Pattern
The reaction of weakly basic anions with H2O tends to make the actual solubility of many salts higher than predicted.
A crystal of calcite (CaCO3), illustrating the phenomenon of double refraction. When a transparent crystal of calcite is placed over a page, we see two images of the letters.
Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example 1. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations as we did in Chapter 15, remembering that the concentration of the pure solid is essentially constant.
Example 17.1.2
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C
Given: K sp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
A Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
B Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
1. A The dissolution equilibrium for Ca3(PO4)2 (Equation 17.1) is shown in the following table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43− ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43−] will be +2x. We can insert these values into the table.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq)
Ca3(PO4)2 [Ca2+] [PO43−]
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.1.2):
\begin{align}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \notag \2.07\times10^{-33}&=108x^5 \notag \1.92\times10^{-35}&=x^5 \notag \1.14\times10^{-7}\textrm{ M}&=x \notag \end{align}
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that
[PO43−] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
2. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water: $\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2} \notag$
Exercise
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature
Answer
1. 1.28 × 10−4 M
2. 3.54 mg
The Ion Product
The ion product (Q)A quantity that has precisely the same form as the solubility product for the dissolution of a sparingly soluble salt, except that the concentrations used are not necessarily equilibrium concentrations. of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibriums in Chapter 15. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.
Note the Pattern
The ion product Q is analogous to the reaction quotient Q for gaseous equilibriums.
As summarized in Figure 17.1.1 , there are three possible conditions for an aqueous solution of an ionic solid:
1. Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
2. Q = Ksp. The solution is saturated and at equilibrium.
3. Q > Ksp. The solution is supersaturated, and ionic solid will precipitate.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Example 17.1.3
We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10−10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that NaCl is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
A Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
B Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
C Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution:
A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble. The equation for the precipitation of BaSO4 is as follows:
$BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)} \notag$
The solubility product expression is as follows:
Ksp = [Ba2+][SO42−] = 1.08×10−10
B To solve this problem, we must first calculate the ion product—Q = [Ba2+][SO42−]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of BaCl2 solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
$\textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+}$
$[\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+}$
Similarly, the concentration of SO42− after mixing is the total number of moles of SO42− in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
$\textrm{moles SO}_4^{2-}=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-}$
$[\mathrm{SO_4^{2-}}]=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-}$
We can now calculate Q:
Q = [Ba2+][SO42−] = (2.9×10−4)(1.8×10−4) = 5.2×10−8
C We now compare Q with the Ksp. If Q > Ksp, then BaSO4 will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that BaSO4 will precipitate when the two solutions are mixed. In fact, BaSO4 will continue to precipitate until the system reaches equilibrium, which occurs when [Ba2+][SO42−] = Ksp = 1.08 × 10−10.
Exercise
The solubility product of calcium fluoride (CaF2) is 3.45 × 10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10−5 M solution of Ca(NO3)2, will CaF2 precipitate?
Answer: yes (Q = 4.7 × 10−11 > Ksp)
The Common Ion Effect and Solubility
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect discussed in Section 16.6 : adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.The exceptions generally involve the formation of complex ions, which is discussed in Section 17.3 .
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains 3 × (1.14 × 10−7 M) = 3.42 × 10−7 M Ca2+ and 2 × (1.14 × 10−7 M) = 2.28 × 10−7 M PO43−, according to the stoichiometry shown in Equation 17.1.1 (neglecting hydrolysis to form HPO42− as described in Chapter 16). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation 17.1.1 to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43− until Q = Ksp.
Note the Pattern
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example 17.1.4
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
A Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
B Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution:
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43−] is +2x. We can insert these values into the table.
Ca3(PO4)2(s) = 3Ca2+(aq) + 2PO43−(aq)
Ca3(PO4)2 [Ca2+] [PO43−]
initial pure solid 0.20 0
change +3x +2x
final pure solid 0.20 + 3x 2x
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43−]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\begin{align}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33} \notag \x^2&=6.5\times10^{-32} \notag \x&=2.5\times10^{-16}\textrm{ M} \notag \end{align}
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example 2—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer: 2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
Summary
The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion.
Key Takeaway
• The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium.
Conceptual Problems
1. Write an expression for Ksp for each salt.
1. AgI
2. CaF2
3. PbCl2
4. Ag2CrO4
2. Some species are not represented in a solubility product expression. Why?
3. Describe the differences between Q and Ksp.
4. How can an ion product be used to determine whether a solution is saturated?
5. When using Ksp to directly compare the solubilities of compounds, why is it important to compare only the Ksp values of salts that have the same stoichiometry?
6. Describe the effect of a common ion on the solubility of a salt. Is this effect similar to the common ion effect found in buffers? Explain your answer.
7. Explain why the presence of MgCl2 decreases the molar solubility of the sparingly soluble salt MgCO3.
Answers
1. Ksp = [Ag+][I]
2. Ksp = [Ca2+][F]2
3. Ksp = [Pb2+][Cl]2
4. Ksp = [Ag+]2[CrO42−]
1. For a 1:1 salt, the molar solubility is simply $\sqrt{K_{\textrm{sp}}}$ ; for a 2:1 salt, the molar solubility is $\sqrt[3]{K_{\textrm{sp}}/4}$> Consequently, the magnitudes of Ksp can be correlated with molar solubility only if the salts have the same stoichiometry.
2. Because of the common ion effect. Adding a soluble Mg2+ salt increases [Mg2+] in solution, and Le Chatelier’s principle predicts that this will shift the solubility equilibrium of MgCO3 to the left, decreasing its solubility.
Numerical Problems
1. Predict the molar solubility of each compound using the Ksp values given in Table E3.
1. Cd(IO3)2
2. AgCN
3. HgI2
2. Predict the molar solubility of each compound using the Ksp values given.
1. Li3PO4: 2.37 × 10−11
2. Ca(IO3)2: 6.47 × 10−6
3. Y(IO3)3: 1.12 × 10−10
3. A student prepared 750 mL of a saturated solution of silver sulfate (Ag2SO4). How many grams of Ag2SO4 does the solution contain? Ksp = 1.20 × 10−5.
4. Given the Ksp values in Table 17.1.1 and Table E3, predict the molar concentration of each species in a saturated aqueous solution.
1. silver bromide
2. lead oxalate
3. iron(II) carbonate
4. silver phosphate
5. copper(I) cyanide
5. Given the Ksp values in Table 17.1.1 and Table E3, predict the molar concentration of each species in a saturated aqueous solution.
1. copper(I) chloride
2. lanthanum(III) iodate
3. magnesium phosphate
4. silver chromate
5. strontium sulfate
6. Silicon dioxide, the most common binary compound of silicon and oxygen, constitutes approximately 60% of Earth’s crust. Under certain conditions, this compound can react with water to form silicic acid, which can be written as either H4SiO4 or Si(OH)4. Write a balanced chemical equation for the dissolution of SiO2 in basic solution. Write an equilibrium constant expression for the reaction.
7. The Ksp of Mg(OH)2 is 5.61 × 10−12. If you tried to dissolve 24.0 mg of Mg(OH)2 in 250 mL of water and then filtered the solution and dried the remaining solid, what would you predict to be the mass of the undissolved solid? You discover that only 1.0 mg remains undissolved. Explain the difference between your expected value and the actual value.
8. The Ksp of lithium carbonate is 8.15 × 10−4. If 2.34 g of the salt is stirred with 500 mL of water and any undissolved solid is filtered from the solution and dried, what do you predict to be the mass of the solid? You discover that all of your sample dissolves. Explain the difference between your predicted value and the actual value.
9. You have calculated that 24.6 mg of BaSO4 will dissolve in 1.0 L of water at 25°C. After adding your calculated amount to 1.0 L of water and stirring for several hours, you notice that the solution contains undissolved solid. After carefully filtering the solution and drying the solid, you find that 22.1 mg did not dissolve. According to your measurements, what is the Ksp of barium sulfate?
10. In a saturated silver chromate solution, the molar solubility of chromate is 6.54 × 10−5. What is the Ksp?
11. A saturated lead(II) chloride solution has a chloride concentration of 3.24 × 10−2 mol/L. What is the Ksp?
12. From the solubility data given, calculate Ksp for each compound.
1. AgI: 2.89 × 10−7 g/100 mL
2. SrF2: 1.22 × 10−2 g/100 mL
3. Pb(OH)2: 78 mg/500 mL
4. BiAsO4: 14.4 mg/2.0 L
13. From the solubility data given, calculate Ksp for each compound.
1. BaCO3: 10.0 mg/500 mL
2. CaF2: 3.50 mg/200 mL
3. Mn(OH)2: 6.30 × 10−4 g/300 mL
4. Ag2S: 1.60 × 10−13 mg/100 mL
14. Given the following solubilities, calculate Ksp for each compound.
1. BaCO3: 7.00 × 10−5 mol/L
2. CaF2: 1.70 mg/100 mL
3. Pb(IO3)2: 2.30 mg/100 mL
4. SrC2O4: 1.58 × 10−7mol/L
15. Given the following solubilities, calculate Ksp for each compound.
1. Ag2SO4: 4.2 × 10−1 g/100 mL
2. SrSO4: 1.5 × 10−3 g/100 mL
3. CdC2O4: 6.0 × 10−3 g/100 mL
4. Ba(IO3)2: 3.96 × 10−2 g/100 mL
16. The Ksp of the phosphate fertilizer CaHPO4·2H2O is 2.7 × 10−7 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 3.0 L of water at this temperature?
17. The Ksp of zinc carbonate monohydrate is 5.5 × 10−11 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 2.0 L of water at this temperature?
18. Silver nitrate eye drops were formerly administered to newborn infants to guard against eye infections contracted during birth. Although silver nitrate is highly water soluble, silver sulfate has a Ksp of 1.20 × 10−5 at 25°C. If you add 25.0 mL of 0.015 M AgNO3 to 150 mL of 2.8 × 10−3 M Na2SO4, will you get a precipitate? If so, what will its mass be?
19. Use the data in Table E3 to predict whether precipitation will occur when each pair of solutions is mixed.
1. 150 mL of 0.142 M Ba(NO3)2 with 200 mL of 0.089 M NaF
2. 250 mL of 0.079 M K2CrO4 with 175 mL of 0.087 M CaCl2
3. 300 mL of 0.109 M MgCl2 with 230 mL of 0.073 M Na2(C2O4)
20. What is the maximum volume of 0.048 M Pb(NO3)2 that can be added to 250 mL of 0.10 M NaSCN before precipitation occurs? Ksp = 2.0 × 10−5 for Pb(SCN)2.
21. Given 300 mL of a solution that is 0.056 M in lithium nitrate, what mass of solid sodium carbonate can be added before precipitation occurs (assuming that the volume of solution does not change after adding the solid)? Ksp = 8.15 × 10−4 for Li2CO3.
22. Given the information in the following table, calculate the molar solubility of each sparingly soluble salt in 0.95 M MgCl2.
Saturated Solution K sp
MgCO3·3H2O 2.4 × 10−6
Mg(OH)2 5.6 × 10−12
Mg3(PO4)2 1.04 × 10−24
Answers
1. 1.84 × 10−3 M
2. 7.73 × 10−9 M
3. 1.9 × 10−10 M
1. 3.37 g
1. 4.15 × 10−4 M
2. 7.26 × 10−4 M
3. 6.26 × 10−6 M
4. 6.54 × 10−5 M
5. 5.86 × 10−4 M
2. 22.4 mg; a secondary reaction occurs, where OH from the dissociation of the salt reacts with H+ from the dissociation of water. This reaction causes further dissociation of the salt (Le Chatelier’s principle).
3. 1.2 × 10−10
4. 1.70 × 10−5
1. 8.8 × 10−6
2. 6.7 × 10−9
3. 9.0 × 10−8
4. 2.16 × 10−9
5. 7.4 × 10−6 M; 2.1 mg
6. Precipitation will occur in all cases.
7. 8.27 g
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.01%3A_Introduction.txt |
Learning Objectives
• To understand the factors that determine the solubility of ionic compounds.
The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH. The first two situations are described in this section, the formation of complex ions is discussed in Section 17.3 , and changes in pH are discussed in Section 17.4.
Ion-Pair Formation
An ion pairA cation and an anion that are in intimate contact in solution rather than separated by solvent. An ion pair can be viewed as a species that is intermediate between the ionic solid and the completely dissociated ions in solution. consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent (Figure 17.2.1). The ions in an ion pair are held together by the same attractive electrostatic forces that we discussed in Chapter 4 for ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently).
As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
$\mathrm{CaSO_4(s)}\rightleftharpoons\mathrm{Ca^{2+}}\cdot\underset{\textrm{ion pair}}{\mathrm{SO_4^{2-}(aq)}}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \tag{17.2.1}$
The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions.
Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured Ksp for calcium sulfate is 4.93 × 10−5 at 25°C. The solubility of CaSO4 should be 7.02 × 10−3 M if the only equilibrium involved were as follows:
$CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2−}_{4(aq)} \tag{17.2.2}$
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10−2 M, almost twice the value predicted from its Ksp. The reason for the discrepancy is that the concentration of ion pairs in a saturated CaSO4 solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M2+ and M3+ ions, such as Ca2+ and La3+, and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li+. We therefore expect a saturated solution of CaSO4 to contain a high concentration of ion pairs and its solubility to be greater than predicted from its Ksp.
Note the Pattern
The formation of ion pairs increases the solubility of a salt.
Incomplete Dissociation
A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. (For more information about weak organic acids, see Chapter 16 ). Although strong acids (HA) dissociate completely into their constituent ions (H+ and A) in water, weak acids such as carboxylic acids do not (Ka = 1.5 × 10−5). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH3CO2H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C6H5CO2H), with Ka = 6.25 × 10−5. Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
$C_6H_5CO_2H_{(s)} \rightleftharpoons C_6H_5CO_2H_{(aq)} \rightleftharpoons C_6H_5CO^−_{2(aq)} + H^+_{(aq)} \tag{17.2.3}$
In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H+] = [C6H5CO2] = 1.4 × 10−3 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10−2 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—C6H5CO2H(aq)—and only about 5% is present as the dissociated ions (Figure 17.2.2).
Note the Pattern
Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute.
Although ion pairs, such as Ca2+·SO42−, and undissociated electrolytes, such as C6H5CO2H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte.
Summary
There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes in pH. An ion pair is held together by electrostatic attractive forces between the cation and the anion, whereas incomplete dissociation results from intramolecular forces, such as polar covalent O–H bonds.
Key Takeaway
• Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility.
Conceptual Problems
1. Do you expect the actual molar solubility of LaPO4 to be greater than, the same as, or less than the value calculated from its Ksp? Explain your reasoning.
2. Do you expect the difference between the calculated molar solubility and the actual molar solubility of Ca3(PO4)2 to be greater than or less than the difference in the solubilities of Mg3(PO4)2? Why?
3. Write chemical equations to describe the interactions in a solution that contains Mg(OH)2, which forms ion pairs, and in one that contains propanoic acid (CH3CH2CO2H), which forms a hydrated neutral molecule.
4. Draw representations of Ca(IO3)2 in solution
1. as an ionic solid.
2. in the form of ion pairs.
3. as discrete ions.
Numerical Problem
1. Ferric phosphate has a molar solubility of 5.44 × 10−16 in 1.82 M Na3PO4. Predict its Ksp. The actual Ksp is 1.3 × 10−22. Explain this discrepancy.
Answer
1. 9.90 × 10−16; the solubility is much higher than predicted by Ksp due to the formation of ion pairs (and/or phosphate complexes) in the sodium phosphate solution.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.03%3A_Factors_That_Affect_Solubility.txt |
Learning Objectives
• To be introduced to complex ions, including ligands.
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
The Formation Constant
The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows:
\begin{align}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\hspace{5mm}K_1 \ \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}\hspace{3mm}K_2 \ \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}\hspace{3mm}K_3 \ \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}}\hspace{3mm}K_4 \end{align} \label{17.3.1}
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $NH_3$ ligands, not six.
$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants.
Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32 ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42 ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of Cu(NO3)2•6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{17.3.2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{17.3.2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,
\begin{align}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times10^{-14}\end{align}
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise $1$
The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6].
Answer
1.2 × 10−5 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}$
with
$K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}$
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}$
with
$K_f = 2.9 \times 10^{13} \label{17.3.5b}$
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide (HgI2) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer
1. 1.9 × 10−10 M
2. 1.4 M
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$Ca^{2+}_{(aq)} + O_3POPO^{4−}_{4(aq)} \rightleftharpoons [Ca(O_3POPO_3)]^{2−}_{(aq)} \label{17.3.7a}$
with
$K_f = 4\times 10^4\label{17.3.7b}$
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $2$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Figure $2$: An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues.
Summary
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.04%3A_The_Formation_of_Complex_Ions.txt |
Learning Objectives
• To understand why the solubility of many compounds depends on pH.
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH.
The Effect of Acid–Base Equilibriums on the Solubility of Salts
We begin our discussion by examining the effect of pH on the solubility of a representative salt, M+A, where A is the conjugate base of the weak acid HA. When the salt dissolves in water, the following reaction occurs:
$MA_{(s)} \rightleftharpoons M^+_{(aq)} + A^−_{(aq)} \tag{17.4.1a}$
with
$K_{sp} = [M^+][A^−] \tag{17.4.1b}$
The anion can also react with water in a hydrolysis reaction:
$A^−_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^−_{(aq)} + HA_{(aq)} \tag{17.4.2}$
Because of the reaction described in Equation 17.4.2, the predicted solubility of a sparingly soluble salt that has a basic anion such as S2−, PO43−, or CO32− is increased, as described in Section 17.1. If instead a strong acid is added to the solution, the added H+ will react essentially completely with A to form HA. This reaction decreases [A], which decreases the magnitude of the ion product (Q = [M+][A]). According to Le Chatelier’s principle, more MA will dissolve until Q = Ksp. Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)2 is relatively insoluble in water:
$Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^−_{(aq)} \tag{17.4.3a}$
with
$K_{sp} = 5.61 \times 10^{−12} \tag{17.4.3b}$
When acid is added to a saturated solution that contains excess solid Mg(OH)2, the following reaction occurs, removing OH from solution:
$H^+_{(aq)} + OH^−_{(aq)} \rightarrow H_2O_{(l)} \tag{17.4.5}$
The overall equation for the reaction of Mg(OH)2 with acid is thus
$Mg(OH)_{2(s)} + 2H^+_{(aq)} \rightleftharpoons Mg^{2+}_{(aq)} + 2H_2O_{(l)} \tag{17.4.6}$
As more acid is added to a suspension of Mg(OH)2, the equilibrium shown in Equation 17.4.6 is driven to the right, so more Mg(OH)2 dissolves.
Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF2 is a sparingly soluble salt:
$CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^−_{(aq)} \tag{17.4.7a}$
with
$K_{sp} = 3.45 \times 10^{−11} \tag{17.4.7b}$
When strong acid is added to a saturated solution of CaF2, the following reaction occurs:
$H^+_{(aq)} + F^−_{(aq)} \rightleftharpoons HF_{(aq)} \tag{17.4.8}$
Because the forward reaction decreases the fluoride ion concentration, more CaF2 dissolves to relieve the stress on the system. The net reaction of CaF2 with strong acid is thus
$CaF_{2(s)} + 2H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + 2HF_{(aq)} \tag{17.4.9}$
Example 7 shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt.
Note the Pattern
Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution.
Example 17.4.1
Lead oxalate (PbC2O4), lead iodide (PbI2), and lead sulfate (PbSO4) are all rather insoluble, with Ksp values of 4.8 × 10−10, 9.8 × 10−9, and 2.53 × 10−8, respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities?
Given: Ksp values for three compounds
Asked for: relative solubilities in acid solution
Strategy:
Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions.
Solution:
The solubility equilibriums for the three salts are as follows:
$PbC_2O_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + C_2O^{2−}_{4(aq) \notag }$
$PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^−_{(aq)} \notag$
$PbSO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + SO^{2−}_{4(aq)} \notag$
The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI2 will not greatly affect its solubility; the acid will simply dissociate to form H+(aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO2CCO2H), which is a weak diprotic acid (pKa1 = 1.23 and pKa2 = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions:
$C_2O^{2−}_{4(aq)} + H^+_{(aq)} \rightarrow HO_2CCO^−_{2(aq)} \notag$
$HO_2CCO^−_{2(aq)} + H^+_{(aq)} \rightarrow HO_2CCO_2H_{(aq)} \notag$
These reactions will decrease [C2O42−], causing more lead oxalate to dissolve to relieve the stress on the system.The pKa of HSO4 (1.99) is similar in magnitude to the pKa1 of oxalic acid, so adding a strong acid to a saturated solution of PbSO4 will result in the following reaction:
$SO^{2-}_{4(aq)} + H^+_{(aq)} \rightleftharpoons HSO^-_{4(aq)} \notag$
Because HSO4 has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO4 will be significantly less than for PbC2O4.
Exercise
Which of the following insoluble salts—AgCl, Ag2CO3, Ag3PO4, and/or AgBr—will be substantially more soluble in 1.0 M HNO3 than in pure water?
Answer: Ag2CO3 and Ag3PO4
Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility equilibriums (part (a) in Figure 17.6). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows:
$CO_{2(aq)} + H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + HCO^−_{3(aq)} \tag{17.4.10}$
$HCO^−_{3(aq)} \rightleftharpoons H+(aq) + CO^{2−}_{3(aq)} \tag{17.4.11}$
$Ca^{2+}_{(aq)} + CO^{2−}_{3(aq)} \rightleftharpoons CaCO_{3(s)} \tag{17.4.12}$
Limestone deposits that form caves consist primarily of CaCO3 from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO3 in CO2-rich water rises toward Earth’s surface or is otherwise heated, CO2 gas is released as the water warms. CaCO3 then precipitates from the solution according to the following equation (part (b) in Figure 17.4.1):
$Ca^{2+}_{(aq)} + 2HCO^−_{3(aq)} \rightleftharpoons CaCO_{3(s)} + CO_{2(g)} + H_2O_{(l)} \tag{17.4.13}$
The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated.
When groundwater-containing atmospheric CO2 (Equation 17.4.10 and Equation 17.4.11) finds its way into microscopic cracks in the limestone deposits, CaCO3 dissolves in the acidic solution in the reverse direction of Equation 17.4.13. The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in Equation 17.4.13 to shift to the right. A circular layer of solid CaCO3 is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China (Figure 17.4.2).
Acidic, Basic, and Amphoteric Oxides and Hydroxides
One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxidesAn oxide that reacts with water to produce a basic solution or dissolves readily in aqueous acid. and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxidesAn oxide that reacts with water to produce an acidic solution or dissolves in aqueous base. or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. As shown in Figure 17.4.3 , there is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:
$Cs_2O_{(s)} + H_2O_{(l)} \rightarrow 2Cs^+_{(aq)} + 2OH^−_{(aq)} \tag{17.4.14}$
$SO_{3(g)} + H_2O_{(l)} \rightarrow H_2SO_{4(aq)} \tag{17.4.15}$
Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed!
Note the Pattern
Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions.
The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete Mn+ cations and O2− anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: Eδ+–Oδ−. The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO3, which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO42−):
$MoO_{3(s)} + 2OH^−_{(aq)} \rightarrow MoO^{2−}_{4(aq)} + H_2O_{(l)} \tag{17.4.16}$
As shown in Figure 17.4.3, there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called amphoteric oxidesAn oxide that can dissolve in acid to produce water and dissolve in base to produce a soluble complex. (from the Greek ampho, meaning “both,” as in amphiprotic, which was defined in Section 16.1). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in Figure 17.4.4 , for example, mixing the amphoteric oxide Cr(OH)3 (also written as Cr2O3·3H2O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH)3 to dissolve to give a bright violet solution of Cr3+(aq), which contains the [Cr(H2O)6]3+ ion, whereas adding strong base gives a green solution of the [Cr(OH)4] ion. The chemical equations for the reactions are as follows:
$\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \tag{17.4.17}$
$\mathrm{Cr(OH)_3(s)}+\mathrm{OH^-(aq)}\rightarrow\underset{\textrm{green}}{\mathrm{[Cr(OH)_4]^-}}\mathrm{(aq)}\tag{17.4.18}$
Figure 17.4.4 Chromium(III) Hydroxide [Cr(OH)3 or Cr2O3·3H2O] Is an Example of an Amphoteric OxideAll three beakers originally contained a suspension of brownish purple Cr(OH)3(s) (center). When concentrated acid (6 M H2SO4) was added to the beaker on the left, Cr(OH)3 dissolved to produce violet [Cr(H2O)6]3+ ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH)3 to dissolve, producing green [Cr(OH)4]ions.
Example 17.4.2
Aluminum hydroxide, written as either Al(OH)3 or Al2O3·3H2O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base.
Given: amphoteric compound
Asked for: dissolution reactions in acid and base
Strategy:
Using Equation 17.4.17 and Equation 17.4.18 as a guide, write the dissolution reactions in acid and base solutions.
Solution:
1. An acid donates protons to hydroxide to give water and the hydrated metal ion, so aluminum hydroxide, which contains three OH ions per Al, needs three H+ ions: $Al(OH)_{3(s)} + 3H^+_{(aq)} \rightarrow Al^{3+}_{(aq)} + 3H_2O_{(l)} \notag$
In aqueous solution, Al3+ forms the complex ion [Al(H2O)6]3+.
2. In basic solution, OH is added to the compound to produce a soluble and stable poly(hydroxo) complex: $Al(OH)_{3(s)} + OH^−_{(aq)} \rightarrow [Al(OH)_4]^−_{(aq)} \notag$
Exercise
Copper(II) hydroxide, written as either Cu(OH)2 or CuO·H2O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base.
Answer
$Cu(OH)_{2(s)} + 2H^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2H_2O_{(l)} \notag$
$Cu(OH)_{2(s)} + 2OH^−_{(aq)} \rightarrow [Cu(OH)_4]^2_{−(aq)} \notag$
Selective Precipitation Using pH
Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate.
Note the Pattern
The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations.
Suppose, for example, we have a solution that contains 1.0 mM Zn2+ and 1.0 mM Cd2+ and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilibriums can be written as follows:
$ZnS_{(s)} \rightleftharpoons Zn^{2+}_{(aq)} + S^{2−}_{(aq)} \tag{17.4.19a}$
with
$K_{sp}= 1.6 \times 10^{−24} \tag{17.4.19b}$
and
$CdS_{(s)} \rightleftharpoons Cd^{2+}_{(aq)} + S^{2−}_{(aq)} \tag{17.4.20a}$
with
$K_{sp} = 8.0 \times 10^{−27} \tag{17.4.20b}$
Because the S2− ion is quite basic and reacts extensively with water to give HS and OH, the solubility equilibriums are more accurately written as MS(s) = M2+(aq) +HS-(aq) + OH-(aq) rather than MS(s) = M2+(aq) +S2-(aq) Here we use the simpler form involving S2−, which is justified because we take the reaction of S2− with water into account later in the solution, arriving at the same answer using either equilibrium equation.
The sulfide concentrations needed to cause ZnS and CdS to precipitate are as follows:
$K_{sp} = [Zn^{2+}][S^{2−}] \tag{17.4.21a}$
$1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\tag{17.4.21b}$
$1.6 \times 10^{−21}\; M = [S^{2−}]\tag{17.4.21c}$
and
$K_{sp} = [Cd^{2+}][S^{2−}] \tag{17.4.22a}$
$8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\tag{17.4.22b}$
$8.0 \times 10^{−24}\; M = [S^{2−}] \tag{17.4.22c}$
Thus sulfide concentrations between 1.6 × 10−21 M and 8.0 × 10−24 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H2S contains 0.10 M H2S at 20°C. The pKa1 for H2S is 6.97, and pKa2 corresponding to the formation of [S2−] is 12.90. The equations for these reactions are as follows:
$H_2S_{(aq)} \rightleftharpoons H^+_{(aq)} + HS^−_{(aq)} \tag{17.4.24a}$
with
$pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \tag{17.4.24b}$
$HS^−_{(aq)} \rightleftharpoons H^+_{(aq)} + S^{2−}_{(aq)} \tag{17.4.24c}$
with
$pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \tag{17.4.24d}$
We can show that the concentration of S2− is 1.3 × 10−13 by comparing Ka1 and Ka2 and recognizing that the contribution to [H+] from the dissociation of HS is negligible compared with [H+] from the dissociation of H2S. Thus substituting 0.10 M in the equation for Ka1 for the concentration of H2S, which is essentially constant regardless of the pH, gives the following:
$K_{\textrm{a1}}=1.1\times10^{-7}=\dfrac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]}=\dfrac{x^2}{0.10\textrm{ M}} \x=1.1\times10^{-4}\textrm{ M}=[\mathrm{H^+}]=[\mathrm{HS^-}] \tag{17.4.5}$
Substituting this value for [H+] and [HS] into the equation for Ka2,
$K_{\textrm{a2}}=1.3\times10^{-13}=\dfrac{[\mathrm{H^+}][\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}=\dfrac{(1.1\times10^{-4}\textrm{ M})x}{1.1\times10^{-4}\textrm{ M}}=x=[\mathrm{S^{2-}}] \notag$
Although [S2−] in an H2S solution is very low (1.3 × 10−13 M), bubbling H2S through the solution until it is saturated would precipitate both metal ions because the concentration of S2− would then be much greater than 1.6 × 10−21 M. Thus we must adjust [S2−] to stay within the desired range. The most direct way to do this is to adjust [H+] by adding acid to the H2S solution (recall Le Chatelier's principle), thereby driving the equilibrium in Equation 17.4.244 to the left. The overall equation for the dissociation of H2S is as follows:
$H_2S_{(aq)} \rightleftharpoons 2H^+_{(aq)} + S^{2−}_{(aq)} \tag{17.4.26}$
Now we can use the equilibrium constant K for the overall reaction, which is the product of Ka1 and Ka2, and the concentration of H2S in a saturated solution to calculate the H+ concentration needed to produce [S2−] of 1.6 × 10−21 M:
$K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \tag{17.4.27}$
\begin{align}\mathrm{[H^+]^2}=\dfrac{K[\mathrm{H_2S}]}{[\mathrm{S^{2-}}]}=\dfrac{(1.4\times10^{-20})(\textrm{0.10 M})}{1.6\times10^{-21}\textrm{ M}}&=0.88 \notag \ [\mathrm{H^+}]&=0.94 \notag \end{align} \tag{17.4.28}
Thus adding a strong acid such as HCl to make the solution 0.94 M in H+ will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H2S.
Example 17.4.3
A solution contains 0.010 M Ca2+ and 0.010 M La3+. What concentration of HCl is needed to precipitate La2(C2O4)3·9H2O but not Ca(C2O4)·H2O if the concentration of oxalic acid is 1.0 M? Ksp values are 2.32 × 10−9 for Ca(C2O4) and 2.5 × 10−27 for La2(C2O4)3; pKa1 = 1.25 and pKa2 = 3.81 for oxalic acid.
Given: concentrations of cations, Ksp values, and concentration and pKa values for oxalic acid
Asked for: concentration of HCl needed for selective precipitation of La2(C2O4)3
Strategy:
A Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La2(C2O4)3·9H2O.
B Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox2−] needed to precipitate La2(C2O4)3·9H2O into the overall equation for the dissociation of oxalic acid to calculate the required [H+].
Solution:
A Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox2− for oxalate, we write the solubility product expression for calcium oxalate as follows:
$K_{sp} = [Ca^{2+}][ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \notag$
$[ox^{2−}] = 2.32 \times 10^{−7}\; M \notag$
The expression for lanthanum oxalate is as follows:
$K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \notag$
$[ox^{2−}] = 2.9 \times 10^{−8}\; M \notag$
Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between 2.9 × 10−8 M and 2.32 × 10−7 M.
B To prevent Ca2+ from precipitating as calcium oxalate, we must add enough H+ to give a maximum oxalate concentration of 2.32 × 10−7 M. We can calculate the required [H+] by using the overall equation for the dissociation of oxalic acid to oxalate:
$HO_2CCO_2H_{(aq)} \rightleftharpoons 2H^+_{(aq)} + C_2O^{2−}_{4(aq)} \notag$
K = Ka1Ka2 = (10−1.25)(10−3.81) = 10−5.06 = 8.7×10−6
Substituting the desired oxalate concentration into the equilibrium constant expression,
\begin{align}8.7\times10^{-6}=\dfrac{[\mathrm{H^+}]^2[\mathrm{ox^{2-}}]}{[\mathrm{HO_2CCO_2H}]} &=\dfrac{[\mathrm{H^+}]^2(2.32\times10^{-7})}{1.0} \notag \ [\mathrm{H^+}] &=\textrm{6.1 M} \notag \end{align} \notag
Thus adding enough HCl to give [H+] = 6.1 M will cause only La2(C2O4)3·9H2O to precipitate from the solution.
Exercise
A solution contains 0.015 M Fe2+ and 0.015 M Pb2+. What concentration of acid is needed to ensure that Pb2+ precipitates as PbS in a saturated solution of H2S, but Fe2+ does not precipitate as FeS? Ksp values are 6.3 × 10−18 for FeS and 8.0 × 10−28 for PbS.
Answer: 0.018 M H+
Summary
The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation.
Key Takeaway
• The anion in sparingly soluble salts is often the conjugate base of a weak acid that may become protonated in solution, so the solubility of simple oxides and sulfides, both strong bases, often depends on pH.
Conceptual Problems
1. Which of the following will show the greatest increase in solubility if 1 M HNO3 is used instead of distilled water? Explain your reasoning.
1. CuCl2
2. K[Pb(OH)3]
3. Ba(CH3CO2)2
4. CaCO3
2. Of the compounds Sn(CH3CO2)2 and SnS, one is soluble in dilute HCl and the other is soluble only in hot, concentrated HCl. Which is which? Provide a reasonable explanation.
3. Where in the periodic table do you expect to find elements that form basic oxides? Where do you expect to find elements that form acidic oxides?
4. Because water can autoionize, it reacts with oxides either as a base (as OH) or as an acid (as H3O+). Do you expect oxides of elements in high oxidation states to be more acidic (reacting with OH) or more basic (reacting with H3O+) than the corresponding oxides in low oxidation states? Why?
5. Given solid samples of CrO, Cr2O3, and CrO3, which would you expect to be the most acidic (reacts most readily with OH)? Which would be the most basic (reacts most readily with H3O+)? Why?
6. Which of these elements—Be, B, Al, N, Se, In, Tl, Pb—do you expect to form an amphoteric oxide? Why?
Numerical Problems
1. A 1.0 L solution contains 1.98 M Al(NO3)3. What are [OH] and [H+]? What pH is required to precipitate the cation as Al(OH)3? Ksp = 1.3 × 10−33 and Ka = 1.05 × 10−5 for the hydrated Al3+ ion.
2. A 1.0 L solution contains 2.03 M CoCl2. What is [H+]? What pH is required to precipitate the cation as Co(OH)2? Ksp = 5.92 × 10−15 and Ka = 1.26 × 10−9 for the hydrated Co2+ ion.
3. Given 100 mL of a solution that contains 0.80 mM Ag+ and 0.80 mM Cu+, can the two metals be separated by selective precipitation as the insoluble bromide salts by adding 10 mL of an 8.0 mM solution of KBr? Ksp values are 6.27 × 10−9 for CuBr and 5.35 × 10−13 for AgBr. What maximum [Br] will separate the ions?
4. Given 100 mL of a solution that is 1.5 mM in Tl+, Zn2+, and Ni2+, which ions can be separated from solution by adding 5.0 mL of a 12.0 mM solution of Na2C2O4?
Precipitate K sp
Tl2C2O4 2 × 10−4
ZnC2O4·2H2O 1.38 × 10−9
NiC2O4 4 × 10−10
How many milliliters of 12.0 mM Na2C2O4 should be added to separate Tl+ and Zn2+ from Ni2+?
Answers
1. [H+] = 4.56 × 10−3; [OH] = 2.19 × 10−12; pH = 2.94
2. No; both metal ions will precipitate; AgBr will precipitate as Br is added, and CuBr will begin to precipitate at [Br] = 8.6 × 10−6 M.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.05%3A_Solubility_and_pH.txt |
Learning Objectives
• To know how to separate metal ions by selective precipitation.
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysisA procedure for determining the identity of metal ions present in a mixture that does not include information about their amounts., a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts).
The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure 17.5.1.
Group 1: Insoluble Chlorides
Most metal chloride salts are soluble in water; only Ag+, Pb2+, and Hg22+ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M HCl, thereby causing AgCl, PbCl2, and/or Hg2Cl2 to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation.
Group 2: Acid-Insoluble Sulfides
Next, the acidic solution is saturated with H2S gas. Only those metal ions that form very insoluble sulfides, such as As3+, Bi3+, Cd2+, Cu2+, Hg2+, Sb3+, and Sn2+, precipitate as their sulfide salts under these acidic conditions. All others, such as Fe2+ and Zn2+, remain in solution. Once again, the precipitates are collected by filtration or centrifugation.
Group 3: Base-Insoluble Sulfides (and Hydroxides)
Ammonia or NaOH is now added to the solution until it is basic, and then (NH4)2S is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions Co2+, Fe2+, Mn2+, Ni2+, and Zn2+ precipitate as their sulfides, and the trivalent metal ions Al3+ and Cr3+ precipitate as their hydroxides: Al(OH)3 and Cr(OH)3. If the mixture contains Fe3+, sulfide reduces the cation to Fe2+, which precipitates as FeS.
Group 4: Insoluble Carbonates or Phosphates
The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When Na2CO3 is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding (NH4)2HPO4 causes the same metal ions to precipitate as insoluble phosphates.
Group 5: Alkali Metals
At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+). We now take a second sample from the original solution and add a small amount of NaOH to neutralize the ammonium ion and produce NH3. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. As discussed in Chapter 2, the other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present.
Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing Ag+, Pb2+, and Hg22+, are all quite insoluble in water. Because PbCl2 is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any PbCl2 present. Isolating the solution and adding a small amount of Na2CrO4 solution to it will produce a bright yellow precipitate of PbCrO4 if Pb2+ was in the original sample (Figure 17.5.2).
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any AgCl because Ag+ forms a stable complex with ammonia: [Ag(NH3)2]+. In addition, Hg2Cl2 disproportionates in ammonia (2Hg22+ → Hg + Hg2+) to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any AgCl because Ag+ forms a stable complex with ammonia: [Ag(NH3)2]+. In addition, Hg2Cl2 disproportionates in ammonia (2Hg22+ → Hg + Hg2+) to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
$Hg_2Cl_{2(s)} + 2NH_{3(aq)} \rightarrow Hg_{(l)} + Hg(NH_2)Cl_{(s)} + NH^+_{4(aq)} + Cl^−_{(aq)} \tag{17.39}$
Any silver ion in the solution is then detected by adding HCl, which reverses the reaction and gives a precipitate of white AgCl that slowly darkens when exposed to light:
$[Ag(NH_3)_2]^+_{(aq)} + 2H^+_{(aq)} + Cl^−_{(aq)} \rightarrow AgCl_{(s)} + 2NH^+_{4(aq)} \tag{17.40}$
Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups.
Summary
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
Key Takeaway
• Several common metal cations can be identified in a solution using selective precipitation.
Conceptual Problem
1. Given a solution that contains a mixture of NaCl, CuCl2, and ZnCl2, propose a method for separating the metal ions.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.06%3A_Qualitative_Analysis_Using_Selective_Precipitation.txt |
Learning Objectives
• To understand the relationships between work, heat, and energy.
• To become familiar with the concept of PV work.
We begin our discussion of thermodynamics by reviewing some important terms introduced in Chapter 9. First, we need to distinguish between a system and its surroundings. A system is that part of the universe in which we are interested, such as a mixture of gases in a glass bulb or a solution of substances in a flask. The surroundings are everything else—the rest of the universe. We can therefore state the following:
$system + surroundings = universe \tag{18.1.1}$
A closed system, such as the contents of a sealed jar, cannot exchange matter with its surroundings, whereas an open system can; in this case, we can convert a closed system (the jar) to an open system by removing the jar’s lid.
In Chapter 9, we also introduced the concept of a state function A property of a system whose magnitude depends on only the present state of the system, not its previous history., a property of a system that depends on only the present state of the system, not its history. Thus a change in a state function depends on only the difference between the initial and final states, not the pathway used to go from one to the other. To help understand the concept of a state function, imagine a person hiking up a mountain (Figure 18.1.1). If the person is well trained and fit, he or she may be able to climb almost vertically to the top (path A), whereas another less athletic person may choose a path that winds gradually to the top (path B). If both hikers start from the same point at the base of the mountain and end up at the same point at the top, their net change in altitude will be the same regardless of the path chosen. Hence altitude is a state function. On the other hand, a person may or may not carry a heavy pack and may climb in hot weather or cold. These conditions would influence changes in the hiker’s fatigue level, which depends on the path taken and the conditions experienced. Fatigue, therefore, is not a state function. Thermodynamics is generally concerned with state functions and does not deal with how the change between the initial state and final state occurs.
The Connections among Work, Heat, and Energy
The internal energy (E)A state function that is the sum of the kinetic and potential energies of all a system’s components. of a system is the sum of the potential energy and the kinetic energy of all the components; internal energy is a state function. Although a closed system cannot exchange matter with its surroundings, it can exchange energy with its surroundings in two ways: by doing work or by releasing or absorbing heat—the flow of thermal energy. Work and heat are therefore two distinct ways of changing the internal energy of a system. We defined work (w) in Chapter 9 as a force F acting through a distance d:
$w=Fd \tag{18.1.2}$
Because work occurs only when an object, such as a person, or a substance, such as water, moves against an opposing force, work requires that a system and its surroundings be connected. In contrast, the flow of heat, the transfer of energy due to differences in temperature between two objects, represents a thermal connection between a system and its surroundings. Thus doing work causes a physical displacement, whereas the flow of heat causes a temperature change. The units of work and heat must be the same because both processes result in the transfer of energy. In the SI system, those units are joules (J), the same unit used for energy. There is no difference between an energy change brought about by doing work on a system and an equal energy change brought about by heating it.
The connections among work, heat, and energy were first described by Benjamin Thompson (1753–1814), an American-born scientist who was also known as Count Rumford. While supervising the manufacture of cannons, Rumford recognized the relationship between the amount of work required to drill out a cannon and the temperature of the water used to cool it during the drilling process (Figure 18.1.2). At that time, it was generally thought that heat and work were separate and unrelated phenomena. Hence Rumford’s ideas were not widely accepted until many years later, after his findings had been corroborated in other laboratories.
PV Work
As we saw in Chapter 9, there are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work, focusing on the work done during changes in the pressure or the volume of a gas. To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions.
Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure Pint and initial volume Vi (Figure 18.1.3). If Pext = Pint, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston (Pext) is less than Pint, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume (Vf) will be greater than Vi. If Pext > Pint, then the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area A, the external pressure exerted by the piston is, by definition, the force per unit area: Pext = F/A. The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height (V = Ah). Rearranging to give F = PextA and defining the distance the piston moves (d) as Δh, we can calculate the magnitude of the work performed by the piston by substituting into Equation 18.1.2:
$w=Fd=P_{ext}A\Delta h \tag{18.1.3}$
The change in the volume of the cylinder (ΔV) as the piston moves a distance d is ΔV = AΔh, as shown in Figure 18.1.4. The work performed is thus
$w=P_{ext}\Delta V \tag{18.1.4}$
The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
$w=\left ( \dfrac{F}{A} \right ) \Delta V= \dfrac{newton}{m^{2}\times m^{3}}$
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
$R= \dfrac{0.08206\;L\cdot atm}{mol\cdot K} =\dfrac{8.314\;J}{mol\cdot K} \notag$
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J. (For more information on the ideal gas law, see Chapter 10.)
Whether work is defined as having a positive sign or a negative sign is a matter of convention. In Chapter 9, we defined heat flow from a system to its surroundings as negative. Using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings.This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, Equation 18.4 must be written with a negative sign to describe PV work done by the system as negative:
$w= P_{ext}\Delta V \tag{18.1.5}$
The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings.
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining Figure 18.1.5, in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from (V1, P1) to (V2, P2), which means that work is not a state function.
Note the Pattern
Internal energy is a state function, whereas work is not.
Example 18.1.1
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Asked for: work done
Strategy:
A Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio.
B Use Equation 18.1.5 to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution:
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
$w= -P_{ext}\Delta V=-\left ( 40.0\;atm \right )\left ( 0.400\;L-0.0400\;L \right )=14.4\; L\cdot atm$
Converting from liter-atmospheres to joules,
$w= -P_{ext}\Delta V=-\left (14.4\; \cancel{L\cdot atm} \right )\left [ 101.3J/\left ( \cancel{L\cdot atm} \right ) \right ]=-1.46\times 10^{3}\;J$
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
Exercise
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?
Answer: −0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
Summary
Thermodynamics is the study of the interrelationships among heat, work, and the energy content of a system at equilibrium. The sum of the potential energy and the kinetic energy of all the components of a system is the internal energy (E) of the system, which is a state function. When the pressure or the volume of a gas is changed, any mechanical work done is called PV work. Work done by a system on its surroundings is given a negative value, whereas work done on a system by its surroundings has a positive value.
Key Takeaway
• Internal energy is a state function that is the sum of the potential and kinetic energy of the system, whereas work is not a state function.
Key Equation
Definition of PV work
Equation 18.1.5: w = −PextΔV
Conceptual Problems
1. Thermodynamics focuses on the energetics of the reactants and products and provides information about the composition of the reaction system at equilibrium. What information on reaction systems is not provided by thermodynamics?
2. Given a system in which a substance can produce either of two possible products, A → B or A → C, which of the following can be predicted using chemical thermodynamics?
1. At equilibrium, the concentration of product C is greater than the concentration of product B.
2. Product C forms more quickly than product B.
3. The reaction A → C is exothermic.
4. Low-energy intermediates are formed in the reaction A → B.
5. The reaction A → C is spontaneous.
3. In what two ways can a closed system exchange energy with its surroundings? Are these two processes path dependent or path independent?
4. A microwave oven operates by providing enough energy to rotate water molecules, which produces heat. Can the change in the internal energy of a cup of water heated in a microwave oven be described as a state function? Can the heat produced be described as a state function?
Answers
1. Thermodynamics tells us nothing about the rate at which reactants are converted to products.
2. heat and work; path dependent
Numerical Problems
1. Calculate the work done in joules in each process.
1. compressing 12.8 L of hydrogen gas at an external pressure of 1.00 atm to 8.4 L at a constant temperature
2. expanding 21.9 L of oxygen gas at an external pressure of 0.71 atm to 23.7 L at a constant temperature
2. How much work in joules is done when oxygen is compressed from a volume of 22.8 L and an external pressure of 1.20 atm to 12.0 L at a constant temperature? Was work done by the system or the surroundings?
3. Champagne is bottled at a CO2 pressure of about 5 atm. What is the force on the cork if its cross-sectional area is 2.0 cm2? How much work is done if a 2.0 g cork flies a distance of 8.2 ft straight into the air when the cork is popped? Was work done by the system or the surroundings?
4. One mole of water is converted to steam at 1.00 atm pressure and 100°C. Assuming ideal behavior, what is the change in volume when the water is converted from a liquid to a gas? If this transformation took place in a cylinder with a piston, how much work could be done by vaporizing the water at 1.00 atm? Is work done by the system or the surroundings?
5. Acceleration due to gravity on the earth’s surface is 9.8 m/s2. How much work is done by a 175 lb person going over Niagara Falls (approximately 520 ft high) in a barrel that weighs 145 lb?
6. Recall that force can be expressed as mass times acceleration (F = ma). Acceleration due to gravity on the earth’s surface is 9.8 m/s2.
1. What is the gravitational force on a person who weighs 52 kg?
2. How much work is done if the person leaps from a burning building out of a window that is 20 m above the ground?
3. If the person lands on a large rescue cushion fitted with a pressure-release valve that maintains an internal pressure of 1.5 atm, how much air is forced out of the cushion?
7. A gas is allowed to expand from a volume of 2.3 L to a volume of 5.8 L. During the process, 460 J of heat is transferred from the surroundings to the gas.
1. How much work has been done if the gas expands against a vacuum?
2. How much work has been done if the gas expands against a pressure of 1.3 atm?
3. What is the change in the internal energy of the system?
8. One mole of an ideal gas is allowed to expand from an initial volume of 0.62 L to a final volume of 1.00 L at constant temperature against a constant external pressure of 1.0 atm. How much work has been done?
1. −230 kJ
1. 0 J
2. −460 J
3. 0 J
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/18%3A_Chemical_Thermodynamics/18.01%3A_Thermodynamics_and_Work.txt |
Learning Objectives
• To calculate changes in internal energy.
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamicsThe energy of the universe is constant: ΔEuniverse = ΔEsystem +ΔEsurroundings = 0., which states that the energy of the universe is constant. Using Equation 18.1.1, we can express this law mathematically as follows:
$\Delta E_{univ}= \Delta E_{sys}+\Delta E_{surr}=0 \tag{18.2.1}$
$\Delta E_{sys}= -\Delta E_{surr}$
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the CO2/H2O product mixture is less than that of the isooctane/O2 reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):
$\Delta E_{sys}= q + w \tag{18.2.2}$
Although q and w are not state functions on their own, their sum (ΔEsys) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed, as discussed in Section 18.5. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
Note the Pattern
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
Note the Pattern
Although q and w are not state functions, their sum (ΔEsys) is independent of the path taken and therefore is a state function.
Example 18.2.1
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔE) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
A Determine the sign of q to use in Equation 18.2.2.
B From Equation 18.1.5, calculate w from the values given. Substitute this value into Equation 18.2.2 to calculate ΔE.
Solution:
A From Equation 18.2.2, we know that ΔE = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so w must be positive. From Equation 18.1.5,
$w= -P_{ext}\Delta V=-\left ( 8.0\;\cancel{atm} \right )\left ( 0.0500\;\cancel{L}-0.400\;\cancel{L} \right )\left ( \dfrac{101.3\;J}{\cancel{L}\cdot \cancel{atm}} \right )=284\; J$
Thus
$\Delta E=q+w=-140\;J+284\;J=144\;J$
In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔE) of the gas in joules?
Answer: −216 J
Note the Pattern
By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Enthalpy
To further understand the relationship between heat flow (q) and the resulting change in internal energy (ΔE), we can look at two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. We will assume that PV work is the only kind of work possible for the system, so we can substitute its definition from Equation 18.1.5 into Equation 18.7 to obtain the following:
$\Delta E=q-p\Delta V \tag{18.2.3}$
where the subscripts have been deleted.
If the reaction occurs in a closed vessel, the volume of the system is fixed, and ΔV is zero. Under these conditions, the heat flow (often given the symbol qv to indicate constant volume) must equal ΔE:
$\underset{constant\; volume}{q_{v}=\Delta E} \tag{18.2.4}$
No PV work can be done, and the change in the internal energy of the system is equal to the amount of heat transferred from the system to the surroundings or vice versa.
Many chemical reactions are not, however, carried out in sealed containers at constant volume but in open containers at a more or less constant pressure of about 1 atm. The heat flow under these conditions is given the symbol qp to indicate constant pressure. Replacing q in Equation 18.2.3 by qp and rearranging to solve for qp,
$\underset{constant\; pressure}{q_{p}=\Delta E+P\Delta V} \tag{18.2.5}$
Thus, at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done, as we stated in Chapter 9.
Because conditions of constant pressure are so important in chemistry, a new state function called enthalpy (H)A state function that is the sum of the system’s internal energy E and the product of its pressure P and volume V. H=E+PV is defined as H = E + PV. At constant pressure, the change in the enthalpy of a system is as follows:
$\Delta H=\Delta E+\Delta \left ( PV \right )= \Delta E+P\Delta \left ( V \right ) \tag{18.2.6}$
Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH = qp. This expression is consistent with our definition of enthalpy in Section Section 9.1, where we stated that enthalpy is the heat absorbed or produced during any process that occurs at constant pressure.
Note the Pattern
At constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH = qp.
Example 18.2.2
The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔE for the melting of ice at 0.0°C. (For more information on enthalpy, see Section 9.2.)
Given: enthalpy of fusion for ice, pressure, and molar volumes of ice and water
Asked for: ΔH and ΔE for ice melting at 0.0°C
Strategy:
A Determine the sign of q and set this value equal to ΔH.
B Calculate Δ(PV) from the information given.
C Determine ΔE by substituting the calculated values into Equation 18.2.6.
Solution:
A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = qp = ΔH = 6.01 kJ.
B To find ΔE using Equation 18.2.6, we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so
$\Delta PV=P \Delta V = P \left ( V_{f}-V \right ) =\left ( 1.00\;atm \right ) \left ( 0.0180\;L-0.0197\; lL \right )$
$=\left (-1.7\times 10{-3}\;\cancel{L\cdot atm} \right )\left ( 101.3\;J/\cancel{L\cdot atm} \right )$
C Substituting the calculated values of ΔH and PΔV into Equation 18.2.6,
$\Delta E = \Delta H - P \Delta V = -6010 \; J - \left (-0.0017\;J \right ) = 6010 \; J= 6.01 \; kJ$
Exercise
At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔE for the conversion of C (graphite) to C (diamond) under these conditions.
Answer: ΔH = 1.85 kJ/mol; ΔE = 1.85 kJ/mol
The Relationship between ΔH and ΔE
If ΔH for a reaction is known, we can use the change in the enthalpy of the system (Equation 18.2.6) to calculate its change in internal energy. When a reaction involves only solids, liquids, liquid solutions, or any combination of these, the volume does not change appreciably (ΔV = 0). Under these conditions, we can simplify Equation 18.2.6 to ΔH = ΔE. If gases are involved, however, ΔH and ΔE can differ significantly. We can calculate ΔE from the measured value of ΔH by using the right side of Equation 18.11 together with the ideal gas law, PV = nRT. Recognizing that Δ(PV) = Δ(nRT), we can rewrite Equation 18.2.6 as follows:
$\Delta H= \Delta E+ \Delta \left (PV \right )= \Delta E+ \Delta \left (nRT \right ) \tag{18.2.7}$
At constant temperature, Δ(nRT) = RTΔn, where Δn is the difference between the final and initial numbers of moles of gas. Thus
$\Delta E= \Delta H - RT \Delta n \tag{18.2.8}$
For reactions that result in a net production of gas, Δn > 0, so ΔE < ΔH. Conversely, endothermic reactions (ΔH > 0) that result in a net consumption of gas have Δn < 0 and ΔE > ΔH. The relationship between ΔH and ΔE for systems involving gases is illustrated in Example 4.
Note the Pattern
For reactions that result in a net production of gas, ΔE < ΔH. For endothermic reactions that result in a net consumption of gas, ΔE > ΔH.
Example 18.2.3
The combustion of graphite to produce carbon dioxide is described by the equation C (graphite, s) + O2(g) → CO2(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔE for the reaction?
Given: balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant
Asked for: ΔE
Strategy:
A Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.
B Substitute this value and the data given into Equation 18.2.8 to obtain ΔE.
Solution:
A In this reaction, 1 mol of gas (CO2) is produced, and 1 mol of gas (O2) is consumed. Thus Δn = 1 − 1 = 0.
B Substituting this calculated value and the given values into Equation 18.2.8,
$\Delta E=\Delta H - RT\Delta n =\left (-393.5\;kJ/mol\right )-\left [ 8.314\;J/mol\cdot K \right ]\left ( 298\;K \right )\left ( 0 \right )$
$=\left (-393.5\;kJ/mol\right )-\left ( 0 \;J/mol \right )=-393.5\;kJ$
To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite.
Exercise
Calculate ΔE for the conversion of oxygen gas to ozone at 298 K: 3O2(g) → 2O3(g). The value of ΔH for the reaction is 285.4 kJ.
Answer: 288 kJ
As the exercise in Example 4 illustrates, the magnitudes of ΔH and ΔE for reactions that involve gases are generally rather similar, even when there is a net production or consumption of gases.
Summary
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (E) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔE.
Key Takeaway
• Enthalpy is a state function, and the change in enthalpy of a system is equal to the sum of the change in the internal energy of the system and the PV work done.
Key Equations
Internal energy change
Equation 18.2.2: ΔEsys = q + w
Enthalpy change
Equation 18.2.6: ΔH = ΔE + Δ(PV)
Relationship between Δ H and Δ E for an ideal gas
Equation 18.2.8: ΔE = ΔHRTΔn
Conceptual Problems
1. Describe how a swinging pendulum that slows with time illustrates the first law of thermodynamics.
2. When air is pumped into a bicycle tire, the air is compressed. Assuming that the volume is constant, express the change in internal energy in terms of q and w.
3. What is the relationship between enthalpy and internal energy for a reaction that occurs at constant pressure?
4. An intrepid scientist placed an unknown salt in a small amount of water. All the salt dissolved in the water, and the temperature of the solution dropped several degrees.
1. What is the sign of the enthalpy change for this reaction?
2. Assuming the heat capacity of the solution is the same as that of pure water, how would the scientist calculate the molar enthalpy change?
3. Propose an explanation for the decrease in temperature.
5. For years, chemists and physicists focused on enthalpy changes as a way to measure the spontaneity of a reaction. What arguments would you use to convince them not to use this method?
6. What is the relationship between enthalpy and internal energy for a reaction that occurs at constant volume?
7. The enthalpy of combustionHcomb) is defined thermodynamically as the enthalpy change for complete oxidation. The complete oxidation of hydrocarbons is represented by the following general equation: hydrocarbon + O2(g) → CO2(g) + H2O(g). Enthalpies of combustion from reactions like this one can be measured experimentally with a high degree of precision. It has been found that the less stable the reactant, the more heat is evolved, so the more negative the value of ΔHcomb. In each pair of hydrocarbons, which member do you expect to have the greater (more negative) heat of combustion? Justify your answers.
1. cyclopropane or cyclopentane
2. butane or 2-methylpropane
3. hexane or cyclohexane
8. Using a structural argument, explain why the trans isomer of 2-butene is more stable than the cis isomer. The enthalpies of formation of cis- and trans-2-butene are −7.1 kJ/mol and −11.4 kJ/mol, respectively.
9. Using structural arguments, explain why cyclopropane has a positive ΔHf° (12.7 kJ/mol), whereas cyclopentane has a negative ΔHf° (−18.4 kJ/mol). (Hint: consider bond angles.)
Answers
1. At constant pressure, ΔH = ΔE + PΔV.
2. With bond angles of 60°, cyclopropane is highly strained, causing it to be less stable than cyclopentane, which has nearly ideal tetrahedral geometry at each carbon atom.
Numerical Problems
1. A block of CO2 weighing 15 g evaporates in a 5.0 L container at 25°C. How much work has been done if the gas is allowed to expand against an external pressure of 0.98 atm under isothermal conditions? The enthalpy of sublimation of CO2 is 25.1 kJ/mol. What is the change in internal energy (kJ/mol) for the sublimation of CO2 under these conditions?
2. Zinc and HCl react according to the following equation:
Zn(s) + 2HCl(aq) → Zn2+(aq) + 2Cl(aq) + H2(g)
When 3.00 g of zinc metal is added to a dilute HCl solution at 1.00 atm and 25°C, and this reaction is allowed to go to completion at constant pressure, 6.99 kJ of heat must be removed to return the final solution to its original temperature. What are the values of q and w, and what is the change in internal energy?
3. Acetylene torches, used industrially to cut and weld metals, reach flame temperatures as high as 3000°C. The combustion reaction is as follows:
$2C_{2}H_{2}\left ( g \right )+5O_{2}\left ( g \right ) \rightarrow 4CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )$\;\;\; \Delta H=-2599\;kJ \)
Calculate the amount of work done against a pressure of 1.0 atm when 4.0 mol of acetylene are allowed to react with 10 mol of O2 at 1.0 atm at 20°C. What is the change in internal energy for the reaction?
4. When iron dissolves in 1.00 M aqueous HCl, the products are FeCl2(aq) and hydrogen gas. Calculate the work done if 30 g of Fe react with excess hydrochloric acid in a closed vessel at 20°C. How much work is done if the reaction takes place in an open vessel with an external pressure of 1.0 atm?
Answer
1. −350 J; 8.2 kJ
Contributors
• Anonymous
Modified by Joshua B. Halpern
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Learning Objectives
• To understand the relationship between internal energy and entropy.
The first law of thermodynamics governs changes in the state function we have called internal energy ($U$). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔHsoln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure $1$.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S), a thermodynamic property of all substances that is proportional to their degree of "disorder". In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Entropy
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (ΔS > 0) or a decrease in entropy (ΔS < 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: ΔS = Sf − Si.
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
We can illustrate the concepts of microstates and entropy using a deck of playing cards, as shown in Figure $2$. In any new deck, the 52 cards are arranged by four suits, with each suit arranged in descending order. If the cards are shuffled, however, there are approximately 1068 different ways they might be arranged, which corresponds to 1068 different microscopic states. The entropy of an ordered new deck of cards is therefore low, whereas the entropy of a randomly shuffled deck is high. Card games assign a higher value to a hand that has a low degree of disorder. In games such as five-card poker, only 4 of the 2,598,960 different possible hands, or microstates, contain the highly ordered and valued arrangement of cards called a royal flush, almost 1.1 million hands contain one pair, and more than 1.3 million hands are completely disordered and therefore have no value. Because the last two arrangements are far more probable than the first, the value of a poker hand is inversely proportional to its entropy.
We can see how to calculate these kinds of probabilities for a chemical system by considering the possible arrangements of a sample of four gas molecules in a two-bulb container (Figure $3$). There are five possible arrangements: all four molecules in the left bulb (I); three molecules in the left bulb and one in the right bulb (II); two molecules in each bulb (III); one molecule in the left bulb and three molecules in the right bulb (IV); and four molecules in the right bulb (V). If we assign a different color to each molecule to keep track of it for this discussion (remember, however, that in reality the molecules are indistinguishable from one another), we can see that there are 16 different ways the four molecules can be distributed in the bulbs, each corresponding to a particular microstate. As shown in Figure $3$, arrangement I is associated with a single microstate, as is arrangement V, so each arrangement has a probability of 1/16. Arrangements II and IV each have a probability of 4/16 because each can exist in four microstates. Similarly, six different microstates can occur as arrangement III, making the probability of this arrangement 6/16. Thus the arrangement that we would expect to encounter, with half the gas molecules in each bulb, is the most probable arrangement. The others are not impossible but simply less likely.
There are 16 different ways to distribute four gas molecules between the bulbs, with each distribution corresponding to a particular microstate. Arrangements I and V each produce a single microstate with a probability of 1/16. This particular arrangement is so improbable that it is likely not observed. Arrangements II and IV each produce four microstates, with a probability of 4/16. Arrangement III, with half the gas molecules in each bulb, has a probability of 6/16. It is the one encompassing the most microstates, so it is the most probable.
Instead of four molecules of gas, let’s now consider 1 L of an ideal gas at standard temperature and pressure (STP), which contains 2.69 × 1022 molecules (6.022 × 1023 molecules/22.4 L). If we allow the sample of gas to expand into a second 1 L container, the probability of finding all 2.69 × 1022 molecules in one container and none in the other at any given time is extremely small, approximately $\frac{2}{2.69 \times 10^{22}}$. The probability of such an occurrence is effectively zero. Although nothing prevents the molecules in the gas sample from occupying only one of the two bulbs, that particular arrangement is so improbable that it is never actually observed. The probability of arrangements with essentially equal numbers of molecules in each bulb is quite high, however, because there are many equivalent microstates in which the molecules are distributed equally. Hence a macroscopic sample of a gas occupies all of the space available to it, simply because this is the most probable arrangement.
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure $4$, the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔSsoln > 0.
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Example $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
2. 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
1. Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
2. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Exercise $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm or a sample of 2 mol of NH3(g) at 25°C and 1 atm
Answer a
1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
Answer a
a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm (more molecules of gas are present)
Video Solution
Reversible and Irreversible Changes
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (Pext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = - PextΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev. Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So
$ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1}$
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev.
In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (qrev) to define entropy quantitatively.
The Relationship between Internal Energy and Entropy
Because the quantity of heat transferred (qrev) is directly proportional to the absolute temperature of an object (T) (qrev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ qrev). Combining these relationships for any reversible process,
$q_{\textrm{rev}}=T\Delta S\;\textrm{ and }\;\Delta S=\dfrac{q_{\textrm{rev}}}{T} \label{Eq2}$
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is wrev = −PΔV, we can express Equation $\ref{Eq1}$ as follows:
\begin{align} ΔU &= q_{rev} + w_{rev} \[4pt] &= TΔS − PΔV \label{Eq3} \end{align}
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the $PV$ work done.
To illustrate the use of Equation $\ref{Eq2}$ and Equation $\ref{Eq3}$, we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its $T$ constant (Figure $5$). The internal energy of the gas does not change because the temperature of the gas does not change; that is, $ΔU = 0$ and $q_{rev} = −w_{rev}$. During expansion, ΔV > 0, so the gas performs work on its surroundings:
$w_{rev} = −PΔV < 0.$
According to Equation $\ref{Eq3}$, this means that qrev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore ΔSsys = +qrev/T, and the entropy change of the surroundings is
$ΔS_{surr} = −\dfrac{q_{rev}}{T}.$
The corresponding change in entropy of the universe is then as follows:
\begin{align*} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \[4pt] &= \dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \[4pt] &= 0 \label{Eq4} \end{align*}
Thus no change in ΔSuniv has occurred.
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure $6$. The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation $\ref{Eq2}$, we see that the entropy of fusion of ice can be written as follows:
$\Delta S_{\textrm{fus}}=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{\Delta H_{\textrm{fus}}}{T} \label{Eq5}$
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case, ΔSfus = (6.01 kJ/mol)/(273 K) = 22.0 J/(mol•K) = ΔSsys. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so ΔSsurr = qrev/T = −(6.01 kJ/mol)/(273 K) = −22.0 J/(mol•K). Once again, we see that the entropy of the universe does not change:
ΔSuniv = ΔSsys + ΔSsurr = 22.0 J/(mol•K) − 22.0 J/(mol•K) = 0
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The Second Law of Thermodynamics
The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔScold = q/Tcold. Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔShot = −q/Thot, where Tcold and Thot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
$\Delta S_{\textrm{univ}}=\Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}}=\dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6}$
The numerators on the right side of Equation $\ref{Eq6}$ are the same in magnitude but opposite in sign. Whether ΔSuniv is positive or negative depends on the relative magnitudes of the denominators. By definition, Thot > Tcold, so −q/Thot must be less than q/Tcold, and ΔSuniv must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔSuniv is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔSuniv is negative will not occur as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure $7$).
Example $2$: Tin Pest
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to have plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and may have disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
1. What is ΔS for this process?
2. Which is the more highly ordered form of tin—white or gray?
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy:
Use Equation $\ref{Eq2}$ to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution
1. We know from Equation $\ref{Eq2}$ that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{(-2.1\;\mathrm{kJ/mol})(1000\;\mathrm{J/kJ})}{\textrm{286.4 K}}=-7.3\;\mathrm{J/(mol\cdot K)}$
1. The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Video $1$: Time lapse tin pest reaction.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C)
Exercise $2$
Elemental sulfur exists in two forms: an orthorhombic form (Sα), which is stable below 95.3°C, and a monoclinic form (Sβ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.
1. What is ΔS for this process?
2. Which is the more highly ordered form of sulfur—Sα or Sβ?
Answer a
1.09 J/(mol•K)
Answer b
Sα
Summary
For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases. $\Delta S=\frac{q_{\textrm{rev}}}{T}$
A measure of the disorder of a system is its entropy (S), a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/18%3A_Chemical_Thermodynamics/18.03%3A_The_Second_Law_of_Thermodynamics.txt |
Learning Objectives
• To use thermodynamic cycles to calculate changes in entropy.
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $1$). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Third Law of Thermodynamics
The entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (Cp) as a function of temperature and then plotting the quantity Cp/T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Calculating ΔS from Standard Molar Entropy Values
One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol•K)] and 298 K.
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol•K), whereas S° for water vapor is 188.8 J/(mol•K). Likewise, S° is 260.7 J/(mol•K) for gaseous I2 and 116.1 J/(mol•K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Gases Liquids Solids
Substance S° [J/(mol•K)] Substance S° [J/(mol•K)] Substance S° [J/(mol•K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
Note
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol•K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol•K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure.
Note
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $1$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
Example $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table $1$. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
$\mathrm{C_8H_{18}(l)}+\dfrac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(g)}$
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\ &=515.3\;\mathrm{J/K}\end{align}
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer: −361.1 J/K
Calculating ΔS from Thermodynamic Cycles
We can also calculate a change in entropy using a thermodynamic cycle. As you learned previously, the molar heat capacity (Cp) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, Cv is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in Figure $2$ is approximately proportional to the heat capacity of the substance.
Recall that the entropy change (ΔS) is related to heat flow (qrev) by ΔS = qrev/T. Because qrev = nCpΔT at constant pressure or nCvΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T1 to T2 is as follows:
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=nC_\textrm p\dfrac{\Delta T}{T}\hspace{4mm}(\textrm{constant pressure})$
As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in Chapter 11.
$\Delta S=nC_\textrm p\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant pressure}) \tag{18.20}$
Similarly,
$Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \tag{18.21}$
Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample.
We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in Figure $3$): an orthorhombic form with a highly ordered structure (Sα) and a less-ordered monoclinic form (Sβ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from Sα to Sβ can be described by the thermodynamic cycle shown in part (b) in Figure $3$, in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to Sβ (−ΔSfus(β) = ΔS3 in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS3 can be calculated from the overall entropy change (ΔSt) for the Sα–Sβ transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of Sα and Sβ, ΔHfus(α), and the melting point of Sα.)
If we know the melting point of Sα (Tm = 115.2°C = 388.4 K) and ΔSt for the overall phase transition [calculated to be 1.09 J/(mol•K) in the exercise in Example 6], we can calculate ΔS3 from the values given in part (b) in Figure $3$ where Cp(α) = 22.70 J/mol•K and Cp(β) = 24.77 J/mol•K (subscripts on ΔS refer to steps in the cycle):
\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 \ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right) \ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right) \ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right) \ &=[1.194\;\mathrm{J/(mol\cdot K)}]+[4.434\;\mathrm{J/(mol\cdot K)}]+\Delta S_3+[-1.303\;\mathrm{J/(mol\cdot K)}]\end{align}
Solving for ΔS3 gives a value of −3.24 J/(mol•K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS3 is negative.
Summary
• Entropy changes can be calculated using the “products minus reactants” rule or from a combination of heat capacity measurements and measured values of enthalpies of fusion or vaporization.
The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/18%3A_Chemical_Thermodynamics/18.04%3A_Entropy_Changes_and_the_Third_Law_of_Thermodynamics.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Free Energy and the Direction of Spontaneous Reactions
The Gibbs free energy ($G$), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
$G = H − TS \label{Eq1}$
Because it is a combination of state functions, $G$ is also a state function.
J. Willard Gibbs (1839–1903)
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce Equation $\ref{Eq2}$ to $ΔG = q − q_{rev}$. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T}$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align}
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the disorder of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of ΔS (Equation $\ref{Eq3}$), we know that for 1 mol of water,
\begin{align*} \Delta S_{\textrm{vap}}&=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \[4pt] &=\textrm{108.96 J/K} \end{align*}
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\begin{align*}\Delta G_{100^\circ\textrm C}&=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{0 J}\end{align*}
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
\begin{align*}\Delta G_{110^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \[4pt] &=-\textrm{1091 J}\end{align*}
At 110°C, $ΔG < 0$, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate $ΔG$ for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\begin{align*}\Delta G_{90^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{1088 J}\end{align*}
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Relating Enthalpy and Entropy changes under Equilibrium Conditions
$ΔG = 0$ only if $ΔH = TΔS$.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Eq2}$), setting $ΔG = 0$, and solving for $T$,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus $ΔG = 0$ at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, $ΔG$ is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, $ΔG$ is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the $ΔH$ and $TΔS$ terms vary with temperature for the vaporization of water. When the two lines cross, $ΔG = 0$, and $ΔH = TΔS$.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see the previous section). Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $1$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $1$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach leaf cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10-20
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$\ce{ H2(g) + O2(g) \rightleftharpoons H2O2(l)}\nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know $ΔH^o$, $ΔS^o$, and $T$. We are given $ΔH^o$, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \[4pt]&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*}
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, $ΔS^o$ is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}]\nonumber \[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \[4pt] &=-120.31\textrm{ kJ/mol}\nonumber \end{align*}
The negative value of $ΔG^o$ indicates that the reaction is spontaneous as written. Because $ΔS^o$ and $ΔH^o$ for this reaction have the same sign, the sign of $ΔG^o$ depends on the relative magnitudes of the $ΔH^o$ and $TΔS^o$ terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable $ΔS^o$ term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l)\nonumber .$
Is the reaction spontaneous as written at 25°C?
Hint
At 25°C, the standard enthalpy change ($ΔH^o$) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are
• S°(N2H4) = 121.2 J/(mol•K),
• S°(N2) = 191.6 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Answer
149.5 kJ/mol
no, not spontaneous
Video Solution
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $\Delta G^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$\Delta G^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
The "Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated ΔGf values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\ce{C8H_{18}(l) + 25/2 O2 (g) \rightarrow 8CO2(g) + 9H2O(l)}\nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\[4pt]&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}})\nonumber \end{align*}
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Video Solution
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \}\nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align}\nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})]\nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2)\nonumber\end{align}\nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2)\nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ\nonumber \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt]&=21.7\textrm{ kJ (per mole of N}_2) \end{align*}
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 9.5.3 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \[4pt] \Delta H^\circ &=T\Delta S^\circ \[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*}
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example $3$, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Video Solution
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change:
$ΔG = ΔH − TΔS\nonumber$
• Standard free-energy change:
$ΔG° = ΔH° − TΔS°\nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/18%3A_Chemical_Thermodynamics/18.05%3A_Free_Energy.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously: ΔSuniv > 0, ΔGsys < 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. (For more information on the reaction quotient and the equilibrium constant, see Chapter 15.) Recall that if Q < K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q > K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. Table 18.6.1 summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. The relationship between ΔSuniv and ΔGsys was described in Section 18.5. In this section, we explore the relationship between the standard free energy of reaction (ΔG°) and the equilibrium constant (K).
Table 18.6.1 Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
*Spontaneous in the reverse direction.
Free Energy and the Equilibrium Constant
Because ΔH° and ΔS° determine the magnitude of ΔG° (Equation 18.5.5), and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. As you learned in Section 18.5, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. Using higher math, the general relationship can be shown as follows:
$\Delta G= V \Delta P -S \Delta T \tag{18.6.1}$
If a reaction is carried out at constant temperature (ΔT = 0), then Equation 18.29 simplifies to
$\Delta G= V \Delta P \tag{18.6.2}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important. What Equation 18.6.2 tells us is how the free energy changes as the pressure changes at constant temperature.
Assuming ideal gas behavior, we can replace the V in Equation 18.6.2 by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express ΔG in terms of the initial and final pressures (Pi and Pf, respectively) as in Equation 18.4.1.
$\Delta G = \left (\dfrac{nRT}{P} \right ) \Delta P= nRT \dfrac{\Delta P}{P} = nRT ln \left (\dfrac{P_{f}}{P_{i}} \right )\tag{18.6.4}$
To actually show the final step one has to consider infinitesmal changes in both the free energy and the pressure and then integrate both sides of the equation using calculus, which is beyond the scope of this course. If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
$G - G^{o} = nRT ln\;P$
This can be rearranged as follows:
$G = G^{o} + nRT ln\;P\tag{18.6.5}$
As you will soon discover, Equation 18.6.5 allows us to relate ΔG° and Kp. Any relationship that is true for Kp must also be true for K because Kp and K are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA + bB \rightleftharpoons cC + dD \tag{18.6.6}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
$\Delta G = \sum mG_{products} - \sum mG_{reactants} =\left ( cG_{C}+dG_{D} \right )- \left ( aG_{A}+bG_{B} \right )\tag{18.6.7}$
Substituting Equation 18.6.5 for each term into Equation 18.6.7,
$\Delta G = \left [\left (cG_{C}+cRTln\;P_{C} \right ) + \left (dG_{D}+dRTln\;P_{D} \right ) \right ] - \left [\left (aG_{A}+aRTln\;P_{A} \right ) + \left (bG_{B}+bRTln\;P_{B} \right ) \right ]$
Combining terms gives the following relationship between ΔG and the reaction quotient Q:
$\Delta G = \Delta G^{o}+ RTln\left ( \dfrac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}} \right ) = \Delta G^{o}+ RTln\;Q \tag{18.6.8}$
where ΔG° indicates that all reactants and products are in their standard states. In Chapter 15, you learned that for gases Q = Kp at equilibrium, and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and Kp for gases as follows:
$0 = \Delta G^{o}+ RTln\;K_{P} \tag{18.6.9}$
$\Delta G^{o} = - RTln\;K_{P}$
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then Kp = 1, and neither reactants nor products are favored: the system is at equilibrium.
Note the Pattern
For a spontaneous process under standard conditions, Keq and Kp are greater than 1.
Example 18.6.1
In Example 18.5.3, we calculated that ΔG° = −32.7 kJ/mol of N2 for the reaction N2(g)+3H2(g ⇌ 2NH3(g) This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: P(N2) = 2.00 atm, P(H2) = 7.00 atm, P(NH3) = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
A Using the values given and Equation 18.6.8, calculate Q.
B Substitute the values of ΔG° and Q into Equation 18.6.8 to obtain ΔG for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation 18.6.8. Substituting the partial pressures given, we can calculate Q:
$Q=\dfrac{P_{NH_{3}}^{2}}{P_{N_{2}}P_{H_{2}}^{3}} =\dfrac{\left ( 0.021 \right )^{2}}{\left ( 2.00 \right )\left ( 7.00 \right )^{3}}=6.4\times 10^{-7}$
B Substituting the values of ΔG° and Q into Equation 18.6.8,
$\Delta G=\Delta G^{o}+RTln\;Q = -32.7\;kJ+\left [ \left (8.314\;\cancel{J}\cancel{K} \right )\left ( 373\;\cancel{K} \right )\left ( 1\;kJ/1000\;\cancel{kJ} \right ) ln \left ( 6.4\times 10^{-7} \right )\right ]$
$=-32.7\; kJ +\left ( -44\; kJ \right ) = -77 \; kJ/mol\;N_{2}$
Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise
Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P(NO) = 0.0100 atm, P(O2) = 0.200 atm, and P(NO2) = 1.00 × 10−4 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer: −92.9 kJ/mol of O2; the reaction is spontaneous to the right as written, so products are favored.
Example 18.6.2
Calculate Kp for the reaction of H2 with N2 to give NH3 at 25°C. As calculated in Example 18.5.3 , ΔG° for this reaction is −32.7 kJ/mol of N2.
Given: balanced chemical equation from Example 10, ΔG°, and temperature
Asked for: K p
Strategy:
Substitute values for ΔG° and T (in kelvins) into Equation 18.6.9 to calculate Kp, the equilibrium constant for the formation of ammonia.
Solution:
In Example 18.5.3 , we used tabulated values of ΔG° to calculate ΔG° for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation 18.6.9,
$\Delta G^{o}=-RTln\:K_{P}$
$\dfrac{- \Delta G^{o}}{RT}=ln\:K_{P}$
Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
$ln\:K_{P}= -\dfrac{\left ( -32.7\;\cancel{kJ} \right )\left ( 1000\;\cancel{J}/\cancel{kJ} \right )}{8.314 \;\cancel{J}/\cancel{K}\left ( 298\; \cancel{K} \right )}=13.2$
$K_{P}= 5.4\times 10^{5}$
Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in Chapter 15, however, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O2.
Answer: 2.2 × 1012
Although Kp is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of Kp and K in Chapter 15 and showed that they are related:
$K_{P}= K\left ( RT \right )^{\Delta n} \tag{18.6.10}$
where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so Kp = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation 18.36 can be written in a more general form:
$\Delta G^{o}= -RT ln\; K \tag{18.6.11}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation 18.38 for the difference between Kp and K.Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation 18.26 and Equation 18.6.11 provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
$\Delta G^{o}= \Delta H^{o} -T \Delta S^{o}= -RT ln\; K \tag{18.6.12}$
Notice that K becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
Note the Pattern
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible.
Temperature Dependence of the Equilibrium Constant
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation 18.6.12, which can be rearranged as follows:
$ln\; K=\dfrac{\Delta H^{o}}{RT} + \dfrac{\Delta S^{o}}{R} \tag{18.6.13}$
Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation 18.6.13 agrees with the qualitative predictions made by applying Le Chatelier’s principle, which we discussed in Chapter 15. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation 18.6.13 also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K1 and K2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively. Applying Equation 18.6.13 gives the following relationship at each temperature:
$ln\; K_{1}=\dfrac{\Delta H^{o}}{RT_{1}} + \dfrac{\Delta S^{o}}{R}$
$ln\; K_{2}=\dfrac{\Delta H^{o}}{RT_{2}} + \dfrac{\Delta S^{o}}{R}$
Subtracting ln K1 from ln K2,
$ln\; K_{2}-ln\; K_{1}= ln\dfrac{K_{2}}{K_{1}}=\dfrac{\Delta H^{o}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right ) \tag{18.6.13}$
Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K1) allow us to calculate the value of the equilibrium constant at any other temperature (K2), assuming that ΔH° and ΔS° are independent of temperature.
Example 18.6.3
The equilibrium constant for the formation of NH3 from H2 and N2 at 25°C was calculated to be Kp = 5.4 × 105 in Example 13. What is Kp at 500°C? (Use the data from Example 10.)
Given: balanced chemical equation, ΔH°, initial and final T, and Kp at 25°C
Asked for: Kp at 500°C
Strategy:
Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation 18.41 to obtain K2, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation 18.6.13 we obtain the following:
$= \dfrac{\left ( -91.8\;\cancel{kJ} \right )\left ( 1000\;\cancel{J}/\cancel{kJ} \right )}{8.314\;\cancel{J}/\cancel{K} }\left ( \dfrac{1}{298\;\cancel{K}}- \dfrac{1}{773\;\cancel{K}}\right )= -22.8$
$\dfrac{K_{2}}{K_{1}}= 1.3 \pm \times 10^{-10}$
$K_{2}= \left ( 5.4\times 10^{5} \right )\left ( 1.3 \pm \times 10^{-10} \right )=7.0\times 10^{-5}$
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise
In the exercise in Example 13, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the ΔHf° values in the exercise in Example 10 to calculate Kp for this reaction at 1000°C.
Answer: 5.6 × 10−4
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or Kp > 1, and products are favored over reactants. If ΔG° > 0, then K or Kp < 1, and reactants are favored over products. If ΔG° = 0, then K or Kp = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Key Takeaway
• The change in free energy of a reaction can be expressed in terms of the standard free-energy change and the equilibrium constant K or Kp and indicates whether a reaction will occur spontaneously under a given set of conditions.
Key Equations
Relationship between standard free-energy change and equilibrium constant
Equation 18.6.9: $\Delta G^{o}=-RTln\;K$
Temperature dependence of equilibrium constant
Equation 18.6.11: $ln\;K = \dfrac{-\Delta H^{o}}{RT}+\dfrac{-\Delta S^{o}}{R}$
Calculation of K at second temperature
Equation 18.6.11: $ln\;\dfrac {K_{2}}{K_{1}} = \dfrac{-\Delta H^{o}}{R}\left ( \dfrac{1}{T_{2}}- \dfrac{1}{T_{1}}\right )$
Conceptual Problems
1. Do you expect products or reactants to dominate at equilibrium in a reaction for which ΔG° is equal to
1. 1.4 kJ/mol?
2. 105 kJ/mol?
3. −34 kJ/mol?
2. The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds?
3. What happens to the change in free energy of the reaction N2(g) + 3F2(g) → 2NF3(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids?
4. Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions?
Numerical Problems
1. Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO2 and H2, as shown in the equation CO(g)+H2O(g) ⇌CO2(g)+H2(g), for which ΔH° = −41.0 kJ/mol and ΔS° = −42.3 J cal/(mol·K) at 25°C and 1 atm.
1. What is ΔG° for this reaction?
2. What is ΔG if the gases have the following partial pressures: P(CO) = 1.3 atm, P(H2O) = 0.8 atm, P(CO2) = 2.0 atm, and P(H2) = 1.3 atm?
3. What is ΔG if the temperature is increased to 150°C assuming no change in pressure?
2. Methane and water react to form carbon monoxide and hydrogen according to the equation CH4(g) + H2O(g) ⇌CO(g) + 3H2(g)
1. What is the standard free energy change for this reaction?
2. What is Kp for this reaction?
3. What is the carbon monoxide pressure if 1.3 atm of methane reacts with 0.8 atm of water, producing 1.8 atm of hydrogen gas?
4. What is the hydrogen gas pressure if 2.0 atm of methane is allowed to react with 1.1 atm of water?
5. At what temperature does the reaction become spontaneous?
3. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. CCl4(g) + 6 H2O(l) ⇌ CO2(g) +4HCl(aq); ΔG° = −377 kJ/mol
2. Xe(g) + 2F2(g) ⇌ XeF4(s); ΔH° = −66.3 kJ/mol, ΔS° = −102.3 J/(mol·K)
3. PCl3(g) + S ⇌ PSCl3(l); ΔGf°(PCl3) = −272.4 kJ/mol, ΔGf°(PSCl3) = −363.2 kJ/mol
4. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. 2KClO3(s) ⇌ 2KCl(s) + 3O2(g); ΔG° = −225.8 kJ/mol
2. CoCl2(s) + 6 H2O(l) ⇌6CoCl26 H2O(s); ΔHrxn°= −352 kJ/mol, ΔSrxn°= −899 J/(mol·K)
3. PCl3(g) + O2(g) ⇌ 2POCl3(g); (PCl3); ΔGf°(PCl3)= −272.4 kJ/mol, ΔGf°(POCl3) = −558.5 kJ/mol
5. The gas-phase decomposition of N2O4 to NO2 is an equilibrium reaction with Kp = 4.66 × 10−3. Calculate the standard free-energy change for the equilibrium reaction between N2O4 and NO2.
6. The standard free-energy change for the dissolution K4Fe(CN)6H2O(s) ⇌ 4K+(aq) + Fe(CN)64-(aq) + H2O(l) is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C?
7. Ammonia reacts with water in liquid ammonia solution (am) according to the equation
NH3(g) + H2O(am) ⇌ NH4+(am) + OH-(am) The change in enthalpy for this reaction is 21 kJ/mol, and ΔS° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)?
8. At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba2+] = [CO32−] = 5.08 × 10−5 M. Determine ΔG° for the dissolution of BaCO3.
9. Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is Pb3(PO4)2(s) + 4H+(aq) ⇌ 3Pb2+(aq) + 2H2PO4-(aq) for which log K = −1.80. What is ΔG° for this reaction?
10. The conversion of butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol.
1. What is the change in entropy for this conversion?
2. Based on structural arguments, are the sign and magnitude of the entropy change what you would expect? Why?
3. What is the equilibrium constant for this reaction?
11. The reaction of CaCO3(s) to produce CaO(s) and CO2(g) has an equilibrium constant at 25°C of 2 × 10−23. Values of ΔHf° are as follows: CaCO3, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO2, −393.5 kJ/mol.
1. What is ΔG° for this reaction?
2. What is the equilibrium constant at 900°C?
3. What is the partial pressure of CO2(g) in equilibrium with CaO and CaCO3 at this temperature?
4. Are reactants or products favored at the lower temperature? at the higher temperature?
12. In acidic soils, dissolved Al3+ undergoes a complex formation reaction with SO42− to form [AlSO4+]. The equilibrium constant at 25°C for the reaction Al3+(aq) + SO42-(aq) ⇌ AlSO4+(aq) is 1585.
1. What is ΔG° for this reaction?
2. How does this value compare with ΔG° for the reaction Al3+(aq) + F-(aq) ⇌ AlF2+(aq) for which K = 107 at 25°C?
3. Which is the better ligand to use to trap Al3+ from the soil?
Answers
1. −28.4 kJ/mol
2. −26.1 kJ/mol
3. −19.9 kJ/mol
1. 1.21 × 1066; equilibrium lies far to the right.
2. 1.89 × 106; equilibrium lies to the right.
3. 5.28 × 1016; equilibrium lies far to the right.
1. 13.3 kJ/mol
2. 5.1 × 10−21
3. 10.3 kJ/mol
1. 129.5 kJ/mol
2. 6
3. 6.0 atm
4. Products are favored at high T; reactants are favored at low T.
Contributors
• Anonymous
Modified by Joshua B. Halpern
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Learning Objectives
• To understand the differences between the information that thermodynamics and kinetics provide about a system.
Because thermodynamics deals with state functions, it can be used to describe the overall properties, behavior, and equilibrium composition of a system. It is not concerned with the particular pathway by which physical or chemical changes occur, however, so it cannot address the rate at which a particular process will occur. Although thermodynamics provides a significant constraint on what can occur during a reaction process, it does not describe the detailed steps of what actually occurs on an atomic or a molecular level.
Note
Thermodynamics focuses on the energetics of the products and the reactants, whereas kinetics focuses on the pathway from reactants to products.
Table 18.4 gives the numerical values of the equilibrium constant (K) that correspond to various approximate values of ΔG°. Note that ΔG° ≥ +10 kJ/mol or ΔG° ≤ −10 kJ/mol ensures that an equilibrium lies essentially all the way to the left or to the right, respectively, under standard conditions, corresponding to a reactant-to-product ratio of approximately 10,000:1 (or vice versa). Only if ΔG° is quite small (±10 kJ/mol) are significant amounts of both products and reactants present at equilibrium. Most reactions that we encounter have equilibrium constants substantially greater or less than 1, with the equilibrium strongly favoring either products or reactants. In many cases, we will encounter reactions that are strongly favored by thermodynamics but do not occur at a measurable rate. In contrast, we may encounter reactions that are not thermodynamically favored under standard conditions but nonetheless do occur under certain nonstandard conditions.
Table 18.4 The Relationship between K and ΔG° at 25°C
ΔG° (kJ/mol) K Physical Significance
500 3 × 10−88 For all intents and purposes, the reaction does not proceed in the forward direction: only reactants are present at equilibrium.
100 3 × 10−18
10 2 × 10−2 Both forward and reverse reactions occur: significant amounts of both products and reactants are present at equilibrium.
0 1
−10 60
−100 3 × 1017 For all intents and purposes, the forward reaction proceeds to completion: only products are present at equilibrium.
−500 4 × 1087
A typical challenge in industrial processes is a reaction that has a large negative value of ΔG° and hence a large value of K but that is too slow to be practically useful. In such cases, mixing the reactants results in only a physical mixture, not a chemical reaction. An example is the reaction of carbon tetrachloride with water to produce carbon dioxide and hydrochloric acid, for which ΔG° is −232 kJ/mol:
$CCl_4(l)+2H_2O(l) \rightleftharpoons CO_2(g)+2HCl(g)\tag{18.42}$
The value of K for this reaction is 5 × 1040 at 25°C, yet when $CCl_4$ and water are shaken vigorously at 25°C, nothing happens: the two immiscible liquids form separate layers, with the denser $CCl_4$ on the bottom. In comparison, the analogous reaction of SiCl4 with water to give SiO2 and HCl, which has a similarly large equilibrium constant, occurs almost explosively. Although the two reactions have comparable thermodynamics, they have very different kinetics!
There are also many reactions for which ΔG° << 0 but that do not occur as written because another possible reaction occurs more rapidly. For example, consider the reaction of lead sulfide with hydrogen peroxide. One possible reaction is as follows:
$PbS(s) +4H_2O_2(l) \rightleftharpoons PbO2(s)+SO_2(g)+4H_2O(l)\tag{18.43}$
for which ΔG° is −886 kJ/mol and K is 10161. Yet when lead sulfide is mixed with hydrogen peroxide, the ensuing vigorous reaction does not produce $PbO_2$ and $SO_2$. Instead, the reaction that actually occurs is as follows:
$PbS(s) +4H_2O_2(l) \rightleftharpoons PbSO_4(s)+4H_2O(l)$
This reaction has a ΔG° value of −1181 kJ/mol, within the same order of magnitude as the reaction in Equation 18.43, but it occurs much more rapidly.
Now consider reactions with ΔG° > 0. Thermodynamically, such reactions do not occur spontaneously under standard conditions. Nonetheless, these reactions can be made to occur under nonstandard conditions. An example is the reduction of chromium(III) chloride by hydrogen gas:
$\mathrm{CrCl_3(s)}+\dfrac{1}{2}\mathrm{H_2(g)}\rightleftharpoons \mathrm{CrCl_2(s)}+\mathrm{HCl(g)}\tag{18.45}$
At 25°C, ΔG° = 35 kJ/mol and Kp = 7 × 10−7. However, at 730°C, ΔG° = −52 kJ/mol and Kp = 5 × 102; at this elevated temperature, the reaction is a convenient way of preparing chromium(II) chloride in the laboratory. Moreover, removing HCl gas from the system drives the reaction to completion, as predicted by Le Chatelier’s principle. Although the reaction is not thermodynamically spontaneous under standard conditions, it becomes spontaneous under nonstandard conditions.
There are also cases in which a compound whose formation appears to be thermodynamically prohibited can be prepared using a different reaction. The reaction for the preparation of chlorine monoxide from its component elements, for example, is as follows:
$\dfrac{1}{2}\mathrm{O_2(g)}+\mathrm{Cl_2(g)}\rightleftharpoons\mathrm{Cl_2O(g)} \tag{18.46}$
for which ΔGf is 97.9 kJ/mol. The large positive value of ΔGf for this reaction indicates that mixtures of chlorine and oxygen do not react to any extent to form Cl2O. Nonetheless, Cl2O is easily prepared using the reaction
$\mathrm{HgO(s)}+\mathrm{2Cl_2(g)}\rightleftharpoons\mathrm{Cl_2O(g)}+\mathrm{HgCl_2(s)}\tag{18.47}$
which has a ΔG° of −22.2 kJ/mol and a Kp of approximately 1 × 104.
Finally, the ΔG° values for some reactions are so positive that the only way to make them proceed in the desired direction is to supply external energy, often in the form of electricity. Consider, for example, the formation of metallic lithium from molten lithium chloride:
$\mathrm{LiCl(l)}\rightleftharpoons\mathrm{Li(l)}+\dfrac{1}{2}\mathrm{Cl_2(g)} \tag{18.48}$
Even at 1000°C, ΔG is very positive (324 kJ/mol), and there is no obvious way to obtain lithium metal using a different reaction. Hence in the industrial preparation of metallic lithium, electrical energy is used to drive the reaction to the right.
Note
A reaction that does not occur under standard conditions can be made to occur under nonstandard conditions, such as by driving the reaction to completion using Le Chatelier’s principle or by providing external energy.
Often reactions that are not thermodynamically spontaneous under standard conditions can be made to occur spontaneously if coupled, or connected, in some way to another reaction for which ΔG° << 0. Because the overall value of ΔG° for a series of reactions is the sum of the ΔG° values for the individual reactions, virtually any unfavorable reaction can be made to occur by chemically coupling it to a sufficiently favorable reaction or reactions. In the preparation of chlorine monoxide from mercuric oxide and chlorine (Equation 18.47), we have already encountered one example of this phenomenon of coupled reactions, although we did not describe it as such at the time. We can see how the chemical coupling works if we write Equation 18.47 as the sum of three separate reactions:
\begin{align} \frac{1}{2}\mathrm{O_2(g)}+\mathrm{Cl_2(g)}&\rightleftharpoons\mathrm{Cl_2O(g)} &\quad \Delta G^\circ &=97.9\textrm{ kJ/mol }\tag{1} \ \mathrm{HgO(s)}&\rightleftharpoons\mathrm{Hg(l)}+\frac{1}{2}\mathrm{O_2(g)} & \Delta G^\circ &=58.5\textrm{ kJ/mol }\tag{2}\ \mathrm{Hg(l)}+\mathrm{Cl_2(g)}&\rightleftharpoons\mathrm{HgCl_2(s)} &\quad \Delta G^\circ &=-178.6\textrm{ kJ/mol }\tag{3} \ \mathrm{HgO(s)}+\mathrm{2Cl_2(g)}&\rightleftharpoons\mathrm{Cl_2O(g)}+\mathrm{HgCl_2(s)} &\quad \Delta G^\circ_{\textrm{rxn}} &=-22.2\textrm{ kJ/mol } \ \end{align}
Comparing the ΔG° values for the three reactions shows that reaction 3 is so energetically favorable that it more than compensates for the other two energetically unfavorable reactions. Hence the overall reaction is indeed thermodynamically spontaneous as written.
Note
By coupling reactions, a reaction that is thermodynamically nonspontaneous can be made spontaneous.
Example 15
Bronze Age metallurgists were accomplished practical chemists who unknowingly used coupled reactions to obtain metals from their ores. Realizing that different ores of the same metal required different treatments, they heated copper oxide ore in the presence of charcoal (carbon) to obtain copper metal, whereas they pumped air into the reaction system if the ore was copper sulfide. Assume that a particular copper ore consists of pure cuprous oxide (Cu2O). Using the ΔGf values given for each, calculate
1. ΔG° and Kp for the decomposition of $Cu_2O$ to metallic copper and oxygen gas [ΔGf(Cu2O)=−146.0 kJ/mol].
2. ΔG° and Kp for the reaction of $Cu_2O$ with carbon to produce metallic copper and carbon monoxide [ΔGf (CO)=−137.2 kJ/mol].
Given: reactants and products, ΔGf values for Cu2O and CO, and temperature
Asked for: ΔG° and Kp for the formation of metallic copper from Cu2O in the absence and presence of carbon
Strategy:
1. Write the balanced equilibrium equation for each reaction. Using the “products minus reactants” rule, calculate ΔG° for the reaction.
2. Substitute appropriate values into Equation 18.36 to obtain Kp.
Solution
A The chemical equation for the decomposition of cuprous oxide is as follows:
$\mathrm{Cu_2O(s)}\rightleftharpoons\mathrm{2Cu(s)}+\frac{1}{2}\mathrm{O_2(g)}$
The substances on the right side of this equation are pure elements in their standard states, so their ΔGf values are zero. ΔG° for the reaction is therefore
\begin{align}\Delta G^\circ &=\left[2\Delta G^\circ_\textrm f(\mathrm{Cu})+\frac{1}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right]-\Delta G^\circ_\textrm f(\mathrm{Cu_2O})\ &=\left[(\textrm{2 mol})(\textrm{0 kJ/mol})+\left ( \frac{1}{2}\textrm{ mol} \right )(\textrm{0 kJ/mol})\right]-[(\textrm{1 mol})(-\textrm{146.0 kJ/mol})]\ &= \textrm{146.0 kJ}\end{align}
B Rearranging and substituting the appropriate values into Equation 18.36,
\begin{align}\ln K_\textrm p &=-\dfrac{\Delta G^\circ}{RT}=-\dfrac{(\textrm{146.0 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298.15 K})}=-58.90\ K_\textrm p &=2.6\times10^{-26}\end{align}
This is a very small number, indicating that Cu2O does not spontaneously decompose to a significant extent at room temperature.
A The O2 produced in the decomposition of Cu2O can react with carbon to form CO:
$\frac{1}{2}\mathrm{O_2(g)}+\mathrm{C(s)}\rightleftharpoons\mathrm{CO(g)}$
Because ΔG° for this reaction is equal to ΔGf for CO (−137.2 kJ/mol), it is energetically more feasible to produce metallic copper from cuprous oxide by coupling the two reactions:
\begin{align} \mathrm{Cu_2O(s)}&\rightleftharpoons\mathrm{2Cu(s)}+\frac{1}{2}\mathrm{O_2(g)} & \Delta G^\circ &=146.0\textrm{ kJ/mol }\ \frac{1}{2}\mathrm{O_2(g)}+\mathrm{C(s)}&\rightleftharpoons\mathrm{CO(g)} &\quad \Delta G^\circ &=-137.2\textrm{ kJ/mol } \ \mathrm{Cu_2O(s)}+\mathrm{C(s)}&\rightleftharpoons\mathrm{2Cu(s)}+\mathrm{CO(g)} &\quad \Delta G^\circ_{\textrm{f}} &=8.8\textrm{ kJ/mol } \end{align}
B We can find the corresponding value of Kp:
\begin{align}\ln K_\textrm p &=-\dfrac{\Delta G^\circ}{RT}=-\dfrac{(\textrm{8.8 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298.15 K})}=-3.6\ K_\textrm p &=0.03\end{align}
Although this value is still less than 1, indicating that reactants are favored over products at room temperature, it is about 24 orders of magnitude greater than Kp for the production of copper metal in the absence of carbon. Because both ΔH° and ΔS° are positive for this reaction, it becomes thermodynamically feasible at slightly elevated temperatures (greater than about 80°C). At temperatures of a few hundred degrees Celsius, the reaction occurs spontaneously, proceeding smoothly and rapidly to the right as written and producing metallic copper and carbon monoxide from cuprous oxide and carbon.
Exercise 1
Use the ΔGf values given to calculate ΔG° and Kp for each reaction.
1. the decomposition of cuprous sulfide to copper metal and elemental sulfur [ΔGf (Cu2S)=−86.2 kJ/mol]
2. the reaction of cuprous sulfide with oxygen gas to produce sulfur dioxide and copper metal [ΔGf [SO2(g)] = −300.1 kJ/mol]
Answer:
1. ΔG° = 86.2 kJ/mol; Kp = 7.90 × 10−16
2. ΔG° = −213.9 kJ/mol; Kp = 2.99 × 1037
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Learning Objectives
• To understand the importance of thermodynamics in biochemical systems.
In a thermodynamic sense, a living cell can be viewed as a low-entropy system that is not in equilibrium with its surroundings and is capable of replicating itself. A constant input of energy is needed to maintain the cell’s highly organized structure, its wide array of precisely folded biomolecules, and its intricate system of thousands of chemical reactions. A cell also needs energy to synthesize complex molecules from simple precursors (e.g., to make proteins from amino acids), create and maintain differences in the concentrations of various substances inside and outside the cell, and do mechanical work (e.g., muscle contraction). In this section, we examine the nature of the energy flow between a cell and its environment as well as some of the chemical strategies cells use to extract energy from their surroundings and store that energy.
Energy Flow between a Cell and Its Surroundings
One implication of the first and second laws of thermodynamics is that any closed system must eventually reach equilibrium. With no external input, a clock will run down, a battery will lose its charge, and a mixture of an aqueous acid and an aqueous base will achieve a uniform intermediate pH value. In contrast, a cell is an open system that can exchange matter with its surroundings as well as absorb energy from its environment in the form of heat or light. Cells use the energy obtained in these ways to maintain the nonequilibrium state that is essential for life.
Because cells are open systems, they cannot be described using the concepts of classical thermodynamics that we have discussed in this chapter, which have focused on reversible processes occurring in closed chemical systems that can exchange energy—but not matter—with their surroundings. Consequently, a relatively new subdiscipline called nonequilibrium thermodynamics has been developed to quantitatively describe open systems such as living cells.
Because a cell cannot violate the second law of thermodynamics, the only way it can maintain a low-entropy, nonequilibrium state characterized by a high degree of structural organization is to increase the entropy of its surroundings. A cell releases some of the energy that it obtains from its environment as heat that is transferred to its surroundings, thereby resulting in an increase in Ssurr (Figure 18.8.1). As long as ΔSsurr is positive and greater than ΔSsys, the entropy of the universe increases, so the second law of thermodynamics is not violated. Releasing heat to the surroundings is necessary but not sufficient for life: the release of energy must be coupled to processes that increase the degree of order within a cell. For example, a wood fire releases heat to its surroundings, but unless energy from the burning wood is also used to do work, there is no increase in order of any portion of the universe.
Note the Pattern
Any organism in equilibrium with its environment is dead.
Extracting Energy from the Environment
Although organisms employ a wide range of specific strategies to obtain the energy they need to live and reproduce, they can generally be divided into two categories: organisms are either phototrophs (from the Greek photos, meaning “light,” and trophos, meaning “feeder”), whose energy source is sunlight, or chemotrophs, whose energy source is chemical compounds, usually obtained by consuming or breaking down other organisms. Phototrophs, such as plants, algae, and photosynthetic bacteria, use the radiant energy of the sun directly, converting water and carbon dioxide to energy-rich organic compounds, whereas chemotrophs, such as animals, fungi, and many nonphotosynthetic bacteria, obtain energy-rich organic compounds from their environment. Regardless of the nature of their energy and carbon sources, all organisms use oxidation–reduction, or redox, reactions to drive the synthesis of complex biomolecules. Organisms that can use only O2 as the oxidant (a group that includes most animals) are aerobic organisms that cannot survive in the absence of O2. Many organisms that use other oxidants (such as SO42−, NO3, or CO32−) or oxidized organic compounds can live only in the absence of O2, which is a deadly poison for them; such species are called anaerobic organisms.
The fundamental reaction by which all green plants and algae obtain energy from sunlight is photosynthesisThe fundamental reaction by which all green plants and algae obtain energy from sunlight in which CO2 is photochemically reduced to a carbon compound such as glucose. Oxygen in water is concurrently oxidized to O2, the photochemical reduction of CO2 to a carbon compound such as glucose. Concurrently, oxygen in water is oxidized to O2 (recall that hν is energy from light):
$6CO_{2} + 6H_{2}O \xrightarrow[photosynthesis]{h\nu} C_{6}H_{12}O_{6} +6O_{2} \tag{18.8.1}$
This reaction is not a spontaneous process as written, so energy from sunlight is used to drive the reaction. Photosynthesis is critical to life on Earth; it produces all the oxygen in our atmosphere.
In many ways, chemotrophs are more diverse than phototrophs because the nature of both the reductant (the nutrient) and the oxidant can vary. The most familiar chemotrophic strategy uses compounds such as glucose as the reductant and molecular oxygen as the oxidant in a process called respirationA process by which chemotrophs obtain energy from their environment; the overall reaction of respiration is the reverse of photosynthesis. Respiration is the combustion of a carbon compound such as glucose to CO2 and water. The overall reaction of respiration is the reverse of photosynthesis:
$C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O \tag{18.8.2}$
An alternative strategy uses fermentationA process used by some chemotrophs to obtain energy from their environment; a chemical reaction in which both the oxidant and the reductant are organic compounds. reactions, in which an organic compound is simultaneously oxidized and reduced. Common examples are alcohol fermentation, used in making wine, beer, and bread, and lactic acid fermentation, used in making yogurt:
$C_{6}H_{12}O_{6} \rightarrow \underset{alcohol fermentation}{2CO_{2} + 2CH_{3}CH_{2}OH} \tag{18.8.3}$
$C_{6}H_{12}O_{6} \rightarrow \underset{lactic acid fermentation}{ 2CH_{3}CH\left ( OH \right )CO_{2}H} \tag{18.8.4}$
In these reactions, some of the carbon atoms of glucose are oxidized, while others are reduced. Recall that a reaction in which a single species is both oxidized and reduced is called a disproportionation reaction.
The Role of NADH and ATP in Metabolism
Regardless of the identity of the substances from which an organism obtains energy, the energy must be released in very small increments if it is to be useful to the cell. Otherwise, the temperature of the cell would rise to lethal levels. Cells store part of the energy that is released as ATP (adenosine triphosphate), a compound that is the universal energy currency of all living organisms (Figure 18.8.2).
Most organisms use several intermediate species to shuttle electrons between the terminal reductant (such as glucose) and the terminal oxidant (such as O2). In virtually all cases, an intermediate species oxidizes the energy-rich reduced compound, and the now-reduced intermediate then migrates to another site where it is oxidized. The most important of these electron-carrying intermediates is NAD+ (nicotinamide adenine dinucleotide; Figure 18.19), whose reduced form, formally containing H, is NADH (reduced nicotinamide adenine dinucleotide). The reduction of NAD+ to NADH can be written as follows:
$NAD^{+} + H^{+} \rightarrow NADH \tag{18.8.5}$
Most organisms use NAD+ to oxidize energy-rich nutrients such as glucose to CO2 and water; NADH is then oxidized to NAD+ using an oxidant such as O2. During oxidation, a fraction of the energy obtained from the oxidation of the nutrient is stored as ATP. The phosphoric acid anhydride bonds in ATP can then be hydrolyzed by water, releasing energy and forming ADP (adenosine diphosphate). It is through this sequence of reactions that energy from the oxidation of nutrients is made available to cells. Thus ATP has a central role in metabolism: it is synthesized by the oxidation of nutrients, and its energy is then used by cells to drive synthetic reactions and perform work (Figure 18.8.4).
Under standard conditions in biochemical reactions, all reactants are present in aqueous concentrations of 1 M at a pressure of 1 atm. For H+, this corresponds to a pH of zero, but very little biochemistry occurs at pH = 0. For biochemical reactions, chemists have therefore defined a new standard state in which the H+ concentration is 1 × 10−7 M (pH 7.0), and all other reactants and products are present in their usual standard-state conditions (1 M or 1 atm). The free-energy change and corresponding equilibrium constant for a reaction under these new standard conditions are denoted by the addition of a prime sign (′) to the conventional symbol: ΔG°′ and K′. If protons do not participate in a biological reaction, then ΔG°′ = ΔG°. Otherwise, the relationship between ΔG°′ and ΔG° is as follows:
${\Delta G^{o}}'= \Delta G^{o} +RTln\left ( 10^{-7} \right )^{n} \tag{18.8.6}$
where ΔG°′ and ΔG° are in kilojoules per mole and n is the number of protons produced in the reaction. At 298 K, this simplifies to
${\Delta G^{o}}'= \Delta G^{o} -39.96n \tag{18.8.7}$
Thus any reaction that involves the release of protons is thermodynamically more favorable at pH 7 than at pH 0.
The chemical equation that corresponds to the hydrolysis of ATP to ADP and phosphate is as follows:
$ATP^{4-} + H_{2}O \rightleftharpoons \rightarrow ATP^{3-} + HPO_{4}^{2-} + H^{+} \tag{18.8.8}$
This reaction has a ΔG°′ of −34.54 kJ/mol, but under typical physiological (or biochemical) conditions, the actual value of ΔG′ for the hydrolysis of ATP is about −50 kJ/mol. Organisms use this energy to drive reactions that are energetically uphill, thereby coupling the reactions to the hydrolysis of ATP. One example is found in the biochemical pathway of glycolysis, in which the 6-carbon sugar glucose (C6H12O6) is split into two 3-carbon fragments that are then used as the fuel for the cell. Initially, a phosphate group is added to glucose to form a phosphate ester, glucose-6-phosphate (abbreviated glucose-6-P), in a reaction analogous to that of an alcohol and phosphoric acid:
$glucose\left ( aq \right ) + HPO_{4}^{2-}\left ( aq \right ) \rightleftharpoons glucose-6-P^{2-}\left ( aq \right ) + H_{2}O\left ( l \right ) \tag{18.8.9}$
$ROH\left ( aq \right ) + HOPO_{3}^{2-}\left ( aq \right ) \rightleftharpoons ROPO_{3}^{2-}\left ( aq \right ) + H_{2}O\left ( l \right )$
Due to its electrical charge, the phosphate ester is unable to escape from the cell by diffusing back through the membrane surrounding the cell, ensuring that it remains available for further reactions. For the reaction in Equation 18.57, ΔG° is 17.8 kJ/mol and K is 7.6 × 10−4, indicating that the equilibrium lies far to the left. To force this reaction to occur as written, it is coupled to a thermodynamically favorable reaction, the hydrolysis of ATP to ADP:
$\begin{matrix} glucose + \cancel{HPO_{4}^{2-}} & \rightleftharpoons glucose-6-P^{2-} + \cancel{H_{2}O} & {\Delta G^{o}}' & =17.8\; kJ/mol & K_{1}\ & & & & \ ATP^{4-} + \cancel{H_{2}O} & \rightleftharpoons ATP^{3-} + \cancel{HPO_{4}^{2-}} + H^{+} & {\Delta G^{o}}' & =-34.54 \; kJ/mol & K_{2} \ & & & & \ glucose + ATP^{4-} & \rightleftharpoons glucose-6-P^{2-} + ATP^{3-} + H^{+} & {\Delta G^{o}}' & =-16.7 \; kJ/mol & K_{3} \end{matrix}$
Thus the formation of glucose-6-phosphate is thermodynamically spontaneous if ATP is used as the source of phosphate.
Note the Pattern
Organisms use energy from the hydrolysis of ATP to drive reactions that are thermodynamically nonspontaneous.
The formation of glucose-6-phosphate is only one of many examples of how cells use ATP to drive an otherwise nonspontaneous biochemical reaction. Under nonstandard physiological conditions, each ATP hydrolyzed actually results in approximately a 108 increase in the magnitude of the equilibrium constant, compared with the equilibrium constant of the reaction in the absence of ATP. Thus a reaction in which two ATP molecules are converted to ADP increases K by about 1016, three ATP molecules by 1024, and so forth. Virtually any energetically unfavorable reaction or sequence of reactions can be made to occur spontaneously by coupling it to the hydrolysis of a sufficiently large number of ATP molecules.
Energy Storage in Cells
Although all organisms use ATP as the immediate free-energy source in biochemical reactions, ATP is not an efficient form in which to store energy on a long-term basis. If the caloric intake of an average resting adult human were stored as ATP, two-thirds of the body weight would have to consist of ATP. Instead, a typical 70 kg adult human has a total of only about 50 g of both ATP and ADP, and, far from being used for long-term storage, each molecule of ATP is turned over about 860 times per day. The entire ATP supply would be exhausted in less than 2 minutes if it were not continuously regenerated.
How does the body store energy for the eventual production of ATP? Three primary means are as sugars, proteins, and fats. The combustion of sugars and proteins yields about 17 kJ of energy per gram, whereas the combustion of fats yields more than twice as much energy per gram, about 39 kJ/g. Moreover, sugars and proteins are hydrophilic and contain about 2 g of water per gram of fuel, even in very concentrated form. In contrast, fats are hydrophobic and can be stored in essentially anhydrous form. As a result, organisms can store about six times more energy per gram as fats than in any other form. A typical 70 kg adult human has about 170 kJ of energy in the form of glucose circulating in the blood, about 2600 kJ of energy stored in the muscles and liver as glycogen (a polymeric form of glucose), about 100,000 kJ stored in the form of protein (primarily muscle tissue), and almost 500,000 kJ in the form of fats (Figure 18.21). Thus fats constitute by far the greatest energy reserve, while accounting for only about 12 kg of the 70 kg body mass. To store this amount of energy in the form of sugars would require a total body mass of about 144 kg, more than half of which would be sugar.
Example 18.8.1
Glucose is one form in which the body stores energy.
1. Calculate ΔG°′ for the respiration of glucose to CO2 and H2O using these values of ΔGf°: −910.4 kJ/mol for glucose, −394.4 kJ/mol for CO2(g), and −237.1 kJ/mol for H2O(l).
2. Assuming 50% efficiency in the conversion of the released energy to ATP, how many molecules of ATP can be synthesized by the combustion of one molecule of glucose? At 298.15 K, ΔG°′ for the hydrolysis of ATP is −34.54 kJ/mol.
Given: balanced chemical equation (Equation 18.50), values of ΔGf°, conversion efficiency, and ΔG°′ for hydrolysis of ATP
Asked for: ΔG°′ for the combustion reaction and the number of molecules of ATP that can be synthesized
Strategy:
A Using the “products minus reactants” rule, calculate ΔGrxn° for the respiration reaction.
B Multiply the calculated value of ΔGrxn° by the efficiency to obtain the number of kilojoules available for ATP synthesis. Then divide this value by ΔG°′ for the hydrolysis of ATP to find the maximum number of ATP molecules that can be synthesized.
Solution:
1. A Protons are not released or consumed in the reaction, so ΔG°′ = ΔG°. We begin by using the balanced chemical equation in Equation 18.50: C6H12O6 + 6O2 → 6CO2 + 6H2O
From the given values of ΔGf° (remember that ΔGf° is zero for an element such as O2 in its standard state), we can calculate ΔGrxn°
$\Delta G_{rxn}^{o}=\sum m \Delta G_{f}^{o}\left ( products \right )-\sum n \Delta G_{f}^{o}\left ( reactants- \right )$
$=\left [ 6\left ( -394.4\; kJ/mol \right )+ 6\left ( -237.1\; kJ/mol \right )\right ]-\left [ \left ( 910.4 \; kJ/mol \right ) + 0 \; kJ/mol\right ]$
$=-2879 \; kJ/mol\; of \; glucose$
2. B If we assume that only 50% of the available energy is used, then about 1440 kJ/mol of glucose is available for ATP synthesis. The value of ΔG°′ for the hydrolysis of ATP under biochemical conditions is −34.54 kJ/mol, so in principle an organism could synthesize $\dfrac{1440\; \cancel{kJ/mol\; glucose}}{34.54 \cancel{kJ/mol\; ATP}}=41.7 \approx 42\; ATP/glucose$
Most aerobic organisms actually synthesize about 32 molecules of ATP per molecule of glucose, for an efficiency of about 45%.
Exercise
Some bacteria synthesize methane using the following redox reaction:
CO2(g) + 4H2(g) → CH4(g) + 2H2O(g)
1. Calculate ΔG° for this reaction using values of ΔGf° in Table T1.
2. Calculate how many ATP molecules could be synthesized per mol of CO2 reduced if the efficiency of the process were 100%.
Answer
1. −86.6 kJ/mol CO2
2. 2.5 ATP/mol CO2
Summary
A living cell is a system that is not in equilibrium with its surroundings; it requires a constant input of energy to maintain its nonequilibrium state. Cells maintain a low-entropy state by increasing the entropy of their surroundings. Aerobic organisms cannot survive in the absence of O2, whereas anaerobic organisms can live only in the absence of O2. Green plants and algae are phototrophs, which extract energy from the environment through a process called photosynthesis, the photochemical reduction of CO2 to a reduced carbon compound. Other species, called chemotrophs, extract energy from chemical compounds. One of the main processes chemotrophs use to obtain energy is respiration, which is the reverse of photosynthesis. Alternatively, some chemotrophs obtain energy by fermentation, in which an organic compound is both the oxidant and the reductant. Intermediates used by organisms to shuttle electrons between the reductant and the oxidant include NAD+ and NADH. Energy from the oxidation of nutrients is made available to cells through the synthesis of ATP, the energy currency of the cell. Its energy is used by cells to synthesize substances through coupled reactions and to perform work. The body stores energy as sugars, protein, or fats before using it to produce ATP.
Conceptual Problem
1. The tricarboxylic acid (TCA) cycle in aerobic organisms is one of four pathways responsible for the stepwise oxidation of organic intermediates. The final reaction in the TCA cycle has ΔG° = 29.7 kJ/mol, so it should not occur spontaneously. Suggest an explanation for why this reaction proceeds in the forward direction in living cells.
Answer
1. It is coupled to another reaction that is spontaneous, which drives this reaction forward (Le Chatelier’s principle).
Contributors
• Anonymous
Modified by Joshua B. Halpern
Thumbnail from Wikimedia | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/18%3A_Chemical_Thermodynamics/18.08%3A_Thermodynamics_and_Life.txt |
In oxidation–reduction (redox) reactions, electrons are transferred from one species (the reductant) to another (the oxidant). This transfer of electrons provides a means for converting chemical energy to electrical energy or vice versa. The study of the relationship between electricity and chemical reactions is called electrochemistry (the study of the relationship between electricity and chemical reactions., an area of chemistry we introduced in Chapter 8 and Chapter 9). In this chapter, we describe electrochemical reactions in more depth and explore some of their applications.
In the first three sections, we review redox reactions; describe how they can be used to generate an electrical potential, or voltage; and discuss factors that affect the magnitude of the potential. We then explore the relationships among the electrical potential, the change in free energy, and the equilibrium constant for a redox reaction, which are all measures of the thermodynamic driving force for a reaction. Finally, we examine two kinds of applications of electrochemical principles: (1) those in which a spontaneous reaction is used to provide electricity and (2) those in which electrical energy is used to drive a thermodynamically nonspontaneous reaction. By the end of this chapter, you will understand why different kinds of batteries are used in cars, flashlights, cameras, and portable computers; how rechargeable batteries operate; and why corrosion occurs and how to slow—if not prevent—it. You will also discover how metal objects can be plated with silver or chromium for protection; how silver polish removes tarnish; and how to calculate the amount of electricity needed to produce aluminum, chlorine, copper, and sodium on an industrial scale.
A view from the top of the Statue of Liberty, showing the green patina coating the statue. The patina is formed by corrosion of the copper skin of the statue, which forms a thin layer of an insoluble compound that contains copper(II), sulfate, and hydroxide ions.
Contributors
• Anonymous
Modified by Joshua B. Halpern
19.02: Describing Electrochemical Cells
Learning Objectives
• To distinguish between galvanic and electrolytic cells.
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. As we described in Chapter 7, a redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant (a substance that is capable of donating electrons and in the process is oxidized) is the substance that loses electrons and is oxidized in the process; the oxidant (A substance that is capable of accepting electrons and in the process is reduced) is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions (reactions that represent either the oxidation half or the reduction half of an oxidation–reduction (redox) reaction), one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$Zn\left ( s \right ) + Br_{2}\left ( l \right ) \rightarrow Zn^{2+}\left ( aq \right ) + 2Br^{-}\left ( aq \right ) \tag{19.1.1}$
The half-reactions are as follows:
$reduction\;half\;reaction: Br_{2}\left ( l \right )+2e^{-} \rightarrow 2Br^{-}\left ( aq \right ) \tag{19.1.2}$
$oxidation\;half\;reaction: Zn\left ( s \right ) \rightarrow Zn^{2+}\left ( aq \right ) +2e^{-} \tag{19.1.3}$
Each half-reaction is written to show what is actually occurring in the system; Zn is the reductant in this reaction (it loses electrons), and Br2 is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation 19.1). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
Note the Pattern
In any redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell (an apparatus that generates electricity from a spontaneous oxidation–reduction (redox) reaction or, conversely, uses electricity to drive a nonspontaneous redox reaction).
There are two types of electrochemical cells: galvanic cells and electrolytic cells. A galvanic (voltaic) cell (an electrochemical cell that uses the energy released during a spontaneous oxidation–reduction (redox) reaction (ΔGo) to generate electricity). Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cellAn electrochemical cell that consumes electrical energy from an external source to drive a nonspontaneous (ΔGo) oxidation–reduction (redox) reaction. consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two electrodesA solid metal connected by an electrolyte and an external circuit that provides an electrical connection between systems in an electrochemical cell (galvanic or electrolytic)., which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure 19.1.1). The oxidation half-reaction occurs at one electrode (the anodeOne of two electrodes in an electrochemical cell, it is the site of the oxidation half-reaction.), and the reduction half-reaction occurs at the other (the cathodeOne of two electrodes in an electrochemical cell, it is the site of the reduction half-reaction.). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells. We discuss electrolytic cells in Section 19.7.
Galvanic (Voltaic) Cells
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
$Zn\left ( s \right ) + Cu^{2+}\left (aq \right ) \rightarrow Zn^{2+} \left (aq \right ) + Cu \left ( s \right ) \tag{19.1.4}$
We can cause this reaction to occur by inserting zinc powder into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc dissolves, and a mass of metallic copper forms (Figure 19.1.2). These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
Figure 19.1.2 The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Single Compartment When zinc powder is inserted into a beaker that contains an aqueous solution of copper(II) sulfate, a spontaneous redox reaction occurs: the zinc electrode dissolves to give Zn2+(aq) ions, while Cu2+(aq) ions are simultaneously reduced to metallic copper. The reaction occurs so rapidly that the copper is deposited as very fine particles that appear black, rather than the usual reddish color of copper. The reaction demonstration starts at 3:00 in the video, the chemistry of the reaction is discussed in the first three minutes and at the end
This same reaction can be carried out using the galvanic cell illustrated in part (a) in Figure 19.1.3. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of Cu2+ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of Zn2+ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge (a U-shaped tube inserted into both solutions of a galvanic cell that contains a concentrated liquid or gelled electrolyte and completes the circuit between the anode and the cathode), a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are Na+ or K+ and NO3 or SO42−, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to Zn2+ ions at the zinc electrode (the anode), and Cu2+ ions are reduced to Cu metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of Zn2+ ions in the Zn2+ solution increases; simultaneously, the copper strip gains mass, and the concentration of Cu2+ ions in the Cu2+ solution decreases (part (b) in Figure 19.1.3). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the Zn2+ solution would increase as the zinc metal dissolves, and the total positive charge in the Cu2+ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the Zn2+ solution and a flow of cations into the Cu2+ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A galvanic cell. This galvanic cell illustrates the use of a salt bridge to connect two solutions generating a voltage.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential (Ecell)Related to the energy needed to move a charged particle in an electric field, it is the difference in electrical potential beween two half-reactions. of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (part (a) in Figure 19.1.3 ). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example 1.
Note the Pattern
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
Example 19.1.1
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
3Sn(s) + 2NO3(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
A Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
B From the direction of electron flow, assign each electrode as either positive or negative.
Solution:
1. A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows: reduction: NO3(aq) + 4H+(aq) + 3e → NO(g) + 2H2O(l) oxidation: Sn(s) → Sn2+(aq) + 2e
Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
2. Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
3. B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.
Exercise
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of MnO4 in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of Sn2+ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
2MnO4(aq) + 5Sn2+(aq) + 16H+(aq) → 2Mn2+(aq) + 5Sn4+(aq) + 8H2O(l)
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.
Answer
1. MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l); Sn2+(aq) → Sn4+(aq) + 2e
2. The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
3. The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.
Constructing a Cell Diagram
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the Zn/Cu cell shown in part (a) in Figure 19.1.3 is written as follows:
Figure 19.1.4 A cell diagram includes solution concentrations when they are provided.
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
$Pt\left (s \right )\mid H_{2}\left (g \right )\mid HCl\left (aq \right )\mid AgCl\left (s \right )\mid Ag\left (s \right ) \tag{19.1.5}$
This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
$cathode\;reaction: AgCl\left (s \right ) + e^{-} \rightarrow Ag \left (s \right ) + Cl^{-}\left (aq \right ) \tag{19.1.6}$
$anode\;reaction: \dfrac{1}{2}H_{2}\left (g \right ) \rightarrow H^{+} \left (aq \right ) + e^{-} \tag{19.1.7}$
$overall: AgCl\left (s \right ) +\dfrac{1}{2}H_{2}\left (g \right ) \rightarrow Ag \left (s \right ) + Cl^{-}\left (aq \right ) + H^{+} \left (aq \right ) \tag{19.1.8}$
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Example 19.1.2
Draw a cell diagram for the galvanic cell described in Example 1. The balanced chemical reaction is as follows:
3Sn(s) + 2NO3(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)
Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution:
The anode is the tin strip, and the cathode is the Pt electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus Sn(s)∣Sn2+(aq). We could include H2SO4(aq) with the contents of the anode compartment, but the sulfate ion (as HSO4) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (NO) and the Pt electrode. These are written as HNO3(aq)∣NO(g)∣Pt(s), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
Sn(s)∣Sn2+(aq)∥HNO3(aq)∣NO(g)∣Pt(s)
The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise
Draw a cell diagram for the following reaction, assuming the concentration of Ag+ and Mg2+ are each 1 M:
Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s)
Answer: Mg(s)∣Mg2+(aq, 1 M)∥Ag+(aq, 1 M)∣Ag(s)
Summary
Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential.
Key Takeaway
• A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur.
Conceptual Problems
1. Is 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) an oxidation–reduction reaction? Why or why not?
2. If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?
3. What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?
4. What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?
5. One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K+ and Cl ions do. What would happen if the diffusion rates of the anions and cations differed significantly?
6. It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?
Answer
1. A large difference in cation/anion diffusion rates would increase resistance in the salt bridge and limit electron flow through the circuit.
Numerical Problems
1. Copper(I) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO4 solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of ZnCl2?
2. Consider the following spontaneous redox reaction: NO3(aq) + H+(aq) + SO32−(aq) → SO42−(aq) + HNO2(aq).
1. Write the two half-reactions for this overall reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?
3. Which electrode is negatively charged, and which is positively charged?
3. The reaction Pb(s) + 2VO2+(aq) + 4H+(aq) → Pb2+(aq) + 2V3+(aq) + 2H2O(l) occurs spontaneously.
1. Write the two half-reactions for this redox reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which reaction occurs at the cathode and which occurs at the anode?
3. Which electrode is positively charged, and which is negatively charged?
4. Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)6]3−, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.
5. Sulfate is reduced to HS in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?
6. Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.
1. Pb(s)∣PbSO4(s)∣SO42−(aq)∥Cu2+(aq)∣Cu(s)
2. Hg(l)∣Hg2Cl2(s)∣Cl(aq) ∥ Cd2+(aq)∣Cd(s)
7. For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.
1. Zn(s)∣Zn2+(aq) ∥ H+(aq)∣H2(g), Pt(s)
2. Ag(s)∣AgCl(s)∣Cl(aq) ∥ H+(aq)∣H2(g)∣Pt(s)
3. Pt(s)∣H2(g)∣H+(aq) ∥ Fe2+(aq), Fe3+(aq)∣Pt(s)
8. For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.
1. Ag(s) + Fe3+(aq) → Ag+(aq) + Fe2+(aq)
2. Fe3+(aq) + 1/2H2(g) → Fe2+(aq) + H+(aq)
9. Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
1. 2Cl(aq) + Br2(l) → Cl2(g) + 2Br(aq)
2. 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(l)
3. 2H2O(l) + 2Cl(aq) → H2(g) + Cl2(g) + 2OH(aq)
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
10. Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
1. Co(s) + Fe2+(aq) → Co2+(aq) + Fe(s)
2. O2(g) + 4H+(aq) + 4Fe2+(aq) → 2H2O(l) + 4Fe3+(aq)
3. 6Hg2+(aq) + 2NO3(aq) + 8H+ → 3Hg22+(aq) + 2NO(g) + 4H2O(l)
4. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Answers
1. reduction: SO42−(aq) + 9H+(aq) + 8e → HS(aq) + 4H2O(l) oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3(g) + 30H+(aq) + 24e overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3(g) + 3H+(aq) + 3HS(aq)
1. reduction: 2H+(aq) + 2e → H2(aq); cathode;
oxidation: Zn(s) → Zn2+(aq) + 2e; anode;
overall: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(aq)
2. reduction: AgCl(s) + e → Ag(s) + Cl(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e; anode;
overall: AgCl(s) + H2(g) → 2H+(aq) + Ag(s) + Cl(aq)
3. reduction: Fe3+(aq) + e → Fe2+(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e; anode;
overall: 2Fe3+(aq) + H2(g) → 2H+(aq) + 2Fe2+(aq)
Contributors
• Anonymous
Modified by Joshua B. Halpern
Figure 9.1.2 the reaction between Zn and Cu2+ from YouTube courtesy of UC Berkeley Chemistry and Respect Chemistry
Galvanic Cell demonstration from YouTube courtesy of Angela Stott | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.01%3A_Introduction.txt |
Learning Objectives
• To use redox potentials to predict whether a reaction is spontaneous.
• To balance redox reactions using half-reactions.
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. (For more information on atomic orbitals, see Section 2.5.) Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper (Figure 19.2.1). Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work.
Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure 19.1.3 but instead of copper use a strip of cobalt metal and 1 M Co2+ in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V.
The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potentialThe potential of an electrochemical cell measured under standard conditions (1 M for solutions, 1 atm for gases, and pure solids or pure liquids for other substances) and at a fixed temperature (usually 298 K). (E°cell), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions,Concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for nonideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C.
Note the Pattern
Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system.
Measuring Standard Electrode Potentials
It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured. (This is analogous to measuring absolute enthalpies or free energies. Recall from Chapter 18 that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram:
$Co \left (s \right ) \mid Co^{2+} \left (aq,\;1M \right ) \parallel Cu ^{2+} \left (aq,\;1M \right ) \mid Cu \left (s \right ) \tag{19.2.1}$
This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu2+ in solution at the copper cathode.
All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances. (Standard electrode potentials for various reduction reactions are given in Table P1.) The standard cell potential (E°cell) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum:
$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} \tag{19.2.2}$
In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation 19.2.1, when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell.
Note the Pattern
The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials: E°cell = E°cathodeE°anode.
Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE)The electrode chosen as the reference for all other electrodes, which has been assigned a standard potential of 0 V and consists of a Pt wire in contact with an aqueous solution that contains 1 M H+ in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface. is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H+. The [H+] in solution is in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface (Figure 19.2.2). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
$2H^{+}\left ( aq \right ) + 2e^{-} \rightleftharpoons H_{2}\left ( g \right ) \tag{19.2.3}$
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure 19.2.3 shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn2+ ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn2+, and H+ ions are reduced to H2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows:
$Zn\left (s \right )\mid Zn^{2+}\left (aq \right )\parallel H^{+}\left (aq,\; 1M \right )\mid H_{2}\left (g, 1\; atm \right )\mid Pt\left (s \right ) \tag{19.2.4}$
The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
$cathode: 2H^{+}\left (aq,\; 1M \right ) + 2e^{-} \rightarrow H_{2}\left ( g \right ) \;\;\; E_{cathode}^{o} = 0\;V \tag{19.2.5}$
$anode: Zn\left (s \right )\rightarrow Zn^{2+}\left (aq \right )+2e^{-} \;\;\; E_{anode}^{o} = -0.76\;V \tag{19.2.6}$
$overall: Zn\left (s \right ) + 2H^{+}\left (aq \right )\rightarrow Zn^{2+}\left (aq \right ) + H_{2}\left ( g \right ) \tag{19.2.7}$
$E_{cell}^{o} = E_{cathode}^{o}-E_{anode}^{o} = 0.76$
Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potentialThe potential of a half-reaction measured against the SHE under standard conditions. for that half-reaction.In this example, the standard reduction potential for Zn2+(aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn2+, often called the Zn/Zn2+ redox couple, or the Zn/Zn2+ couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E°anode from E°cathode to obtain E°cell: 0 − (−0.76 V) = 0.76 V.
Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.
Note the Pattern
E° values do not depend on the stoichiometric coefficients for a half-reaction.
To measure the potential of the Cu/Cu2+ couple, we can construct a galvanic cell analogous to the one shown in Figure 19.2.3 but containing a Cu/Cu2+ couple in the sample compartment instead of Zn/Zn2+. When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of E°cell indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn2+ couple. Hence the reactions that occur spontaneously, indicated by a positive E°cell, are the reduction of Cu2+ to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H2 is oxidized to H+ at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu2+/Cu couple on the right:
$Pt\left (s \right ) \mid H_{2}\left (g, 1\; atm \right ) \mid H^{+}\left (aq,\; 1M \right ) \parallel Cu^{2+}\left (aq \right ) \mid Cu\left (s \right ) \tag{19.2.8}$
The half-cell reactions and potentials of the spontaneous reaction are as follows:
$cathode: Cu^{2+}\left (aq \right )+2e^{-} \rightarrow Cu \left (s \right ) \;\;\; E_{cathode}^{o} = 0.34\;V \tag{19.2.9}$
$anode: H_{2}\left ( g \right ) \rightarrow 2H^{+} \left ( aq \right ) + 2e^{-} \;\;\; E_{anode}^{o} = 0\;V \tag{19.2.10}$
$overall: H_{2}\left ( g \right )+ Cu^{2+}\left (aq \right ) \rightarrow Cu\left (s \right ) + 2H^{+}\left (aq \right ) \tag{19.2.11}$
$E_{cell}^{o} = E_{cathode}^{o}-E_{anode}^{o} = 0.34$
Thus the standard electrode potential for the Cu2+/Cu couple is 0.34 V.
Balancing Redox Reactions Using the Half-Reaction Method
In Chapter 8, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other.
We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas:
$Al\left (s \right ) + OH^{-}\left (aq \right ) \rightarrow Al\left (OH \right )_{4}^{-}\left (aq \right ) + H_{2}\left (g \right ) \tag{19.2.12}$
In this reaction, Al(s) is oxidized to Al3+, and H+ in water is reduced to H2 gas, which bubbles through the solution, agitating it and breaking up the clogs.
The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed in Table P1, we find the corresponding half-reactions that describe the reduction of H+ ions in water to H2 and the oxidation of Al to Al3+ in basic solution:
$reduction: \; 2H_{2}O \left (l \right ) + 2e^{-} \rightarrow 2OH^{-}\left (aq \right ) + H_{2}\left (g \right ) \tag{19.2.13}$
$oxidation: \; Al \left (s \right ) + 4OH^{-}\left (aq \right ) \rightarrow Al(OH)_{4}^{-}\left (aq \right ) + 3e^{-} \tag{19.2.14}$
The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution.
In Equation 19.2.13, two H+ ions gain one electron each in the reduction; in Equation 19.2.14, the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation 19.2.13) by 3 and the oxidation half-reaction (Equation 19.2.14) by 2 to give the same number of electrons in both half-reactions:
$reduction: \; 6H_{2}O \left (l \right ) + 6e^{-} \rightarrow 6OH^{-}\left (aq \right ) + 3H_{2}\left (g \right ) \tag{19.2.15}$
$oxidation: \; 2Al \left (s \right ) + 8OH^{-}\left (aq \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right ) + 6e^{-} \tag{19.2.16}$
$6H_{2}O \left (l \right ) + 2Al \left (s \right ) + 8OH^{-}\left (aq \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right )+ 6OH^{-}\left (aq \right ) + 3H_{2}\left (g \right ) \tag{19.2.17}$
Simplifying by canceling substances that appear on both sides of the equation,
$6H_{2}O \left (l \right ) + 2Al \left (s \right ) + 2OH^{-}\left (aq \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right )+ 3H_{2}\left (g \right ) \tag{19.2.18}$
We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced:
$2Al + 8O + 14H = 2Al + 8O + 14H \tag{19.2.19}$
The atoms also balance, so Equation 19.2.18 is a balanced chemical equation for the redox reaction depicted in Equation 19.20.
Note the Pattern
The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction.
We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in Table P1 but instead focus on the atoms whose oxidation states change, as illustrated in the following steps:
Step 1: Write the reduction half-reaction and the oxidation half-reaction.
For the reaction shown in Equation 19.213, hydrogen is reduced from H+ in OH to H2, and aluminum is oxidized from Al0 to Al3+:
$reduction: \; OH^{-}(aq) \rightarrow H_{2}(g) \tag{19.2.20}$
$oxidation: \; Al \left (s \right ) \rightarrow Al(OH)_{4}^{-}\left (aq \right ) \tag{19.2.21}$
Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H2O and balance H atoms by adding H+.
Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction:
$reduction: \; OH^{-}(aq) \rightarrow H_{2}(g)+ H_{2}O\left ( l \right ) \tag{19.2.22}$
$oxidation: \; Al \left (s \right ) + 4H_{2}O\left ( l \right ) \rightarrow Al(OH)_{4}^{-}\left (aq \right ) \tag{19.2.23}$
Balancing H atoms by adding H+, we obtain the following:
$reduction: \; OH^{-}(aq) + 3H^{+}\left (aq \right ) \rightarrow H_{2}(g)+ H_{2}O\left ( l \right )\tag{19.2.24}$
$oxidation: \; Al \left (s \right ) + 4H_{2}O\left ( l \right ) \rightarrow Al(OH)_{4}^{-}\left (aq \right ) + 4H^{+}\left ( aq \right ) \tag{19.2.25}$
We have now balanced the atoms in each half-reaction, but the charges are not balanced.
Step 3: Balance the charges in each half-reaction by adding electrons.
Two electrons are gained in the reduction of H+ ions to H2, and three electrons are lost during the oxidation of Al0 to Al3+:
$reduction: \; OH^{-}(aq) + 3H^{+}\left (aq \right ) + 2e^{-} \rightarrow H_{2}(g)+ H_{2}O\left ( l \right )\tag{19.2.26}$
$oxidation: \; Al \left (s \right ) + 4H_{2}O\left ( l \right ) \rightarrow Al(OH)_{4}^{-}\left (aq \right ) + 4H^{+}\left ( aq \right ) + 3e^{-} \tag{19.2.27}$
Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions.
In this case, we multiply Equation 19.2.26 (the reductive half-reaction) by 3 and Equation 19.2.27 (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions:
$reduction: \; 3OH^{-}(aq) + 9H^{+}\left (aq \right ) + 6e^{-} \rightarrow 3H_{2}(g)+ 3H_{2}O\left ( l \right )\tag{19.2.28}$
$oxidation: \; 2Al \left (s \right ) + 8H_{2}O\left ( l \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right ) + 8H^{+}\left ( aq \right ) + 6e^{-} \tag{19.2.29}$
Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation.
Adding and, in this case, canceling 8H+, 3H2O, and 6e,
$oxidation: \; 2Al \left (s \right ) + 5H_{2}O\left ( l \right ) +3OH^{-}\left (aq \right )+ H^{+}\left ( aq \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right ) + 3H_{2}\left (g \right ) \tag{19.2.30}$
We have three OH and one H+ on the left side. Neutralizing the H+ gives us a total of 5H2O + H2O = 6H2O and leaves 2OH on the left side:
$oxidation: \; 2Al \left (s \right ) + 6H_{2}O\left ( l \right ) + 2OH^{-}\left (aq \right ) \rightarrow 2Al(OH)_{4}^{-}\left (aq \right ) + 3H_{2}\left (g \right ) \tag{19.2.31}$
Step 6: Check to make sure that all atoms and charges are balanced.
Equation 19.2.31 is identical to Equation 19.2.18, obtained using the first method, so the charges and numbers of atoms on each side of the equation balance.
Figure 19.2.4 The Reaction of Dichromate with Iodide The reaction of a yellow solution of sodium dichromate with a colorless solution of sodium iodide produces a deep amber solution that contains a green Cr3+(aq) complex and brown I2(aq) ions.
Example 19.2.1
In acidic solution, the redox reaction of dichromate ion (Cr2O72−) and iodide (I) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green Cr3+(aq) complex and brown I2(aq) ions (Figure 19.2.4.):
$Cr_{2}O_{7}^{2-} \left (aq \right ) + I^{-}\left (aq \right ) \rightarrow Cr^{3+}\left (aq \right ) + 3I_{2}\left (aq \right )$
Balance this equation using half-reactions.
Given: redox reaction and Table P1
Asked for: balanced chemical equation using half-reactions
Strategy:
Follow the steps to balance the redox reaction using the half-reaction method.
Solution:
From the standard electrode potentials listed in Table P1, we find the half-reactions corresponding to the overall reaction:
$reduction: \; Cr_{2}O_{7}^{2-} \left (aq \right ) + 14 H^{+}\left (aq \right ) + 6e^{-} \rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right )$
$oxidation: \; 2I^{-}\left (aq \right ) \rightarrow I_{2}\left (aq \right )+ 2e^{-}$
Balancing the number of electrons by multiplying the oxidation reaction by 3,
$oxidation: \; 6I^{-}\left (aq \right ) \rightarrow 3I_{2}\left (aq \right )+6e^{-}$
oxidation: 6I(aq) → 3I2(aq) + 6e
Adding the two half-reactions and canceling electrons,
$Cr_{2}O_{7}^{2-} \left (aq \right ) + 14 H^{+}\left (aq \right ) + 6I^{-}\left (aq \right ) \rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right ) + 3I_{2}\left (aq \right )$
We must now check to make sure the charges and atoms on each side of the equation balance:
$\begin{matrix} \left ( -2 \right )+14 +\left ( -6 \right ) & = +6\ +6 & = +6\ 2Cr + 7O + 14H + 6I & 2Cr + 7O + 14H + 6I \end{matrix}$
The charges and atoms balance, so our equation is balanced.
We can also use the alternative procedure, which does not require the half-reactions listed in Table P1.
Step 1: Chromium is reduced from Cr6+ in Cr2O72−to Cr3+, and I ions are oxidized to I2. Dividing the reaction into two half-reactions,
$reduction: \; Cr_{2}O_{7}^{2-} \left (aq \right ) \rightarrow Cr^{3+}\left (aq \right )$
$oxidation: \; I^{-}\left (aq \right ) \rightarrow I_{2}\left (aq \right )$
Step 2: Balancing the atoms other than oxygen and hydrogen,
$reduction: \; Cr_{2}O_{7}^{2-} \left (aq \right ) \rightarrow 2Cr^{3+}\left (aq \right )$
$oxidation: \; 2I^{-}\left (aq \right ) \rightarrow I_{2}\left (aq \right )$
We now balance the O atoms by adding H2O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.
Next we balance the H atoms by adding H+ to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction.
$reduction: \; Cr_{2}O_{7}^{2-} \left (aq \right ) \rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right )$
Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr+6 to 2Cr+3) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I2) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge:
$reduction: \; Cr_{2}O_{7}^{2-} \left (aq \right )+ 14 H^{+}\left (aq \right ) + 6e^{-}\rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right )$
Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3:
$oxidation: \; 6I^{-}\left (aq \right ) \rightarrow 3I_{2}\left (aq \right )+6e^{-}$
Step 5: Adding the two half-reactions and canceling substances that appear in both reactions,
$Cr_{2}O_{7}^{2-} \left (aq \right )+ 14 H^{+}\left (aq \right ) + 6I^{-} \rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right ) + 3I_{2}\left (aq \right )$
Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance.
Exercise
Copper is commonly found as the mineral covellite (CuS). The first step in extracting the copper is to dissolve the mineral in nitric acid (HNO3), which oxidizes sulfide to sulfate and reduces nitric acid to NO:
$CuS\left ( s \right ) + HNO_{3}\rightarrow NO\left ( g \right ) + CuSO_{4}$
Balance this equation using the half-reaction method.
Answer: $3CuS\left ( s \right ) +8 HNO_{3} \rightarrow 8NO\left ( g \right ) + 3CuSO_{4} +4H_{2}O\left ( l \right )$
Calculating Standard Cell Potentials
The standard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram:
$Zn\left (s \right )\mid Zn^{2+}\left (aq, 1 M \right )\parallel Cu^{2+}\left (aq, 1 M \right )\mid Cu\left (s \right ) \tag{19.2.32}$
We know the values of E°anode for the reduction of Zn2+ and E°cathode for the reduction of Cu2+, so we can calculate E°cell:
$cathode: \; Cu^{2+}\left (aq, 1 M \right )+2e^{-} \rightarrow Cu\left (s \right )\;\;\; E_{cathode}^{o}=0.34\;V \tag{19.2.33}$
$anode: \; Zn\left (s \right ) \rightarrow Zn^{2+}\left (aq, 1 M \right ) + 2e^{-} \;\;\; E_{anode}^{o}=-0.76\;V \tag{19.2.34}$
$overall \; Zn\left (s \right ) + Cu^{2+}\left (aq, 1 M \right ) \rightarrow Zn^{2+}\left (aq, 1 M \right ) + Cu\left (s \right ) \tag{19.2.35}$
$E_{cell}=E_{cathode}^{o}-E_{anode}=1.10\;V$
This is the same value that is observed experimentally. If the value of E°cell is positive, the reaction will occur spontaneously as written. If the value of E°cell is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction.As we shall see in Section 19.7, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example 4 and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples.
Note the Pattern
A positive E°cell means that the reaction will occur spontaneously as written. A negative E°cell means that the reaction will proceed spontaneously in the opposite direction.
Example 19.2.2
A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl3, and the other contains a piece of nickel immersed in a 1 M solution of NiCl2. The half-reactions that occur when the compartments are connected are as follows:
$cathode: \; Ni^{2+}\left (aq \right ) + 2e^{-} \rightarrow Ni\left (s \right )$
$anode: \; Ga\left (s \right ) \rightarrow Ga^{3+}\left (aq \right ) + 3e^{-}$
If the potential for the oxidation of Ga to Ga3+ is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni2+?
Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions
Asked for: standard electrode potential of reaction occurring at the cathode
Strategy:
A Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction.
B Use Equation 19.2.2 to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions.
Solution:
A We have been given the potential for the oxidation of Ga to Ga3+ under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga3+(aq) + 3e → Ga(s), E°anode = −0.55 V.
B Using the value given for E°cell and the calculated value of E°anode, we can calculate the standard potential for the reduction of Ni2+ to Ni from Equation 19.2.2:
$\begin{matrix} E_{cell} & = &E_{cathode}^{o}-E_{anode}\ 0.27 \; V & = & E_{cathode}^{o}-\left ( -0.55 \; V \right )\ E_{cathode}^{o} & = & -0.28 \; V \end{matrix}$
This is the standard electrode potential for the reaction Ni2+(aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni2+ under standard conditions, we must reverse the sign of E°cathode. Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry.
Exercise
A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate [Hg(CH3CO2)2] and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of MgCl2. When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur:
$cathode: \; Hg^{2+}\left (aq \right ) + 2e^{-} \rightarrow Hg\left (l \right )$
$anode: \; Mg\left (s \right ) \rightarrow Mg^{2+}\left (aq \right ) + 2e^{-}$
If the potential for the oxidation of Mg to Mg2+ is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the anode?
Answer: 0.85 V
Reference Electrodes and Measuring Concentrations
When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrodeThe electrode of a galvanic cell whose potential is related to the concentration of the substance being measured., whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrodeAn electrode in an galvanic cell whose potential is unaffected by the properties of the solution., must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode.
Note the Pattern
The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated.
There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrodeA reference electrode that consists of a silver wire coated with a very thin layer of AgCl and dipped into a chloride ion solution with a fixed concentration., which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows:
$Cl^{-}\left (aq \right ) \mid AgCl\left (s \right ) \mid Ag\left (s \right ) \tag{19.2.36}$
$AgCl\left (s \right ) + e^{-} \rightarrow Ag\left (s \right ) + Cl^{-}\left (aq \right )$
If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE.
A second common reference electrode is the saturated calomel electrode (SCE)A reference electrode that consists of a platinum wire inserted into a moist paste of liquid mercury (calomel; <math display="inline" xml:id="av_1.0-19_m021"><semantics><mrow><msub><mrow><mtext>Hg</mtext></mrow><mn>2</mn></msub><msub><mrow><mtext>Cl</mtext></mrow><mn>2</mn></msub></mrow></semantics>[/itex]) and KCl in an interior cell, which is surrounded by an aqueous KCl solution., which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg2Cl2; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure 19.2.5). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows:
$Pt\left (s \right ) \mid Hg_{2}Cl_{2}\left (s \right ) \mid KCl\left (aq, sat \right ) \tag{19.2.37}$
$Hg_{2}Cl_{2}\left (s \right ) + 2e^{-} \rightarrow 2Hg\left (l \right ) + 2Cl^{-}\left (aq \right ) \tag{19.2.38}$
At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential.
One of the most common uses of electrochemistry is to measure the H+ ion concentration of a solution. A glass electrodeAn electrode used to measure the <math display="inline" xml:id="av_1.0-19_m022"><semantics><mrow><msup><mtext>H</mtext><mo>+</mo></msup></mrow></semantics>[/itex] ion concentration of a solution and consisting of an internal Ag/AgCl electrode immersed in a 1 M HCl solution that is separated from the solution by a very thin glass membrane. is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure 19.2.5). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H+] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H+] as follows (recall that pH = −log[H+]:
$E_{glass} = E' + \left ( 0.0591 \; V \times log\left [H^{+} \right ] \right ) = E'- 0.0591 \; V \times pH \tag{19.2.39}$
The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH.
Ion-selective electrodesAn electrode whose potential depends on only the concentration of a particular species in solution. are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure 19.2.5). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped LaF3 as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table 19.2.1.
Table 19.2.1 Some Species Whose Aqueous Concentrations Can Be Measured Using Electrochemical Methods
Species Type of Sample
H+ laboratory samples, blood, soil, and ground and surface water
NH3/NH4+ wastewater and runoff water
K+ blood, wine, and soil
CO2/HCO3 blood and groundwater
F groundwater, drinking water, and soil
Br grains and plant extracts
I milk and pharmaceuticals
NO3 groundwater, drinking water, soil, and fertilizer
Summary
The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E°cell). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E°cell = E°cathodeE°anode). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. If E°cell is positive, the reaction will occur spontaneously under standard conditions. If E°cell is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°.
Key Takeaways
• Redox reactions can be balanced using the half-reaction method.
• The standard cell potential is a measure of the driving force for the reaction.
Key Equation
Standard cell potential
Equation 19.2.2: E°cell = E°cathodeE°anode
Conceptual Problems
1. Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?
2. List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.
3. What is the relationship between electron flow and the potential energy of valence electrons? If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?
4. If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?
5. Write a cell diagram representing a cell that contains the Ni/Ni2+ couple in one compartment and the SHE in the other compartment. What are the values of E°cathode, E°anode, and E°cell?
6. Explain why E° values are independent of the stoichiometric coefficients in the corresponding half-reaction.
7. Identify the oxidants and the reductants in each redox reaction.
1. Cr(s) + Ni2+(aq) → Cr2+(aq) + Ni(s)
2. Cl2(g) + Sn2+(aq) → 2Cl(aq) + Sn4+(aq)
3. H3AsO4(aq) + 8H+(aq) + 4Zn(s) → AsH3(g) + 4H2O(l) + 4Zn2+(aq)
4. 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(l)
8. Identify the oxidants and the reductants in each redox reaction.
1. Br2(l) + 2I(aq) → 2Br(aq) + I2(s)
2. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
3. H+(aq) + 2MnO4(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4(aq)
4. IO3(aq) + 5I(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
9. All reference electrodes must conform to certain requirements. List the requirements and explain their significance.
10. For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?
1. measuring the potential of a Cl/Cl2 couple
2. measuring the pH of a solution
3. measuring the potential of a MnO4−/Mn2+ couple
Answers
1. $Ni\left (s \right )\mid Ni^{2+}\left (aq \right ) \parallel H^{+} \left (aq, 1\;M \right ) \mid H_{2}\left ( g, 1 \; atm \right )\mid Pt\left (s \right )$
$\begin{matrix} E_{anode}^{o} & \;\;\; & Ni^{2+}\left (aq \right ) + 2e^{-} \rightarrow Ni\left (s \right ) & \;\;-0.257 \; V \ E_{cathode}^{o} & \;\;\;& 2H^{+}\left (aq \right )+2e^{-} \rightarrow H_{2}\left ( g,\right )& & \;\;0.000 \; V \ E_{cell}^{o} & \;\;\;& 2H^{+}\left (aq \right )+ Ni\left (s \right )\rightarrow H_{2}\left ( g,\right )+ Ni^{2+}\left (aq \right ) & \;\;0.257 \; V \end{matrix}$
1. oxidant: Ni2+(aq); reductant: Cr(s)
2. oxidant: Cl2(g); reductant: Sn2+(aq)
3. oxidant: H3AsO4(aq); reductant: Zn(s)
4. oxidant: NO2(g); reductant: NO2(g)
Numerical Problems
1. Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction:
H2(g) + Cu2+(aq) → 2H+(aq) + Cu(s).
2. Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g).
3. Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.
1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq)
2. Br2(aq) + Fe2+(aq) → 2Br(aq) + Fe3+(aq)
3. Fe3+(aq) + Cd(s) → Fe2+(aq) + Cd2+(aq)
4. Balance each reaction and calculate the standard reduction potential for each. Be sure to include the physical state of each product and reactant.
1. Cu+(aq) + Ag+(aq) → Cu2+(aq) + Ag(s)
2. Sn(s) + Fe3+(aq) → Sn2+(aq) + Fe2+(aq)
3. Mg(s) + Br2(l) → 2Br(aq) + Mg2+(aq)
5. Write a balanced chemical equation for each redox reaction.
1. H2PO2(aq) + SbO2(aq) → HPO32−(aq) + Sb(s) in basic solution
2. HNO2(aq) + I(aq) → NO(g) + I2(s) in acidic solution
3. N2O(g) + ClO(aq) → Cl(aq) + NO2(aq) in basic solution
4. Br2(l) → Br(aq) + BrO3(aq) in basic solution
5. Cl(CH2)2OH(aq) + K2Cr2O7(aq) → ClCH2CO2H(aq) + Cr3+(aq) in acidic solution
6. Write a balanced chemical equation for each redox reaction.
1. I(aq) + HClO2(aq) → IO3(aq) + Cl2(g) in acidic solution
2. Cr2+(aq) + O2(g) → Cr3+(aq) + H2O(l) in acidic solution
3. CrO2(aq) + ClO(aq) → CrO42−(aq) + Cl(aq) in basic solution
4. S(s) + HNO2(aq) → H2SO3(aq) + N2O(g) in acidic solution
5. F(CH2)2OH(aq) + K2Cr2O7(aq) → FCH2CO2H(aq) + Cr3+(aq) in acidic solution
7. The standard cell potential for the oxidation of Pb to Pb2+ with the concomitant reduction of Cu+ to Cu is 0.39 V. You know that E° for the Pb2+/Pb couple is −0.13 V. What is E° for the Cu+/Cu couple?
8. You have built a galvanic cell similar to the one in Figure 19.2.3 using an iron nail, a solution of FeCl2, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is Ecell?
9. Carbon is used to reduce iron ore to metallic iron. The overall reaction is as follows:
2Fe2O3·xH2O(s) + 3C(s) → 4Fe(l) + 3CO2(g) + 2xH2O(g)
Write the two half-reactions for this overall reaction.
10. Will each reaction occur spontaneously under standard conditions?
1. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
2. Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq)
11. Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.
1. Se(s) + Br2(l) → H2SeO3(aq) + Br(aq)
2. NO3(aq) + S(s) → HNO2(aq) + H2SO3(aq)
3. Fe3+(aq) + Cr3+(aq) → Fe2+(aq) + Cr2O72−(aq)
12. Calculate E°cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?
13. If you place Zn-coated (galvanized) tacks in a glass and add an aqueous solution of iodine, the brown color of the iodine solution fades to a pale yellow. What has happened? Write the two half-reactions and the overall balanced chemical equation for this reaction. What is E°cell?
14. Your lab partner wants to recover solid silver from silver chloride by using a 1.0 M solution of HCl and 1 atm H2 under standard conditions. Will this plan work?
Answers
1. Pt(s)∣H2(g, 1 atm) | H+(aq, 1M)∥Cu2+(aq)∣Cu(s)
1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq); E° = 1.358 V
2. Br2(l) + 2Fe2+(aq) → 2Br(aq) + 2Fe3+(aq); E° = 0.316 V
3. 2Fe3+(aq) + Cd(s) → 2Fe2+(aq) + Cd2+(aq); E° = 1.174 V
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.03%3A_Standard_Potentials.txt |
Learning Objectives
• To know how to predict the relative strengths of various oxidants and reductants.
We can use the procedure described in Section 19.2 to measure the standard potentials for a wide variety of chemical substances, some of which are listed in Table 19.3.1 (Table P1 contains a more extensive listing.) These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table 19.3.1 . All reactants that lie above the SHE in the table are stronger oxidants than H+, and all those that lie below the SHE are weaker. The strongest oxidant in the table is F2, with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element.
Table 19.3.1 Standard Potentials for Selected Reduction Half-Reactions at 25°C
Half-Reaction E° (V)
F2(g) + 2e→ 2F(aq) 2.87
H2O2(aq) + 2H+(aq) + 2e → 2H2O(l) 1.78
Ce4+(aq) + e → Ce3+(aq) 1.72
PbO2(s) + HSO4(aq) + 3H+(aq) + 2e → PbSO4(s) + 2H2O(l) 1.69
Cl2(g) + 2e → 2Cl(aq) 1.36
Cr2O72−(aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l) 1.23
O2(g) + 4H+(aq) + 4e → 2H2O(l) 1.23
MnO2(s) + 4H+(aq) + 2e → Mn2+(aq) + 2H2O(l) 1.22
Br2(aq) + 2e → 2Br(aq) 1.09
NO3(aq) + 3H+(aq) + 2e → HNO2(aq) + H2O(l) 0.93
Ag+(aq) + e → Ag(s) 0.80
Fe3+(aq) + e → Fe2+(aq) 0.77
H2SeO3(aq) + 4H+ + 4e → Se(s) + 3H2O(l) 0.74
O2(g) + 2H+(aq) + 2e → H2O2(aq) 0.70
MnO4(aq) + 2H2O(l) + 3e → MnO2(s) + 4OH(aq) 0.60
MnO42−(aq) + 2H2O(l) + 2e → MnO2(s) + 4OH(aq) 0.60
I2(s) + 2e → 2I(aq) 0.54
H2SO3(aq) + 4H+(aq) + 4e → S(s) + 3H2O(l) 0.45
O2(g) + 2H2O(l) + 4e → 4OH(aq) 0.40
Cu2+(aq) + 2e → Cu(s) 0.34
AgCl(s) + e → Ag(s) + Cl(aq) 0.22
Cu2+(aq) + e → Cu+(aq) 0.15
Sn4+(aq) + 2e → Sn2+(aq) 0.15
2H+(aq) + 2e → H2(g) 0.00
Sn2+(aq) + 2e → Sn(s) −0.14
2SO42−(aq) + 4H+(aq) + 2e → S2O62−(aq) + 2H2O(l) −0.22
Ni2+(aq) + 2e → Ni(s) −0.26
PbSO4(s) + 2e → Pb(s) + SO42−(aq) −0.36
Cd2+(aq) + 2e → Cd(s) −0.40
Cr3+(aq) + e → Cr2+(aq) −0.41
Fe2+(aq) + 2e → Fe(s) −0.45
Ag2S(s) + 2e → 2Ag(s) + S2−(aq) −0.69
Zn2+(aq) + 2e → Zn(s) −0.76
Al3+(aq) + 3e → Al(s) −1.662
Be2+(aq) + 2e → Be(s) −1.85
Li+(aq) + e → Li(s) −3.04
Similarly, all species in Table 19.3.1 that lie below H2 are stronger reductants than H2, and those that lie above H2 are weaker. The strongest reductant in the table is thus metallic lithium, with a standard electrode potential of −3.04 V. This fact might be surprising because cesium, not lithium, is the least electronegative element. The apparent anomaly can be explained by the fact that electrode potentials are measured in aqueous solution, where intermolecular interactions are important, whereas ionization potentials and electron affinities are measured in the gas phase. Due to its small size, the Li+ ion is stabilized in aqueous solution by strong electrostatic interactions with the negative dipole end of water molecules. These interactions result in a significantly greater ΔHhydration for Li+ compared with Cs+. Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution.
Note the Pattern
Species in Table 19.3.1 that lie below H2 are stronger reductants (more easily oxidized) than H2. Species that lie above H2 are stronger oxidants.
Because the half-reactions shown in Table 19.3.1 are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E°cell > 0), as illustrated in Example 5.
Example 19.3.1
The black tarnish that forms on silver objects is primarily Ag2S. The half-reaction for reversing the tarnishing process is as follows:
$AgS\left (s \right ) 2e^{-}\rightarrow Ag\left (s \right )+ S^{2-}\left (aq \right ) \;\;\; E^{o}=-0.69\; V$
1. Referring to Table 19.3.1 , predict which species—H2O2(aq), Zn(s), I(aq), Sn2+(aq)—can reduce Ag2S to Ag under standard conditions.
2. Of these species—H2O2(aq), Zn(s), I(aq), Sn2+(aq), identify which is the strongest reducing agent in aqueous solution and thus the best candidate for a commercial product.
3. From the data in Table 19.3.1 , suggest an alternative reducing agent that is readily available, inexpensive, and possibly more effective at removing tarnish.
Given: reduction half-reaction, standard electrode potential, and list of possible reductants
Asked for: reductants for Ag2S, strongest reductant, and potential reducing agent for removing tarnish
Strategy:
A From their positions in Table 19.3.1, decide which species can reduce Ag2S. Determine which species is the strongest reductant.
B Use Table 19.3.1 to identify a reductant for Ag2S that is a common household product.
Solution:
We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag2S/Ag couple in Table 19.3.1 or (2) compare E° for each species with E° for the Ag2S/Ag couple (−0.69 V).
1. A The species in Table 19.3.1 are arranged from top to bottom in order of increasing reducing strength. Of the four species given in the problem, I(aq), Sn2+(aq), and H2O2(aq) lie above Ag2S, and one [Zn(s)] lies below it. We can therefore conclude that Zn(s) can reduce Ag2S(s) under standard conditions, whereas I(aq), Sn2+(aq), and H2O2(aq) cannot. Sn2+(aq) and H2O2(aq) appear twice in the table: on the left side (oxidant) in one half-reaction and on the right side (reductant) in another.
2. The strongest reductant is Zn(s), the species on the right side of the half-reaction that lies closer to the bottom of Table 19.3.1 than the half-reactions involving I(aq), Sn2+(aq), and H2O2(aq). (Commercial products that use a piece of zinc are often marketed as a “miracle product” for removing tarnish from silver. All that is required is to add warm water and salt for electrical conductivity.)
3. B Of the reductants that lie below Zn(s) in Table 19.3.1 , and therefore are stronger reductants, only one is commonly available in household products: Al(s), which is sold as aluminum foil for wrapping foods.
Exercise
Refer to Table 19.3.1 to predict
1. which species—Sn4+(aq), Cl(aq), Ag+(aq), Cr3+(aq), and/or H2O2(aq)—can oxidize MnO2(s) to MNO4 under standard conditions.
2. which species—Sn4+(aq), Cl(aq), Ag+(aq), Cr3+(aq), and/or H2O2(aq)—is the strongest oxidizing agent in aqueous solution.
Answer
1. Ag+(aq); H2O2(aq)
2. H2O2(aq)
Example 19.3.2
Use the data in Table 19.3.1 to determine whether each reaction is likely to occur spontaneously under standard conditions:
1. Sn(s) + Be2+(aq) → Sn2+(aq) + Be(s)
2. MnO2(s) + H2O2(aq) + 2H+(aq) → O2(g) + Mn2+(aq) + 2H2O(l)
Given: redox reaction and list of standard electrode potentials (Table 19.3.1)
Asked for: reaction spontaneity
Strategy:
A Identify the half-reactions in each equation. Using Table 19.3.1, determine the standard potentials for the half-reactions in the appropriate direction.
B Use Equation 19.2.2 to calculate the standard cell potential for the overall reaction. From this value, determine whether the overall reaction is spontaneous.
Solution:
1. A Metallic tin is oxidized to Sn2+(aq), and Be2+(aq) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.85 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table 19.3.1.
B Adding the two half-reactions gives the overall reaction:
$\begin{matrix} cathode: & MnO_{2}\left ( s \right )+4H^{+}\left ( aq \right )+ 2e^{-}\rightarrow Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & E_{cathode}^{o}=1.22\; V \ anode: & H_{2}O_{2}\left ( aq \right )\rightarrow O_{2}\left( g \right ) + 2H^{+}\left ( aq \right ) + 2e^{-}& E_{anode}^{o}=1.22\; V \ overall: & MnO_{2}\left ( s \right )+2H^{+}\left ( aq \right )+ H_{2}O_{2}\left ( aq \right )\rightarrow & E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\ & O_{2}\left( g \right )+ Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & -0.53 \; V \end{matrix}$
The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot be used to reduce Be2+ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn2+) by metallic beryllium, which has a positive value of E°cell, will occur spontaneously.
2. A MnO2 is the oxidant (Mn4+ is reduced to Mn2+), while H2O2 is the reductant (O2− is oxidized to O2). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table 19.2.
B The two half-reactions and their corresponding potentials are as follows:
$\begin{matrix} cathode: & MnO_{2}\left ( s \right )+4H^{+}\left ( aq \right )+ 2e^{-}\rightarrow Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & E_{cathode}^{o}=1.22\; V \ anode: & H_{2}O_{2}\left ( aq \right )\rightarrow O_{2}\left( g \right ) + 2H^{+}\left ( aq \right ) + 2e^{-}& E_{anode}^{o}=1.22\; V \ overall: & MnO_{2}\left ( s \right )+2H^{+}\left ( aq \right )+ H_{2}O_{2}\left ( aq \right )\rightarrow & E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\ & O_{2}\left( g \right )+ Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & -0.53 \; V \end{matrix}$
The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO2, and oxygen gas will evolve from the solution.
Exercise
Use the data in Table 19.3.2 to determine whether each reaction is likely to occur spontaneously under standard conditions:
1. 2Ce4+(aq) + 2Cl(aq) → 2Ce3+(aq) + Cl2(g)
2. 4MnO2(s) + 3O2(g) + 4OH(aq) → 4MnO4(aq) + 2H2O
Answer
1. spontaneous (E°cell = 0.36 V)
2. nonspontaneous (E°cell = −0.20 V)
Although the sign of E°cell tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics, as described in Section 19.4.
Summary
The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase.
Key Takeaway
• The relative strengths of various oxidants and reductants can be predicted using E° values.
Conceptual Problems
1. The order of electrode potentials cannot always be predicted by ionization potentials and electron affinities. Why? Do you expect sodium metal to have a higher or a lower electrode potential than predicted from its ionization potential? What is its approximate electrode potential?
2. Without referring to tabulated data, of Br2/Br, Ca2+/Ca, O2/OH, and Al3+/Al, which would you expect to have the least negative electrode potential and which the most negative? Why?
3. Because of the sulfur-containing amino acids present in egg whites, eating eggs with a silver fork will tarnish the fork. As a chemist, you have all kinds of interesting cleaning products in your cabinet, including a 1 M solution of oxalic acid (H2C2O4). Would you choose this solution to clean the fork that you have tarnished from eating scrambled eggs?
4. The electrode potential for the reaction Cu2+(aq) + 2e → Cu(s) is 0.34 V under standard conditions. Is the potential for the oxidation of 0.5 mol of Cu equal to −0.34/2 V? Explain your answer.
Answer
1. No; E° = −0.691 V for Ag2S(s) + 2e → Ag(s) + S2−(aq), which is too negative for Ag2S to be spontaneously reduced by oxalic acid [E° = 0.49 V for 2CO2(g) + 2H+(aq) + 2e → H2C2O4(aq)]
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.04%3A_Comparing_Strengths_of_Oxidants_and_Reductants.txt |
Learning Objectives
• To understand the relationship between cell potential and the equilibrium constant.
• To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of Co(s) with Ni2+(aq) to form Ni(s) and Co2+(aq) occurs spontaneously, but if we reduce the concentration of Ni2+ by a factor of 100, so that [Ni2+] is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential and Free Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C)The SI unit of measure for the number of electrons that pass a given point in 1 second; it is defined as 6.25×1018 e/s and relates electron potential (in volts) to energy (in joules): 1 J/1 V = 1 C., an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A)The fundamental SI unit of electric current; it is defined as the flow of 1 C/s past a given point: 1A = 1 C/s.; 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{1\;J}{1\;V}= 1\;C=1\;A\cdot s \tag{19.4.1}$
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F)The charge on 1 mol of electrons; it is obtained by multiplying the charge on the electron by Avogadro’s number., named after the English physicist and chemist Michael Faraday (1791–1867):
$F= \left ( 1.60218 \times 10^{-19}\;C \right )\dfrac{6.02214\times 10^{23}}{1\;mole} \tag{19.4.2}$
$F= 9.64855 \times 10^{4}\;C/mole \simeq 96,486 C/\left (mol \; e^{-} \right )$
The total charge transferred from the reductant to the oxidant is therefore nF, where n is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell (wmax) is equal to the product of the cell potential (Ecell) and the total charge transferred during the reaction (nF):
$w_{max} = nFE_{cell} \tag{19.4.3}$
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
As you learned in Chapter 18, the change in free energy (ΔG) is also a measure of the maximum amount of work that can be performed during a chemical process (ΔG = wmax). Consequently, there must be a relationship between the potential of an electrochemical cell and ΔG, the most important thermodynamic quantity discussed in Chapter 18. This relationship is as follows:
$\Delta G = -nFE_{cell} \tag{19.4.4}$
A spontaneous redox reaction is therefore characterized by a negative value of ΔG and a positive value of Ecell, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and E°cell is as follows:
$\Delta G^{o} = -nFE^{o}_{cell} \tag{19.4.5}$
Note the Pattern
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example 19.4.1
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table 19.3.1, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: ΔG° for the reaction and spontaneity
Strategy:
A From the relevant half-reactions and the corresponding values of E°, write the overall reaction and calculate E°cell using Equation 19.2.2.
B Determine the number of electrons transferred in the overall reaction. Then use Equation 19.4.5 to calculate ΔG°. If ΔG° is negative, then the reaction is spontaneous.
Solution:
A As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of E°. From Table 19.3.1, we can find the reduction and oxidation half-reactions and corresponding E° values:
$\begin{matrix} cathode: & Cr_{2}O_{7}^{2-} \left (aq \right )+ 14 H^{+}\left (aq \right ) + 6e^{-}\rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right ) & E_{cathode}^{o}=1.23\;V \ anode: & 2Br^{-}\left (aq \right ) \rightarrow Br_{2}\left (aq \right )+2e^{-} & E_{anode}^{o}=1.09\;V \end{matrix}$
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of E° is not affected:
$\begin{matrix} cathode: & Cr_{2}O_{7}^{2-} \left (aq \right )+ 14 H^{+}\left (aq \right ) + 6e^{-}\rightarrow 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right ) & E_{cathode}^{o}=1.23\;V \ anode: & 6Br^{-}\left (aq \right ) \rightarrow 3Br_{2}\left (aq \right )+6e^{-} & E_{anode}^{o}=1.09\;V \ overall: & Cr_{2}O_{7}^{2-} \left (aq \right )+ 14 H^{+}\left (aq \right ) + 6BrI^{-} \rightarrow & E_{cell}^{o}=0.14\;V \ & 2Cr^{3+}\left (aq \right ) + 7H_{2}O \left (l \right ) + 3Br_{2}\left (aq \right )& \end{matrix}$
B We can now calculate ΔG° using Equation 19.4.10 Because six electrons are transferred in the overall reaction, the value of n is 6:
$\begin{matrix} \Delta G^{o} &= -nFE_{cell}^{o} & =\left ( 6 \; \cancel{mol} \right )\left ( 96,468\;J/\left ( \cancel{V}\cdot \cancel{mol} \right ) \right )\left ( 0.14 \; \cancel{V} \right ) \ & & -8.1\times 10^{4}\;J\ & & -81 \; kJ/mol \; Cr_{2}O_{7} \end{matrix}$
Thus ΔG° is −81 kJ for the reaction as written, and the reaction is spontaneous.
Exercise
Use the data in Table 19.3.1 to calculate ΔG° for the reduction of ferric ion by iodide:
2Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(s)
Is the reaction spontaneous?
Answer: −44 kJ/mol I2; yes
Potentials for the Sums of Half-Reactions
Although Table 19.3.1 and Table P1 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of Fe3+(aq) to Fe(s) is not listed in the table, but two related reductions are given:
$Fe^{3+}\left ( aq \right ) + e^{-} \rightarrow Fe^{2+}\left ( aq \right ) \;\;\; E^{o}=0.77\;V \tag{19.4.6}$
$Fe^{2+}\left ( aq \right ) + 2e^{-} \rightarrow Fe\left ( aq \right ) \;\;\; E^{o}=-0.45\;V \tag{19.4.7}$
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because E° is not a state function. However, because ΔG° is a state function, the sum of the ΔG° values for the individual reactions gives us ΔG° for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. To obtain the value of E° for the overall half-reaction, we first must add the values of ΔG° (= −nFE°) for each individual half-reaction to obtain ΔG° for the overall half-reaction:
$Fe^{3+}\left ( aq \right ) + e^{-} \rightarrow \cancel{Fe^{2+}\left ( aq \right )} \;\;\; \Delta G^{o}=\left (-1 \right )F \left (0.77\;V \right ) \tag{19.4.8}$
$\cancel{Fe^{2+}\left ( aq \right )} + 2e^{-} \rightarrow Fe\left ( s \right ) \;\;\; \Delta G^{o}= \left (-2 \right )F \left (-0.457\;V \right ) \notag{}$
$Fe^{3+}\left ( aq \right ) + e^{-} \rightarrow Fe\left ( s \right ) \;\;\; \Delta G^{o}=\left (-1 \right )F \left (0.77\;V \right )+ \left (-2 \right )F \left (-0.457\;V \right ) \notag{}$
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta G^{o}=F \left [\left (-0.77\;V \right )+ \left (-2 \right )\left (-0.457\;V \right ) \right ] =F\left ( 0.13\;V \right ) \tag{19.4.9}$
Three electrons (n = 3) are transferred in the overall reaction (Equation 19.4.15), so substituting into Equation 19.4.10 and solving for E° gives the following:
$\Delta G^{o}=-nFE_{cell}^{o}$
$F\left ( 0.13\;V \right )=-3 F E_{cell}^{o}$
$E_{cell}^{o}=-\dfrac{0.13\;V}{3}= -0.043\;V$
This value of E° is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Note the Pattern
Values of E° for half-reactions cannot be added to give E° for the sum of the half-reactions; only values of ΔG° = −nFE°cell for half-reactions can be added.
The Relationship between Cell Potential and the Equilibrium Constant
We can use the relationship between ΔG° and the equilibrium constant K, defined in Chapter 18, to obtain a relationship between E°cell and K. Recall that for a general reaction of the type aA + bB → cC + dD, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta G^{o}=-RTln\;K \tag{19.4.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation 19.4.9), we can write
$-nFE_{cell}^{o}=-RTln\;K \tag{19.4.11}$
Rearranging this equation,
$E_{cell}^{o}=\left (\dfrac{RT}{nF} \right )ln\;K \tag{19.4.12}$
For T = 298 K, Equation 19.4.12 can be simplified as follows:
$E_{cell}^{o}=\left (\dfrac{RT}{nF} \right )ln\;K =\left [ \dfrac{8.314\;\cancel{J}/\left ( \cancel{mol}\cdot \cancel{K} \right )\left ( 298\;\cancel{K} \right )}{n\left [ 96,486\; \cancel{J}/\left ( V \cdot \cancel{mol} \right ) \right ]} \right ]2.303\;log \; K = \left ( \dfrac{0.0591}{n} \right )log\; K \tag{19.4.13}$
Thus E°cell is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of E°cell and vice versa.
Example 19.4.2
Use the data in Table 19.3.1 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: K
Strategy:
A Write the relevant half-reactions and potentials. From these, obtain the overall reaction and E°cell.
B Determine the number of electrons transferred in the overall reaction. Use Equation 19.4.13 to solve for log K and then K.
Solution:
A The relevant half-reactions and potentials from Table 19.3.1 are as follows:
$\begin{matrix} cathode: & PbO_{2} \left (s \right )+ SO_{4}^{2-}\left (aq \right ) + 4H^{+}\left (aq \right )+ 2e^{-}\rightarrow PbSO_{4}\left (s \right ) + 2H_{2}O \left (l \right ) & E_{cathode}^{o}= 1.69 \;V \ anode: & Pb\left (s \right )+ SO_{4}^{2-}\left (aq \right )\rightarrow PbSO_{4}\left (s \right )+2e^{-} & E_{anode}^{o}=0.36 \;V \ overall: & Pb\left (s \right )PbO_{2} \left (s \right )+ 2SO_{4}^{2-}\left (aq \right ) + 4H^{+}\left (aq \right )\rightarrow 2PbSO_{4}\left (s \right )+ 2H_{2}O \left (l \right )& E_{cell}^{o}=2.05\;V \end{matrix}$
B Two electrons are transferred in the overall reaction, so n = 2. Solving Equation 19.4.13 for log K and inserting the values of n and E°,
$ln\;K=\dfrac{nE^{o}}{0.0591\;V}=\dfrac{2\left ( 2.05\;\cancel{V} \right )}{0.0591\;\cancel{V}}=69.37$
$K=2.3\times 10^{69}$
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise
Use the data in Table 19.3.1 to calculate the equilibrium constant for the reaction of Sn2+(aq) with oxygen to produce Sn4+(aq) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$2Sn^{2+}\left ( aq \right )+O_{2}\left ( g \right )+4H^{+}\left ( aq \right )\rightleftharpoons 2Sn^{4+}\left ( aq \right )+2H_{2}O\left ( l \right )$
Answer: 1.2 × 1073
Figure 19.4.1 summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between ΔG and the reaction quotient Q developed in Chapter 18 .
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall from Chapter 18 that the actual free-energy change for a reaction under nonstandard conditions, ΔG, is given as follows:
$\Delta G =\Delta G^{o}+RT\;ln\;Q \tag{19.4.14}$
We also know that ΔG = −nFEcell and ΔG° = −nFE°cell. Substituting these expressions into Equation 19.4.14, we obtain
$-nFE_{cell} =-nFE_{cell}^{o}+RT\;ln\;Q \tag{19.4.15}$
Dividing both sides of this equation by −nF,
$E_{cell} =E_{cell}^{o}-\dfrac{RT}{nF}\;ln\;Q \tag{19.4.16}$
Equation 19.4.16 is called the Nernst equationAn equation for calculating cell potentials Ecell under nonstandard conditions; it can be used to determine the direction of spontaneous reaction for any redox reaction under an conditions:Ecell=</mo><msubsup><mi>E°−(RT/nF)lnQ, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium (ΔG = 0), Equation 19.4.16 reduces to Equation 19.4.12 because Q = K, and there is no net transfer of electrons (i.e., Ecell = 0).
Substituting the values of the constants into Equation 19.4.16 with T = 298 K and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in Q):
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{F}\;ln\;Q \tag{19.4.17}$
Note the Pattern
The Nernst equation can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation 19.4.17 allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation 19.4.17 that the cell potential changes by 0.0591/n V for each 10-fold change in the value of Q because log 10 = 1.
Example 19.4.3
In the exercise in Example 6, you determined that the following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which you now know means that ΔG° < 0):
$2Ce^{4+}\left ( aq \right )+2Cl^{-}\left ( aq \right )\rightarrow 2Ce^{3+}\left ( aq \right )+ Cl_{2}\left ( g \right )$
Calculate E for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl] = 0.0030 M, Cl2= 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution:
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation 19.4.17 instead of 19.46. The overall reaction involves the net transfer of two electrons:
2Ce4+(aq) + 2e → 2Ce3+(aq)
2Cl(aq) → Cl2(g) + 2e
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation 19.4.17,
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{F}\;log\;Q$
$\;\;\; =0.25\;V-\dfrac{0.0591\;V}{2}\;logn\;\left (\dfrac{\left [ Ce^{3+} \right ]^{2}P_{Cl_{2}}}{\left [ Ce^{4+} \right ]^{2}\left [ Cl^{-} \right ]^{2}} \right )$
$\;\;\; =0.25\;V-\left [ \left ( 0.0296\;V \right ) \left ( 8.37 \right )\right ]=0.00\;V$
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture.
Exercise
In the exercise in Example 6, you determined that molecular oxygen will not oxidize MnO2 to permanganate via the reaction
$4MnO_{2}\left ( s \right )+3O_{2}\left ( g \right )+4OH^{-}\left ( aq \right ) \rightarrow 4MnO_{4}^{-}\left ( aq \right )+ 2H_{2}O\left ( l \right )$ \;\;\; E_{cell}^{o}=-0.20\;V \)
Calculate Ecell for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, P(O2) = 0.20 atm, [MNO4] = 1.0 × 10−4 M, and T = 25°C.
Answer: Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell discussed in Section 19.2 allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
$Zn\left ( s \right )+Cu^{2+}\left ( aq \right ) \rightarrow Zn^{2+}\left ( aq \right )+ Cu\left ( s \right )$ \;\;\; E_{cell}^{o}=1.10\;V \tag{19.4.18}\)
The reaction quotient is therefore Q = [Zn2+]/[Cu2+]. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation 19.4.17:
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \dfrac{\left [ Zn^{2+} \right ]}{\left [ Cu^{2+} \right ]} \tag{19.4.19}$
$E_{cell} =1.10\;V-\dfrac{0.0591\;V}{2}\;log \dfrac{1.0\times 10^{-6}}{1.0} = 1.28\;V$
Thus the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with log Q over this range is linear with a slope of −0.0591/n, as illustrated in Figure 19.4.2. As the reaction proceeds still further, Q continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of Q when Ecell = 0 is calculated as follows:
$E_{ cell} =E_{cell}^{o} - \dfrac{0.0591\;V}{n} \; log \; Q = 0 \tag{19.4.20}$
$E^{o}=\dfrac{0.0591\;V}{n}\;log \; Q$
$log \; Q )=\dfrac{E^{o} \; n}{0.0591\;V}=\dfrac{\left ( 1.10\;\cancel{V} \right )\left ( 2 \right )}{0.0591\;\cancel{V}}=37.23$
$Q =10^{37.23}=1.7\times 10^{37}$
Recall that at equilibrium, Q = K. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$Ag\left ( s \right )\mid Ag^{+}\left ( aq,\; 0.010\; M \right )\parallel Ag^{+}\left ( aq,\; 1.0 \; M \right )\mid Ag\left ( s \right ) \tag{19.4.21}$
$cathode: \;\; Ag^{+}\left ( aq,\; 1.0 \; M \right )+e^{-}\rightarrow Ag\left ( s \right ) \tag{19.4.22}$
$anode: \;\; Ag\left ( s \right )\rightarrow Ag^{+}\left ( aq,\; 0.010 \; M \right )+e^{-} \tag{19.4.23}$
$overall: \;\; Ag^{+}\left ( aq,\; 1.0 \; M \right )\rightarrow Ag^{+}\left ( aq,\; 0.010 \; M \right ) \tag{19.4.24}$
As the reaction progresses, the concentration of Ag+ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the Ag+ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of Ag(s) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log Q = 0-\left ( \dfrac{0.0591\;V}{1} \right )log\dfrac{0.010}{1.0} = 0.12 \tag{19.4.25}$
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cellAn electrochemical cell in which the anode and the cathode compartments are identical except for the concentration of a reactant.. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example 19.4.4
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
A Write the overall reaction that occurs in the cell.
B Determine the number of electrons transferr
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \;Q = 0-\left ( \dfrac{0.0591\;V}{2} \right )log\dfrac{5.2\times 10^{-2}}{2.0} = 0.047$
ed. Substitute this value into the Nernst equation to calculate the voltage.
Solution:
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42−) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
Mn2+(aq, 2.0 M) → Mn2+(aq, 5.2 × 10−2 M)
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation 19.4.17:
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with P(O2) = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water, O2(g) + 4H+(aq) + 4e → 2H2O(l), what will be the potential when the circuit is closed?
Answer: 0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products (Ksp) of sparingly soluble substances. As you learned in Chapter 17, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure 19.4.3, which is designed to measure the solubility product of silver chloride: Ksp = [Ag+][Cl]. In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl] and substituting: [Ag+] = Ksp/[Cl] = Ksp/1.0 = Ksp. The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
$E_{cell} = 0-\left ( \dfrac{0.0591\;V}{1} \right )log\left (\dfrac{\left [Ag^{+} \right ]_{dilute}}{\left [Ag^{+} \right ]_{concentrated}} \right ) = -0.0591 \; V\; log\left ( \dfrac{K_{sp}}{1.0} \right )=-0.0591\; V \tag{19.4.26}$
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation 19.4.26 for Ksp,
$log\;K_{sp} = \dfrac{-E_{cell}}{0.0591\;V } = \dfrac{-0.580\; \cancel{V}}{0.0591\;\cancel{V}} =-9.81 \tag{19.4.27}$
$K_{sp}=1.5\times 10^{-10}$
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example 19.4.5
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure 19.4.3, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: K sp
Strategy:
A From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
B Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation 19.4.26 and solve for Ksp.
Solution:
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of Pb2+ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
$\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=K_{sp}$
$\left [ Pb^{2+} \right ]=\dfrac{K_{sp}}{\left [ SO_{4}^{2-} \right ]}=\dfrac{K_{sp}}{1.0 \; M}= K_{sp}$
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)
so
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \;Q = \(0.230\;V=0-\left ( \dfrac{0.0591\;V}{2} \right )log \left (\dfrac{\left [ Pb^{2+} \right ]_{dilute}}{\left [ Pb^{2+} \right ]_{concentrated}} \right ) = -0.0296\; V\; log\left ( \dfrac{K_{sp}}{1.0} \right )$
$-7.77=log \; K_{sp}$
$1.7\times 10^{-8} = K_{sp}$
Exercise
A concentration cell similar to the one described in Example 11 contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer: 5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example 11, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated in Example 12.
Example 19.4.6
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment (Figure 19.4.3). The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
Zn(s)∣Zn2+(aq, 1.0 M) ∥ H+(aq, ? M)∣H2(g, 1.0 atm)∣Pt(s)
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
A Write the overall cell reaction.
B Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution:
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table 19.3.1):
$Zn\left ( s \right )+ 2H^{+}\left ( aq \right )\rightarrow Zn^{2+}\left ( aq \right )+H_{2}\left ( g \right )\;\;\; E_{o}=0.76\;V$
B By substituting the given values into the simplified Nernst equation (Equation 19.64), we can calculate [H+] under nonstandard conditions:
$E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{n}\;log \left ( \dfrac{\left [ Zn^{2+} \right ]P_{H_{2}}}{\left [ H^{+} \right ]^{2}} \right ) \(0.26\;V=0.76 \; V -\left ( \dfrac{0.0591\;V}{2} \right )log \left ( \dfrac{\left ( 1.0 \right )\left ( 1.0 \right )}{\left [ H^{+} \right ]^{2}} \right ) = -0.0296\; V\; log\left ( \dfrac{K_{sp}}{1.0} \right )$
$16.9=log\left ( \dfrac{1}{\left [ H^{+} \right ]^{2}} \right )=log\left [ H^{+} \right ]^{-2} =-2\;log\left [ H^{+} \right ]$
$8.46 = -log\left [ H^{+} \right ]$
$8.5 = pH$
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows”
Pb(s) ∣Pb2+(aq, ? M)∥H+(aq), 1.0 M∣O2(g, 1.0 atm)∣Pt(s)
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table 19.7.1 and Table T1 to determine the concentration of Pb2+ in the groundwater.
Answer: 1.2 × 10−9 M
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species.
Key Takeaway
• The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution.
Key Equations
Charge on a mole of electrons (faraday)
Equation 19.4.2: F ≈ 96,486 J/(V·mol)
Maximum work from an electrochemical cell
Equation 19.4.3: wmax = −nFEcell
Relationship between Δ G ° and Δ E °
Equation 19.4.5: ΔG° = −nFE°cell
Relationship between Δ G ° and K for a redox reaction
Equation 19.4.10: ΔG° = −RT ln K
Relationship between Δ E ° and K for a redox reaction at 25°C
Equation 19.4.12: $E_{cell}^{o}=\left (\dfrac{RT}{nF} \right ) ln\;K$
Equation 19.4.13: $E_{cell}^{o}= \left ( \dfrac{0.0591}{n} \right )log\; K$
Relationship between Δ G ° and Q
Equation 19.4.14: ΔG = ΔG° + RT ln Q
Relationship between E cell and Q at 25°C
Equation 19.4.17: $E_{cell} =E_{cell}^{o}-\dfrac{0.0591\;V}{F}\;ln\;Q$
Conceptual Problems
1. State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn2+(aq) + 2e—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.
2. What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?
3. In the equation wmax = −nFE°cell, which quantities are extensive properties and which are intensive properties?
4. For any spontaneous redox reaction, E is positive. Use thermodynamic arguments to explain why this is true.
5. State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.
6. Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?
7. Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?
8. Blood analyzers, which measure pH, P(CO2) and P(O2) are frequently used in clinical emergencies. For example, blood P(CO2) is measured with a pH electrode covered with a plastic membrane that is permeable to CO2. Based on your knowledge of how electrodes function, explain how such an electrode might work. Hint: CO2(g) + H2O(l) → HCO3(aq) + H+(aq).
9. Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?
10. Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.
Answers
1. extensive: wmax and n; intensive: E°cell
2. Gold is highly resistant to corrosion because of its very positive reduction potential.
Numerical Problems
1. The chemical equation for the combustion of butane is as follows:
$C_{4}H{10}\left ( g \right )+\dfrac{13}{2}O_{2}\left ( g \right )\rightarrow 2CO_{2}\left ( g \right )+5H_{2}O\left ( g \right )$
This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?
2. How many electrons are transferred during the reaction Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg2Cl2 spontaneous? Calculate ΔG° for this reaction.
3. For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?
4. Explain why the sum of the potentials for the half-reactions Sn2+(aq) + 2e → Sn(s) and Sn4+(aq) + 2e → Sn2+(aq) does not equal the potential for the reaction Sn4+(aq) + 4e → Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.
5. Based on Table 19.3.1 and Table P1, do you agree with the proposed potentials for the following half-reactions? Why or why not?
1. Cu2+(aq) + 2e → Cu(s), E° = 0.68 V
2. Ce4+(aq) + 4e → Ce(s), E° = −0.62 V
6. For each reaction, calculate E°cell and then determine ΔG°. Indicate whether each reaction is spontaneous.
1. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
2. K2S2O6(aq) + I2(s) → 2KI(aq) + 2K2SO4(aq)
3. Sn(s) + CuSO4(aq) → Cu(s) + SnSO4(aq)
7. What is the standard change in free energy for the reaction between Ca2+ and Na(s) to give Ca(s) and Na+? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?
8. In acidic solution, permanganate (MnO4) oxidizes Cl to chlorine gas, and MnO4 is reduced to Mn2+(aq).
1. Write the balanced chemical equation for this reaction.
2. Determine E°cell.
3. Calculate the equilibrium constant.
9. Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.
Titrant (mL) E (mV)
2.00 50
6.00 100
9.00 255
10.00 960
11.00 1325
12.00 1625
14.00 1875
1. Write the balanced chemical equation for the oxidation of Fe2+ by Ce4+.
2. Plot the data and then locate the endpoint.
3. How many millimoles of Fe2+ did the solution being titrated originally contain?
10. The standard electrode potential (E°) for the half-reaction Ni2+(aq) + 2e → Ni(s) is −0.257 V. What pH is needed for this reaction to take place in the presence of 1.00 atm H2(g) as the reductant if [Ni2+] is 1.00 M?
11. The reduction of Mn(VII) to Mn(s) by H2(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:
1. MnO4(aq) + e → MnO42−(aq); E° = +0.56 V (purple → dark green)
2. MnO42−(aq) + 2e + 4H+(aq) → MnO2(s); E° = +2.26 V (dark green → dark brown solid)
3. MnO2(s) + e + 4H+(aq) → Mn3+(aq); E° = +0.95 V (dark brown solid → red-violet)
4. Mn3+(aq) + e → Mn2+(aq); E° = +1.51 V (red-violet → pale pink)
5. Mn2+(aq) + 2e → Mn(s); E° = −1.18 V (pale pink → colorless)
1. Is the reduction of MnO4 to Mn3+(aq) by H2(g) spontaneous under standard conditions? What is E°cell?
2. Is the reduction of Mn3+(aq) to Mn(s) by H2(g) spontaneous under standard conditions? What is E°cell?
12. Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:
Mn3+(aq) + e → Mn2+(aq); E° = +1.51 V
Mn3+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + e; E° = +0.95 V
1. What is E° for the disproportionation reaction?
2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
3. How could you prevent the disproportionation reaction from occurring?
13. For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?
14. The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction NAD+ + H+ + 2e → NADH; E° = −0.32 V.
1. Would NADH be able to reduce acetate to pyruvate?
2. Would NADH be able to reduce pyruvate to lactate?
3. What potential is needed to convert acetate to lactate?
acetate + CO2 + 2H+ + 2e→ pyruvate + H2O; E° = +0.70 V
pyruvare + 2H+ +2e→ lactate; E° = -0.185 V
15. Given the following biologically relevant half-reactions, will FAD (flavin adenine dinucleotide), a molecule used to transfer electrons whose reduced form is FADH2, be an effective oxidant for the conversion of acetaldehyde to acetate at pH 4.00?
acetate + 2H+ + 2e→ acetaldehyde + H2O; E° = -0.58 V
FAD + 2H+ + 2e→ FADH2; E° = -0.18 V
16. Ideally, any half-reaction with E° > 1.23 V will oxidize water as a result of the half-reaction O2(g) + 4H+(aq) + 4e → 2H2O(l).
1. Will FeO42− oxidize water if the half-reaction for the reduction of Fe(VI) → Fe(III) is FeO42−(aq) + 8H+(aq) + 3e → Fe3+(aq) + 4H2O; E° = 1.9 V?
2. What is the highest pH at which this reaction will proceed spontaneously if [Fe3+] = [FeO42−] = 1.0 M and <math display="inline" xml:id="av_1.0-19_m077"><semantics><mrow><msub><mi>P</mi><mrow><msub><mtext>O</mtext><mn>2</mn></msub></mrow></msub></mrow></semantics>[/itex] = 1.0 atm?
17. Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction
O2(g) + 4H+(aq) + 4e → 2H2O(l).
1. Will aqueous acidic KMnO4 evolve oxygen with the formation of MnO2?
2. At pH 14.00, what is E° for the oxidation of water by aqueous KMnO4 (1 M) with the formation of MnO2?
3. At pH 14.00, will water be oxidized if you are trying to form MnO2 from MnO42− via the reactio
2MnO42−(aq) + 2H2O(l) → 2MnO2(s) + O2(g) + 4OH(aq)?
18. Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CN as a complexing agent for Mn3+/Mn2+?
Mn3+(aq) + e → Mn2+(aq); E° = +1.51 V
Mn(CN)63-(aq) + e → Mn(CN)64-(aq); E° = -0.24V
19. You have constructed a cell with zinc and lead amalgam electrodes described by the cell diagram Zn(Hg)(s)∣Zn(NO3)2(aq)∥Pb(NO3)2(aq)∣Pb(Hg)(s). If you vary the concentration of Zn(NO3)2 and measure the potential at different concentrations, you obtain the following data:
Zn(NO3)2 (M) Ecell (V)
0.0005 0.7398
0.002 0.7221
0.01 0.7014
1. Write the half-reactions that occur in this cell.
2. What is the overall redox reaction?
3. What is E°cell? What is ΔG° for the overall reaction?
4. What is the equilibrium constant for this redox reaction?
20. Hydrogen gas reduces Ni2+ according to the following reaction: Ni2+(aq) + H2(g) → Ni(s) + 2H+(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.
1. What is K for this redox reaction?
2. Is this reaction likely to occur?
3. What conditions can be changed to increase the likelihood that the reaction will occur as written?
4. Is the reaction more likely to occur at higher or lower pH?
21. The silver–silver bromide electrode has a standard potential of 0.07133 V. What is Ksp of AgBr?
Answers
1. 6e; E°cell = 1.813 V; the reaction is spontaneous; ΔG° = −525 kJ/mol Al.
2. yes; E° = 0.40 V
1. yes; E° = 0.45 V
2. 0.194 V
3. yes; E° = 0.20 V
Contributors
• Anonymous
Modified by Joshua B. Halpern
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Learning Objectives
• To learn how commercial galvanic cells work.
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell)A galvanic cell (or series of galvanic cells) that contains all the reactants needed to produce electricity. is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cellA galvanic cell that requires a constant external supply of one or more reactants to generate electricity. is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
Batteries
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure 19.4.3). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
Leclanché Dry Cell
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cellA battery consisting of an electrolyte that is an acidic water-based paste containing MnO2, NH4Cl, ZnCl2, graphite, and starch. is actually a “wet cell”: the electrolyte is an acidic water-based paste containing MnO2, NH4Cl, ZnCl2, graphite, and starch (part (a) in Figure 19.5.1). The half-reactions at the anode and the cathode can be summarized as follows:
$cathode:\; 2MnO_{2}\left (s \right ) + 2NH_{4}^{+}\left ( aq \right ) + 2e^{-} \rightarrow Mn_{2}O_{3}\left ( s \right ) + 2NH_{3}\left (aq \right ) + H_{2}O\left (l \right ) \tag{19.5.1}$
$anode:\; Zn\left (s \right ) \rightarrow Zn^{2+}\left ( aq \right ) + 2e^{-} \tag{19.5.2}$
The Zn2+ ions formed by the oxidation of Zn(s) at the anode react with NH3 formed at the cathode and Cl ions present in solution, so the overall cell reaction is as follows:
$overall:\; 2MnO_{2}\left (s \right ) + 2NH_{4}Cl\left ( aq \right ) + Zn\left (s \right ) \rightarrow Mn_{2}O_{3}\left ( s \right ) + Zn\left (NH_{3} \right )_{2}Cl_{2}\left (s \right ) + H_{2}O\left (l \right ) \tag{19.5.13}$
The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the MnO2 that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the Zn anode reacts spontaneously with NH4Cl in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline batteryA battery that consists of a Leclanché cell adapted to operate under alkaline (basic) conditions. is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
$cathode:\; 2MnO_{2}\left (s \right ) + H_{2}O \left ( l \right ) + 2e^{-} \rightarrow Mn_{2}O_{3}\left ( s \right ) + 2OH^{-}\left (aq \right ) \tag{19.5.4}$
$anode:\; Zn\left (s \right ) + 2OH^{-}\left (aq \right ) \rightarrow ZnO \left ( S \right ) + H_{2}O \left ( l \right ) + 2e^{-} \tag{19.5.5}$
$overall:\; 2MnO_{2}\left (s \right ) + Zn\left (s \right ) \rightarrow Mn_{2}O_{3}\left ( s \right ) + ZnO \left (s \right ) \tag{19.5.6}$
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Button Batteries
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either HgO or Ag2O as the oxidant rather than MnO2 (part (b) in Figure 19.13). The cathode and overall reactions and cell output for these two types of button batteries are as follows:
$cathode \left ( Hg \right ):\; HgO\left (s \right ) + H_{2}O \left ( l \right ) + 2e^{-} \rightarrow Hg\left ( l \right ) + 2OH^{-}\left (aq \right ) \tag{19.5.7}$
$overall \left ( Hg \right ):\; HgO\left (s \right ) + Zn\left (s \right ) \rightarrow Hg\left (l \right ) + ZnO \left (s \right ) \;\;\; E_{cell}=1.35 \; V \tag{19.5.8}$
$cathode \left ( Ag \right ):\; Ag_{2}O\left (s \right ) + H_{2}O \left ( l \right ) + 2e^{-} \rightarrow 2Ag\left ( s \right ) + 2OH^{-}\left (aq \right ) \tag{19.5.9}$
$overall \left ( Ag \right ):\; Ag_{2}O\left (s \right ) + Zn\left (s \right ) \rightarrow 2Ag\left (s \right ) + ZnO \left (s \right ) \;\;\; E_{cell}=1.6 \; V \tag{19.5.10}$
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as Hg and Ag.
Lithium–Iodine Battery
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries.
One of the few commercially successful water-free batteries is the lithium–iodine batteryA battery that consists of an anode of lithium metal and a cathode containing a solid complex of I2, with a layer of solid LiI in between that allows the diffusion of Li+ ions. The anode is lithium metal, and the cathode is a solid complex of I2. Separating them is a layer of solid LiI, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
$cathode:\; I_{2}\left (s \right ) + 2e^{-} \rightarrow 2I^{-}\left ( LiI \right ) \tag{19.5.11}$
$anode:\; 2Li\left (s \right ) \rightarrow 2Li^{+}\left ( LiI \right )+ 2e^{-} \tag{19.5.12}$
$overall:\; 2Li\left (s \right ) + I_{2}\left (s \right ) \rightarrow 2LiI\left (s \right ) \;\;\; E_{cell}=3.5 \; V \tag{19.5.13}$
Cardiac pacemaker. An x-ray of a patient showing the location and size of a pacemaker powered by a lithium–iodine battery.
As shown in part (c) in Figure 19.5.1, a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using TiS2, for example, for the cathode.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
Nickel–Cadmium (NiCad) Battery
The nickel–cadmiumA type of battery that consists of a water-based cell with a cadmium anode and a highly oxidized nickel cathode., or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure 19.5.2, the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
The electrode reactions during the discharge of a NiCad battery are as follows:
$cathode:\; 2NiO\left ( OH \right )\left (s \right ) + 2H_{2}O\left ( l \right ) + 2e^{-} \rightarrow 2Ni\left ( OH \right )_{2}\left (s \right )+ 2OH^{-}\left ( aq \right ) \tag{19.5.14}$
$anode:\; Cd \left (s \right )+ 2OH^{-}\left ( aq \right )\rightarrow Cd\left ( OH \right )_{2} \left ( s \right )+ 2e^{-} \tag{19.5.15}$
$overall:\; Cd \left (s \right ) + 2NiO\left ( OH \right )\left (s \right )+2H_{2}O\left ( l \right ) \rightarrow Cd\left ( OH \right )_{2} \left ( s \right ) + 2Ni\left ( OH \right )_{2}\left (s \right )\;\;\; E_{cell}=1.4 \; V \tag{19.5.16}$
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
$NiO\left ( OH \right )\left ( s \right )+MH \rightarrow Ni\left ( OH \right )_{2}\left ( s \right ) + M\left ( s \right )$
The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
Lead–Acid (Lead Storage) Battery
The lead–acid batteryA battery consisting of a plate or grid of spongy lead metal, a cathode containing powdered PbO2, and an electrolyte that is usually an aqueous solution of H2SO4 is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells. As shown in Figure 19.5.3, the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide (PbO2). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M H2SO4). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
$cathode:\; 2PbO_{2}\left (s \right ) + HSO_{4}^{-}\left ( aq \right ) + 3H^{+}\left ( aq \right ) +2e^{-} \rightarrow PbSO_{4}\left (s \right )+ 2H_{2}O\left ( l \right ) \;\;\; E_{cathode}=1.685 \; V\tag{19.5.17}$
$anode:\; PbS \left (s \right )+ HSO_{4}^{-}\left ( aq \right ) \rightarrow PbSO_{4}\left (s \right )+ H^{+}\left ( aq \right ) + 2e^{-} \;\;\; E_{anode}=-0.356 \; V\tag{19.5.18}$
$overall:\; 2PbO_{2}\left (s \right ) + 2HSO_{4}^{-}\left ( aq \right ) + 2H^{+}\left ( aq \right ) \rightarrow 2bSO_{4}\left (s \right )+ 2H_{2}O\left ( l \right ) \;\;\; E_{cell}=2.041 \; V \tag{19.5.19}$
As the cell is discharged, a powder of PbSO4 forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte.
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and PbSO4 is converted back to metallic lead and PbO2. If the battery is recharged too vigorously, however, electrolysis of water can occur, resulting in the evolution of potentially explosive hydrogen gas. (For more information on electrolysis, see Section 19.7.) The gas bubbles formed in this way can dislodge some of the PbSO4 or PbO2 particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
Fuel Cells
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in Figure 19.5.4 . The electrode reactions are as follows:
$cathode:\; O_{2} \left (g \right )+ 4H^{+} +4e^{-} \rightarrow 2H_{2}O\left ( g \right ) \tag{19.5.20}$
$anode:\; 2H_{2} \left (g \right ) \rightarrow 4H^{+} + 4e^{-} \tag{19.5.21}$
$overall:\; 2H_{2} \left (g \right ) + O_{2}\left ( g \right ) \rightarrow 2H_{2}O\left ( g \right ) \tag{19.5.22}$
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of O2(g) at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of O2.
Summary
A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction.
Key Takeaway
• Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass.
Conceptual Problems
1. What advantage is there to using an alkaline battery rather than a Leclanché dry cell?
2. Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?
3. What type of battery would you use for each application and why?
1. powering an electric motor scooter
2. a backup battery for a smartphone
3. powering an iPod
4. Why are galvanic cells used as batteries and fuel cells? What is the difference between a battery and a fuel cell? What is the advantage to using highly concentrated or solid reactants in a battery?
Answer
1. lead storage battery
2. lithium–iodine battery
3. NiCad, NiMH, or lithium ion battery (rechargeable)
Numerical Problem
1. This reaction is characteristic of a lead storage battery:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?
Answer
1. [H2SO4] = 3.52 M; E > E°
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.06%3A_Commercial_Galvanic_Cells.txt |
Learning Objectives
• To understand the process of corrosion.
CorrosionA galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Note the Pattern
Corrosion is a galvanic process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide (Fe2O3·xH2O), commonly known as rust, that does not provide a tight protective film (Figure 19.17). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil saturated with oxygen will not rust because of the absence of water.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
$cathode:\; O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) +4 e^{-} \rightarrow 2H_{2}O\left ( l \right ) \;\;\; E^{o}=1.23 \; V \tag{19.6.1}$
$anode:\; Fe\left ( s \right )\rightarrow Fe^{2+} \left ( aq \right ) +2e^{-} \;\;\; E^{o}=-0.24 \; V \tag{19.6.2}$
$overall:\; 2Fe\left ( s \right ) + O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) \rightarrow 2Fe^{2+} \left ( aq \right ) + 2H_{2}O \left ( l \right ) \;\;\; E^{o}=1.68 \; V \tag{19.6.3}$
The Fe2+ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing Fe3+, as represented in the following equation:
$4Fe^{2+} \left ( aq \right ) + O_{2}\left ( g \right ) + \left ( 2+4x \right )H_{2}O \left ( l \right )\rightarrow 2Fe_{2}O_{3}\cdot x H_{2}O +4H^{+} \tag{19.6.4}$
The sign and magnitude of E° for the corrosion process (Equation 19.6.3) indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3 provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure 19.6.2).
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process that will be discussed in Section 19.7. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually made of steel coated with a thin layer of tin. Neither chromium nor tin is intrinsically resistant to corrosion, but both form protective oxide coatings.
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rat that Fe is more easily oxidized than Sn. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure 19.6.3). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as Zn (E° = −0.76 V for Zn2+ + 2e → Zn) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
$cathode:\; O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) +4 e^{-} \rightarrow 2H_{2}O\left ( l \right ) \tag{19.6.5}$
$anode:\; Zn\left ( s \right )\rightarrow Zn^{2+} \left ( aq \right ) +2e^{-} \;\;\; E^{o}=-0.24 \; V \tag{19.6.6}$
$overall:\; 2Zn\left ( s \right ) + O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) \rightarrow 2Zn^{2+} \left ( aq \right )+ 2H_{2}O \left ( l \right ) \ \tag{19.6.7}$
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans. In a similar strategy, sacrificial electrodesAn electrode containing a more reactive metal that is attached to a metal object to inhibit that object’s corrosion. using magnesium, for example, are used to protect underground tanks or pipes (Figure 19.20). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example 19.6.1
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
1. If the boat is immersed in seawater, what corrosion reaction will occur? What is E°cell?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, E°cell, and preventive measures
Strategy:
A Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate E°cell.
B Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution:
1. A According to Table 19.3.1, both copper and tin are less active metals than iron (i.e., they have higher positive values of E° than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which O2 is reduced, and the iron screws will act as anodes at which iron dissolves:
$cathode:\; O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) +4 e^{-} \rightarrow 2H_{2}O\left ( l \right ) \;\;\; E_{cathode}^{o} = 1.23 \; V$
$anode:\; Fe\left ( s \right )\rightarrow Fe^{2+} \left ( aq \right ) + 2e^{-} \;\;\; E_{anode}^{o}=-0.45 \; V$
$overall:\; 2Fe\left ( s \right ) + O_{2}\left ( g \right )+ 4H^{+}\left ( aq \right ) \rightarrow 2Fe^{2+} \left ( aq \right )+ 2H_{2}O \left ( l \right ) \;\; E_{cell}^{o}=1.68 \; V$
Over time, the iron screws will dissolve, and the boat will fall apart.
2. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation 19.102).
Exercise
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
1. Do you accept his proposal?
2. What else should you have the plumber do while at your home?
Answer
1. Not unless you plan to sell the house very soon because the Cu/Fe pipe joints will lead to rapid corrosion.
2. Any existing Pb/Fe joints should be examined carefully for corrosion of the iron pipes due to the Pb–Fe junction; the less active Pb will have served as the cathode for the reduction of O2, promoting oxidation of the more active Fe nearby.
Summary
The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it.
Key Takeaway
• Corrosion is a galvanic process that can be prevented using cathodic protection.
Conceptual Problems
1. Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?
2. What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?
3. Why is it important for automobile manufacturers to apply paint to the metal surface of a car? Why is this process particularly important for vehicles in northern climates, where salt is used on icy roads?
Answer
1. Paint keeps oxygen and water from coming into direct contact with the metal, which prevents corrosion. Paint is more necessary because salt is an electrolyte that increases the conductivity of water and facilitates the flow of electric current between anodic and cathodic sites.
Contributors
• Anonymous
Modified by Joshua B. Halpern
Thumbnail from Wikimedia | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.07%3A_Corrosion.txt |
Learning Objectives
• To understand electrolysis and describe it quantitatively.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysisAn electrochemical process in which an external voltage is applied to an electrolytic cell to drive a nonspontaneous reaction., occurs: an external voltage is applied to drive a nonspontaneous reaction (Figure 19.1.1). In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
Note the Pattern
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.Electrolytic Cells
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a 1 M Cd2+ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (part (a) in Figure 19.7.19). The overall reaction is as follows:
$Cd\left ( s \right ) +Cu^{2+}\left ( aq \right )\rightarrow Cu\left ( s \right ) +Cd^{2+}\left ( aq \right ) \tag{19.7.1}$
This reaction is thermodynamically spntaneous as written (
$\Delta G^{o}=nFE_{cell}^{o}= \left ( 2\; \cancel{mol \; e^{-}} \right )\left [ 96,486 \;J/\left ( \cancel{V}\cdot \cancel{mol} \right ) \right ]\left ( 0.74 \; \cancel{V} \right ) =-140 \; kJ/mol \tag{19.7.2}$
In this direction, the system is acting as a galvanic cell.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (part (b) in Figure 19.21). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
$cathode:\; Cd^{2+}\left ( aq \right )+ 2e^{-} \rightarrow Cd\left ( s \right ) \;\;\; E_{cathode}^{o} = -0.40 \; V \tag{19.7.3}$
$anode:\; Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right ) + 2e^{-} \;\;\; E_{anode}^{o}= 0.34 \; V \tag{19.7.4}$
$overall:\; Cd^{2+}\left ( aq \right ) + Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right )+ Cd \left ( s \right ) \;\; E_{cell}^{o}=-0.74 \; V \tag{19.7.5}$
Because E°cell < 0, the overall reaction—the reduction of Cd2+ by Cu—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table 19.7.1.
Table 19.7.1 Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
E cell > 0 < 0
Electrode Process
anode oxidation oxidation
cathode reduction reduction
Sign of Electrode
anode +
cathode +
Electrolytic Reactions
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten NaCl, for example, and an electrical potential is applied, Cl is oxidized at the anode, and Na+ is reduced at the cathode. The overall reaction is as follows:
$2NaCl\left ( l \right ) \rightarrow Na\left ( l \right ) +Cl_{2}\left ( g \right ) \tag{19.7.6}$
This is the reverse of the formation of NaCl from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of NaCl (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten NaCl in a Downs cell (Figure 19.7.2). In this specialized cell, CaCl2 (melting point = 772°C) is first added to the NaCl to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
$2Al_{2}O_{3}\left ( l \right ) + 3C\left ( s \right ) \rightarrow 4Al\left ( l \right ) + 3CO_{2}\left ( g \right ) \tag{19.7.7}$
Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
1. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table 19.3.1 and Table P1, which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions.
2. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl2 is electrolyzed, Cl is oxidized because it is the only anion present, but either Na+ or Ca2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al3+), but there are three species that can be oxidized: C, O2−, and F.
In the Hall–Heroult process, C is oxidized instead of O2− or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
Example 19.7.1
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
A List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 3.3.9, determine which species will be reduced and which species will be oxidized.
B Identify the products that will form at each electrode.
Solution:
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer: Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (Figure 19.7.3). The reactions that occur are as follows:
$cathode:\; 2H^{+}\left ( aq \right )+ 2e^{-} \rightarrow H_{2}\left ( g \right ) \;\;\; E_{cathode}^{o} = 0 \; V \tag{19.7.8}$
$anode:\; 2H_{2}O\left ( l \right ) \rightarrow O_{2}\left ( g \right ) + 4H^{+} \left ( g \right ) + 4e^{-} \;\;\; E_{anode}^{o}= 1.23 \; V \tag{19.7.9}$
$overall:\; 2H_{2}O\left ( l \right )\rightarrow O_{2}\left ( g \right )+ 2H_{2}\left ( g \right ) \;\; E_{cell}^{o}=-1.23 \; V \tag{19.7.10}$
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH] = 1.0 × 10−7. Assuming that P(O2) = P(H2) = 1 atm, we can use the standard potentials and Equation 19.4.17 to calculate E for the overall reaction:
$E_{cell}= E_{cell}^{o}-\left ( \dfrac{0.0591 \; V}{n} \right )log\;\left ( P_{O_{2}}P_{H_{2}}^{2} \right ) \tag{19.7.11}$
$E_{cell}= -1.23 \; V -\left ( \dfrac{0.0591 \; V}{4} \right )log\;\left ( 1 \right ) = -1.23 \; V$
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltageThe voltage that must be applied in electrolysis in addition to the calculated (theoretical) value to overcome factors such as a high activation energy and the formation of bubbles on a surface., represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3, SO42−, PO43−, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
Note the Pattern
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
Electroplating
In a process called electroplatingA process in which a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis., a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure 19.7.4.
The half-reactions in electroplating a fork, for example, with silver are as follows:
$cathode\left ( fork \right ):\; Ag^{+}\left ( aq \right )+ e^{-} \rightarrow Ag\left ( s \right ) \;\;\; E_{cathode}^{o} = 0.80 \; V \tag{19.7.12}$
$anode\left ( silver bar \right ):\; Ag\left ( s \right ) \rightarrow Ag^{+} \left ( aq \right ) + e^{-} \;\;\; E_{anode}^{o}= 0.80 \; V \tag{19.7.13}$
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because E°cell = 0 V, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
Quantitative Considerations
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction Ag+(aq) + e → Ag(s), 1 mol of electrons reduces 1 mol of Ag+ to Ag metal. In contrast, in the reaction Cu2+(aq) + 2e → Cu(s), 1 mol of electrons reduces only 0.5 mol of Cu2+ to Cu metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,486 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (C) transferred is the product of the current (A) and the time (t, in seconds):
$C=A\times t \tag{19.7.14}$
The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of CuSO4 for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
$charge=\left (0.60 \;A \right )\left ( 6.0 \; \cancel{min} \right ) \left ( 60 \; s/\cancel{min} \right ) =220 A \cdot s \tag{19.7.15}$
The number of moles of electrons transferred to Cu2+ is therefore
$moles \; e^{-}= \dfrac{220 \; \cancel{C}}{96,486 \; \cancel{C}/mol}=2.3\times 10^{-3} \; mol \; e^{-} \tag{19.7.16}$
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
Example 19.7.2
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
A Calculate the number of moles of metal corresponding to the given mass transferred.
B Write the reaction and determine the number of moles of electrons required for the electroplating process.
C Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution:
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
$moles \; Ag= \dfrac{2.0 \; \cancel{g}}{107.868 \; \cancel{g}/mol}=1.85\times 10^{-2} \; mol \; Ag$
B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
$Coulombs=\left ( 1.85\times 10^{-2} \; \cancel{mol} \right )\left ( 96,486 \; C/\cancel{mol} \right ) =1.78 \times 10^{3} \; C$
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
$amperes=\dfrac{1.78 \times 10^{3} \; C}{ \left ( 12.0 \; \cancel{h} \right ) \left ( 60 \; \cancel{min}/ \cancel{h} \right ) \left ( 60 \; s/\cancel{min} \right ) }$
$\;\;\;\; =4.12\times 10^{-2} \; C/s = 4.12\times 10^{-2} \; A$
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer: 5.8 h
Summary
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. A Downs cell is used to produce sodium metal from a mixture of salts, and the Hall–Heroult process is used to produce aluminum commercially. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.
Key Takeaways
• In electrolysis, an external voltage is applied to drive a nonspontaneous reaction.
• The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred.
Key Equation
Relationship of charge, current and time
Equation 19.116: C = A × t
Conceptual Problems
1. Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?
2. How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?
3. Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?
4. Two solutions, one containing Fe(NO3)2·6H2O and the other containing the same molar concentration of Fe(NO3)3·6H2O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer.
Numerical Problems
1. The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?
1. AlCl3
2. MgCl2
3. FeCl3
2. Electrolysis is the most direct way of recovering a metal from its ores. However, the Na+(aq)/Na(s), Mg2+(aq)/Mg(s), and Al3+(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?
3. What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl2 is electrolyzed using a current of 12.4 A for 1.0 h?
4. What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO3)2 solution for 1.5 h.
5. What mass of PbO2 is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery?
6. Electrolysis of Cr3+(aq) produces Cr2+(aq). If you had 500 mL of a 0.15 M solution of Cr3+(aq), how long would it take to reduce the Cr3+ to Cr2+ using a 0.158 A current?
7. Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
1. AgNO3
2. RbI
8. Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
1. MgBr2
2. Hg(CH3CO2)2
3. Al2(SO4)3
Answers
1. 5.2 L
1. cathode: Ag(s); anode: O2(g);
2. cathode: H2(g); anode: I2(s)
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.08%3A_Electrolysis.txt |
• 20.1: Introduction
• 20.2: Overview of Periodic Trends
The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms.
• 20.3: The Chemistry of Hydrogen
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell.
• 20.4: The Alkali Metals (Group 1)
The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M+ ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts.
• 20.5: The Alkaline Earth Metals (Group 2)
Group 2 elements almost exclusively form ionic compounds containing the M2+ ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations.
• 20.6: The s-Block Elements in Biology
• 20.E: Periodic Trends and the s-Block Elements (Exercises)
20: Periodic Trends and the s-Block Elements
In previous chapters, we used the principles of chemical bonding, thermodynamics, and kinetics to provide a conceptual framework for understanding the chemistry of the elements. Beginning in Chapter 21, we use the periodic table to guide our discussion of the properties and reactions of the elements and the synthesis and uses of some of their commercially important compounds. We begin this chapter with a review of periodic trends as an introduction, and then we describe the chemistry of hydrogen and the other s-block elements. In Chapter 22, we consider the chemistry of the p-block elements; Chapter 23 presents the transition metals, in which the d-subshell is being filled. In this chapter, you will learn why potassium chloride is used as a substitute for sodium chloride in a low-sodium diet, why cesium is used as a photosensor, why the heating elements in electric ranges are coated with magnesium oxide, and why exposure to a radioactive isotope of strontium is more dangerous for children than for adults.
Flame tests. Heating a compound in a very hot flame results in the formation of its component atoms in electronically excited states. When an excited atom decays to the ground state, it emits light. Each element emits light at characteristic frequencies. Flame tests are used to identify many elements based on the color of light emitted in the visible region of the electromagnetic spectrum. As shown here, sodium compounds produce an intense yellow light, whereas potassium compounds produce a crimson color.
Contributors
• Anonymous
Modified by Joshua B. Halpern
Video by Sully Science from YouTube https://www.youtube.com/watch?v=jJvS4uc4TbU | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.01%3A_Introduction.txt |
Learning Objectives
• To know important periodic trends in several atomic properties.
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries.
The most important periodic trends in atomic properties are summarized in Figure \(1\). Recall that these trends are based on periodic variations in a single fundamental property, the effective nuclear charge (Zeff), which increases from left to right and from top to bottom in the periodic table.
The diagonal line in Figure \(1\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations.
Unique Chemistry of the Lightest Elements
The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms.
In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group.
Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF3 forms only the four-coordinate, tetrahedral BF4 ion, whereas under the same conditions AlF3 forms the six-coordinate, octahedral AlF63− ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H2O).
Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF6 have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results.
One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N2 contains an N≡N bond, but each phosphorus atom in tetrahedral P4 forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements.
Another important trend to note in main group chemistry is the chemical similarity between the lightest element of one group and the element immediately below and to the right of it in the next group, a phenomenon known as the diagonal effect (Figure \(2\)) There are, for example, significant similarities between the chemistry of Li and Mg, Be and Al, and B and Si. Both BeCl2 and AlCl3 have substantial covalent character, so they are somewhat soluble in nonpolar organic solvents. In contrast, although Mg and Be are in the same group, MgCl2 behaves like a typical ionic halide due to the lower electronegativity and larger size of magnesium.
The Inert-Pair Effect
The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths.
In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.
The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2electrons, as shown in Table \(1\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.
Table \(1\): Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
Element Electron Configuration I1 (kJ/mol) I1 + I2 + I3 (kJ/mol) Average M–Cl Bond Energy (kJ/mol)
B [He] 2s22p1 801 6828 536
Al [Ne] 3s23p1 578 5139 494
Ga [Ar] 3d104s24p1 579 5521 481
In [Kr] 4d105s2p1 558 5083 439
Tl [Xe] 4f145d106s2p1 589 5439 373
Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).
Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(1\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.
The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.
Example \(1\)
Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +1 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of another group will exhibit chemistry most similar to that of Al.
Given: positions of elements in the periodic table
Asked for: classification, oxidation-state stability, and chemical reactivity
Strategy:
From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.
Solution
1. Group 13 spans the diagonal line separating the metals from the nonmetals. Although Al and B both lie on the diagonal line, only B is a semimetal; the heavier elements are metals.
2. All five elements in group 13 have an ns2np1 valence electron configuration, so they are expected to form ions with a +3 charge from the loss of all valence electrons. The inert-pair effect should be most important for the heaviest element (Tl), so it is most likely to form compounds in an oxidation state that is lower by 2. Thus the +1 oxidation state is predicted to be most important for thallium.
3. Among the main group elements, the lightest member of each group exhibits unique chemistry because of its small size resulting in a high concentration of charge, energetically unavailable d orbitals, and a tendency to form multiple bonds. In group 13, we predict that the chemistry of boron will be quite different from that of its heavier congeners.
4. Within the s and p blocks, similarities between elements in different groups are most marked between the lightest member of one group and the element of the next group immediately below and to the right of it. These elements exhibit similar electronegativities and charge-to-radius ratios. Because Al is the second member of group 13, we predict that its chemistry will be most similar to that of Be, the lightest member of group 2.
Exercise \(1\)
Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +2 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of group 14 will be chemically most similar to a group 15 element.
Answer
1. nonmetal: C; semimetals: Si and Ge; metals: Sn and Pb
2. Pb is most stable as M2+.
3. C is most different.
4. C and P are most similar in chemistry.
Summary
The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms. The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.02%3A_Overview_of_Periodic_Trends.txt |
Learning Objectives
• To describe the physical and chemical properties of hydrogen and predict its reactivity.
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s1 electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in Group 17 because the addition of a single electron to a hydrogen atom completes its valence shell.
Although hydrogen has an ns1 electron configuration, its chemistry does not resemble that of the Group 1 metals.
Isotopes of Hydrogen
Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table $1$ "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is protium (1H or H), followed by deuterium (2H or D), which has an additional neutron. The rarest isotope of hydrogen is tritium (3H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to 3He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes.
Table $1$: The Isotopes of Hydrogen
Protium Deuterium Tritium
symbol $\mathrm{_1^1H}$ $\mathrm{_1^2H}$ $\mathrm{_1^3H}$
neutrons 0 1 2
mass (amu) 1.00783 2.0140 3.01605
abundance (%) 99.9885 0.0115 ~10−17
half-life (years) 12.32
boiling point of X2 (K) 20.28 23.67 25
melting point/boiling point of X2O (°C) 0.0/100.0 3.8/101.4 4.5/?
The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H2, D2, and T2 differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction.
Harold Urey (1893–1981)
Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium (2H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth.
Because the normal boiling point of D2O is 101.4°C (compared to 100.0°C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D+) has twice the mass of a proton (H+), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H2 over D2 or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D2O (1.1044 g/cm3 at 25°C) is greater than that of H2O (0.99978 g/cm3). Heavy water was an important constituent of early nuclear reactors.
Because deuterons diffuse so much more slowly, D2O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D+ is so slow compared with that of H+ because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions.
Bonding in Hydrogen and Hydrogen-Containing Compounds
The 1s1 electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure $1$):
1. Losing its electron to form a proton (H+) with an empty 1s orbital. The proton is a Lewis acid that can accept a pair of electrons from another atom to form an electron-pair bond. In the acid–base reactions, the proton always binds to a lone pair of electrons on an atom in another molecule to form a polar covalent bond. If the lone pair of electrons belongs to an oxygen atom of a water molecule, the result is the hydronium ion (H3O+).
2. Accepting an electron to form a hydride ion (H), which has a filled 1s2 orbital. Hydrogen reacts with relatively electropositive metals, such as the alkali metals (group 1) and alkaline earth metals (group 2), to form ionic hydrides, which contain metal cations and H ions.
3. Sharing its electron with an electron on another atom to form an electron-pair bond. With a half-filled 1s1 orbital, the hydrogen atom can interact with singly occupied orbitals on other atoms to form either a covalent or a polar covalent electron-pair bond, depending on the electronegativity of the other atom.
Hydrogen can also act as a bridge between two atoms. One familiar example is the hydrogen bond, an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure $2$). An example of this kind of interaction is the hydrogen bonding network found in water (Figure $2$). Hydrogen can also form a three-center bond (or electron-deficient bond), in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds.
Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp3 orbitals on two boron atoms (Figure $3$). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures.
Hydrogen can lose its electron to form H+, accept an electron to form H, share its electron, hydrogen bond, or form a three-center bond.
Synthesis, Reactions, and Compounds of Hydrogen
The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure $4$). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K).
The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid:
$M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}$
Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base:
$\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}$
Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D2O:
$MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}$
On an industrial scale, H2 is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H2 known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst:
$\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}$
Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H2(g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities.
Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework.
Summary and Key Takeaway
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H+), a hydride ion (H), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.03%3A_The_Chemistry_of_Hydrogen.txt |
Learning Objectives
1. To describe how the alkali metals are isolated.
2. To be familiar with the reactions, compounds, and complexes of the alkali metals.
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
Humphry Davy (1778–1829)
Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl3), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH3) and hydrogen telluride (H2Te), both of which are highly toxic.
Robert Wilhelm Bunsen (1811–1899)
Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line).
Preparation of the Alkali Metals
Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides:
$\mathrm{LiCl(l)}\rightarrow\mathrm{Li(l)}+\frac{1}{2}\mathrm{Cl_2(g)} \label{21.15}$
In practice, CaCl2 is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O2 and water because Li reacts with nitrogen gas to form lithium nitride (Li3N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl2. In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg:
$2RbOH_{(s)} + Mg_{(s)} \rightarrow 2Rb_{(l)} + Mg(OH)_{2(s)} \label{21.6}$
Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates.
Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al3+ from the ore; basic precipitation to remove Al3+ from the mixture as Al(OH)3; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. Figure $1$ illustrates the isolation of liquid lithium from a lithium silicate ore by this process.
General Properties of the Alkali Metals
Various properties of the group 1 elements are summarized in Table $1$. In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns2 electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C).
Table $1$: Selected Properties of the Group 1 Elements
Lithium Sodium Potassium Rubidium Cesium Francium
*The values cited are for four-coordinate ions except for Rb+ and Cs+, whose values are given for the six-coordinate ion.
atomic symbol Li Na K Rb Cs Fr
atomic number 3 11 19 37 55 87
atomic mass 6.94 22.99 39.10 85.47 132.91 223
valence electron configuration 2s1 3s1 4s1 5s1 6s1 7s1
melting point/boiling point (°C) 180.5/1342 97.8/883 63.5/759 39.3/688 28.5/671 27/—
density (g/cm3) at 25°C 0.534 0.97 0.89 1.53 1.93
atomic radius (pm) 167 190 243 265 298
first ionization energy (kJ/mol) 520 496 419 403 376 393
most common oxidation state +1 +1 +1 +1 +1 +1
ionic radius (pm)* 76 102 138 152 167
electron affinity (kJ/mol) −60 −53 −48 −47 −46
electronegativity 1.0 0.9 0.8 0.8 0.8 0.7
standard electrode potential (E°, V) −3.04 −2.71 −2.93 −2.98 −3.03
product of reaction with O2 Li2O Na2O2 KO2 RbO2 CsO2
type of oxide basic basic basic basic basic
product of reaction with N2 Li3N none none none none
product of reaction with X2 LiX NaX KX RbX CsX
product of reaction with H2 LiH NaH KH RbH CsH
The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (Table $1$). Because Li+ is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li+ more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends.
Reactions and Compounds of the Alkali Metals
All alkali metals are electropositive elements with an ns1 valence electron configuration, forming the monocation (M+) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M+ ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds.
All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where $X$ is a halogen:
$2M_{(s)} + X_{2(s, l, g)} \rightarrow 2M^+X^−_{(s)} \label{21.7}$
Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te:
$2M_{(s)} + Y_{(s)} \rightarrow M_2Y_{(s)} \label{21.8}$
When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na2Sn, where n = 2–6). For example, Na2S3 contains the S32− ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S3) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine (Figure $2$).
Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li2O (white), Na2O2 (pale yellow), KO2 (orange), RbO2 (brown), and CsO2 (orange). Only Li2O has the stoichiometry expected for a substance that contains two M+ cations and one O2− ion. In contrast, Na2O2 contains the O22− (peroxide) anion plus two Na+ cations. The other three salts, with stoichiometry MO2, contain the M+ cation and the O2 (superoxide) ion. Because O2− is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li+). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na+ cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO2).
The chemistry of the alkali metals is largely that of ionic compounds containing the M+ ions.
The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na2O2 is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na2O2 and KO2 are used to purify and regenerate the air by removing the CO2 produced by respiration and replacing it with O2. Both compounds react with CO2 in a redox reaction in which O22− or O2 is simultaneously oxidized and reduced, producing the metal carbonate and O2:
$2Na_2O_{2(s)} + 2CO_{2(g)} \rightarrow 2Na_2CO_{3(s)} + O_{2(g)} \label{21.9}$
$4KO_{2(s)} + 2CO_{2(g)} \rightarrow 2K_2CO_{3(s)} + 3O_{2(g)} \label{21.10}$
The presence of water vapor, the other product of respiration, makes KO2 even more effective at removing CO2 because potassium bicarbonate, rather than potassium carbonate, is formed:
$4KO_{2(s)} + 4CO_{2(g)} + 2H_2O_{(g)} \rightarrow 4KHCO_{3(s)} + 3O_{2(g)} \label{21.11}$
Notice that 4 mol of CO2 are removed in this reaction, rather than 2 mol in Equation 21.10.
Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li3N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K+ cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As):
$12M_{(s)} + Z_{4(s)} \rightarrow 4M_3Z_{(s)} \label{21.12}$
Because of lattice energies, only lithium forms a stable oxide and nitride.
The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K4Si4 whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M2C2 (where M is Li or Na):
$2M_{(s)} + 2C_{(s)} \rightarrow M_2C_{2(s)} \label{21.13}$
The same compounds can be obtained by reacting the metal with acetylene (C2H2). In this reaction, the metal is again oxidized, and hydrogen is reduced:
$2M_{(s)} + C_2H_{2(g)} \rightarrow M_2C_{2(s)} + H_{2(g)} \label{21.14}$
The acetylide ion (C22−), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq).
The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called graphite intercalation compounds (part (a) in Figure $3$). The stoichiometries of these compounds include MC60 and MC48, which are black/gray; MC36 and MC24, which are blue; and MC8, which is bronze (part (b) in Figure $3$). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K+C8.
All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M+H):
$2M_{(s)} + H_{2(g)} \rightarrow 2MH_{(s)} \label{21.15a}$
All are also capable of reducing water to produce hydrogen gas:
$\mathrm{M(s)}+\mathrm{H_2O(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{MOH(aq)} \label{21.16}$
Although lithium reacts rather slowly with water, sodium reacts quite vigorously (Figure $4$), and the heavier alkali metals (K, Rb, and Cs) react so vigorously that they invariably explode. This trend, which is not consistent with the relative magnitudes of the reduction potentials of the elements, serves as another example of the complex interplay of different forces and phenomena—in this case, kinetics and thermodynamics. Although the driving force for the reaction is greatest for lithium, the heavier metals have lower melting points. The heat liberated by the reaction causes them to melt, and the larger surface area of the liquid metal in contact with water greatly accelerates the reaction rate.
Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water:
$2MOH_{(aq)} + H_2SO_{4(aq)} \rightarrow M_2SO_{4(aq)} + 2H_2O_{(l)} \label{21.17}$
$MOH_{(aq)} + HNO_{3(aq)} \rightarrow MNO_{3(aq)} + H_2O_{(l)} \label{21.18}$
Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH3CO2Na) by reacting sodium hydroxide and acetic acid:
$CH_3CO_2H_{(aq)} + NaOH_{(s)} \rightarrow CH_3CO_2Na_{(aq)} + H_2O_{(l)} \label{21.19}$
Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH3(CH2)14CO2H] and stearic acid [CH3(CH2)16CO2H]. Lithium salts, such as lithium stearate [CH3(CH2)14CO2Li], are used as additives in motor oils and greases.
Complexes of the Alkali Metals
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes. Complex formation is most significant for the smallest cation (Li+) and decreases with increasing radius. In aqueous solution, for example, Li+ forms the tetrahedral [Li(H2O)4]+ complex. In contrast, the larger alkali metal cations form octahedral [M(H2O)6]+ complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li+ and Na+ ions (such as Na2SO4) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration.
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases.
Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in Chapter 13, crown ethers are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li+, whereas 18-crown-6 forms the strongest complexes with K+.
Cryptands are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in Figure 13.7). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO4 in nonpolar organic solvents.
Liquid Ammonia Solutions
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2) rather than hydroxide:
$\mathrm{M(s)}+\mathrm{NH_3(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{M^+(am)}+\mathrm{NH_2^-(am)} \label{21.20}$
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation 21.20 tends to be rather slow. In many cases, the alkali metal amide salt (MNH2) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e, NH3), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure $5$) and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M+), the neutral metal atom (M), metal dimers (M2), and the metal anion (M). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M+NH2 and hydrogen gas (Equation 21.20). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry.
Organometallic Compounds of the Group 1 Elements
Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called organometallic compounds. The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH3), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi)n, where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in Figure $6$, where each triangular face of the Li4 tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH3 group is using a single pair of electrons in an sp3 hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes. Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text.
• The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components.
• Organosodium and organopotassium compounds are more ionic than organolithium compounds. They contain discrete M+ and R ions and are insoluble or only sparingly soluble in nonpolar solvents.
Uses of the Alkali Metals
Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response.
Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na2CO3, used in the manufacture of glass; K2O, used in porcelain glazes; and Na4SiO4, used in detergents.
Several other alkali metal compounds are also important. For example, Li2CO3 is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder.
$1$
For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case.
1. For a reaction that requires a strong base in a solution of tetrahydrofuran (THF), would you use LiOH or CsOH?
2. To extinguish a fire caused by burning lithium metal, would you use water, CO2, N2 gas, or sand (SiO2)?
3. Both LiNO3and CsNO3 are highly soluble in acetone (2-propanone). Which of these alkali metal salts would you use to precipitate I from an acetone solution?
Given: application and selected alkali metals
Asked for: appropriate metal for each application
Strategy:
Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application.
Solution
1. Both LiOH and CsOH are ionic compounds that contain the hydroxide anion. Li+, however, is much smaller than Cs+, so the Li+ cation will be more effectively solvated by the oxygen of THF with its lone pairs of electrons. This difference will have two effects: (1) LiOH is likely to be much more soluble than CsOH in the nonpolar solvent, which could be a significant advantage, and (2) the solvated Li+ ions are less likely to form tight ion pairs with the OH ions in the relatively nonpolar solution, making the OH more basic and thus more reactive. Thus LiOH is the better choice.
2. Lithium is a potent reductant that reacts with water to form LiOH and H2 gas, so adding a source of hydrogen such as water to a lithium fire is likely to produce an explosion. Lithium also reacts with oxygen and nitrogen in the air to form Li2O and Li3N, respectively, so we would not expect nitrogen to extinguish a lithium fire. Because CO2 is a gaseous molecule that contains carbon in its highest accessible oxidation state (+4), adding CO2 to a strong reductant such as Li should result in a vigorous redox reaction. Thus water, N2, and CO2 are all unsuitable choices for extinguishing a lithium fire. In contrast, sand is primarily SiO2, which is a network solid that is not readily reduced. Smothering a lithium fire with sand is therefore the best choice.
3. The salt with the smaller cation has the higher lattice energy, and high lattice energies tend to decrease the solubility of a salt. However, the solvation energy of the cation is also important in determining solubility, and small cations tend to have higher solvation energies. Recall that high solvation energies tend to increase the solubility of ionic substances. Thus CsI should be the least soluble of the alkali metal iodides, and LiI the most soluble. Consequently, CsNO3 is the better choice.
$1$
Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application.
1. drying agent for an organic solvent—Li2SO4 or Rb2SO4
2. removing trace amounts of N2 from highly purified Ar gas—Li, K, or Cs
3. reacting with an alkyl halide (formula RX) to prepare an organometallic compound (formula MR)—Li or K
Answer
1. Li2SO4
2. Li
3. Li
$2$
Predict the products of each reaction and then balance each chemical equation.
1. Na(s) + O2(g) →
2. Li2O(s) + H2O(l) →
3. K(s) + CH3OH(l) →
4. Li(s) + CH3Cl(l) →
5. Li3N(s) + KCl(s) →
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
A Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction.
B If a reaction is predicted to occur, balance the chemical equation.
Solution
1. A Sodium is a reductant, and oxygen is an oxidant, so a redox reaction is most likely. We expect an electron to be transferred from Na (thus forming Na+) to O2. We now need to determine whether the reduced product is a superoxide (O2), peroxide (O22−), or oxide (O2−). Under normal reaction conditions, the product of the reaction of an alkali metal with oxygen depends on the identity of the metal. Because of differences in lattice energy, Li produces the oxide (Li2O), the heavier metals (K, Rb, Cs) produce the superoxide (MO2), and Na produces the peroxide (Na2O2).
B The balanced chemical equation is 2Na(s) + O2(g) → Na2O2(s).
1. A Li2O is an ionic salt that contains the oxide ion (O2−), which is the completely deprotonated form of water and thus is expected to be a strong base. The other reactant, water, is both a weak acid and a weak base, so we can predict that an acid–base reaction will occur.
B The balanced chemical equation is Li2O(s) + H2O(l) → 2LiOH(aq).
1. A Potassium is a reductant, whereas methanol is both a weak acid and a weak base (similar to water). A weak acid produces H+, which can act as an oxidant by accepting an electron to form $\frac{1}{2}\mathrm{H_2}$. This reaction, therefore, is an acid dissociation that is driven to completion by a reduction of the protons as they are released.
B The balanced chemical equation is as follows: $\mathrm{K(s)}+\mathrm{CH_3OH(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{CH_3OK(soln)}$.
1. A One of the reactants is an alkali metal, a potent reductant, and the other is an alkyl halide. Any compound that contains a carbon–halogen bond can, in principle, be reduced, releasing a halide ion and forming an organometallic compound. That outcome seems likely in this case because organolithium compounds are among the most stable organometallic compounds known.
B Two moles of lithium are required to balance the equation: 2Li(s) + CH3Cl(l) → LiCl(s) + CH3Li(soln).
1. A Lithium nitride and potassium chloride are largely ionic compounds. The nitride ion (N3−) is a very strong base because it is the fully deprotonated form of ammonia, a weak acid. An acid–base reaction requires an acid as well as a base, however, and KCl is not acidic. What about a redox reaction? Both substances contain ions that have closed-shell valence electron configurations. The nitride ion could act as a reductant by donating electrons to an oxidant and forming N2. KCl is not an oxidant, however, and a redox reaction requires an oxidant as well as a reductant.
B We conclude that the two substances will not react with each other.
$2$
Predict the products of each reaction and balance each chemical equation.
1. K(s) + N2(g) →
2. Li3N(s) + H2O(l) →
3. Na(s) + (CH3)2NH(soln) →
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li →
Answer
1. no reaction
2. Li3N(s) + 3H2O(l) → NH3(aq) + 3LiOH(aq)
3. $\mathrm{Na(s)}+\mathrm{(CH_3)_2NH(soln)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{Na[(CH_3)_2N](soln)}$
4. C6H5Li(soln) + D2O(l) → C6H5D(l) + LiOD(soln)
5. CH3CH2Cl(soln) + 2Li → CH3CH2Li(soln) + LiCl(soln)
Summary
The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M+ ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH2). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.04%3A_The_Alkali_Metals_%28Group_1%29.txt |
Learning Objectives
• To describe how to isolate the alkaline earth metals.
• To be familiar with the reactions, compounds, and complexes of the alkaline earth metals.
Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium ($M_2CO_3$). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides.
Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium:
$\mathrm{BeCl_2(s)}+\mathrm{2K(s)}\xrightarrow\Delta\mathrm{Be(s)}+\mathrm{2KCl(s)} \label{Eq1}$
Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure $RaCl_2$. Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses.
Preparation of the Alkaline Earth Metals
The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium:
$CaCl_{2\;(l)} \rightarrow Ca_{(l)} + Cl_{2\;(g)} \label{Eq2}$
The group 2 metal chlorides are obtained from a variety of sources. For example, $BeCl_2$ is produced by reacting $HCl$ with beryllia ($BeO$), which is obtained from the semiprecious stone beryl $[Be_3Al_2(SiO_3)_6]$.
Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO3·MgCO3) with an inexpensive iron/silicon alloy at 1150°C. Initially $CO_2$ is released, leaving behind a mixture of $CaO$ and MgO; Mg2+ is then reduced:
$2CaO·MgO_{(s)} + Fe/Si_{(s)} \rightarrow 2Mg(l) + Ca_2SiO_{4\;(s)} + Fe(s) \label{Eq3}$
An early source of magnesium was an ore called magnesite ($MgCO_3$) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite ($SrCO_3$) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble.
General Properties of the Alkaline Earth Metals
Several important properties of the alkaline earth metals are summarized in Table $1$. Although many of these properties are similar to those of the alkali metals (Table $1$), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns2 for the alkaline earth metals versus ns1 for the alkali metals).
Table $1$: Selected Properties of the Group 2 Elements
Beryllium Magnesium Calcium Strontium Barium Radium
*The values cited are for six-coordinate ions except for Be2+, for which the value for the four-coordinate ion is given.
atomic symbol Be Mg Ca Sr Ba Ra
atomic number 4 12 20 38 56 88
atomic mass 9.01 24.31 40.08 87.62 137.33 226
valence electron configuration 2s2 3s2 4s2 5s2 6s2 7s2
melting point/boiling point (°C) 1287/2471 650/1090 842/1484 777/1382 727/1897 700/—
density (g/cm3) at 25°C 1.85 1.74 1.54 2.64 3.62 ~5
atomic radius (pm) 112 145 194 219 253
first ionization energy (kJ/mol) 900 738 590 549 503
most common oxidation state +2 +2 +2 +2 +2 +2
ionic radius (pm)* 45 72 100 118 135
electron affinity (kJ/mol) ≥ 0 ≥ 0 −2 −5 −14
electronegativity 1.6 1.3 1.0 1.0 0.9 0.9
standard electrode potential (E°, V) −1.85 −2.37 −2.87 −2.90 −2.91 −2.8
product of reaction with O2 BeO MgO CaO SrO BaO2
type of oxide amphoteric weakly basic basic basic basic
product of reaction with N2 none Mg3N2 Ca3N2 Sr3N2 Ba3N2
product of reaction with X2 BeX2 MgX2 CaX2 SrX2 BaX2
product of reaction with H2 none MgH2 CaH2 SrH2 BaH2
As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns2 valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance.
One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy.
Reactions and Compounds of the Alkaline Earth Metals
With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M2+ ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent. Some compounds of Mg2+ also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2.
The group 2 elements almost exclusively form ionic compounds containing M2+ ions.
All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX2). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M2+ cation and two X anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl2. These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons:
$BeCl_{2(s)} + 2(CH_3CH_2)_2O_{(l)} \rightarrow BeCl_2[O(CH_2CH_3)_2]_{2(soln)} \label{Eq4}$
Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds.
The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO2) because the larger O22− ion is better able to separate the large Ba2+ ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts:
$\mathrm{MCO_3(s)}\xrightarrow\Delta\mathrm{MO(s)}+\mathrm{CO_2(g)} \label{Eq5}$
The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Yn2−) are formed.
In the reverse of Equation $\ref{Eq5}$, the oxides of Ca, Sr, and Ba react with CO2 to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH)2:
$MO_{(s)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} + 2OH^−_{(aq)} \label{Eq6}$
and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges.
The carbonates of the alkaline earth metals also react with aqueous acid to give CO2 and H2O:
$MCO_{3(s)} + 2H^+_{(aq)} \rightarrow M^{2+}_{(aq)} + CO_{2(g)} + H_2O_{(l)} \label{Eq7}$
The reaction in Equation $\ref{Eq7}$ is the basis of antacids that contain MCO3, which is used to neutralize excess stomach acid.
The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N2 to form the nitride (M3N2), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M2+ and N3− ions is apparently sufficient to overcome the chemical inertness of the N2 molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M3Z2.
Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements.
When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC2. The most important alkaline earth carbide is calcium carbide (CaC2), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be2C, which formally contains the C4− ion (although the compound is covalent). Consistent with this formulation, reaction of Be2C with water or aqueous acid produces methane:
$Be_2C_{(s)} + 4H_2O_{(l)} \rightarrow 2Be(OH)_{2(s)} + CH_{4(g)} \label{Eq8}$
Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH2 can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH2). The hydrides of the heavier alkaline earth metals are ionic, but both BeH2 and MgH2 have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas:
$CaH_{2(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(s)} + 2H_{2(g)} \label{Eq9}$
Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size.
The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease.
Complexes of the Alkaline Earth Metals
Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be2+) and decreases rapidly with the increasing radius of the metal ion.
The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals.
The chemistry of Be2+ is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be2+ salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H2O)4]2+ ion. Because of its high charge-to-radius ratio, the Be2+ ion polarizes coordinated water molecules, thereby increasing their acidity:
$[Be(H_2O)_4]^{2+}_{(aq)} \rightarrow [Be(H_2O)_3(OH)]^+_{(aq)} + H^+_{(aq)} \label{Eq10}$
Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH)4]2−. Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF4]2−. Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases (Equation $\ref{Eq4}$).
The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg2+ and Ca2+). Thus aqueous solutions of Mg2+ contain the octahedral [Mg(H2O)6]2+ ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in Section 21.3.
Organometallic Compounds Containing Group 2 Elements
Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called Grignard reagents, after Victor Grignard (1871–1935), the French chemist who discovered them. Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines.
Uses of the Alkaline Earth Metals
Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 105 tn per year). Its low density (1.74 g/cm3 compared with 7.87 g/cm3 for iron and 2.70 g/cm3 for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (Figure $1$). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation:
$TiCl_{4\;(l)} + 2Mg(s) \rightarrow Ti_{(s)} + 2MgCl_{2\;(s)} \label{11}$
The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil.
Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl2 is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO3 is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO3 (Equation $\ref{Eq5}$), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO4 in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks.
Example $1$
For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as Ksp values (Table 17.1), lattice energies (Table 8.1), and band-gap energies.
1. To neutralize excess stomach acid that causes indigestion, would you use BeCO3, CaCO3, or BaCO3?
2. To remove CO2 from the atmosphere in a space capsule, would you use MgO, CaO, or BaO?
3. As a component of the alloy in an automotive spark plug electrode, would you use Be, Ca, or Ba?
Given: application and selected alkaline earth metals
Asked for: most appropriate substance for each application
Strategy:
Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use.
Solution
1. All the alkaline earth carbonates will neutralize an acidic solution by Equation $\ref{Eq7}$. Because beryllium and its salts are toxic, however, BeCO3 cannot be used as an antacid. Of the remaining choices, CaCO3 is somewhat more soluble than BaCO3 (according to the Ksp values in Table 17.1), suggesting that it will act more rapidly. Moreover, the formula mass of CaCO3 is 100.1 amu, whereas that of BaCO3 is almost twice as large. Therefore, neutralizing a given amount of acid would require twice the mass of BaCO3 compared with CaCO3. Furthermore, reaction of BaCO3 with acid produces a solution containing Ba2+ ions, which are toxic. (Ba2+ is a stimulant that can cause ventricular fibrillation of the heart.) Finally, CaCO3 is produced on a vast scale, so CaCO3 is likely to be significantly less expensive than any barium compound. Consequently, CaCO3 is the best choice for an antacid.
2. This application involves reacting CO2 with an alkaline earth oxide to form the carbonate, which is the reverse of the thermal decomposition reaction in which MCO3 decomposes to CO2 and the metal oxide MO (Equation $\ref{Eq5}$). Owing to their higher lattice energies, the smallest alkaline earth metals should form the most stable oxides. Hence their carbonates should decompose at the lowest temperatures, as is observed (BeCO3 decomposes at 100°C; BaCO3 at 1360°C). If the carbonate with the smallest alkaline earth metal decomposes most readily, we would expect the reverse reaction (formation of a carbonate) to occur most readily with the largest metal cation (Ba2+). Hence BaO is the best choice.
3. The alloy in a spark plug electrode must release electrons and promote their flow across the gap between the electrodes at high temperatures. Of the three metals listed, Ba has the lowest ionization energy and thus releases electrons most readily. Heating a barium-containing alloy to high temperatures will cause some ionization to occur, providing the initial step in forming a spark.
Exercise $1$
Which of the indicated alkaline earth metals or their compounds is most appropriate for each application?
1. drying agent for removing water from the atmosphere—CaCl2, MgSO4, or BaF2
2. removal of scale deposits (largely CaCO3) in water pipes—HCl(aq) or H2SO4(aq)
3. removal of traces of N2 from purified argon gas—Be, Ca, or Ba
Answer
1. MgSO4
2. HCl
3. Ba
Example $2$
Predict the products of each reaction and then balance each chemical equation.
1. CaO(s) + HCl(g) →
2. MgO(s) + excess OH(aq) →
3. $\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta$
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation.
Solution
1. A Gaseous HCl is an acid, and CaO is a basic oxide that contains the O2− ion. This is therefore an acid–base reaction that produces CaCl2 and H2O.
B The balanced chemical equation is $CaO_{(s)} + 2HCl_{(g)} → CaCl_{2(aq)} + H_2O_{(l)}$
1. A Magnesium oxide is a basic oxide, so it can either react with water to give a basic solution or dissolve in an acidic solution. Hydroxide ion is also a base. Because we have two bases but no acid, an acid–base reaction is impossible. A redox reaction is not likely because MgO is neither a good oxidant nor a good reductant.
B We conclude that no reaction occurs.
1. A Because CaH2 contains the hydride ion (H), it is a good reductant. It is also a strong base because H ions can react with H+ ions to form H2. Titanium oxide (TiO2) is a metal oxide that contains the metal in its highest oxidation state (+4 for a group 4 metal); it can act as an oxidant by accepting electrons. We therefore predict that a redox reaction will occur, in which H is oxidized and Ti4+ is reduced. The most probable reduction product is metallic titanium, but what is the oxidation product? Oxygen must appear in the products, and both CaO and H2O are stable compounds. The +1 oxidation state of hydrogen in H2O is a sign that an oxidation has occurred (2H → 2H+ + 4e).
B The balanced chemical equation is
$\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta\mathrm{Ti(s)}+\mathrm{CaO(s)}+\mathrm{H_2O(l)}$
We could also write the products as Ti(s) + Ca(OH)2(s).
Exercise $2$
Predict the products of each reaction and then balance each chemical equation.
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta$
2. BaCl2(aq) + Na2SO4(aq) →
3. BeO(s) + OH(aq) + H2O(l) →
Answer
1. $\mathrm{BeCl_2(s)}+\mathrm{Mg(s)}\xrightarrow\Delta\mathrm{Be(s)}+ \mathrm{MgCl_2(s)}$
2. BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
3. BeO(s) + 2OH(aq) + H2O(l) → [Be(OH)4]2−(aq)
Summary
Group 2 elements almost exclusively form ionic compounds containing the M2+ ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO2 to form carbonates, which in turn react with acid to produce CO2 and H2O. Except for Be, all the alkaline earth metals react with N2 to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.05%3A_The_Alkaline_Earth_Metals_%28Group_2%29.txt |
Learning Objectives
• To be familiar with the roles of the s-block elements in biology.
The s-block elements play important roles in biological systems. Covalent hydrides, for example, are the building blocks of organic compounds, and other compounds and ions containing s-block elements are found in tissues and cellular fluids. In this section, we describe some ways in which biology depends on the properties of the group 1 and group 2 elements.
Covalent Hydrides
There are three major classes of hydrides—covalent, ionic, and metallic—but only covalent hydrides occur in living cells and have any biochemical significance. As you learned in Chapter 3 , carbon and hydrogen have similar electronegativities, and the C–H bonds in organic molecules are strong and essentially nonpolar. Little acid–base chemistry is involved in the cleavage or formation of these bonds. In contrast, because hydrogen is less electronegative than oxygen and nitrogen (symbolized by Z), the H–Z bond in the hydrides of these elements is polarized (Hδ+–Zδ−). Consequently, the hydrogen atoms in these H–Z bonds are relatively acidic. Moreover, S–H bonds are relatively weak due to poor s orbital overlap, so they are readily cleaved to give a proton. Hydrides in which H is bonded to O, N, or S atoms are therefore polar, hydrophilic molecules that form hydrogen bonds. They also undergo acid–base reactions by transferring a proton.
Note the Pattern
Covalent hydrides in which H is bonded to O, N, or S atoms are polar and hydrophilic, form hydrogen bonds, and transfer a proton in their acid-base reactions.
Hydrogen bonds are crucial in biochemistry, in part because they help hold proteins in their biologically active folded structures. Hydrogen bonds also connect the two intertwining strands of DNA (deoxyribonucleic acid), the substance that contains the genetic code for all organisms. (For more information on DNA, see Section 23.6 .) Because hydrogen bonds are easier to break than the covalent bonds that form the individual DNA strands, the two intertwined strands can be separated to give intact single strands, which is essential for the duplication of genetic information.
In addition to the importance of hydrogen bonds in biochemical molecules, the extensive hydrogen-bonding network in water is one of the keys to the existence of life on our planet. Based on its molecular mass, water should be a gas at room temperature (20°C), but the strong intermolecular interactions in liquid water greatly increase its boiling point. Hydrogen bonding also produces the relatively open molecular arrangement found in ice, which causes ice to be less dense than water. Because ice floats on the surface of water, it creates an insulating layer that allows aquatic organisms to survive during cold winter months.
These same strong intermolecular hydrogen-bonding interactions are also responsible for the high heat capacity of water and its high heat of fusion. A great deal of energy must be removed from water for it to freeze. Consequently, as noted in Chapter 9, large bodies of water act as “thermal buffers” that have a stabilizing effect on the climate of adjacent land areas. Perhaps the most striking example of this effect is the fact that humans can live comfortably at very high latitudes. For example, palm trees grow in southern England at the same latitude (51°N) as the southern end of frigid Hudson Bay and northern Newfoundland in North America, areas known more for their moose populations than for their tropical vegetation. Warm water from the Gulf Stream current in the Atlantic Ocean flows clockwise from the tropical climate at the equator past the eastern coast of the United States and then turns toward England, where heat stored in the water is released. The temperate climate of Europe is largely attributable to the thermal properties of water.
Note the Pattern
Strong intermolecular hydrogen-bonding interactions are responsible for the high heat capacity of water and its high heat of fusion.
Macrominerals
The members of group 1 and group 2 that are present in the largest amounts in organisms are sodium, potassium, magnesium, and calcium, all of which form monatomic cations with a charge of +1 (group 1, M+) or +2 (group 2, M2+). Biologically, these elements can be classified as macrominerals.
For example, calcium is found in the form of relatively insoluble calcium salts that are used as structural materials in many organisms. Hydroxyapatite [Ca5(PO4)3OH] is the major component of bones, calcium carbonate (CaCO3) is the major component of the shells of mollusks and the eggs of birds and reptiles, and calcium oxalate (CaO2CCO2) is found in many plants. Because calcium and strontium have similar sizes and charge-to-radius ratios, small quantities of strontium are always found in bone and other calcium-containing structural materials. Normally this is not a problem because the Sr2+ ions occupy sites that would otherwise be occupied by Ca2+ ions. When trace amounts of radioactive 90Sr are released into the atmosphere from nuclear weapons tests or a nuclear accident, however, the radioactive strontium eventually reaches the ground, where it is taken up by plants that are consumed by dairy cattle. The isotope then becomes concentrated in cow’s milk, along with calcium. Because radioactive strontium coprecipitates with calcium in the hydroxyapatite that surrounds the bone marrow (where white blood cells are produced), children, who typically ingest more cow’s milk than adults, are at substantially increased risk for leukemia, a type of cancer characterized by the overproduction of white blood cells.
Ion Transport
The Na+, K+, Mg2+, and Ca2+ ions are important components of intracellular and extracellular fluids. Both Na+ and Ca2+ are found primarily in extracellular fluids, such as blood plasma, whereas K+ and Mg2+ are found primarily in intracellular fluids. Substantial inputs of energy are required to establish and maintain these concentration gradients and prevent the system from reaching equilibrium. Thus energy is needed to transport each ion across the cell membrane toward the side with the higher concentration. The biological machines that are responsible for the selective transport of these metal ions are complex assemblies of proteins called ion pumpsA complex assembly of proteins that selectively transport ions across cell membranes by their high affinity for ions of a certain charge and radius.. Ion pumps recognize and discriminate between metal ions in the same way that crown ethers and cryptands do, with a high affinity for ions of a certain charge and radius.
Defects in the ion pumps or their control mechanisms can result in major health problems. For example, cystic fibrosis, the most common inherited disease in the United States, is caused by a defect in the transport system (in this case, chloride ions). Similarly, in many cases, hypertension, or high blood pressure, is thought to be due to defective Na+ uptake and/or excretion. If too much Na+ is absorbed from the diet (or if too little is excreted), water diffuses from tissues into the blood to dilute the solution, thereby decreasing the osmotic pressure in the circulatory system. The increased volume increases the blood pressure, and ruptured arteries called aneurysms can result, often in the brain. Because high blood pressure causes other medical problems as well, it is one of the most important biomedical disorders in modern society.
For patients who suffer from hypertension, low-sodium diets that use NaCl substitutes, such as KCl, are often prescribed. Although KCl and NaCl give similar flavors to foods, the K+ is not readily taken up by the highly specific Na+-uptake system. This approach to controlling hypertension is controversial, however, because direct correlations between dietary Na+ content and blood pressure are difficult to demonstrate in the general population. More important, recent observations indicate that high blood pressure may correlate more closely with inadequate intake of calcium in the diet than with excessive sodium levels. This finding is important because the typical “low-sodium” diet is also low in good sources of calcium, such as dairy products.
Some of the most important biological functions of the group 1 and group 2 metals are due to small changes in the cellular concentrations of the metal ion. The transmission of nerve impulses, for example, is accompanied by an increased flux of Na+ ions into a nerve cell. Similarly, the binding of various hormones to specific receptors on the surface of a cell leads to a rapid influx of Ca2+ ions; the resulting sudden rise in the intracellular Ca2+ concentration triggers other events, such as muscle contraction, the release of neurotransmitters, enzyme activation, or the secretion of other hormones.
Within cells, K+ and Mg2+ often activate particular enzymes by binding to specific, negatively charged sites in the enzyme structure. Chlorophyll, the green pigment used by all plants to absorb light and drive the process of photosynthesis, contains magnesium. During photosynthesis, CO2 is reduced to form sugars such as glucose. The structure of the central portion of a chlorophyll molecule resembles a crown ether (part (a) in Figure 13.2.4) with four five-member nitrogen-containing rings linked together to form a large ring that provides a “hole” the proper size to tightly bind Mg2+.
The structure of the central core of chlorophyll, a magnesium complex present in all photosynthetic tissues. Note the resemblance to the crown ether complexes discussed in Chapter 13.
Ionophores
Because the health of cells depends on maintaining the proper levels of cations in intracellular fluids, any change that affects the normal flux of metal ions across cell membranes could well cause an organism to die. Molecules that facilitate the transport of metal ions across membranes are generally called ionophoresA molecule that facilitates the transport of metal ions across membranes. (ion plus phore from the Greek phorein, meaning “to carry”). Many ionophores are potent antibiotics that can kill or inhibit the growth of bacteria. An example is valinomycin, a cyclic molecule with a central cavity lined with oxygen atoms (part (a) in Figure 20.5.1) that is similar to the cavity of a crown ether (part (a) in Figure 13.2.4). Like a crown ether, valinomycin is highly selective: its affinity for K+ is about 1000 times greater than that for Na+. By increasing the flux of K+ ions into cells, valinomycin disrupts the normal K+ gradient across a cell membrane, thereby killing the cell (part (b) in Figure 20.5.1).
Example 20.5.1
A common way to study the function of a metal ion in biology is to replace the naturally occurring metal with one whose reactivity can be traced by spectroscopic methods. The substitute metal ion must bind to the same site as the naturally occurring ion, and it must have a similar (or greater) affinity for that site, as indicated by its charge density. Arrange the following ions in order of increasing effectiveness as a replacement for Ca2+, which has an ionic radius of 100 pm (numbers in parentheses are ionic radii): Na+ (102 pm), Eu2+ (117 pm), Sr2+ (118 pm), F (133 pm), Pb2+ (119 pm), and La3+ (103 pm). Explain your reasoning.
Given: ions and ionic radii
Asked for: suitability as replacement for Ca2+
Strategy:
Use periodic trends to arrange the ions from least effective to most effective as a replacement for Ca2+.
Solution:
The most important properties in determining the affinity of a biological molecule for a metal ion are the size and charge-to-radius ratio of the metal ion. Of the possible Ca2+ replacements listed, the F ion has the opposite charge, so it should have no affinity for a Ca2+-binding site. Na+ is approximately the right size, but with a +1 charge it will bind much more weakly than Ca2+. Although Eu2+, Sr2+, and Pb2+ are all a little larger than Ca2+, they are probably similar enough in size and charge to bind. Based on its ionic radius, Eu2+ should bind most tightly of the three. La3+ is nearly the same size as Ca2+ and more highly charged. With a higher charge-to-radius ratio and a similar size, La3+ should bind tightly to a Ca2+ site and be the most effective replacement for Ca2+. The order is F << Na+ << Pb2+ ~ Sr2+ ~ Eu2+ < La3+.
Exercise
The ionic radius of K+ is 138 pm. Arrange the following ions in order of increasing affinity for a K+-binding site in an enzyme (numbers in parentheses are ionic radii): Na+ (102 pm), Rb+ (152 pm), Ba2+ (135 pm), Cl (181 pm), and Tl+ (150 pm).
Answer Cl << Na+ < Tl+ ~ Rb+ < Ba2+
Summary
Covalent hydrides in which hydrogen is bonded to oxygen, nitrogen, or sulfur are polar, hydrophilic molecules that form hydrogen bonds and undergo acid–base reactions. Hydrogen-bonding interactions are crucial in stabilizing the structure of proteins and DNA and allow genetic information to be duplicated. The hydrogen-bonding interactions in water and ice also allow life to exist on our planet. The group 1 and group 2 metals present in organisms are macrominerals, which are important components of intracellular and extracellular fluids. Small changes in the cellular concentration of a metal ion can have a significant impact on biological functions. Metal ions are selectively transported across cell membranes by ion pumps, which bind ions based on their charge and radius. Ionophores, many of which are potent antibiotics, facilitate the transport of metal ions across membranes.
Key Takeaway
• Among their many roles in biology, the s-block elements allow genetic information to be duplicated and are important components of intracellular and extracellular fluids.
Conceptual Problems
1. Explain the thermochemical properties of water in terms of its intermolecular bonding interactions. How does this affect global climate patterns?
2. Of the three classes of hydrides, which is (are) biochemically significant? How do you account for this?
3. Many proteins are degraded and become nonfunctional when heated higher than a certain temperature, even though the individual protein molecules do not undergo a distinct chemical change. Propose an explanation for this observation.
4. Los Angeles has moderate weather throughout the year, with average temperatures between 57°F and 70°F. In contrast, Palm Springs, which is just 100 miles inland, has average temperatures between 55°F and 95°F. Explain the difference in the average temperature ranges between the two cities.
5. Although all group 1 ions have the same charge (+1), Na+ and K+ ions are selectively transported across cell membranes. What strategy do organisms employ to discriminate between these two cations?
Structure and Reactivity
1. A 0.156 g sample of a chloride salt of an alkaline earth metal is dissolved in enough water to make 20.5 mL of solution. If this solution has an osmotic pressure of 2.68 atm at 25°C, what is the identity of the alkaline earth metal?
2. The thermal buffering capacity of water is one of the reasons the human body is capable of withstanding a wide range of temperatures. How much heat (in kilojoules) is required to raise the temperature of a 70.0 kg human from 37.0°C to 38.0°C? Assume that 70% of the mass of the body is water and that body fluids have the same specific heat as water.
3. During illness, body temperature can increase by more than 2°C. One piece of folklore is that you should “feed a fever.” Using the data in Table 8.7.2, how many fried chicken drumsticks would a 70.0 kg person need to eat to generate a 2.0°C change in body temperature? Assume the following: there is complete conversion of the caloric content of the chicken to thermal energy, 70% of the mass of the body is solely due to water, and body fluids have the same specific heat as water.
4. Hydrogen bonding is partly responsible for the high enthalpy of vaporization of water (ΔHvap = 43.99 kJ/mol at 25°C), which contributes to cooling the body during exercise. Assume that a 50.0 kg runner produces 20.0 g of perspiration during a race, and all the perspiration is converted to water vapor at 37.0°C. What amount of heat (in joules) is removed from the runner’s skin if the perspiration consists of only water?
1. Ba
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.06%3A_The_s-Block_Elements_in_Biology.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: Principles, Patterns, and Applications" by Bruce A. Averill and Patricia Eldredge. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Problems
1. List three physical properties that are important in describing the behavior of the main group elements.
2. Arrange K, Cs, Sr, Ca, Ba, and Li in order of
1. increasing ionization energy.
2. increasing atomic size.
3. increasing electronegativity.
1. Arrange Rb, H, Be, Na, Cs, and Ca in order of
1. decreasing atomic size.
2. decreasing magnitude of electron affinity.
1. Which periodic trends are affected by Zeff? Based on the positions of the elements in the periodic table, which element would you expect to have the highest Zeff? the lowest Zeff?
1. Compare the properties of the metals and nonmetals with regard to their electronegativities and preferred oxidation states.
1. Of Ca, Br, Li, N, Zr, Ar, Sr, and S, which elements have a greater tendency to form positive ions than negative ions?
1. Arrange As, O, Ca, Sn, Be, and Sb in order of decreasing metallic character.
1. Give three reasons the chemistry of the second-period elements is generally not representative of their groups as a whole.
1. Compare the second-period elements and their heavier congeners with regard to
1. magnitude of electron affinity.
2. coordination number.
3. the solubility of the halides in nonpolar solvents.
1. The heavier main group elements tend to form extended sigma-bonded structures rather than multiple bonds to other atoms. Give a reasonable explanation for this tendency.
1. What is the diagonal effect? How does it explain the similarity in chemistry between, for example, boron and silicon?
1. Although many of the properties of the second- and third-period elements in a group are quite different, one property is similar. Which one?
1. Two elements are effective additives to solid rocket propellant: beryllium and one other element that has similar chemistry. Based on the position of beryllium in the periodic table, identify the second element.
1. Give two reasons for the inert-pair effect. How would this phenomenon explain why Sn2+ is a better reducing agent than Pb2+?
1. Explain the following trend in electron affinities: Al (−41.8 kJ/mol), Si (−134.1 kJ/mol), P (−72.0 kJ/mol), and S (−200.4 kJ/mol).
1. Using orbital energy arguments, explain why electron configurations with more than four electron pairs around the central atom are not observed for second-period elements.
Answers
1. Cs > Rb > Ca > Na > Be > H
2. H > Na > Rb > Cs > Ca > Be
1. Ca > Be > Sn > Sb > As > O
1. aluminum
1. The magnitude of electron affinity increases from left to right in a period due to the increase in Zeff; P has a lower electron affinity than expected due to its half-filled 3p shell, which requires the added electron to enter an already occupied 3p orbital.
Structure and Reactivity
1. The following table lists the valences, coordination numbers, and ionic radii for a series of cations. Which would you substitute for K+ in a crystalline lattice? Explain your answer.
Metal Charge Coordination Number Ionic Radius (pm)
Li +1 4 76
Na +1 6 102
K +1 6 138
Mg +2 6 72
Ca +2 6 100
Sr +2 6 118
Answer
1. Sr2+; it is the ion with the radius closest to that of K+. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.E%3A_Periodic_Trends_and_the_s-Block_Elements_%28Exercises%29.txt |
• 21.1: The Elements of Group 13
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the prop
• 21.2: The Elements of Group 14
Group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead.
• 21.3: The Elements of Group 15 (The Pnicogens)
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter p
• 21.4: The Elements of Group 16 (The Chalcogens)
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements.
• 21.5: The Elements of Group 17 (The Halogens)
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X− ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group.
• 21.6: The Elements of Group 18 (The Noble Gases)
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases.
• 21.E: The p-Block Elements (Exercises)
21: The p-Block Elements
Learning Objectives
• To understand the trends in properties and the reactivity of the group 13 elements.
Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B2O3 and AlCl3, respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13.
Group 13 elements are never found in nature in their free state.
Preparation and General Properties of the Group 13 Elements
As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores.
Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds (Figure $1$) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH)3], which is then dehydrated to the oxide (B2O3). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure:
$\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}$
$\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}$
Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl3 with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B2H6):
$\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}$
$B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}$
The reaction shown in Equation $\ref{Eq3}$ is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware.
In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al2O3, are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH)4]. Neutralization of the resulting solution with gaseous CO2 results in the precipitation of Al(OH)3:
$2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}$
Thermal dehydration of Al(OH)3 produces Al2O3, and metallic aluminum is obtained by the electrolytic reduction of Al2O3 using the Hall–Heroult process. Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass.
The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO2. Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As).
Table $1$ summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns2np1 valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids.
Table $1$: Selected Properties of the Group 13 Elements
Property Boron Aluminum* Gallium Indium Thallium
*This is the name used in the United States; the rest of the world inserts an extra i and calls it aluminium.
The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate ions in the most common oxidation state, except for Al3+, for which the value for the four-coordinate ion is given. The B3+ ion is not a known species; the radius cited is an estimated four-coordinate value.
§X is Cl, Br, or I. Reaction with F2 gives the trifluorides (MF3) for all group 13 elements.
atomic symbol B Al Ga In Tl
atomic number 5 13 31 49 81
atomic mass (amu) 10.81 26.98 69.72 114.82 204.38
valence electron configuration 2s22p1 3s23p1 4s24p1 5s25p1 6s26p1
melting point/boiling point (°C) 2075/4000 660/2519 29.7/2204 156.6/2072 304/1473
density (g/cm3) at 25°C 2.34 2.70 5.91 7.31 11.8
atomic radius (pm) 87 118 136 156 156
first ionization energy (kJ/mol) 801 578 579 558 589
most common oxidation state +3 +3 +3 +3 +1
ionic radius (pm) −25 54 62 80 162
electron affinity (kJ/mol) −27 −42 −40 −39 −37
electronegativity 2.0 1.6 1.8 1.8 1.8
standard reduction potential (E°, V) −0.87 −1.66 −0.55 −0.34 +0.741 of M3+(aq)
product of reaction with O2 B2O3 Al2O3 Ga2O3 In2O3 Tl2O
type of oxide acidic amphoteric amphoteric amphoteric basic
product of reaction with N2 BN AlN GaN InN none
product of reaction with X2§ BX3 Al2X6 Ga2X6 In2X6 TlX
Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids.
In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.
Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding.
Reactions and Compounds of Boron
Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms.
Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding.
The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B12 icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure $3$). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B4C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation $\ref{Eq22.6}$); boron oxide (B2O3), which is formed when boron is heated with excess oxygen (Equation $\ref{Eq22.7}$); and the boron trihalides (BX3), which are formed by heating boron with excess halogen (Equation $\ref{Eq22.8}$).
$\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}$
$\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}$
$\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}$
As is typical of elements lying near the dividing line between metals and nonmetals, many compounds of boron are amphoteric, dissolving in either acid or base.
Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B3N3 rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B2O3) contains layers of trigonal planar BO3 groups (analogous to BX3) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO2, to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass.
At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB12 and CaB6—are shown in Figure $4$. Because metal-rich borides such as ZrB2 and TiB2 are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles.
Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B2H6), which contains two bridging hydrogen atoms (part (a) in Figure $5$. An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B12H122− ion, which has a polyhedral structure similar to the icosahedral B12 unit of elemental boron, with a single hydrogen atom bonded to each boron atom.
A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy.
The enthalpy of combustion of diborane (B2H6) is −2165 kJ/mol, one of the highest values known:
$B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}$
Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B2O3 proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity.
Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH4 ion are reductants, whereas electron-deficient species such as B2H6 act as oxidants.
Example $1$
For each reaction, explain why the given products form.
1. B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l)
2. BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
3. $\mathrm{2BI_3(s)}+\mathrm{3H_2(g)}\xrightarrow{\Delta}\frac{1}{6}\mathrm{B_{12}(s)}+\mathrm{6HI(g)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Molecular oxygen is an oxidant. If the other reactant is a potential reductant, we expect that a redox reaction will occur. Although B2H6 contains boron in its highest oxidation state (+3), it also contains hydrogen in the −1 oxidation state (the hydride ion). Because hydride is a strong reductant, a redox reaction will probably occur. We expect that H will be oxidized to H+ and O2 will be reduced to O2−, but what are the actual products? A reasonable guess is B2O3 and H2O, both stable compounds.
2. Neither BCl3 nor water is a powerful oxidant or reductant, so a redox reaction is unlikely; a hydrolysis reaction is more probable. Nonmetal halides are acidic and react with water to form a solution of the hydrohalic acid and a nonmetal oxide or hydroxide. In this case, the most probable boron-containing product is boric acid [B(OH)3].
3. We normally expect a boron trihalide to behave like a Lewis acid. In this case, however, the other reactant is elemental hydrogen, which usually acts as a reductant. The iodine atoms in BI3 are in the lowest accessible oxidation state (−1), and boron is in the +3 oxidation state. Consequently, we can write a redox reaction in which hydrogen is oxidized and boron is reduced. Because compounds of boron in lower oxidation states are rare, we expect that boron will be reduced to elemental boron. The other product of the reaction must therefore be HI.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow$
3. $\mathrm{B_2O_3(s)}+\mathrm{Ca(s)}\xrightarrow{\Delta}$
Answer
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}\mathrm{2B(OH)_3(s)}+\mathrm{6H_2(g)}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow\textrm{no reaction}$
3. $\mathrm{6B_2O_3(s)}+18\mathrm{Ca(s)}\xrightarrow{\Delta}\mathrm{B_{12}(s)}+\mathrm{18CaO(s)}$
Reactions and Compounds of the Heavier Group 13 Elements
All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry:
$2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}$
The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI3, it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl+ salt of the triiodide ion (I3). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state.
Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl3 and TlBr3, are more covalent in character and form halogen-bridged dimers (part (b) in Figure $4$). Although the structure of these dimers is similar to that of diborane (B2H6), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct:
$Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}$
In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide ($M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}$
In a related reaction, Al2(SO4)3 is used to clarify drinking water by the precipitation of hydrated Al(OH)3, which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H2O)6]3+.
Of the group 13 halides, only the fluorides behave as typical ionic compounds.
Like boron (Equation $\ref{Eq22.7}$), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M2O3), although Tl2O3 is unstable:
$\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}$
Aluminum oxide (Al2O3), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al3+ ions in crystalline alumina with Cr3+ ions forms the gemstone ruby, whereas replacing Al3+ with a mixture of Fe2+, Fe3+, and Ti4+ produces blue sapphires. The gallium oxide compound MgGa2O4 gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH)4 ions.
Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct.
Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M2Y3. However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S2−, Se2−, and Te2−, thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl2Y. Only aluminum, like boron, reacts directly with N2 (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs1−x−yPxNy) is used in the displays of calculators and digital watches.
All group 13 oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric.
Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH3 is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH3 is unstable at room temperature.
Complexes of Group 13 Elements
Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H2O)63+], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H2O)5(OH)2+ ion:
$[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}$
Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{2Al(s)} + \mathrm{Fe_2O_3(s)}\xrightarrow{\Delta}\mathrm{2Fe(l)} + \mathrm{Al_2O_3(s)}$
2. $\mathrm{2Ga(s)} + \mathrm{6H_2O(l)}+ \mathrm{2OH^-(aq)}\xrightarrow{\Delta}\mathrm{3H_2(g)} + \mathrm{2Ga(OH)^-_4(aq)}$
3. $\mathrm{In_2Cl_6(s)}\xrightarrow{\mathrm{H_2O(l)}}\mathrm{2In^{3+}(aq)}+\mathrm{6Cl^-(aq)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Aluminum is an active metal and a powerful reductant, and Fe2O3 contains Fe(III), a potential oxidant. Hence a redox reaction is probable, producing metallic Fe and Al2O3. Because Al is a main group element that lies above Fe, which is a transition element, it should be a more active metal than Fe. Thus the reaction should proceed to the right. In fact, this is the thermite reaction, which is so vigorous that it produces molten Fe and can be used for welding.
2. Gallium lies immediately below aluminum in the periodic table and is amphoteric, so it will dissolve in either acid or base to produce hydrogen gas. Because gallium is similar to aluminum in many of its properties, we predict that gallium will dissolve in the strong base.
3. The metallic character of the group 13 elements increases with increasing atomic number. Indium trichloride should therefore behave like a typical metal halide, dissolving in water to form the hydrated cation.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. LiH(s) + Al2Cl6(soln)→
2. Al2O3(s) + OH(aq)→
3. Al(s) + N2(g) $\xrightarrow{\Delta}$
4. Ga2Cl6(soln) + Cl(soln)→
Answer
1. 8LiH(s) + Al2Cl6(soln)→2LiAlH4(soln) + 6LiCl(s)
2. Al2O3(s) + 2OH(aq) + 3H2O(l) → 2Al(OH)4(aq)
3. 2Al(s) + N2(g) $\xrightarrow{\Delta}$ 2AlN(s)
4. Ga2Cl6(soln) + 2Cl(soln) → 2GaCl4(soln)
Summary
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Zeff that arises from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.01%3A_The_Elements_of_Group_13.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 14 elements.
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure $1$). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater.
Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.)
Although the first glasses were prepared from silica (silicon oxide, SiO2) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na2SiF6 with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country.
Preparation and General Properties of the Group 14 Elements
The natural abundance of the group 14 elements varies tremendously. Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite:
$\mathrm{SiO_2(s)}+\mathrm{3C(s)}\xrightarrow{\Delta}\mathrm{SiC(s)}+\mathrm{2CO(g)} \label{$1$}$
$\mathrm{SiC(s)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{C(graphite)} \label{$2$}$
One allotrope of carbon, diamond, is metastable under normal conditions, with a ΔG°f of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm3 versus 2.2 g/cm3). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices.
The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (part (a) in Figure $2$). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds.
Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl2 to give SiCl4, followed by the fractional distillation of the impure SiCl4 and reduction with H2:
$\mathrm{SiCl_4(l)}+\mathrm{2H_2(g)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{4HCl(g)} \label{$3$}$
Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (part (b) in Figure $2$).
In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices.
Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity:
$\mathrm{PbS(s)}+\frac{3}{2}\mathrm{O_2(g)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\mathrm{SO_2(g)} \label{$4$}$
$\mathrm{PbO(s)}+\mathrm{C(s)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO(g)} \label{$5$}$
or
$\mathrm{PbO(s)}+\mathrm{CO(g)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO_2(g)} \label{$6$}$
By far the single largest use of lead is in lead storage batteries. The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Zeff, the relative stability of the +2 oxidation state increases smoothly from carbon to lead.
The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14.
Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO2 is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO2 is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO2 molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms.
The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH2 group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH2 group into the silicon analogue of n-hexane (Si6H14) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead.
In Table $1$ "Selected Properties of the Group 14 Elements" we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge.
Table $1$: Selected Properties of the Group 14 Elements
Property Carbon Silicon Germanium Tin Lead
*The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated.
X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element.
atomic symbol C Si Ge Sn Pb
atomic number 6 14 32 50 82
atomic mass (amu) 12.01 28.09 72.64 118.71 207.2
valence electron configuration* 2s22p2 3s23p2 4s24p2 5s25p2 6s26p2
melting point/boiling point (°C) 4489 (at 10.3 MPa)/3825 1414/3265 939/2833 232/2602 327/1749
density (g/cm3) at 25°C 2.2 (graphite), 3.51 (diamond) 2.33 5.32 7.27(white) 11.30
atomic radius (pm) 77 (diamond) 111 125 145 154
first ionization energy (kJ/mol) 1087 787 762 709 716
most common oxidation state +4 +4 +4 +4 +4
ionic radius (pm) ≈29 ≈40 53 69 77.5
electron affinity (kJ/mol) −122 −134 −119 −107 −35
electronegativity 2.6 1.9 2.0 2.0 1.8
standard reduction potential (E°, V) (for EO2 → E in acidic solution) 0.21 −0.86 −0.18 −0.12 0.79
product of reaction with O2 CO2, CO SiO2 GeO2 SnO2 PbO
type of oxide acidic (CO2) acidic neutral (CO) amphoteric amphoteric amphoteric
product of reaction with N2 none Si3N4 none Sn3N4 none
product of reaction with X2 CX4 SiX4 GeX4 SnX4 PbX2
product of reaction with H2 CH4 none none none none
The group 14 elements follow the same pattern as the group 13 elements in their periodic properties.
Reactions and Compounds of Carbon
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
$CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{$7$}$
The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
$\mathrm{HCO_2H(l)}\xrightarrow{\mathrm{H_2SO_4(l)}}\mathrm{CO(g)}+\mathrm{H_3O^+(aq)}+\mathrm{HSO^-_4}\label{$8$}$
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
$\mathrm{CO(g)}+\mathrm{Cl_2(g)}\xrightarrow{\Delta}\textrm{Cl}_2\textrm{C=O(g)}\label{$9$}$
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
$\mathrm{C(s)}+\mathrm{2S(g)}\xrightarrow{\Delta}\mathrm{CS_2(g)}\label{$1$0}$
The selenium analog CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
$\pi$ bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C4− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation $\ref{Eq1}$) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
Example $1$
For each reaction, explain why the given product forms.
1. CO(g) + Cl2(g) → Cl2C=O(g)
2. CO(g) + BF3(g) → F3B:C≡O(g)
3. Sr(s) + 2C(s) $\xrightarrow{\Delta}$ SrC2(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Because the carbon in CO is in an intermediate oxidation state (+2), CO can be either a reductant or an oxidant; it is also a Lewis base. The other reactant (Cl2) is an oxidant, so we expect a redox reaction to occur in which the carbon of CO is further oxidized. Because Cl2 is a two-electron oxidant and the carbon atom of CO can be oxidized by two electrons to the +4 oxidation state, the product is phosgene (Cl2C=O).
2. Unlike Cl2, BF3 is not a good oxidant, even though it contains boron in its highest oxidation state (+3). Nor can BF3 behave like a reductant. Like any other species with only six valence electrons, however, it is certainly a Lewis acid. Hence an acid–base reaction is the most likely alternative, especially because we know that CO can use the lone pair of electrons on carbon to act as a Lewis base. The most probable reaction is therefore the formation of a Lewis acid–base adduct.
3. Typically, both reactants behave like reductants. Unless one of them can also behave like an oxidant, no reaction will occur. We know that Sr is an active metal because it lies far to the left in the periodic table and that it is more electropositive than carbon. Carbon is a nonmetal with a significantly higher electronegativity; it is therefore more likely to accept electrons in a redox reaction. We conclude, therefore, that Sr will be oxidized, and C will be reduced. Carbon forms ionic carbides with active metals, so the reaction will produce a species formally containing either C4− or C22−. Those that contain C4− usually involve small, highly charged metal ions, so Sr2+ will produce the acetylide (SrC2) instead.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$
2. C(s) + H2O(l) →
3. NaHCO3(s) + H2SO4(aq) →
Answer
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$ CO2(g)
2. C(s) + H2O(l) → no reaction
3. NaHCO3(s) + H2SO4(aq) → CO2(g) + NaHSO4(aq) + H2O(l)
Reactions and Compounds of the Heavier Group 14 Elements
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Video $1$: Time lapse tin pest reaction.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62− ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
$MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{$1$1}$
In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure $3$). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44− unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure $4$). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na+ ions inside the zeolite cavities are displaced by the more highly charged Mg2+ and Ca2+ ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
$3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{$1$2}$
Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4− ion. They react with aqueous acid to form silicon hydrides such as SiH4:
$Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{$1$3}$
Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure $5$). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Example $2$
For each reaction, explain why the given products form.
1. Pb(s) + Cl2(g) → PbCl2(s)
2. Mg2Si(s) + 4H2O(l) → SiH4(g) + 2Mg(OH)2(s)
3. GeO2(s) + 4OH(aq) → GeO44−(aq) + 2H2O(l)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Lead is a metal, and chlorine is a nonmetal that is a strong oxidant. Thus we can expect a redox reaction to occur in which the metal acts as a reductant. Although lead can form compounds in the +2 and +4 oxidation states, Pb4+ is a potent oxidant (the inert-pair effect). Because lead prefers the +2 oxidation state and chlorine is a weaker oxidant than fluorine, we expect PbCl2 to be the product.
2. This is the reaction of water with a metal silicide, which formally contains the Si4− ion. Water can act as either an acid or a base. Because the other compound is a base, we expect an acid–base reaction to occur in which water acts as an acid. Because Mg2Si contains Si in its lowest possible oxidation state, however, an oxidation–reduction reaction is also a possibility. But water is a relatively weak oxidant, so an acid–base reaction is more likely. The acid (H2O) transfers a proton to the base (Si4−), which can accept four protons to form SiH4. Proton transfer from water produces the OH ion, which will combine with Mg2+ to give magnesium hydroxide.
3. We expect germanium dioxide (GeO2) to be amphoteric because of the position of germanium in the periodic table. It should dissolve in strong aqueous base to give an anionic species analogous to silicate.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. PbO2(s) $\xrightarrow{\Delta}$
2. GeCl4(s) + H2O(l) →
3. Sn(s) + HCl(aq) →
Answer
1. $\mathrm{PbO_2(s)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\frac{1}{2}\mathrm{O_2(g)}$
2. GeCl4(s) + 2H2O(l) → GeO2(s) + 4HCl(aq)
3. Sn(s) + 2HCl(aq) → Sn2+(aq) + H2(g) + 2Cl(aq)
Summary
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22− (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
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Learning Objectives
• To understand the trends in properties and reactivity of the group 15 elements: the pnicogens.
Like the group 14 elements, the lightest member of group 15, nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores. Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb2S3), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (Figure $1$). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells.
In the form of its yellow sulfide ore, orpiment (As2S3), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting.
Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!).
Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry.
Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO2), noticed that a gas remained when CO2 was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO3 (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals.
Preparation and General Properties of the Group 15 Elements
Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide:
$\mathrm{2NaN_3(s)\xrightarrow{\Delta}2Na(l)+3N_2(g)} \label{Eq1}$
In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO3 and NaNO3 in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it.
The other three pnicogens are much less abundant: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M2S3), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals.
In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (Table $1$). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns2np3 valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH3 and PH3), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases.
Table $1$: Selected Properties of the Group 15 Elements
Property Nitrogen Phosphorus Arsenic Antimony Bismuth
*The configuration shown does not include filled d and f subshells. For white phosphorus. For gray arsenic. §The values cited are for six-coordinate ions in the indicated oxidation states. The N5+, P5+, and As5+ ions are not known species. ||The chemical form of the elements in these oxidation states varies considerably. For N, the reaction is NO3 + 3H+ + 2e HNO2 + H2O; for P and As, it is $\ce{H3EO4 + 2H^{+} + 2e^{−} → H3EO3 + H2O}$; and for Sb it is $\ce{Sb2O5 + 4e^{-} + 10H^{+} → 2Sb^{3+} + 5H2O}$.
atomic symbol N P As Sb Bi
atomic number 7 15 33 51 83
atomic mass (amu) 14.01 30.97 74.92 121.76 209.98
valence electron configuration* 2s22p3 3s23p3 4s24p3 5s25p3 6s26p3
melting point/boiling point (°C) −210/−196 44.15/281c 817 (at 3.70 MPa)/603 (sublimes) 631/1587 271/1564
density (g/cm3) at 25°C 1.15 (g/L) 1.82 5.75 6.68 9.79
atomic radius (pm) 56 98 114 133 143
first ionization energy (kJ/mol) 1402 1012 945 831 703
common oxidation state(s) −3 to +5 +5, +3, −3 +5, +3 +5, +3 +3
ionic radius (pm)§ 146 (−3), 16 (+3) 212 (−3), 44 (+3) 58 (+3) 76 (+3), 60 (+5) 103 (+3)
electron affinity (kJ/mol) 0 −72 −78 −101 −91
electronegativity 3.0 2.2 2.2 2.1 1.9
standard reduction potential (E°, V) (EV → EIII in acidic solution)|| +0.93 −0.28 +0.56 +0.65
product of reaction with O2 NO2, NO P4O6, P4O10 As4O6 Sb2O5 Bi2O3
type of oxide acidic (NO2), neutral (NO, N2O) acidic acidic amphoteric basic
product of reaction with N2 none none none none
product of reaction with X2 none PX3, PX5 AsF5, AsX3 SbF5, SbCl5, SbBr3, SbI3 BiF5, BiX3
product of reaction with H2 none none none none none
In group 15, the stability of the +5 oxidation state decreases from P to Bi.
Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases.
Reactions and Compounds of Nitrogen
Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp3 hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N3).
Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N2, whose N≡N bond is so strong (DN≡N = 942 kJ/mol) compared with the N–N and N=N bonds (DN–N = 167 kJ/mol; DN=N = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N2. In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive.
Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. At higher temperatures, however, N2 reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances.
Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N2 reacts with H2 to form ammonia, with O2 to form a mixture of NO and NO2, and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX3) are known. Except for NF3, all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH3 rather than N2. Both nitrogen monoxide (NO) and nitrogen dioxide (NO2) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO2 reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO2) and nitric acid (HNO3):
$\ce{2NO2(g) + H2O(l) \rightarrow HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms N2O (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with CO2 and can be represented as N=N+=O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows:
Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements.
At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as Li3N and Ca3N2. These compounds consist of ionic lattices formed by Mn+ and N3− ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity.
Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as Si3N4 and BN, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials.
Ammonia (NH3) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an O2 atmosphere:
$\ce{4NH3(g) + 3O2(g) \rightarrow 2N2(g) + 6H2O(g)} \label{Eq3}$
About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer.
Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid (HN3), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine (N2H4) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers.
B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials.
Example $1$
For each reaction, explain why the given products form when the reactants are heated.
1. Sr(s) + N2O(g) $\xrightarrow{\Delta}$ SrO(s) + N2(g)
2. NH4NO2(s) $\xrightarrow{\Delta}$ N2(g) + 2H2O(g)
3. Pb(NO3)2(s) $\xrightarrow{\Delta}$ PbO2(s) + 2NO2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. As an alkali metal, strontium is a strong reductant. If the other reactant can act as an oxidant, then a redox reaction will occur. Nitrous oxide contains nitrogen in a low oxidation state (+1), so we would not normally consider it an oxidant. Nitrous oxide is, however, thermodynamically unstable (ΔH°f > 0 and ΔG°f > 0), and it can be reduced to N2, which is a stable species. Consequently, we predict that a redox reaction will occur.
2. When a substance is heated, a decomposition reaction probably will occur, which often involves the release of stable gases. In this case, ammonium nitrite contains nitrogen in two different oxidation states (−3 and +3), so an internal redox reaction is a possibility. Due to its thermodynamic stability, N2 is the probable nitrogen-containing product, whereas we predict that H and O will combine to form H2O.
3. Again, this is probably a thermal decomposition reaction. If one element is in an usually high oxidation state and another in a low oxidation state, a redox reaction will probably occur. Lead nitrate contains the Pb2+ cation and the nitrate anion, which contains nitrogen in its highest possible oxidation state (+5). Hence nitrogen can be reduced, and we know that lead can be oxidized to the +4 oxidation state. Consequently, it is likely that lead(II) nitrate will decompose to lead(IV) oxide and nitrogen dioxide when heated. Even though PbO2 is a powerful oxidant, the release of a gas such as NO2 can often drive an otherwise unfavorable reaction to completion (Le Chatelier’s principle). Note, however, that PbO2 will probably decompose to PbO at high temperatures.
Exercise 23.3.1
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$
2. NH4NO3(s) $\xrightarrow{\Delta}$
3. Sr(s) + N2(g) →
Answer
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$ no reaction
2. NH4NO3(s) $\xrightarrow{\Delta}$ N2O(g) + 2H2O(g)
3. 3Sr(s) + N2(g) → Sr3N2(s)
Reactions and Compounds of the Heavier Pnicogens
Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P4 tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors.
As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group.
The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O2, whereas arsenic burns in pure O2 if ignited, and antimony and bismuth react with O2 only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO3. Only bismuth behaves like a metal, dissolving in HNO3 to give the hydrated Bi3+ cation.
The reactivity of the heavier group 15 elements decreases as we go down the column.
The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp3 or d2sp3 hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl5) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H3AsO4). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base:
$\ce{AsF5(soln) + F^{−}(soln) \rightarrow AsF^{−}6(soln)} \label{Eq4}$
In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth.
Except for BiF3, which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, $\ce{H3PO3}$ and $\ce{H3AsO3}$, where E is P or As:
$\ce{EX3(l) + 3H2O(l) \rightarrow H3EO3(aq) + 3HX(aq)} \label{Eq5}$
Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers.
Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C.
With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P4O6 and As4O6 have cage structures formed by inserting an oxygen atom into each edge of the P4 or As4 tetrahedron (part (a) in Figure $2$), and they behave like typical nonmetal oxides. For example, P4O6 reacts with water to form phosphorous acid (H3PO3). Consistent with its position between the nonmetal and metallic oxides, Sb4O6 is amphoteric, dissolving in either acid or base. In contrast, Bi2O3 behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi3+ ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E4O10 in the +5 oxidation state (part (b) in Figure $2$. In contrast, Bi2O5 is so unstable that there is no absolute proof it exists.
The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P4S3 (part (c) in Figure $2$), to layered or ribbon structures, such as Sb2S3 and Bi2S3, which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M4P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP15) are lower melting and less thermally stable because they contain catenated Pn units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth.
The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{Bi(s) +\frac{3}{2}Br(l)\rightarrow BiBr_3(s)}$
2. 2(CH3)3As(l) + O2(g) → 2(CH3)3As=O(s)
3. PBr3(l) + 3H2O(l) → H3PO3(aq) + 3HBr(aq)
4. As(s) + Ga(s) $\xrightarrow{\Delta}$ GaAs(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Bromine is an oxidant, and bismuth is a metal that can be oxidized. Hence a redox reaction is likely to occur. To identify the product, recall that bismuth can form compounds in either the +3 or +5 oxidation state. The heaviest pnicogen, bismuth is rather difficult to oxidize to the +5 oxidation state because of the inert-pair effect. Hence the product will probably be bismuth(III) bromide.
2. Trimethylarsine, with a lone pair of electrons on the arsenic atom, can act as either a Lewis base or a reductant. If arsenic is oxidized by two electrons, then oxygen must be reduced, most probably by two electrons to the −2 oxidation state. Because As(V) forms strong bonds to oxygen due to π bonding, the expected product is (CH3)3As=O.
3. Phosphorus tribromide is a typical nonmetal halide. We expect it to react with water to produce an oxoacid of P(III) and the corresponding hydrohalic acid.Because of the strength of the P=O bond, phosphorous acid (H3PO3) is actually HP(O)(OH)2, which contains a P=O bond and a P–H bond.
4. Gallium is a metal with a strong tendency to act as a reductant and form compounds in the +3 oxidation state. In contrast, arsenic is a semimetal. It can act as a reductant to form compounds in the +3 or +5 oxidation state, or it can act as an oxidant, accepting electrons to form compounds in the −3 oxidation state. If a reaction occurs, then a binary compound will probably form with a 1:1 ratio of the elements. GaAs is an example of a III-V compound, many of which are used in the electronics industry.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. PCl5(s) + H2O(l) →
2. Bi2O5(s) $\xrightarrow{\Delta}$
3. Ca3P2(s) + H+(aq) →
4. NaNH2(s) + PH3(soln) →
Answer
1. PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
2. Bi2O5(s) $\xrightarrow{\Delta}$ Bi2O3(s) + O2(g)
3. Ca3P2(s) + 6H+(aq) → 2PH3(g) + 3Ca2+(aq)
4. NaNH2(s) + PH3(soln) → NaPH2(s) + NH3(soln)
Summary
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp3 or d2sp3 hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.03%3A_The_Elements_of_Group_15_%28The_Pnicogens%29.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 16 elements: the chalcogens.
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element.
Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility.
Group 16 is the first group in the p block with no stable metallic elements.
Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics.
Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices.
Jöns Jakob Berzelius (1779–1848)
Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium).
The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium.
Preparation and General Properties of the Group 16 Elements
Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2:
$\mathrm{2KClO_3(s)\overset{MnO_2(s)}{\underset{\Delta}\rightleftharpoons}2KCl(s)+3O_2(g)} \label{22.4.1}$
Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure $1$). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2).
Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.”
With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure $1$. As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions.
Table $1$: Selected Properties of the Group 16 Elements
Property Oxygen Sulfur Selenium Tellurium Polonium
*The configuration shown does not include filled d and f subshells.
The values cited for the hexacations are for six-coordinate ions and are only estimated values.
atomic mass (amu) 16.00 32.07 78.96 127.60 209
atomic number 8 16 34 52 84
atomic radius (pm) 48 88 103 123 135
atomic symbol O S Se Te Po
density (g/cm3) at 25°C 1.31 (g/L) 2.07 4.81 6.24 9.20
electron affinity (kJ/mol) −141 −200 −195 −190 −180
electronegativity 3.4 2.6 2.6 2.1 2.0
first ionization energy (kJ/mol) 1314 1000 941 869 812
ionic radius (pm) 140 (−2) 184 (−2), 29 (+6) 198 (−2), 42 (+6) 221 (−2), 56 (+6) 230 (−2), 97 (+4)
melting point/boiling point (°C) −219/−183 115/445 221/685 450/988 254/962
normal oxidation state(s) −2 +6, +4, −2 +6, +4, −2 +6, +4, −2 +2 (+4)
product of reaction with H2 H2O H2S H2Se none none
product of reaction with N2 NO, NO2 none none none none
product of reaction with O2 SO2 SeO2 TeO2 PoO2
product of reaction with X2 O2F2 SF6, S2Cl2, S2Br2 SeF6, SeX4 TeF6, TeX4 PoF4, PoCl2, PoBr2
standard reduction potential (E°, V) (E0 → H2E in acidic solution) +1.23 +0.14 −0.40 −0.79 −1.00
type of oxide acidic acidic amphoteric basic
valence electron configuration* 2s22p4 3s23p4 4s24p4 5s25p4 6s26p4
Reactions and Compounds of Oxygen
As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure:
$2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)} \;\;\; ΔG^o = −119\; kJ/mol \label{1}$
As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds.
Despite the strength of the O=O bond ($D_\mathrm{O_2}$ = 494 kJ/mol), $O_2$ is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in Table $2$. With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens.
Table $2$: Some Properties of O2 and Related Diatomic Species
Species Bond Order Number of Unpaired e O–O Distance (pm)*
*Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175.
O2+ 2.5 1 112
O2 2 2 121
O2 1.5 1 133
O22− 1 0 149
Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid:
$H_2O_{(l)} + SO_{3(g)} \rightarrow H_2SO_{4(aq)} \label{2}$
The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect:
$Al_2O_{3(s)} + 6H^+_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3H_2O_{(l)} \label{3}$
$Al_2O_{3(s)} + 2OH^−_{(aq)} + 3H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} \label{4}$
Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric.
Example $1$
For each reaction, explain why the given products form.
1. Ga2O3(s) + 2OH(aq) + 3H2O(l) → 2Ga(OH)4(aq)
2. 3H2O2(aq) + 2MnO4(aq) + 2H+(aq) → 3O2(g) + 2MnO2(s) + 4H2O(l)
3. KNO3(s) $\xrightarrow{\Delta}$ KNO(s) + O2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Gallium is a metal. We expect the oxides of metallic elements to be basic and therefore not to react with aqueous base. A close look at the periodic table, however, shows that gallium is close to the diagonal line of semimetals. Moreover, aluminum, the element immediately above gallium in group 13, is amphoteric. Consequently, we predict that gallium will behave like aluminum (Equation $\ref{4}$).
2. Hydrogen peroxide is an oxidant that can accept two electrons per molecule to give two molecules of water. With a strong oxidant, however, H2O2 can also act as a reductant, losing two electrons (and two protons) to produce O2. Because the other reactant is permanganate, which is a potent oxidant, the only possible reaction is a redox reaction in which permanganate is the oxidant and hydrogen peroxide is the reductant. Recall that reducing permanganate often gives MnO2, an insoluble brown solid. Reducing MnO4 to MnO2 is a three-electron reduction, whereas the oxidation of H2O2 to O2 is a two-electron oxidation.
3. This is a thermal decomposition reaction. Because KNO3 contains nitrogen in its highest oxidation state (+5) and oxygen in its lowest oxidation state (−2), a redox reaction is likely. Oxidation of the oxygen in nitrate to atomic oxygen is a two-electron process per oxygen atom. Nitrogen is likely to accept two electrons because oxoanions of nitrogen are known only in the +5 (NO3) and +3 (NO2) oxidation states.
Exercise $2$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. SiO2(s) + H+(aq) →
2. NO(g) + O2(g) →
3. SO3(g) + H2O(l) →
4. H2O2(aq) + I(aq) →
Answer
1. SiO2(s) + H+(aq) → no reaction
2. 2NO(g) + O2(g) → 2NO2(g)
3. SO3(g) + H2O(l) → H2SO4(aq)
4. H2O2(aq) + 2I(aq) → I2(aq) + 2OH(aq)
Reactions and Compounds of the Heavier Chalcogens
Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column.
Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (Figure $2$), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group.
As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion.
Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group.
Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure $2$). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known.
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate.
The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid.
The stability of the highest oxidation state of the chalcogens decreases down the column.
Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5.
The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group.
Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous.
Example $2$
For each reaction, explain why the given product forms or no reaction occurs.
1. SO2(g) + Cl2(g) → SO2Cl2(l)
2. SF6(g) + H2O(l) → no reaction
3. 2Se(s) + Cl2(g) → Se2Cl2(l)
Given: balanced chemical equations
Asked for: why the given products (or no products) form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs.
Solution
1. One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds.
1. Sulfur hexafluoride is a nonmetallic halide. Such compounds normally react vigorously with water to produce an oxoacid of the nonmetal and the corresponding hydrohalic acid. In this case, however, we have a highly stable species, presumably because all of sulfur’s available orbitals are bonding orbitals. Thus SF6 is not likely to react with water.
2. Here we have the reaction of a chalcogen with a halogen. The halogen is a good oxidant, so we can anticipate that a redox reaction will occur. Only fluorine is capable of oxidizing the chalcogens to a +6 oxidation state, so we must decide between SeCl4 and Se2Cl2 as the product. The stoichiometry of the reaction determines which of the two is obtained: SeCl4 or Se2Cl2.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. Te(s) + Na(s) $\xrightarrow{\Delta}$
2. SF4(g) + H2O(l) →
3. CH3SeSeCH3(soln) + K(s) →
4. Li2Se(s) + H+(aq) →
Answer
1. Te(s) + 2Na(s) → Na2Te(s)
2. SF4(g) + 3H2O(l) → H2SO3(aq) + 4HF(aq)
3. CH3SeSeCH3(soln) + 2K(s) → 2KCH3Se(soln)
4. Li2Se(s) + 2H+(aq) → H2Se(g) + 2Li+(aq)
Summary
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.04%3A_The_Elements_of_Group_16_%28The_Chalcogens%29.txt |
Learning Objectives
• To understand the periodic trends and reactivity of the group 17 elements: the halogens.
Because the halogens are highly reactive, none is found in nature as the free element. Hydrochloric acid, which is a component of aqua regia (a mixture of HCl and HNO3 that dissolves gold), and the mineral fluorspar (CaF2) were well known to alchemists, who used them in their quest for gold. Despite their presence in familiar substances, none of the halogens was even recognized as an element until the 19th century.
Because the halogens are highly reactive, none is found in nature as the free element.
Chlorine was the first halogen to be obtained in pure form. In 1774, Carl Wilhelm Scheele (the codiscoverer of oxygen) produced chlorine by reacting hydrochloric acid with manganese dioxide. Scheele was convinced, however, that the pale green gas he collected over water was a compound of oxygen and hydrochloric acid. In 1811, Scheele’s “compound” was identified as a new element, named from the Greek chloros, meaning “yellowish green” (the same stem as in chlorophyll, the green pigment in plants). That same year, a French industrial chemist, Bernard Courtois, accidentally added too much sulfuric acid to the residue obtained from burned seaweed. A deep purple vapor was released, which had a biting aroma similar to that of Scheele’s “compound.” The purple substance was identified as a new element, named iodine from the Greek iodes, meaning “violet.” Bromine was discovered soon after by a young French chemist, Antoine Jérôme Balard, who isolated a deep red liquid with a strong chlorine-like odor from brine from the salt marshes near Montpellier in southern France. Because many of its properties were intermediate between those of chlorine and iodine, Balard initially thought he had isolated a compound of the two (perhaps ICl). He soon realized, however, that he had discovered a new element, which he named bromine from the Greek bromos, meaning “stench.” Currently, organic chlorine compounds, such as PVC (polyvinylchloride), consume about 70% of the Cl2 produced annually; organobromine compounds are used in much smaller quantities, primarily as fire retardants.
Because of the unique properties of its compounds, fluorine was believed to exist long before it was actually isolated. The mineral fluorspar (now called fluorite [CaF2]) had been used since the 16th century as a “flux,” a low-melting-point substance that could dissolve other minerals and ores. In 1670, a German glass cutter discovered that heating fluorspar with strong acid produced a solution that could etch glass. The solution was later recognized to contain the acid of a new element, which was named fluorine in 1812. Elemental fluorine proved to be very difficult to isolate, however, because both HF and F2 are extraordinarily reactive and toxic. After being poisoned three times while trying to isolate the element, the French chemist Henri Moissan succeeded in 1886 in electrolyzing a sample of KF in anhydrous HF to produce a pale green gas (Figure $1$). For this achievement, among others, Moissan narrowly defeated Mendeleev for the Nobel Prize in Chemistry in 1906. Large amounts of fluorine are now consumed in the production of cryolite (Na3AlF6), a key intermediate in the production of aluminum metal. Fluorine is also found in teeth as fluoroapatite [Ca5(PO4)3F], which is formed by reacting hydroxyapatite [Ca5(PO4)3OH] in tooth enamel with fluoride ions in toothpastes, rinses, and drinking water.
The heaviest halogen is astatine (At), which is continuously produced by natural radioactive decay. All its isotopes are highly radioactive, and the most stable has a half-life of only about 8 h. Consequently, astatine is the least abundant naturally occurring element on Earth, with less than 30 g estimated to be present in Earth’s crust at any one time.
Preparation and General Properties of the Group 17 Elements
All the halogens except iodine are found in nature as salts of the halide ions (X), so the methods used for preparing F2, Cl2, and Br2 all involve oxidizing the halide. Reacting CaF2 with concentrated sulfuric acid produces gaseous hydrogen fluoride:
$CaF_{2(s)} + H_2SO_{4(l)} \rightarrow CaSO_{4(s)} + 2HF_{(g)} \label{1}$
Fluorine is produced by the electrolysis of a 1:1 mixture of HF and K+HF2 at 60–300°C in an apparatus made of Monel, a highly corrosion-resistant nickel–copper alloy:
$KHF_2\cdot HF(l) \xrightarrow{electrolysis}F_2(g) + H_2(g) \label{2}$
Fluorine is one of the most powerful oxidants known, and both F2 and HF are highly corrosive. Consequently, the production, storage, shipping, and handling of these gases pose major technical challenges.
Although chlorine is significantly less abundant than fluorine, elemental chlorine is produced on an enormous scale. Fortunately, large subterranean deposits of rock salt (NaCl) are found around the world (Figure $2$), and seawater consists of about 2% NaCl by mass, providing an almost inexhaustible reserve. Inland salt lakes such as the Dead Sea and the Great Salt Lake are even richer sources, containing about 23% and 8% NaCl by mass, respectively. Chlorine is prepared industrially by the chloralkali process, which uses the following reaction:
$2NaCl_{(aq)} +2H_2O_{(l)} \xrightarrow{electrolysis} 2NaOH(aq) + Cl_{2(g)} + H_{2(g)} \label{3}$
Bromine is much less abundant than fluorine or chlorine, but it is easily recovered from seawater, which contains about 65 mg of Br per liter. Salt lakes and underground brines are even richer sources; for example, the Dead Sea contains 4 g of Br per liter. Iodine is the least abundant of the nonradioactive halogens, and it is a relatively rare element. Because of its low electronegativity, iodine tends to occur in nature in an oxidized form. Hence most commercially important deposits of iodine, such as those in the Chilean desert, are iodate salts such as Ca(IO3)2. The production of iodine from such deposits therefore requires reduction rather than oxidation. The process is typically carried out in two steps: reduction of iodate to iodide with sodium hydrogen sulfite, followed by reaction of iodide with additional iodate:
$2IO^−_{3(aq)} + 6HSO^−_{3(aq)} \rightarrow 2I^−_{(aq)} + 6SO^2−_{4(aq)} + 6H^+_{(aq)} \label{4}$
$5I^−_{(aq)} + IO^−_{3(aq)} + 6H^+_{(aq)} \rightarrow 3I_{2(s)} + 3H_2O_{(l)} \label{5}$
Because the halogens all have ns2np5 electron configurations, their chemistry is dominated by a tendency to accept an additional electron to form the closed-shell ion (X). Only the electron affinity and the bond dissociation energy of fluorine differ significantly from the expected periodic trends shown in Table $1$. Electron–electron repulsion is important in fluorine because of its small atomic volume, making the electron affinity of fluorine less than that of chlorine. Similarly, repulsions between electron pairs on adjacent atoms are responsible for the unexpectedly low F–F bond dissociation energy. (As discussed earlier, this effect is also responsible for the weakness of O–O, N–N, and N–O bonds.)
Electrostatic repulsions between lone pairs of electrons on adjacent atoms cause single bonds between N, O, and F to be weaker than expected.
Table $1$: Selected Properties of the Group 17 Elements
Property Fluorine Chlorine Bromine Iodine Astatine
*The configuration shown does not include filled d and f subshells. The values cited are for the six-coordinate anion (X−).
atomic symbol F Cl Br I At
atomic number 9 17 35 53 85
atomic mass (amu) 19.00 35.45 79.90 126.90 210
valence electron configuration* 2s22p5 3s23p5 4s24p5 5s25p5 6s26p5
melting point/boiling point (°C) −220/−188 −102/−34.0 −7.2/58.8 114/184 302/—
density (g/cm3) at 25°C 1.55 (g/L) 2.90 (g/L) 3.10 4.93
atomic radius (pm) 42 79 94 115 127
first ionization energy (kJ/mol) 1681 1251 1140 1008 926
normal oxidation state(s) −1 −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1, +1
ionic radius (pm) 133 181 196 220
electron affinity (kJ/mol) −328 −349 −325 −295 −270
electronegativity 4.0 3.2 3.0 2.7 2.2
standard reduction potential (E°, V) (X2 → X in basic solution) +2.87 +1.36 +1.07 +0.54 +0.30
dissociation energy of X2(g) (kJ/mol) 158.8 243.6 192.8 151.1 ~80
product of reaction with O2 O2F2 none none none none
type of oxide acidic acidic acidic acidic acidic
product of reaction with N2 none none none none none
product of reaction with H2 HF HCl HBr HI HAt
Because it is the most electronegative element in the periodic table, fluorine forms compounds in only the −1 oxidation state. Notice, however, that all the halogens except astatine have electronegativities greater than 2.5, making their chemistry exclusively that of nonmetals. The halogens all have relatively high ionization energies, but the energy required to remove electrons decreases substantially as we go down the column. Hence the heavier halogens also form compounds in positive oxidation states (+1, +3, +5, and +7), derived by the formal loss of ns and np electrons.
Because ionization energies decrease down the group, the heavier halogens form compounds in positive oxidation states (+1, +3, +5, and +7).
Reactions and Compounds of the Halogens
Fluorine is the most reactive element in the periodic table, forming compounds with every other element except helium, neon, and argon. The reactions of fluorine with most other elements range from vigorous to explosive; only O2, N2, and Kr react slowly. There are three reasons for the high reactivity of fluorine:
1. Because fluorine is so electronegative, it is able to remove or at least share the valence electrons of virtually any other element.
2. Because of its small size, fluorine tends to form very strong bonds with other elements, making its compounds thermodynamically stable.
3. The F–F bond is weak due to repulsion between lone pairs of electrons on adjacent atoms, reducing both the thermodynamic and kinetic barriers to reaction.
With highly electropositive elements, fluorine forms ionic compounds that contain the closed-shell F ion. In contrast, with less electropositive elements (or with metals in very high oxidation states), fluorine forms covalent compounds that contain terminal F atoms, such as SF6. Because of its high electronegativity and 2s22p5 valence electron configuration, fluorine normally participates in only one electron-pair bond. Only a very strong Lewis acid, such as AlF3, can share a lone pair of electrons with a fluoride ion, forming AlF63−.
Oxidative strength decreases down group 17.
The halogens (X2) react with metals (M) according to the general equation
$M_{(s,l)} + nX_{2(s,l,g)} \rightarrow MX_{n(s,l)} \label{6}$
For elements that exhibit multiple oxidation states fluorine tends to produce the highest possible oxidation state and iodine the lowest. For example, vanadium reacts with the halogens to give VF5, VCl4, VBr4, and VI3.
Metal halides in the +1 or +2 oxidation state, such as CaF2, are typically ionic halides, which have high melting points and are often soluble in water. As the oxidation state of the metal increases, so does the covalent character of the halide due to polarization of the M–X bond. With its high electronegativity, fluoride is the least polarizable, and iodide, with the lowest electronegativity, is the most polarizable of the halogens. Halides of small trivalent metal ions such as Al3+ tend to be relatively covalent. For example, AlBr3 is a volatile solid that contains bromide-bridged Al2Br6 molecules. In contrast, the halides of larger trivalent metals, such as the lanthanides, are essentially ionic. For example, indium tribromide (InBr3) and lanthanide tribromide (LnBr3) are all high-melting-point solids that are quite soluble in water.
As the oxidation state of the metal increases, the covalent character of the corresponding metal halides also increases due to polarization of the M–X bond.
All halogens react vigorously with hydrogen to give the hydrogen halides (HX). Because the H–F bond in HF is highly polarized (Hδ+–Fδ−), liquid HF has extensive hydrogen bonds, giving it an unusually high boiling point and a high dielectric constant. As a result, liquid HF is a polar solvent that is similar in some ways to water and liquid ammonia; after a reaction, the products can be recovered simply by evaporating the HF solvent. (Hydrogen fluoride must be handled with extreme caution, however, because contact of HF with skin causes extraordinarily painful burns that are slow to heal.) Because fluoride has a high affinity for silicon, aqueous hydrofluoric acid is used to etch glass, dissolving SiO2 to give solutions of the stable SiF62− ion.
Glass etched with hydrogen flouride.© Thinkstock
Except for fluorine, all the halogens react with water in a disproportionation reaction, where X is Cl, Br, or I:
$X_{2(g,l,s)} + H_2O_{(l)} \rightarrow H^+_{(aq)} + X^−_{(aq)} + HOX_{(aq)} \label{7}$
The most stable oxoacids are the perhalic acids, which contain the halogens in their highest oxidation state (+7). The acid strengths of the oxoacids of the halogens increase with increasing oxidation state, whereas their stability and acid strength decrease down the group. Thus perchloric acid (HOClO3, usually written as HClO4) is a more potent acid and stronger oxidant than perbromic acid. Although all the oxoacids are strong oxidants, some, such as HClO4, react rather slowly at low temperatures. Consequently, mixtures of the halogen oxoacids or oxoanions with organic compounds are potentially explosive if they are heated or even agitated mechanically to initiate the reaction. Because of the danger of explosions, oxoacids and oxoanions of the halogens should never be allowed to come into contact with organic compounds.
Both the acid strength and the oxidizing power of the halogen oxoacids decrease down the group.
The halogens react with one another to produce interhalogen compounds, such as ICl3, BrF5, and IF7. In all cases, the heavier halogen, which has the lower electronegativity, is the central atom. The maximum oxidation state and the number of terminal halogens increase smoothly as the ionization energy of the central halogen decreases and the electronegativity of the terminal halogen increases. Thus depending on conditions, iodine reacts with the other halogens to form IFn (n = 1–7), ICl or ICl3, or IBr, whereas bromine reacts with fluorine to form only BrF, BrF3, and BrF5 but not BrF7. The interhalogen compounds are among the most powerful Lewis acids known, with a strong tendency to react with halide ions to give complexes with higher coordination numbers, such as the IF8 ion:
$IF_{7(l)} + KF_{(s)} \rightarrow KIF_{8(s)} \label{8}$
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7). The interhalogen compounds are also potent oxidants and strong fluorinating agents; contact with organic materials or water can result in an explosion.
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7), but the importance of the higher oxidation states generally decreases down the group.
Example $1$
For each reaction, explain why the given products form.
1. ClF3(g) + Cl2(g) → 3ClF(g)
2. 2KI(s) + 3H2SO4(aq) → I2(aq) + SO2(g) + 2KHSO4(aq) + 2H2O(l)
3. Pb(s) + 2BrF3(l) → PbF4(s) + 2BrF(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. When the reactants have the same element in two different oxidation states, we expect the product to have that element in an intermediate oxidation state. We have Cl3+ and Cl0 as reactants, so a possible product would have Cl in either the +1 or +2 oxidation state. From our discussion, we know that +1 is much more likely. In this case, Cl2 is behaving like a reductant rather than an oxidant.
2. At first glance, this appears to be a simple acid–base reaction, in which sulfuric acid transfers a proton to I to form HI. Recall, however, that I can be oxidized to I2. Sulfuric acid contains sulfur in its highest oxidation state (+6), so it is a good oxidant. In this case, the redox reaction predominates.
3. This is the reaction of a metallic element with a very strong oxidant. Consequently, a redox reaction will occur. The only question is whether lead will be oxidized to Pb(II) or Pb(IV). Because BrF3 is a powerful oxidant and fluorine is able to stabilize high oxidation states of other elements, it is likely that PbF4 will be the product. The two possible reduction products for BrF3 are BrF and Br2. The actual product will likely depend on the ratio of the reactants used. With excess BrF3, we expect the more oxidized product (BrF). With lower ratios of oxidant to lead, we would probably obtain Br2 as the product.
Exercise $1$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. CaCl2(s) + H3PO4(l) →
2. GeO2(s) + HF(aq) →
3. Fe2O3(s) + HCl(g) $\xrightarrow{\Delta}$
4. NaClO2(aq) + Cl2(g) →
Answer
1. CaCl2(s) + H3PO4(l) → 2HCl(g) + Ca(HPO4)(soln)
2. GeO2(s) + 6HF(aq) → GeF62−(aq) + 2H2O(l) + 2H+(aq)
3. Fe2O3(s) + 6HCl(g) $\xrightarrow{\Delta}$ 2FeCl3(s) + 3H2O(g)
4. 2NaClO2(aq) + Cl2(g) → 2ClO2(g) + 2NaCl(aq)
Summary
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group. Fluorine, the most reactive element in the periodic table, has a low F–F bond dissociation energy due to repulsions between lone pairs of electrons on adjacent atoms. Fluorine forms ionic compounds with electropositive elements and covalent compounds with less electropositive elements and metals in high oxidation states. All the halogens react with hydrogen to produce hydrogen halides. Except for F2, all react with water to form oxoacids, including the perhalic acids, which contain the halogens in their highest oxidation state. Halogens also form interhalogen compounds; the heavier halogen, with the lower electronegativity, is the central atom. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/21%3A_The_p-Block_Elements/21.05%3A_The_Elements_of_Group_17_%28The_Halogens%29.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 18 elements: the noble gases.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
J. W. Strutt (Lord Rayleigh) (1842–1919)
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Sir William Ramsay (1852–1916)
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Preparation and General Properties of the Group 18 Elements
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $1$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table $1$: Selected Properties of the Group 18 Elements
Property Helium Neon Argon Krypton Xenon Radon
*The configuration shown does not include filled d and f subshells. This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given.
atomic symbol He Ne Ar Kr Xe Rn
atomic number 2 10 18 36 54 86
atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222
valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6
triple point/boiling point (°C) —/−269 −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62
density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07
atomic radius (pm) 31 38 71 88 108 120
first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037
normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2)
electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0
electronegativity 2.6
product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none
type of oxide acidic
product of reaction with N2 none none none none none none
product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2
product of reaction with H2 none none none none none none
Reactions and Compounds of the Noble Gases
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $1$).
Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved.
Methane Clathrates
In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (Figure $3$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion:
$\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}$
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
$\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}$
The ionization energies of helium, neon, and argon are so high (Table $1$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
$\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}$
$\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}$
The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF8 ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid:
$\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}$
Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3), and XeF82− is isoelectronic with the IF8 ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
Example $1$
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise $1$
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer
$\ce{KXeF7}$; the xenon atom in XeF7 has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
Summary
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
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Application Problems
1. Borax (Na2B4O5(OH)4·8H2O) is used as a flux during welding operations. As brass is heated during welding, for example, borax cleans the surface of Cu2O and prevents further oxidation of the fused metal. Explain why borax is effective at cleaning the surface and preventing surface oxidation.
2. Extensive research is being conducted into using GaAs as a material for computer memory chips. It has been found, for example, that chips made from GaAs are up to 10 times faster than those made from silicon. Propose an explanation for this increase in speed.
3. Cement that has a high content of alumina (Al2O3) is particularly resistant to corrosion, so it is used for structures that must be resistant to seawater and acidic conditions. Why is this material so effective under these service conditions? Failure occurs under prolonged exposure to a hot, wet environment. Why?
4. Aluminum is light and ductile. If you were considering using aluminum rather than steel as a structural material for building a high-speed ferry, what disadvantages would you need to consider in using aluminum for these service conditions?
5. Life on Earth is based on carbon. A possible explanation is that no other element in the periodic table forms compounds that are so diverse in their chemistry and physical properties. Discuss the chemistry of carbon with regard to
1. types of bonding.
2. the states of matter of carbon compounds.
3. the properties of elemental C.
4. the reactivity of elemental C and its compounds.
Then compare B, Al, Si, N, and P with C in terms of these properties.
1. After a traffic accident in which a tanker truck carrying liquid nitrogen overturned, a reporter at the scene warned of a danger to residents in the vicinity of the accident because nitrogen would react with hydrocarbons in the asphalt to produce ammonia gas. Comment on the credibility of this statement.
2. Nitrogen forms a hydride called hydrazoic acid (HN3), which is a colorless, highly toxic, explosive substance that boils at 37°C. The thermal decomposition of one of its salts—NaN3—is used to inflate automotive air bags. The N3 ion is isoelectronic with CO2.
1. Draw the Lewis electron structure of the N3 anion.
2. Write a balanced chemical equation for the thermal decomposition of NaN3.
3. Based on your answer to part (b), propose an explanation for why many people have suffered skin burns when their air bags exploded.
1. Hydrazine (N2H4), a rocket fuel, is a colorless, oily liquid with a melting point of 1.4°C, and it is a powerful reducing agent. The physical properties of hydrazine presumably reflect the presence of multiple hydrogen-bond acceptors and donors within a single molecule. Explain the basis for this statement.
2. Because the N–C bond is almost as strong as the N–H bond, organic analogues of ammonia, hydrazine, and hydroxylamine are stable and numerous. Conceptually at least, they are formed by the successive replacement of H atoms by alkyl or aryl groups. Methylhydrazine and dimethylhydrazine, for example, were used as fuels in the US Apollo space program. They react spontaneously and vigorously with liquid N2O4, thus eliminating the need for an ignition source. Write balanced chemical equations for these reactions and calculate ΔG° for each reaction.
3. In an effort to remove a troublesome stain from a sink, a member of the cleaning staff of a commercial building first used bleach on the stain and then decided to neutralize the bleach with ammonia. What happened? Why?
4. A slow reaction that occurs on the ocean floor is the conversion of carbonate to bicarbonate, which absorbs CO2. Write a balanced chemical equation for this reaction. Silicate sediments play an important role in controlling the pH of seawater. Given the reaction, propose a chemical explanation for this.
5. Marketing surveys have shown that customers prefer to buy a bright red steak rather than a dull gray one. It is known that NO combines with myoglobin to form a bright red NO complex. What would you add to beef during processing to ensure that this reaction occurs and yields the desired appearance?
6. Covalent azides are used as detonators and explosives. Ionic azides, in contrast, are usually much more stable and are used in dyestuffs. Why is there such a difference between these two types of compounds? The N3 ion is considered a pseudohalide. Why?
7. The heads of modern “strike anywhere” matches contain a mixture of a nonvolatile phosphorus sulfide (P4S3) and an oxidizing agent (KClO3), which is ignited by friction when the match is struck against a rough object. Safety matches separate the oxidant and the reductant by putting KClO3 in the head and a paste containing nonvolatile red phosphorus on the match box or cover. Write a balanced chemical equation for the reaction that occurs when a match is rubbed against the abrasive end of a matchbox.
8. Paris green was a common pigment in paints and wallpaper of the Napoleonic era. It is a mixed acetate/arsenite salt of copper with the formula Cu2(OAc)2(AsO3). In damp conditions, certain fungi are able to convert arsenite salts to volatile, toxic organoarsenic compounds. Shortly after his exile in 1815 to the remote island of St. Helena in the southern Atlantic Ocean, Napoleon died. As a forensic scientist investigating the cause of Napoleon’s mysterious death, you notice that the walls of his enclosed bedchamber are covered in green wallpaper. What chemical clues would you look for to determine the cause of his death?
9. Selenium, an element essential to humans, appears to function biologically in an enzyme that destroys peroxides. Why is selenium especially suited for this purpose? Would sulfur or tellurium be as effective? Why or why not?
10. One way to distinguish between fool’s gold (FeS2, or iron pyrite) and real gold is to heat the sample over a fire. If your sample of “gold” were actually fool’s gold, what would happen?
11. Calcium hypochlorite is sold as swimming pool bleach. It is formed by the hydrolysis of Cl2O, which gives only one product, followed by neutralization with lime [Ca(OH)2]. Write balanced chemical equations for these reactions.
12. There is much interest in the superheavy elements beyond Z = 111 because of their potentially unique properties. Predict the valence electron configurations, preferred oxidation states, and products of the reaction with aqueous acid for elements 113 and 115.
13. Zeolites have become increasingly important in chemical engineering. They can be used as desiccants because the dehydrated zeolite absorbs small molecules, such as water. To be retained by the zeolite frame, a molecule must satisfy two conditions. What are they? Why can linear CO2 and tetrahedral CH4 not be held by a typical zeolite, even though they can penetrate it easily?
Answers
1. Three resonance structures for the azide ion may be reasonably drawn:
1. $\mathrm{2NaN_3(s)\xrightarrow{\Delta}2Na(s)+3N_2(g)}$
2. One of the products of the decomposition of sodium azide is elemental sodium, which is highly reactive and can ignite in the presence of water and air.
1. Examine samples of Napoleon’s hair and/or fingernails from museums or collections to determine arsenic concentrations.
1. Upon heating, pyrite will react with oxygen to form SO2(g), which has a pungent smell.
1. Element 113: 5f146d107s27p1, +1, E+(aq); element 115: 5f146d107s27p3, +3, E3+(aq)
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Learning Objectives
• To understand the trends in properties and reactivity of the d-block elements.
The transition metals, groups 3–12 in the periodic table, are generally characterized by partially filled d subshells in the free elements or their cations. (Although the metals of group 12 do not have partially filled d shells, their chemistry is similar in many ways to that of the preceding groups, and we therefore include them in our discussion.) Unlike the s-block and p-block elements, the transition metals exhibit significant horizontal similarities in chemistry in addition to their vertical similarities.
Electronic Structure and Reactivity of the Transition Metals
The valence electron configurations of the first-row transition metals are given in Table \(1\). As we go across the row from left to right, electrons are added to the 3d subshell to neutralize the increase in the positive charge of the nucleus as the atomic number increases. With two important exceptions, the 3d subshell is filled as expected based on the aufbau principle and Hund’s rule. Unexpectedly, however, chromium has a 4s13d5 electron configuration rather than the 4s23d4 configuration predicted by the aufbau principle, and copper is 4s13d10 rather than 4s23d9. In Chapter 7, we attributed these anomalies to the extra stability associated with half-filled subshells. Because the ns and (n − 1)d subshells in these elements are similar in energy, even relatively small effects are enough to produce apparently anomalous electron configurations.
Table \(1\) Valence Electron Configurations of the First-Row Transition Metals
Sc Ti V Cr Mn Fe Co Ni Cu Zn
4s23d1 4s23d2 4s23d3 4s13d5 4s23d5 4s23d6 4s23d7 4s23d8 4s13d10 4s23d10
In the second-row transition metals, electron–electron repulsions within the 4d subshell cause additional irregularities in electron configurations that are not easily predicted. For example, Nb and Tc, with atomic numbers 41 and 43, both have a half-filled 5s subshell, with 5s14d4 and 5s14d6 valence electron configurations, respectively. Further complications occur among the third-row transition metals, in which the 4f, 5d, and 6s orbitals are extremely close in energy. Although La has a 6s25d1 valence electron configuration, the valence electron configuration of the next element—Ce—is 6s25d04f2. From this point through element 71, added electrons enter the 4f subshell, giving rise to the 14 elements known as the lanthanides. After the 4f subshell is filled, the 5d subshell is populated, producing the third row of the transition metals. Next comes the seventh period, where the actinides have three subshells (7s, 6d, and 5f) that are so similar in energy that their electron configurations are even more unpredictable.
As we saw in the s-block and p-block elements, the size of neutral atoms of the d-block elements gradually decreases from left to right across a row, due to an increase in the effective nuclear charge (Zeff) with increasing atomic number. In addition, the atomic radius increases down a group, just as it does in the s and p blocks. Because of the lanthanide contraction, however, the increase in size between the 3d and 4d metals is much greater than between the 4d and 5d metals (Figure \(1\). (For more information on the lanthanides, see Chapter 7, Section 7.3.) The effects of the lanthanide contraction are also observed in ionic radii, which explains why, for example, there is only a slight increase in radius from Mo3+ to W3+.
Figure \(1\) The Metallic Radii of the First-, Second-, and Third-Row Transition Metals
Because of the lanthanide contraction, the second- and third-row transition metals are very similar in size.
As you learned in Chapter 7, electrons in (n − 1)d and (n − 2)f subshells are only moderately effective at shielding the nuclear charge; as a result, the effective nuclear charge experienced by valence electrons in the d-block and f-block elements does not change greatly as the nuclear charge increases across a row. Consequently, the ionization energies of these elements increase very slowly across a given row (Figure 7.10). In addition, as we go from the top left to the bottom right corner of the d block, electronegativities generally increase, densities and electrical and thermal conductivities increase, and enthalpies of hydration of the metal cations decrease in magnitude, as summarized in Figure \(2\). Consistent with this trend, the transition metals become steadily less reactive and more “noble” in character from left to right across a row. The relatively high ionization energies and electronegativities and relatively low enthalpies of hydration are all major factors in the noble character of metals such as Pt and Au.
Figure \(2\) Some Trends in Properties of the Transition Metals
The electronegativity of the elements increases, and the hydration energies of the metal cations decrease in magnitude from left to right and from top to bottom of the d block. As a result, the metals in the lower right corner of the d block are so unreactive that they are often called the “noble metals.”
Trends in Transition Metal Oxidation States
The similarity in ionization energies and the relatively small increase in successive ionization energies lead to the formation of metal ions with the same charge for many of the transition metals. This in turn results in extensive horizontal similarities in chemistry, which are most noticeable for the first-row transition metals and for the lanthanides and actinides. Thus all the first-row transition metals except Sc form stable compounds that contain the 2+ ion, and, due to the small difference between the second and third ionization energies for these elements, all except Zn also form stable compounds that contain the 3+ ion. The relatively small increase in successive ionization energies causes most of the transition metals to exhibit multiple oxidation states separated by a single electron. Manganese, for example, forms compounds in every oxidation state between −3 and +7. Because of the slow but steady increase in ionization potentials across a row, high oxidation states become progressively less stable for the elements on the right side of the d block. The occurrence of multiple oxidation states separated by a single electron causes many, if not most, compounds of the transition metals to be paramagnetic, with one to five unpaired electrons. This behavior is in sharp contrast to that of the p-block elements, where the occurrence of two oxidation states separated by two electrons is common, which makes virtually all compounds of the p-block elements diamagnetic.
Note the Pattern
Due to a small increase in successive ionization energies, most of the transition metals have multiple oxidation states separated by a single electron.
Note the Pattern:
Most compounds of transition metals are paramagnetic, whereas virtually all compounds of the p-block elements are diamagnetic.
The electronegativities of the first-row transition metals increase smoothly from Sc (χ = 1.4) to Cu (χ = 1.9). Thus Sc is a rather active metal, whereas Cu is much less reactive. The steady increase in electronegativity is also reflected in the standard reduction potentials: thus E° for the reaction M2+(aq) + 2e → M0(s) becomes progressively less negative from Ti (E° = −1.63 V) to Cu (E° = +0.34 V). Exceptions to the overall trends are rather common, however, and in many cases, they are attributable to the stability associated with filled and half-filled subshells. For example, the 4s23d10 electron configuration of zinc results in its strong tendency to form the stable Zn2+ ion, with a 3d10 electron configuration, whereas Cu+, which also has a 3d10 electron configuration, is the only stable monocation formed by a first-row transition metal. Similarly, with a half-filled subshell, Mn2+ (3d5) is much more difficult to oxidize than Fe2+ (3d6). The chemistry of manganese is therefore primarily that of the Mn2+ ion, whereas both the Fe2+ and Fe3+ ions are important in the chemistry of iron.
The transition metals form cations by the initial loss of the ns electrons of the metal, even though the ns orbital is lower in energy than the (n − 1)d subshell in the neutral atoms. This apparent contradiction is due to the small difference in energy between the ns and (n − 1)d orbitals, together with screening effects. The loss of one or more electrons reverses the relative energies of the ns and (n − 1)d subshells, making the latter lower in energy. Consequently, all transition-metal cations possess dn valence electron configurations, as shown in Table \(2\) for the 2+ ions of the first-row transition metals.
Note the Pattern
All transition-metal cations have dn electron configurations; the ns electrons are always lost before the (n − 1)d electrons.
Table \(2\) d-Electron Configurations of the Dications of the First-Row Transition Metals
Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
d 2 d 3 d 4 d 5 d 6 d 7 d 8 d 9 d 10
The most common oxidation states of the first-row transition metals are shown in Table \(3\). The second- and third-row transition metals behave similarly but with three important differences:
1. The maximum oxidation states observed for the second- and third-row transition metals in groups 3–8 increase from +3 for Y and La to +8 for Ru and Os, corresponding to the formal loss of all ns and (n − 1)d valence electrons. As we go farther to the right, the maximum oxidation state decreases steadily, reaching +2 for the elements of group 12 (Zn, Cd, and Hg), which corresponds to a filled (n − 1)d subshell.
2. Within a group, higher oxidation states become more stable down the group. For example, the chromate ion ([CrO4]2−) is a powerful oxidant, whereas the tungstate ion ([WO4]2−) is extremely stable and has essentially no tendency to act as an oxidant.
3. Cations of the second- and third-row transition metals in lower oxidation states (+2 and +3) are much more easily oxidized than the corresponding ions of the first-row transition metals. For example, the most stable compounds of chromium are those of Cr(III), but the corresponding Mo(III) and W(III) compounds are highly reactive. In fact, they are often pyrophoric, bursting into flames on contact with atmospheric oxygen. As we shall see, the heavier elements in each group form stable compounds in higher oxidation states that have no analogues with the lightest member of the group.
Note the Pattern
The highest possible oxidation state, corresponding to the formal loss of all valence electrons, becomes increasingly less stable as we go from group 3 to group 8, and it is never observed in later groups.
Note the Pattern
In the transition metals, the stability of higher oxidation states increases down a column.
Table \(3\) Common Oxidation States of the First-Row Transition Metals*
Sc Ti V Cr Mn Fe Co Ni Cu Zn
electronic structure s 2 d 1 s 2 d 2 s 2 d 3 s 1 d 5 s 2 d 5 s 2 d 6 s 2 d 7 s 2 d 8 s 1 d 10 s 2 d 10
oxidation states I I
II II II II II II II II II
III III III III III III III III III
IV IV IV IV IV IV IV
V V V V V
VI VI VI
VII
*The convention of using roman numerals to indicate the oxidation states of a metal is used here.
Binary transition-metal compounds, such as the oxides and sulfides, are usually written with idealized stoichiometries, such as FeO or FeS, but these compounds are usually cation deficient and almost never contain a 1:1 cation:anion ratio. Thus a substance such as ferrous oxide is actually a nonstoichiometric compound with a range of compositions.
The acid–base character of transition-metal oxides depends strongly on the oxidation state of the metal and its ionic radius. Oxides of metals in lower oxidation states (less than or equal to +3) have significant ionic character and tend to be basic. Conversely, oxides of metals in higher oxidation states are more covalent and tend to be acidic, often dissolving in strong base to form oxoanions.
EXAMPLE 1
Two of the group 8 metals (Fe, Ru, and Os) form stable oxides in the +8 oxidation state. Identify these metals; predict the stoichiometry of the oxides; describe the general physical and chemical properties, type of bonding, and physical state of the oxides; and decide whether they are acidic or basic oxides.
Given: group 8 metals
Asked for: identity of metals and expected properties of oxides in +8 oxidation state
Strategy:
Refer to the trends outlined in Figure \(1\), Figure \(1\), Table \(2\), Table \(2\), and Table \(3\) to identify the metals. Decide whether their oxides are covalent or ionic in character, and, based on this, predict the general physical and chemical properties of the oxides.
Solution:
The +8 oxidation state corresponds to a stoichiometry of MO4. Because the heavier transition metals tend to be stable in higher oxidation states, we expect Ru and Os to form the most stable tetroxides. Because oxides of metals in high oxidation states are generally covalent compounds, RuO4 and OsO4 should be volatile solids or liquids that consist of discrete MO4 molecules, which the valence-shell electron-pair repulsion (VSEPR) model predicts to be tetrahedral. Finally, because oxides of transition metals in high oxidation states are usually acidic, RuO4 and OsO4 should dissolve in strong aqueous base to form oxoanions.
Exercise \(1\)
Predict the identity and stoichiometry of the stable group 9 bromide in which the metal has the lowest oxidation state and describe its chemical and physical properties.
Answer
Because the lightest element in the group is most likely to form stable compounds in lower oxidation states, the bromide will be CoBr2. We predict that CoBr2 will be an ionic solid with a relatively high melting point and that it will dissolve in water to give the Co2+(aq) ion.
Summary
The transition metals are characterized by partially filled d subshells in the free elements and cations. The ns and (n − 1)d subshells have similar energies, so small influences can produce electron configurations that do not conform to the general order in which the subshells are filled. In the second- and third-row transition metals, such irregularities can be difficult to predict, particularly for the third row, which has 4f, 5d, and 6s orbitals that are very close in energy. The increase in atomic radius is greater between the 3d and 4d metals than between the 4d and 5d metals because of the lanthanide contraction. Ionization energies and electronegativities increase slowly across a row, as do densities and electrical and thermal conductivities, whereas enthalpies of hydration decrease. Anomalies can be explained by the increased stabilization of half-filled and filled subshells. Transition-metal cations are formed by the initial loss of ns electrons, and many metals can form cations in several oxidation states. Higher oxidation states become progressively less stable across a row and more stable down a column. Oxides of small, highly charged metal ions tend to be acidic, whereas oxides of metals with a low charge-to-radius ratio are basic.
KEY TAKEAWAYS
• Transition metals are characterized by the existence of multiple oxidation states separated by a single electron.
• Most transition-metal compounds are paramagnetic, whereas virtually all compounds of the p-block elements are diamagnetic.
CONCEPTUAL PROBLEMS
1. The transition metals show significant horizontal similarities in chemistry in addition to their vertical similarities, whereas the same cannot be said of the s-block and p-block elements. Explain why this is so.
2. The energy of the d subshell does not change appreciably in a given period. Why? What effect does this have on the ionization potentials of the transition metals? on their electronegativities?
3. Standard reduction potentials vary across the first-row transition metals. What effect does this have on the chemical reactivity of the first-row transition metals? Which two elements in this period are more active than would be expected? Why?
4. Many transition metals are paramagnetic (have unpaired electrons). How does this affect electrical and thermal conductivities across the rows?
5. What is the lanthanide contraction? What effect does it have on the radii of the transition metals of a given group? What effect does it have on the chemistry of the elements in a group?
6. Why are the atomic volumes of the transition elements low compared with the elements of groups 1 and 2? Ir has the highest density of any element in the periodic table (22.65 g/cm3). Why?
7. Of the elements Ti, Ni, Cu, and Cd, which do you predict has the highest electrical conductivity? Why?
8. The chemistry of As is most similar to the chemistry of which transition metal? Where in the periodic table do you find elements with chemistry similar to that of Ge? Explain your answers.
9. The coinage metals (group 11) have significant noble character. In fact, they are less reactive than the elements of group 12. Explain why this is so, referring specifically to their reactivity with mineral acids, electronegativity, and ionization energies. Why are the group 12 elements more reactive?
STRUCTURE AND REACTIVITY
1. Give the valence electron configurations of the 2+ ion for each first-row transition element. Which two ions do you expect to have the most negative E° value? Why?
2. Arrange Ru3+, Cu2+, Zn, Ti4+, Cr3+, and Ni2+ in order of increasing radius.
3. Arrange Pt4+, Hg2+, Fe2+, Zr4+, and Fe3+ in order of decreasing radius.
4. Of Ti2+, V2+, Mn2+, Fe2+, Co2+, Ni2+, and Zn2+, which divalent ion has the smallest ionic radius? Explain your reasoning.
AnswerS
1. Ti2+, 3d2; V2+, 3d3; Cr2+, 3d4; Mn2+, 3d5; Fe2+, 3d6; Co2+, 3d7; Ni2+, 3d8; Cu2+, 3d9; Zn2+, 3d10. Because Zeff increases from left to right, Ti2+ and V2+ will have the most negative reduction potentials (be most difficult to reduce).
2. Hg2+ > Fe2+ > Zr4+ > Fe3+ > Pt4 | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.01%3A_General_Trends_among_the_Transition_Metals.txt |
Learning Objectives
• To use periodic trends to understand the chemistry of the transition metals.
As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member.
Group 3 (Sc, Y, La, and Ac)
As shown in Table $1$, the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns2(n − 1)d1 valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas:
$2M_{(s)} + 6H_2O_{(l)} \rightarrow 2M(OH)_{3(s)} + 3H_{2(g)}\label{Eq1}$
The chemistry of the group 3 metals is almost exclusively that of the M3+ ion; the elements are powerful reductants.
Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M3+(aq).
Table $1$: Some Properties of the Elements of Groups 3, 4, and 5
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
3 Sc 21 4s23d1 1.36 162 1541 2.99
Y 39 5s24d1 1.22 180 1522 4.47
La 57 6s25d1 1.10 183 918 6.15
Ac 89 7s26d1 1.10 188 1051 10.07
4 Ti 22 4s23d2 1.54 147 1668 4.51
Zr 40 5s24d2 1.33 160 1855 6.52
Hf 72 6s25d24f14 1.30 159 2233 13.31
5 V 23 4s23d3 1.63 134 1910 6.00
Nb 41 5s24d3 1.60 146 2477 8.57
Ta 73 6s25d34f14 1.50 146 3017 16.65
The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX3. The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M2O3), which react with H2O or CO2 to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters.
Group 4 (Ti, Zr, and Hf)
Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns2(n − 1)d2 valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti:
$2FeTiO_{3(s)} + 6C_{(s)} + 7Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + 2FeCl_{3(g)} + 6CO_{(g)}\label{Eq2}$
followed by reduction of the tetrachlorides with an active metal such as Mg.
The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states.
In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO2-containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines.
Consistent with the periodic trends shown in Figure 23.2, the group 4 metals become denser, higher melting, and more electropositive down the column (Table $1$). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns2(n − 1)d2 valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti2+ are usually powerful reductants. In fact, the Ti2+(aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas.
Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX4), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl2 is an ionic salt, whereas TiCl4 is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO2) and dichalcogenides (MY2). Industrially, TiO2, which is used as a white pigment in paints, is prepared by reacting TiCl4 with oxygen at high temperatures:
$TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} + 2Cl_{2(g)}\label{Eq3}$
The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure $1$). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH2), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB2), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity.
Group 5 (V, Nb, and Ta)
Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores.
Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V4C3, which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V2O5, an important catalyst for the industrial conversion of SO2 to SO3 in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb3Zr and Nb3Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones.
The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4.
As indicated in Table $1$, the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M2O5), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H2O)6]2+; the blue-green [V(H2O)6]3+ ion is acidic, dissociating to form small amounts of the [V(H2O)5(OH)]2+ ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H2O)4VO]2+, which contains a formal V=O bond (Figure $2$). Consistent with its covalent character, V2O5 is acidic, dissolving in base to give the vanadate ion ([VO4]3−), whereas both Nb2O5 and Ta2O5 are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric.
Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO2]+ ion; V(IV) is the blue vanadyl ion [VO]2+; and V(III) and V(II) exist as the hydrated V3+ (blue-green) and V2+ (violet) ions, respectively.
Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY2), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools.
Group 6 (Cr, Mo, and W)
As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS2) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr2O4), which is oxidized to the soluble [CrO4]2− ion under basic conditions and reduced successively to Cr2O3 and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr2O3 in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO3, which is then reduced with hydrogen to give the metal.
The metals become increasing polarizable across the d block.
Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (Table $2$). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr2+, which is a powerful reductant, to CrO3, a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known.
Table $2$: Some Properties of the Elements of Groups 6 and 7
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
6 Cr 24 4s13d5 1.66 128 1907 7.15
Mo 42 5s14d5 2.16 139 2623 10.20
W 74 6s25d44f14 1.70 139 3422 19.30
7 Mn 25 4s23d5 1.55 127 1246 7.30
Tc 43 5s24d5 2.10 136 2157 11.50
Re 75 6s25d54f14 1.90 137 3186 20.80
The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states.
As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF6 is unstable at temperatures above −100°C, whereas MoF6 and WF6 are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX3 (X is Cl, Br, or I), while molybdenum forms MoCl5, MoBr4, and MoI3, and tungsten gives WCl6, WBr5, and WI4.
Both Mo and W react with oxygen to form the covalent trioxides (MoO3 and WO3), but Cr reacts to form only the so-called sesquioxide (Cr2O3). Chromium will form CrO3, which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO4]2−). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr2O3) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo8O26]4− ion, whose structure is as follows:
An isopolymolybdate cluster. The [Mo8O26]4− ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution.
Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS2; it has a layered structure similar to that of TiS2 (Figure $1$), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS2 and WS2 are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS2 particles.
As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits.
Group 7 (Mn, Tc, and Re)
Continuing across the periodic table, we encounter the group 7 elements (Table $2$). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, 99mTc (m for metastable), has become an important biomedical tool for imaging internal organs. Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline.
All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (Figure $3$). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d5 electron configuration, the aqueous Mn3+ ion, with a 3d4 valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound.
Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn2+(aq) ion is pale pink; Mn(OH)3, which contains Mn(III), is a dark brown solid; MnO2, which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO4]2− and the purple permanganate ion [MnO4], respectively.
Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F2 gives only MnF3, a high-melting, red-purple solid, whereas Re reacts with F2 to give ReF7, a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl2, a weaker oxidant than F2, gives MnCl2 and ReCl6. Reaction of Mn with oxygen forms only Mn3O4, a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe3O4). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M2O7), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn2O7, an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO4]2− is a potent oxidant, whereas [TcO4] and [ReO4] are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS2, as well as more complex heptasulfides (M2S7). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties.
The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced.
Groups 8, 9, and 10
In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert.
The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3.
Group 8 (Fe, Ru, and Os)
The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age.
Group 9 (Co, Rh, and Ir)
Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores.
Group 10 (Ni, Pd, and Pt)
Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold.
Over 2000 years ago, the Bactrian civilization in Western Asia used a 75:25 alloy of copper and nickel for its coins. A modern US nickel has the same composition, but a modern Canadian nickel is nickel-plated steel and contains only 2.5% nickel by mass.
Some properties of the elements in groups 8–10 are summarized in Table $3$. As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state.
Table $3$: Some Properties of the Elements of Groups 8, 9, and 10
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
8 Fe 26 4s23d6 1.83 126 1538 7.87
Ru 44 5s14d7 2.20 134 2334 12.10
Os 76 6s25d64f14 2.20 135 3033 22.59
9 Co 27 4s23d7 1.88 125 1495 8.86
Rh 45 5s14d8 2.28 134 1964 12.40
Ir 77 6s25d74f14 2.20 136 2446 22.50
10 Ni 28 4s23d8 1.91 124 1455 8.90
Pd 46 4d10 2.20 137 1555 12.00
Pt 78 6s25d84f14 2.20 139 1768 21.50
We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF2. In contrast to Fe, Ru and Os form a series of fluorides up to RuF6 and OsF7. The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe2O3, and the mixed-valent Fe3O4 (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO2) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block.
Higher oxidation states become less stable across the d-block, but more stable down a group.
Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S22−), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S2−). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character.
The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe3C), which is used to strengthen steel. At high temperatures, Fe3C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH0.5. Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H2 but essentially impermeable to all other gases, including He. Consequently, diffusion of H2 through Pd is an effective method for separating hydrogen from other gases.
Group 11 (Cu, Ag, and Au)
The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times.
This 1 kg gold nugget was found in Australia; in 2005, it was for sale in Hong Kong at an asking price of more than US\$64,000.
Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe2). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer.
The Chuquicamata copper mine in northern Chile, the world’s largest open-pit copper mine, is 4.3 km long, 3 km wide, and 825 m deep. Each gigantic truck in the foreground (and barely visible in the lower right center) can hold 330 metric tn (330,000 kg) of copper ore.
Some properties of the coinage metals are listed in Table $4$. The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns1(n − 1)d10 valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s25d10 valence electron configuration.
Table $4$: Some Properties of the Elements of Groups 11 and 12
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
11 Cu 29 4s13d10 1.90 128 1085 8.96
Ag 47 5s14d10 1.93 144 962 10.50
Au 79 6s15d104f14 2.40 144 1064 19.30
12 Zn 30 4s23d10 1.65 134 420 7.13
Cd 48 5s24d10 1.69 149 321 8.69
Hg 80 6s25d104f14 1.90 151 −38.8 13.53
All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O2 at high temperatures to produce Cu2O and with sulfur to form Cu2S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag2S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO3 and in hot concentrated H2SO4, while Au dissolves in the 3:1 HCl:HNO3 mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN)2] ions, a reaction that is used to separate gold from its ores.
Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au.
All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au3+ and Cu2+). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF2. In contrast, all the gold trihalides (AuX3) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements.
Group 12 (Zn, Cd, and Hg)
We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries.
As shown in Table $4$, the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns2(n − 1)d10 valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg22+.
The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble.
All the possible group 12 dihalides (MX2) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl2). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d10 subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO3 and H2SO4. All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur.
Example $1$
For each reaction, explain why the indicated products form.
1. TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(aq)
2. WO3(s) + 3C(s) + 3Cl2(g) $\xrightarrow{\Delta}$ WCl6(s) + 3CO(g)
3. Sc2O3(s) + 2OH(aq) + 3H2O(l) → 2[Sc(OH)4](aq)
4. 2KMnO4(aq) + 2H2SO4(l) → Mn2O7(l) + 2KHSO4(soln) + H2O(soln)
5. 4CrCl2(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) + 8Cl(aq) + 2H2O(l)
Given: balanced chemical equation
Asked for: why the indicated products form
Strategy:
Refer to the periodic trends in this section.
Solution:
1. The most stable oxidation state for Ti is +4, and neither reactant is a particularly strong oxidant or reductant; hence a redox reaction is unlikely. Similarly, neither reactant is a particularly strong acid or base, so an acid–base reaction is unlikely. Because TiCl4 contains Ti in a relatively high oxidation state (+4), however, it is likely to be rather covalent in character, with reactivity similar to that of a semimetal halide such as SiCl4. Covalent halides tend to hydrolyze in water to produce the hydrohalic acid and either the oxide of the other element or a species analogous to an oxoacid.
2. This reaction involves the oxide of a group 6 metal in its highest oxidation state (WO3) and two elements, one of which is a reductant (C) and the other an oxidant (Cl2). Consequently, some sort of redox reaction will occur. Carbon can certainly react with chlorine to form CCl4, and WO3 is a potential source of oxygen atoms that can react with carbon to produce CO, which is very stable. If CO is one of the products, then it seems likely that the other product will contain the metal and chlorine. The most likely combination is WCl6 (leaving the oxidation state of the metal unchanged).
3. One of the reactants is a strong base (OH), so an acid–base reaction is likely if the other reactant is acidic. Because oxides like Sc2O3, in which the metal is in an intermediate oxidation state, are often amphoteric, we expect Sc2O3 to dissolve in base to form a soluble hydroxide complex.
4. Concentrated sulfuric acid is both an oxidant and a strong acid that tends to protonate and dehydrate other substances. The permanganate ion already contains manganese in its highest possible oxidation state (+7), so it cannot be oxidized further. A redox reaction is impossible, which leaves an acid–base reaction as the most likely alternative. Sulfuric acid is likely to protonate the terminal oxygen atoms of permanganate, allowing them to be lost as water.
5. Molecular oxygen is an oxidant, so a redox reaction is likely if the other reactant can be oxidized. Because chromous chloride contains chromium in its lowest accessible oxidation state, a redox reaction will occur in which Cr2+ ions are oxidized and O2 is reduced. In the presence of protons, the reduction product of O2 is water, so we need to determine only the identity of the oxidation product of Cr2+. Chromium forms compounds in two common higher oxidation states: the Cr3+ ion, which is the most stable, and the [Cr2O7]2− ion, which is a more powerful oxidant than O2. We therefore predict that the reaction will form Cr3+(aq) and water.
Exercise $1$
Predict the products of each reactions and then balance each chemical equation.
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$
2. Na2Cr2O7(aq) + H2SO4(l) →
3. FeBr2(aq) + O2(g) $\xrightarrow{\mathrm{H^+}}$
4. VBr4(l) + H2O(l) →
5. ZrO2(s) + C(s) + Cl2(g) $\xrightarrow{\Delta}$
Answer
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$ Cr3+(aq) + Fe2+(aq)
2. Na2Cr2O7(aq) + 2H2SO4(l) → 2NaHSO4(soln) + H2O(soln) + 2CrO3(s)
3. 4FeBr2(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l) + 8Br(aq)
4. VBr4(l) + H2O(l) → VO2+(aq) + 4Br(aq) + 2H+(aq)
5. ZrO2(s) + 2C(s) + 2Cl2(g) $\xrightarrow{\Delta}$ ZrCl4(s) + 2CO(g)
Summary
The elements tend to become more polarizable going across the d block and higher oxidation states become less stable; higher oxidation states become more stable going down a group. The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M2O3). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr2O3 is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands.
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.02%3A_A_Brief_Survey_of_Transition-Metal_Chemistry.txt |
Learning Objectives
• To understand how metals are extracted from their ores.
Very few of the transition metals are found in nature as free metals (e.g., gold and silver). Consequently, almost all metallic elements must be isolated from metal oxide or metal sulfide ores. Metallurgy is the set of processes by which metals are extracted from their ores and converted to more useful forms. Metallurgy consists of three general steps:
1. mining the ore,
2. separating and concentrating the metal or the metal-containing compound, and
3. reducing the ore to the metal.
Additional processes are sometimes required to improve the mechanical properties of the metal or increase its purity. Many ores contain relatively low concentrations of the desired metal; for example, copper ores that contain even 1% Cu by mass are considered commercially useful.
After an ore has been mined, the first step in processing is usually to crush it because the rate of chemical reactions increases dramatically with increased surface area. Next, one of three general strategies is used to separate and concentrate the compound(s) of interest: settling and flotation, which are based on differences in density between the desired compound and impurities; pyrometallurgy, which uses chemical reduction at high temperatures; and hydrometallurgy, which employs chemical or electrochemical reduction of an aqueous solution of the metal. Other methods that take advantage of unusual physical or chemical properties of a particular compound may also be used. For example, crystals of magnetite (Fe3O4) are tiny but rather powerful magnets; in fact, magnetite (also known as lodestone) was used to make the first compasses in China during the first century BC. If a crushed ore that contains magnetite is passed through a powerful magnet, the Fe3O4 particles are attracted to the poles of the magnet, allowing them to be easily separated from other minerals.
Metallurgy depends on the separation of a metal compound from its ore and reduction to the metal at high temperature (pyrometallurgy) or in aqueous solution (hydrometallurgy).
Settling and flotation have been used for thousands of years to separate particles of dense metals such as gold, using the technique known as panning, in which a sample of gravel or sand is swirled in water in a shallow metal pan. Because the density of gold (19.3 g/cm3) is so much greater than that of most silicate minerals (about 2.5 g/cm3), silicate particles settle more slowly and can be poured off with the water, leaving dense gold particles on the bottom of the pan. Conversely, in flotation, the compound of interest is made to float on top of a solution. Blowing air through a suspension of the crude ore in a mixture of water and an organic liquid, such as pine tar, produces a “froth” that contains tiny particles of hydrophobic solids, such as metal sulfides, while more hydrophilic oxide minerals remain suspended in the aqueous phase (Figure 23.3.1). To make the separation more efficient, small amounts of an anionic sulfur-containing compound, such as Na+C2H5OCS2, are added; the additive binds to the sulfur-rich surface of the metal sulfide particles and makes the metal sulfide particles even more hydrophobic. The resulting froth is highly enriched in the desired metal sulfide(s), which can be removed simply by skimming. This method works even for compounds as dense as PbS (7.5 g/cm3).
Pyrometallurgy
In pyrometallurgy, an ore is heated with a reductant to obtain the metal. Theoretically, it should be possible to obtain virtually any metal from its ore by using coke, an inexpensive form of crude carbon, as the reductant. An example of such a reaction is as follows:
$\mathrm{CaO(s) + C(s)\xrightarrow{\Delta}Ca(l) + CO(g)} \tag{23.3.1}$
Unfortunately, many of the early transition metals, such as Ti, react with carbon to form stable binary carbides. Consequently, more expensive reductants, such as hydrogen, aluminum, magnesium, or calcium, must be used to obtain these metals. Many metals that occur naturally as sulfides can be obtained by heating the sulfide in air, as shown for lead in the following equation:
$\mathrm{PbS(s) + O_2(g)\xrightarrow{\Delta}Pb(l) + SO_2(g)}\tag{23.3.2}$
The reaction is driven to completion by the formation of SO2, a stable gas.
Pyrometallurgy is also used in the iron and steel industries. The overall reaction for the production of iron in a blast furnace is as follows:
$\mathrm{Fe_2O_3(s) +3C(s)\xrightarrow{\Delta}2Fe(l) +3CO(g)} \tag{23.3.3}$
The actual reductant is CO, which reduces Fe2O3 to give Fe(l) and CO2(g); the CO2 is then reduced back to CO by reaction with excess carbon. As the ore, lime, and coke drop into the furnace (Figure 23.3.2), any silicate minerals in the ore react with the lime to produce a low-melting mixture of calcium silicates called slag, which floats on top of the molten iron. Molten iron is then allowed to run out the bottom of the furnace, leaving the slag behind. Originally, the iron was collected in pools called pigs, which is the origin of the name pig iron.
Iron that is obtained directly from a blast furnace has an undesirably low melting point (about 1100°C instead of 1539°C) because it contains a large amount of dissolved carbon. It contains other impurities (such as Si, S, P, and Mn from contaminants in the iron ore that were also reduced during processing) that must be removed because they make iron brittle and unsuitable for most structural applications. In the Bessemer process, oxygen is blown through the molten pig iron to remove the impurities by selective oxidation because these impurities are more readily oxidized than iron (Figure 23.3.3). In the final stage of this process, small amounts of other metals are added at specific temperatures to produce steel with the desired combination of properties.
Hydrometallurgy
The most selective methods for separating metals from their ores are based on the formation of metal complexes. For example, gold is often found as tiny flakes of the metal, usually in association with quartz or pyrite deposits. In those circumstances, gold is typically extracted by using cyanide leaching, which forms a stable gold–cyanide complex—[Au(CN)2]:
$4Au_{(s)} + 8NaCN_{(aq)} + O_{2(g)} + 2H_2O_{(l)} \rightarrow 4Na[Au(CN)_2]_{(aq)} + 4NaOH_{(aq)} \tag{23.3.4}$
Virtually pure gold can be obtained by adding powdered zinc to the solution:
$Zn_{(s)} + 2[Au(CN)_2]^−_{(aq)} \rightarrow [Zn(CN)_4]^{2−}_{(aq)} + 2Au_{(s)} \tag{23.3.5}$
A related method, which is used to separate Co3+, Ni2+, and Cu+ from Fe, Mn, and Ti, is based on the formation of stable, soluble ammonia complexes of ions of the late transition metals.
Example 23.3.1
Suppose you are working in the chemistry laboratory of a mining company that has discovered a new source of tungsten ore containing about 5% WS2 in a granite matrix (granite is a complex aluminosilicate mineral). You have been asked to outline an economical procedure for isolating WS2 from the ore and then converting it to elemental tungsten in as few steps as possible. What would you recommend?
Given: composition of ore
Asked for: procedure to isolate metal sulfide
Strategy:
Determine which method would be most effective for separating the metal sulfide from the ore. Then determine the best method for reducing the metal to the pure element.
Solution
You need to separate and concentrate the WS2, convert it to a suitable form so it can be reduced to the metal (if necessary), and then carry out the reduction. Because the new ore is a binary metal sulfide, you could take advantage of the hydrophilic nature of most metal sulfides to separate WS2 by froth flotation. Then, because most metal sulfides cannot be reduced directly to the metal using carbon, you will probably need to convert WS2 to an oxide for subsequent reduction. One point to consider is whether the oxide can be reduced using carbon because many transition metals react with carbon to form stable carbides. Here is one possible procedure for producing tungsten from this new ore:
1. Grind the ore and separate WS2 from the silicate matrix by flotation.
2. Convert the crude WS2 to an oxide by roasting in air (because W is in group 6, you anticipate that roasting will yield WO3, the oxide in the highest possible oxidation state). The reaction will also produce SO2, which will have to be removed by scrubbing the exhaust gases to minimize environmental pollution.
3. Reduce the oxide with hydrogen gas at high temperature to avoid carbide formation:
$\mathrm{WO_3(s) +3H_2(g)\xrightarrow{\Delta}W(s) +3H_2O(g)}$
Exercise 23.3.1
Propose an economical procedure for converting a silicate mineral deposit containing BaCO3 to the pure Ba metal.
Answer:
1. Dissolve the sample containing barium carbonate in HCl(aq) to give Ba2+(aq), which will allow the insoluble silicate minerals to be removed by filtration.
2. Precipitate the barium from solution as BaCO3 by adding solid Na2CO3.
3. Dissolve the solid BaCO3 in concentrated HCl and remove the water by evaporation to obtain anhydrous BaCl2.
4. Reduce molten BaCl2 to the metal by electrolysis.
Summary
• A metal is separated from its ore and then isolated by using pyrometallurgy or hydrometallurgy or by taking advantage of unusual chemical or physical properties of a particular compound.
The conversion of metals from their ores to more useful forms is called metallurgy, which consists of three general steps: mining, separation and concentration, and reduction. Settling and flotation are separation methods based on differences in density, whereas pyrometallurgy is based on a chemical reduction at elevated temperatures, and hydrometallurgy uses chemical or electrochemical reduction of an aqueous solution. In pyrometallurgy, a reductant must be used that does not form stable compounds with the metal of interest. In hydrometallurgy, metals are separated via the formation of metal complexes.
Conceptual Problems
1. Coke is a plentiful and inexpensive reductant that is used to isolate metals from their ores. Of Cr, Co, W, Cu, Ni, Os, Fe, Mn, La, and Hf, which cannot be isolated using this reductant? Why?
2. Hydrometallurgy is the preferred method for separating late transition metals from their ores. What types of ligands are most effective in this process?
Answer
1. Coke cannot be used as a reductant for metals that form stable carbides, such as the early transition metals (La, Hf, and W).
Structure and Reactivity
1. Tantalum and niobium are frequently found together in ores. These elements can be separated from other metals present by treatment with a solution of HF. Explain why this is an effective separation technique.
2. A commercially important ore of chromium is chromite (FeCr2O4), which is an analogue of magnetite (Fe3O4). Based on what you know about the oxidation states of iron in magnetite, predict the oxidation states of the metal ions in chromite.
3. Pure vanadium is obtained by reducing VCl4 with H2 or Mg or by reducing V2O5 with Ca. Write a balanced chemical equation for each reaction. Why is carbon not used for the reduction?
4. Manganese is an important additive in steel because of its reactivity with oxygen and sulfur, both of which contribute to brittleness. Predict the products of reacting Mn with these species.
5. The diagram of a blast furnace in Figure 23.3.2 illustrates several important features of the reduction of Fe2O3 to iron. Write a balanced chemical equation for each step of the process described in the figure and give the overall equation for the conversion. Oxygen is blown through the final product to remove impurities. Why does this step not simply reverse the process and produce iron oxides?
6. Metallic Zr is produced by the Kroll method, which uses Na as the reductant. Write a balanced chemical equation for each reaction involved in this process. The product is frequently contaminated with Hf. Propose a feasible method for separating the two elements.
7. The compound Cr2O3 is important commercially; among other things, it is used as a pigment in paint and as a catalyst for the manufacture of butadiene. Write a balanced chemical equation to show how you would produce this compound from
1. chromium metal.
2. ammonium dichromate.
3. CrCl3 in a basic solution.
Answer
\begin{align} &\mathrm{VCl_4(l) +2H_2(g)} \xrightarrow{\Delta} \mathrm{V(s) +4HCl(g)} \ &\mathrm{VCl_4(l) +2Mg(s)} \xrightarrow{\Delta} \mathrm{V(s) +2MgCl_2(g)} \ &\mathrm {V_2O_5(s) +5Ca(s)} \xrightarrow{\Delta} \mathrm{2V(s) +5CaO(s)} \end{align}
Carbon cannot be used as a reductant because vanadium forms stable carbides, such as VC and VC2. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.03%3A_Metallurgy.txt |
Learning Objectives
• To know the most common structures observed for metal complexes.
• To predict the relative stabilities of metal complexes with different ligands.
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more ligands (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions. Metal complexes are so important in biology that we consider the topic separately in Section 23.6.
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: KFe2(CN)6. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as AlF3·3KF, Fe(CN)2·4KCN, and ZnCl2·2CsCl, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should AlF3·3KF exist but not AlF3·4KF or AlF3·2KF? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, CoCl3·6NH3, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3, and CoCl3·3NH3 were all known and had very different properties, but despite all attempts, chemists could not prepare CoCl3·2NH3 or CoCl3·NH3.
The Great Wave Off Kanagawa. The Japanese artist Katsushika Hokusai used Prussian blue to create this famous woodcut.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in Table $1$. The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in Table $1$ showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Alfred Werner (1866–1919)
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table $1$ Werner’s Data on Complexes of Ammonia with PtCl4
Complex Conductivity (ohm−1) Number of Ions per Formula Unit Number of Cl Ions Precipitated by Ag+
PtCl4·6NH3 523 5 4
PtCl4·5NH3 404 4 3
PtCl4·4NH3 299 3 2
PtCl4·3NH3 97 2 1
PtCl4·2NH3 0 0 0
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligands bound to the metal ion. If Pt had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the PtCl4·NH3 adducts by the following reactions, where the metal complex is enclosed in square brackets:
\begin{aligned}
\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{4} & \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{6}\right]^{4+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \
\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{3} & \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{3+}(\mathrm{aq})+3 \mathrm{Cl}^{-}(\mathrm{aq}) \
\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} & \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \
\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right] \mathrm{Cl} & \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \
\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right] & \rightarrow\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]^{0}(\mathrm{aq})
\end{aligned}
Further work showed that the two missing members of the series—[Pt(NH3)Cl5] and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (part (a) in Figure $1$. Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (part (b) in Figure $1$. His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Figure $1$ Complexes with Different Arrangements of the Same Ligands Have Different Properties
The [Co(NH3)4Cl2]+ ion can have two different arrangements of the ligands, which results in different colors: if the two Cl ligands are next to each other, the complex is red (a), but if they are opposite each other, the complex is green (b).
Example $1$
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Solution
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise $1$
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer
square planar, 2; tetrahedral, 1
Structures of Metal Complexes
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model in Chapter 9 because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Note the Pattern
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Coordination Number 2
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2] ion, which is used to extract gold from its ores, as described in Section 23.3. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Coordination Number 3
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3 ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Coordination Number 4
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4] and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
Coordination Number 5
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
Coordination Number 6
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
Coordination Number 7
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
Coordination Number 8
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
Coordination Number 9
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
Stability of Metal Complexes
The thermodynamic stability of a metal complex depends greatly on the properties of the ligand and the metal ion and on the type of bonding. Recall that the metal–ligand interaction is an example of a Lewis acid–base interaction. Lewis bases can be divided into two categories: hard bases, which contain small, relatively nonpolarizable donor atoms (such as N, O, and F), and soft bases, which contain larger, relatively polarizable donor atoms (such as P, S, and Cl). Metal ions with the highest affinities for hard bases are hard acids, whereas metal ions with the highest affinity for soft bases are soft acids. Some examples of hard and soft acids and bases are given in Table $2$. Notice that hard acids are usually cations of electropositive metals; consequently, they are relatively nonpolarizable and have higher charge-to-radius ratios. Conversely, soft acids tend to be cations of less electropositive metals; consequently, they have lower charge-to-radius ratios and are more polarizable. Chemists can predict the relative stabilities of complexes formed by the d-block metals with a remarkable degree of accuracy by using a simple rule: hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
Table $2$ Examples of Hard and Soft Acids and Bases
Acids Bases
hard H+ NH3, RNH2, N2H4
Li+, Na+, K+ H2O, ROH, R2O
Be2+, Mg2+, Ca2+, VO2+ OH, F, Cl, CH3CO2
Al3+, Sc3+, Cr3+ CO32−
Ti4+ PO43−
soft BF3, Al2Cl6, CO2, SO3
Cu+, Ag+, Au+, Tl+, Hg22+ H
Pd2+, Pt2+, Hg2+ CN, SCN, I, RS
GaCl3, GaBr3, GaI3 CO, R2S
Because the interaction between hard acids and hard bases is primarily electrostatic in nature, the stability of complexes involving hard acids and hard bases increases as the positive charge on the metal ion increases and as its radius decreases. For example, the complex of Al3+ (r = 53.5 pm) with four fluoride ligands (AlF4) is about 108 times more stable than InF4, the corresponding fluoride complex of In3+ (r = 80 pm). In general, the stability of complexes of divalent first-row transition metals with a given ligand varies inversely with the radius of the metal ion, as shown in the following series:The inversion in the order at copper is due to the anomalous structure of copper(II) complexes, which will be discussed shortly.
\begin{aligned}
&\text { complex stability } \mathrm{Mn}^{2+}<\mathrm{Fe}^{2+}<\mathrm{Co}^{2+}<\mathrm{Ni}^{2+}<\mathrm{Cu}^{2+}>\mathrm{Zn}^{2+}\
&78 \quad 74.5\
&69\
&73 \quad 74\
&\text { ionic radius }(\mathrm{pm}) \quad 83 \quad 7\
&6\
&\text { - }
\end{aligned}
Because a hard metal interacts with a base in much the same way as a proton, by binding to a lone pair of electrons on the base, the stability of complexes of hard acids with hard bases increases as the ligand becomes more basic. For example, because ammonia is a stronger base than water, metal ions bind preferentially to ammonia. Consequently, adding ammonia to aqueous solutions of many of the first-row transition-metal cations results in the formation of the corresponding ammonia complexes.
In contrast, the interaction between soft metals (such as the second- and third-row transition metals and Cu+) and soft bases is largely covalent in nature. Most soft-metal ions have a filled or nearly filled d subshell, which suggests that metal-to-ligand π bonding is important. Complexes of soft metals with soft bases are therefore much more stable than would be predicted based on electrostatic arguments.
Note the Pattern
Hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
The hard acid–hard base/soft acid–soft base concept also allows us to understand why metals are found in nature in different kinds of ores. Recall from Section 23.2 that most of the first-row transition metals are isolated from oxide ores but that copper and zinc tend to occur naturally in sulfide ores. This is consistent with the increase in the soft character of the metals across the first row of the transition metals from left to right. Recall also that most of the second- and third-row transition metals occur in nature as sulfide ores, consistent with their greater soft character.
Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, three, or several sites, respectively. Ethylenediamine (H2NCH2CH2NH2, often abbreviated as en) and diethylenetriamine (H2NCH2CH2NHCH2CH2NH2, often abbreviated as dien) are examples of a bidentate and a tridentate ligand, respectively, because each nitrogen atom has a lone pair that can be shared with a metal ion. When a bidentate ligand such as ethylenediamine binds to a metal such as Ni2+, a five-membered ring is formed. A metal-containing ring like that shown is called a chelate ring (from the Greek chele, meaning “claw”). Correspondingly, a polydentate ligand is a chelating agent, and complexes that contain polydentate ligands are called chelate complexes.
Experimentally, it is observed that metal complexes of polydentate ligands are significantly more stable than the corresponding complexes of chemically similar monodentate ligands; this increase in stability is called the chelate effect. For example, the complex of Ni2+ with three ethylenediamine ligands, [Ni(en)3]2+, should be chemically similar to the Ni2+ complex with six ammonia ligands, [Ni(NH3)6]2+. In fact, the equilibrium constant for the formation of [Ni(en)3]2+ is almost 10 orders of magnitude larger than the equilibrium constant for the formation of [Ni(NH3)6]2+:
\begin{aligned}
&{\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+6 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K_{\mathrm{f}}=4 \times 10^{8}} \
&{\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+3 \mathrm{en} \quad \rightleftharpoons\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K_{\mathrm{f}}=2 \times 10^{18}}
\end{aligned}
Note the Pattern
Chelate complexes are more stable than the analogous complexes with monodentate ligands.
The stability of a chelate complex depends on the size of the chelate rings. For ligands with a flexible organic backbone like ethylenediamine, complexes that contain five-membered chelate rings, which have almost no strain, are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings. For example, the complex of copper(II) with two ethylenediamine ligands is about 1000 times more stable than the corresponding complex with triethylenediamine (H2NCH2CH2CH2NH2, abbreviated as trien):
$\begin{array}{ll}{\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+2 \mathrm{en}} & \rightleftharpoons\left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & K_{\mathrm{f}}=10^{20} \ {\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+3 \text { trien }} & \rightleftharpoons\left[\mathrm{Cu}(\text { trien })_{2}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & K_{\mathrm{f}}=10^{17}\end{array}$
Example $2$
Arrange [Cr(en)3]3+, [CrCl6]3−, [CrF6]3−, and [Cr(NH3)6]3+ in order of increasing stability.
Given: four Cr(III) complexes
Asked for: relative stabilities
Solution
A Determine the relative basicity of the ligands to identify the most stable complexes.
B Decide whether any complexes are further stabilized by a chelate effect and arrange the complexes in order of increasing stability.
Solution:
A The metal ion is the same in each case: Cr3+. Consequently, we must focus on the properties of the ligands to determine the stabilities of the complexes. Because the stability of a metal complex increases as the basicity of the ligands increases, we need to determine the relative basicity of the four ligands. Our earlier discussion of acid–base properties suggests that ammonia and ethylenediamine, with nitrogen donor atoms, are the most basic ligands. The fluoride ion is a stronger base (it has a higher charge-to-radius ratio) than chloride, so the order of stability expected due to ligand basicity is [CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ ≈ [Cr(en)3]3+.
B Because of the chelate effect, we expect ethylenediamine to form a stronger complex with Cr3+ than ammonia. Consequently, the likely order of increasing stability is [CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ < [Cr(en)3]3+.
Exercise $2$
Arrange [Co(NH3)6]3+, [CoF6]3−, and [Co(en)3]3+ in order of decreasing stability.
Answer
[Co(en)3]3+ > [Co(NH3)6]3+ > [CoF6]3−
Isomers of Metal Complexes
As we discussed earlier in this section, the existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall from Chapter 2 that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. (For more information on isomers in organic compounds, see Chapter 24, Section 24.2.)
Structural Isomers
Isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another are called structural isomers. Isobutane and n-butane are examples of structural isomers. One kind of isomerism consists of two compounds that have the same empirical formula but differ in the number of formula units present in the molecular formula. An example in coordination compounds is two compounds with the empirical formula Pt(NH3)2Cl2. One is a simple square planar platinum(II) complex, Pt(NH3)2Cl2, and the other is an ionic compound that contains the [Pt(NH3)4]2+ cation and the [PtCl4]2− anion, [Pt(NH3)4][PtCl4]. As you might expect, these compounds have very different physical and chemical properties. One arrangement of the Cl and NH3 ligands around the platinum ion in the former gives the anticancer drug cisplatin, whereas the other arrangement has no known biomedical applications.
Geometrical Isomers
Metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal are called geometrical isomers. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
Because there is no way to convert the cis structure to the trans by rotating or flipping the molecule in space, they are fundamentally different arrangements of atoms in space. Probably the best-known examples of cis and trans isomers of an MA2B2 square planar complex are cis-Pt(NH3)2Cl2, also known as cisplatin, and trans-Pt(NH3)2Cl2, which is actually toxic rather than therapeutic.
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs (the testes in males and the ovaries in females), which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin and went on to win an unprecedented seven Tour de France bicycle races.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional):
Example $1$
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution:
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise $1$
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. The stability of metal complexes with first-row transition metals in a +2 oxidation state varies inversely with their ionic radius. Lewis bases can be hard bases, which have small, relatively nonpolarizable donor atoms, or soft bases, with larger, relatively polarizable donor atoms. Hard acids have the highest affinity for hard bases, and soft acids have the highest affinity for soft bases. Soft metals and soft bases form complexes that are more stable than would be predicted based on electrostatic arguments, which suggests that metal-to-ligand π bonding is important. Ligands that are strong bases form the most stable complexes with metal ions that are hard acids. Exceptionally stable complexes are formed by chelates, which are polyatomic ligands with two or more donor atoms; this enhanced stability is known as the chelate effect. Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans.
KEY TAKEAWAYS
• Coordination compounds are a major feature of the chemistry of over half the elements.
• Coordination compounds have important roles as industrial catalysts in controlling reactivity, and they are essential in biochemical processes.
CONCEPTUAL PROBLEMS
1. Give two reasons a metal can bind to only a finite number of ligands. Based on this reasoning, what do you predict is the maximum coordination number of Ti? of Ac?
2. Can a tetrahedral MA2B2 complex form cis and trans isomers? Explain your answer.
3. The group 12 elements are never found in their native (free) form but always in combination with one other element. What element is this? Why? Which of the group 12 elements has the highest affinity for the element you selected?
Answer
1. The group 12 metals are rather soft and prefer to bind to a soft anion such as sulfide rather than to a hard anion like oxide; hence they are usually found in nature as sulfide ores. Because it is the softest of these metals, mercury has the highest affinity for sulfide.
STRUCTURE AND REACTIVITY
1. Complexes of metals in the +6 oxidation state usually contain bonds to which two Lewis bases? Why are these bonds best described as covalent rather than ionic? Do Ca, Sr, and Ba also form covalent bonds with these two Lewis bases, or is their bonding best described as ionic?
2. Cr, Mn, Fe, Co, and Ni form stable CO complexes. In contrast, the earlier transition metals do not form similar stable complexes. Why?
3. The transition metals Cr through Ni form very stable cyanide complexes. Why are these complexes so much more stable than similar compounds formed from the early transition metals?
4. Of Co(en)33+, CoF63−, Co(NH3)63+, and Co(dien)23+, which species do you expect to be the most stable? Why?
5. Of Ca2+, Ti2+, V2+, Mn2+, Fe2+, Co2+, Ni2+, and Zn2+, which divalent metal ions forms the most stable complexes with ligands such as NH3? Why?
6. Match each Lewis base with the metal ions with which it is most likely to form a stable complex:
Lewis bases: NH3, F, RS, OH, and Cl
Metals: Sc3+, Cu+, W6+, Mg2+, V3+, Fe3+, Zr4+, Co2+, Ti4+, Au+, Al3+, and Mn7+
7. Of ReF2, ReCl5, MnF6, Mn2O7, and ReO, which are not likely to exist?
8. Of WF2, CrF6, MoBr6, WI6, CrO3, MoS2, W2S3, and MoH, which are not likely to exist?
AnswerS
1. Metals in the +6 oxidation state are stabilized by oxide (O2−) and fluoride (F). The M−F and M−O bonds are polar covalent due to extreme polarization of the anions by the highly charged metal. Ca, Sr, and Ba can be oxidized only to the dications (M2+), which form ionic oxides and fluorides.
2. Cyanide is a relatively soft base, and the early transition-metal cations are harder acids than the later transition metals.
3. The formation of complexes between NH3 and a divalent cation is largely due to electrostatic interactions between the negative end of the ammonia dipole moment and the positively charged cation. Thus the smallest divalent cations (Ni2+, Zn2+, and Cu2+) will form the most stable complexes with ammonia.
4. Re2+ is a very soft cation, and F and O2− are very hard bases, so ReO and ReF2 are unlikely to exist. MnF6 is also unlikely to exist: although fluoride should stabilize high oxidation states, in this case Mn6+ is probably too small to accommodate six F ions. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.04%3A_Coordination_Compounds.txt |
Learning Objectives
• To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit (Figure 23.4 and Figure 23.5). In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy.
d-Orbital Splittings
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model in Chapter 9, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall from Chapter 6 that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (part (a) in Figure \(1\). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in part (b) in Figure \(1\), the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
Figure Figure \(1\): An Octahedral Arrangement of Six Negative Charges around a Metal Ion Causes the Five d Orbitals to Split into Two Sets with Different Energies
(a) Distributing a charge of −6 uniformly over a spherical surface surrounding a metal ion causes the energy of all five d orbitals to increase due to electrostatic repulsions, but the five d orbitals remain degenerate. Placing a charge of −1 at each vertex of an octahedron causes the d orbitals to split into two groups with different energies: the dx2−y2 and dz2 orbitals increase in energy, while the, dxy, dxz, and dyz orbitals decrease in energy. The average energy of the five d orbitals is the same as for a spherical distribution of a −6 charge, however. Attractive electrostatic interactions between the negatively charged ligands and the positively charged metal ion (far right) cause all five d orbitals to decrease in energy but does not affect the splittings of the orbitals. (b) The two eg orbitals (left) point directly at the six negatively charged ligands, which increases their energy compared with a spherical distribution of negative charge. In contrast, the three t2g orbitals (right) point between the negatively charged ligands, which decreases their energy compared with a spherical distribution of charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energyo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is 2(0.6Δo) + 3(−0.4Δo) = 0.
Note the Pattern
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in part (a) in Figure \(1\).
Electronic Structures of Metal Complexes
We can use the d-orbital energy-level diagram in Figure \(1\) to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure \(2\), for d1d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
Figure Figure \(2\) The Possible Electron Configurations for Octahedral dn Transition-Metal Complexes (n = 1–10)
Two different configurations are possible for octahedral complexes of metals with d4, d5, d6, and d7 configurations; the magnitude of Δo determines which configuration is observed.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall from Chapter 6 that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
Note the Pattern
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table \(1\)".
Table \(1\) Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes Δo (cm−1) Octahedral Complexes Δo (cm−1) Tetrahedral Complexes Δt (cm−1)
[Ti(H2O)6]3+ 20,300 [Fe(CN)6]4− 32,800 VCl4 9010
[V(H2O)6]2+ 12,600 [Fe(CN)6]3− 35,000 [CoCl4]2− 3300
[V(H2O)6]3+ 18,900 [CoF6]3− 13,000 [CoBr4]2− 2900
[CrCl6]3− 13,000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700
[Cr(H2O)6]2+ 13,900 [Co(H2O)6]3+ 27,000
[Cr(H2O)6]3+ 17,400 [Co(NH3)6]3+ 22,900
[Cr(NH3)6]3+ 21,500 [Co(CN)6]3− 34,800
[Cr(CN)6]3− 26,600 [Ni(H2O)6]2+ 8500
Cr(CO)6 34,150 [Ni(NH3)6]2+ 10,800
[MnCl6]4− 7500 [RhCl6]3− 20,400
[Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27,000
[MnCl6]3− 20,000 [Rh(NH3)6]3+ 34,000
[Mn(H2O)6]3+ 21,000 [Rh(CN)6]3− 45,500
[Fe(H2O)6]2+ 10,400 [IrCl6]3− 25,000
[Fe(H2O)6]3+ 14,300 [Ir(NH3)6]3+ 41,000
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Charge on the Metal Ion
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
Principal Quantum Number of the Metal
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point:
[Co(NH3)6]3+[Rh(NH3)6]3+[Ir(NH3)6]3+ΔoΔoΔo=22,900 cm−1=34,100 cm−1=40,000 cm−1
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I because smaller, more localized charges, such as we see for F, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo:
CO≈CN−>strong-field ligandsNO2−>en>NH3>SCN−>H2O>oxalate2−intermediate-field ligands>OH−>F>acetate−>Cl−>Br−>I−weak-field ligands
The values of Δo listed in Table \(1\) illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
Note the Pattern
The largest Δos are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Colors of Transition-Metal Complexes
The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a dd transition (Figure \(3\)). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δo.
Figure \(3\) A dd Transition
In a dd transition, an electron in one of the t2g orbitals of an octahedral complex such as the [Cr(H2O)6]3+ ion absorbs a photon of light with energy equal to Δo, which causes the electron to move to an empty or singly occupied eg orbital.
Recall from Chapter 6 that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel in End-of-Chapter Application Problem 6 in Chapter 6. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Gem-quality crystals of ruby and emerald. The colors of both minerals are due to the presence of small amounts of Cr3+ impurities in octahedral sites in an otherwise colorless metal oxide lattice.
Crystal Field Stabilization Energies
Recall from Chapter 9 that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table \(2\) gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table \(2\) CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
d 0 0
d 1 0.4
d 2 ↿ ↿ 0.8
d 3 ↿ ↿ ↿ 1.2
d 4 ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6
d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0
d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4
d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ 1.8
d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2
d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6
d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Note the Pattern
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Tetragonal and Square Planar Complexes
If two trans ligands in an octahedral complex are either chemically different from the other four, as in the trans-[Co(NH3)4Cl2]+ ion, or at a different distance from the metal than the other four, the result is a tetragonally distorted octahedral complex. The electronic structures of such complexes are best viewed as the result of distorting an octahedral complex. Consider, for example, an octahedral complex such as [Co(NH3)6]3+ and then slowly remove two trans NH3 molecules by moving them away from the metal along the ±z axes, as shown in the top half of Figure \(4\) As the two axial Co–N distances increase simultaneously, the d orbitals that interact most strongly with the two axial ligands will decrease in energy due to a decrease in electrostatic repulsions between electrons in these orbitals and the negative ends of the ligand dipoles. The affected d orbitals are those with a component along the ±z axes—namely, dz2, dxz, and dyz. They will not be affected equally, however. Because the dz2 orbital points directly at the two ligands being removed, its energy will decrease much more rapidly than the energy of the other two, as shown in the bottom half of Figure \(4\). In addition, the positive charge on the metal will increase somewhat as the axial ligands are removed, causing the four remaining in-plane ligands to be more strongly attracted to the metal. This will increase their interactions with the other two d orbitals and increase their energy. Again, the two d orbitals will not be affected equally. Because the dx2−y2 orbital points directly at the four in-plane ligands, its energy will increase more rapidly than the energy of the dxy orbital, which points between the in-plane ligands. If we remove the two axial ligands to an infinite distance, we obtain a square planar complex. The energies of the dz2 and dxy orbitals actually cross as the axial ligands are removed, and the largest orbital spliting in a square planar complex is identical in magnitude to Δo.
Figure \(4\) d-Orbital Splittings for Tetragonal and Square Planar Complexes
Moving the two axial ligands away from the metal ion along the z axis initially gives an elongated octahedral complex (center) and eventually produces a square planar complex (right). As shown below the structures, an axial elongation causes the dz2 dxz and dyz orbitals to decrease in energy and the dx2−y2 and dxy orbitals to increase in energy. As explained in the text, the change in energy is not the same for all five d orbitals. Removing the two axial ligands completely causes the energy of the dz2 orbital to decrease so much that the order of the dz2 and dxy orbitals is reversed.
Tetrahedral Complexes
In a tetrahedral arrangement of four ligands around a metal ion, none of the ligands lies on any of the three coordinate axes (part (a) in Figure \(5\)); consequently, none of the five d orbitals points directly at the ligands. Nonetheless, the dxy, dxz, and dyz orbitals interact more strongly with the ligands than do dx2−y2 and dz2 again resulting in a splitting of the five d orbitals into two sets. The splitting of the energies of the orbitals in a tetrahedral complex (Δt) is much smaller than that for Δo, however, for two reasons. First, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement. Second, there are only four negative charges rather than six, which decreases the electrostatic interactions by one-third if all other factors are equal. It can be shown that, for complexes of the same metal ion with the same charge, the same ligands, and the same M–L distance, Δt=49Δo. The relationship between the splitting of the five d orbitals in octahedral and tetrahedral crystal fields imposed by the same ligands is shown schematically in part (b) in Figure \(5\).
Figure \(5\) d-Orbital Splittings for a Tetrahedral Complex
(a) In a tetrahedral complex, none of the five d orbitals points directly at or between the ligands. (b) Because the dxy, dxz, and dyz orbitals (the t2g orbitals) interact more strongly with the ligands than do the dx2−y2 and dz2 orbitals (the eg orbitals), the order of orbital energies in a tetrahedral complex is the opposite of the order in an octahedral complex.
Note the Pattern
Δt < Δo because of weaker d-orbital–ligand interactions and decreased electrostatic interactions.
Because Δo is so large for the second- and third-row transition metals, all four-coordinate complexes of these metals are square planar due to the much higher CFSE for square planar versus tetrahedral structures. The only exception is for d10 metal ions such as Cd2+, which have zero CFSE and are therefore tetrahedral as predicted by the VSEPR model. Four-coordinate complexes of the first-row transition metals can be either square planar or tetrahedral. The former is favored by strong-field ligands, whereas the latter is favored by weak-field ligands. For example, the [Ni(CN)4]2− ion is square planar, while the [NiCl4]2− ion is tetrahedral.
Example \(1\)
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Solution
Strategy:
A From the number of ligands, determine the coordination number of the compound.
B Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
C Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
D Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution:
1. A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
2. A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise \(1\)
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Mn(H2O)6]2+
2. [PtCl4]2−
Answer
octahedral; high spin;
five square planar; low spin; no unpaired electrons
Consequences of d-Orbital Splitting
The splitting of the d orbitals because of their interaction with the ligands in a complex has important consequences for the chemistry of transition-metal complexes; they can be divided into structural effects and thermodynamic effects. Although the two kinds of effects are interrelated, we will consider them separately.
Structural Effects
There are two major kinds of structural effects: effects on the ionic radius of metal ions with regular octahedral or tetrahedral geometries, and structural distortions that are observed for specific electron configurations.
Ionic Radii
Figure \(6\) is a plot of the ionic radii of the divalent fourth-period metal ions versus atomic number. Only Ca2+(d0), Mn2+ (high-spin d5), and Zn2+ (d10) fall on the smooth curve calculated based on the effective nuclear charge (Zeff), which assumes that the distribution of d electrons is spherically symmetrical. All the other divalent ions fall below this curve because they have asymmetrical distributions of d electrons. (The points shown for Cr2+ and Cu2+ are only estimated values; as you will learn shortly, these two ions do not form any truly octahedral complexes.) To see why an asymmetrical distribution of d electrons makes a metal ion smaller than expected, consider the Ti2+ ion, which has a d2 configuration with both electrons in the t2g orbitals. Because the t2g orbitals are directed between the ligands, the two d electrons are unable to shield the ligands from the nuclear charge. Consequently, the ligands experience a higher effective nuclear charge than expected, the metal–ligand distance is unusually short, and the ionic radius is smaller than expected. If instead the two electrons were distributed uniformly over all five d orbitals, they would be much more effective at screening the ligands from the nuclear charge, making the metal–ligand distances longer and giving a larger ionic radius.
Figure \(6\) The Effect of d-Orbital Splittings on the Radii of the Divalent Ions of the Fourth-Period Metals
Because these radii are based on the structures of octahedral complexes and Cr2+ and Cu2+ do not form truly octahedral complexes, the points for these ions are shown as open circles. The dashed line represents the behavior predicted based on the effects of screening and variation in effective nuclear charge (Zeff), assuming a spherical distribution of the 3d electrons.
A similar effect is observed for the V2+ ion, which has a d3 configuration. Because the three electrons in the t2g orbitals provide essentially no shielding of the ligands from the metal, the ligands experience the full increase of +1 in nuclear charge that occurs in going from Ti2+ to V2+. Consequently, the observed ionic radius of the V2+ ion is significantly smaller than that of the Ti2+ ion.
Skipping the Cr2+ ion for the moment, we next consider the d5 Mn2+ ion. Because the nuclear charge increases by +2 from V2+ to Mn2+, we might expect Mn2+ to be smaller than V2+. The two electrons that are also added from V2+ to Mn2+ occupy the eg orbitals, however, which point directly at the six ligands. Because these electrons are localized directly between the metal ion and the ligands, they are effective at screening the ligands from the increased nuclear charge. As a result, the ionic radius actually increases significantly as we go from V2+ to Mn2+, despite the higher nuclear charge of the latter.
Exactly the same effects are seen in the second half of the first-row transition metals. In the Fe2+, Co2+, and Ni2+ ions, the extra electrons are added successively to the t2g orbitals, resulting in poor shielding of the ligands from the nuclear charge and abnormally small ionic radii. Skipping over Cu2+, we again see that adding the last two electrons causes a significant increase in the ionic radius of Zn2+, despite its higher nuclear charge.
The Jahn–Teller Effect
Because simple octahedral complexes are not known for the Cr2+ and Cu2+ ions, only estimated values for their radii are shown in Figure \(6\). We see in Figure \(7\) that both the Cr2+ and Cu2+ ions have electron configurations with an odd number of electrons in the eg orbitals. Because the single electron (in the case of Cr2+) or the third electron (in the case of Cu2+) can occupy either one of two degenerate eg orbitals, they have what is called a degenerate ground state. The Jahn–Teller theorem states that such non-linear systems are not stable; they will undergo a distortion that makes the complex less symmetrical and splits the degenerate states, which decreases the energy of the system. The distortion and resulting decrease in energy are collectively referred to as the Jahn–Teller effect. Neither the nature of the distortion nor its magnitude is specified, and in fact, they are difficult to predict. In principle, Jahn–Teller distortions are possible for many transition-metal ions; in practice, however, they are observed only for systems with an odd number of electrons in the eg orbitals, such as the Cr2+ and Cu2+ ions.
To see how a geometrical distortion can decrease the energy of such a system, consider an octahedral Cu2+ complex, the [Cu(H2O)6]2+ ion, which has been elongated along the z axis. As indicated in Figure \(7\), this kind of distortion splits both the eg and t2g sets of orbitals. Because the axial ligands interact most strongly with the dz2 orbital, the splitting of the eg set (δ1) is significantly larger than the splitting of the t2g set (δ2), but both δ1 and δ2 are much, much smaller than the Δo. This splitting does not change the center of gravity of the energy within each set, so a Jahn–Teller distortion results in no net change in energy for a filled or half-filled set of orbitals. If, however, the eg set contains one (as in the d4 ions, Cr2+ and Mn3+) or three (as in the d9 ion, Cu2+) electrons, the distortion decreases the energy of the system. For Cu2+, for example, the change in energy after distortion is 2(−δ1/2) + 1(δ1/2) = −δ1/2. For Cu2+ complexes, the observed distortion is always an elongation along the z axis by as much as 50 pm; in fact, many Cu2+ complexes are so distorted that they are effectively square planar. In contrast, the distortion observed for most Cr2+ complexes is a compression along the z axis. In both cases, however, the net effect is the same: the distorted system is more stable than the undistorted system.
Note the Pattern
Jahn–Teller distortions are most important for d9 and high-spin d4 complexes; the distorted system is more stable than the undistorted one.
Figure \(7\)The Jahn–Teller Effect
Increasing the axial metal–ligand distances in an octahedral d9 complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of eg orbitals to split in energy by an amount δ1; δ1 and δ2 are much smaller than Δo. As a result, the distorted system is more stable (lower in energy) than the undistorted complex by δ1/2.
Thermodynamic Effects
As we previously noted, CFSEs can be as large as several hundred kilojoules per mole, which is the same magnitude as the strength of many chemical bonds or the energy change in most chemical reactions. Consequently, CFSEs are important factors in determining the magnitude of hydration energies, lattice energies, and other thermodynamic properties of the transition metals.
Hydration Energies
The hydration energy of a metal ion is defined as the change in enthalpy for the following reaction:
\( M2+(g) + H2O(l) → M2+(aq) \)
Although hydration energies cannot be measured directly, they can be calculated from experimentally measured quantities using thermochemical cycles. As shown in part (a) in Figure \(8\), a plot of the hydration energies of the fourth-period metal dications versus atomic number gives a curve with two valleys. Note the relationship between the plot in part (a) in Figure \(8\)" and the plot of ionic radii in Figure \(6\): the overall shapes are essentially identical, and only the three cations with spherically symmetrical distributions of d electrons (Ca2+, Mn2+, and Zn2+) lie on the dashed lines. In part (a) in Figure \(8\) the dashed line corresponds to hydration energies calculated based solely on electrostatic interactions. Subtracting the CFSE values for the [M(H2O)6]2+ ions from the experimentally determined hydration energies gives the points shown as open circles, which lie very near the calculated curve. Thus CFSEs are primarily responsible for the differences between the measured and calculated values of hydration energies.
Figure \(8\) Thermochemical Effects of d-Orbital Splittings
(a) A plot of the hydration energies of the divalent fourth-period metal ions versus atomic number (solid circles) shows large deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed line). Correcting for CFSE gives the points shown as open circles, which, except for Ti2+ and Cr2+, are close to the calculated values. The apparent deviations for these ions are caused by the fact that solutions of the Ti2+ ion in water are not stable, and Cr2+ does not form truly octahedral complexes. (b) A plot of the lattice energies for the fourth-period metal dichlorides versus atomic number shows similar deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed lines), again illustrating the importance of CFSEs.
Lattice Energies
Values of the lattice energies for the fourth-period metal dichlorides are plotted versus atomic number in part (b) in Figure \(8\). Recall that the lattice energy is defined as the negative of the enthalpy change for the following reaction. Like hydration energies, lattice energies are determined indirectly by using a thermochemical cycle:
\( M2+(g) + 2Cl(g) → MCl2(s) \)
The shape of the lattice-energy curve is essentially the mirror image of the hydration-energy curve in part (a) in Figure 23.19, with only Ca2+, Mn2+, and Zn2+ lying on the smooth curve. It is not surprising that the explanation for the deviations from the curve is exactly the same as for the hydration energy data: all the transition-metal dichlorides, except MnCl2 and ZnCl2, are more stable than expected due to CFSE.
Summary
Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. The crystal field stabilization energy (CFSE) is the additional stabilization of a complex due to placing electrons in the lower-energy set of d orbitals. CFSE explains the unusual curves seen in plots of ionic radii, hydration energies, and lattice energies versus atomic number. The Jahn–Teller theorem states that a non-linear molecule with a spatially degenerate electronic ground state will undergo a geometrical distortion to remove the degeneracy and lower the overall energy of the system.
KEY TAKEAWAY
• Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity.
CONCEPTUAL PROBLEMS
1. Describe crystal field theory in terms of its
1. assumptions regarding metal–ligand interactions.
2. weaknesses and strengths compared with valence bond theory.
2. In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?
3. Will the value of Δo increase or decrease if I ligands are replaced by NO2 ligands? Why?
4. For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?
5. How can CFT explain the color of a transition-metal complex?
STRUCTURE AND REACTIVITY
1. Do strong-field ligands favor a tetrahedral or a square planar structure? Why?
2. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [TiCl6]3−
2. [CoCl4]2−
3. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Cu(NH3)4]2+
2. [Ni(CN)4]2−
4. The ionic radii of V2+, Fe2+, and Zn2+ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.
Answer
1. d9, square planar, neither high nor low spin, single unpaired electron
2. d8, square planar, low spin, no unpaired electrons | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.05%3A_Crystal_Field_Theory.txt |
Learning Objectives
• To become familiar with some of the roles of transition-metal complexes in biological systems.
In this section, we describe several systems that illustrate the roles transition metals play in biological systems. Our goal is for you to understand why the chemical properties of these elements make them essential for life. We begin with a discussion of the strategies organisms use to extract transition metals from their environment. The section continues with a brief discussion of the use of transition metals in reactions that involve the transfer of electrons, reactions of small molecules such as O2, Lewis-acid catalysis, and the generation of reactive organic radicals.
Uptake and Storage of Transition Metals
There are three possible dietary levels for any essential element: deficient, optimal, and toxic, in order of increasing concentration in the diet. If the concentration of an essential element in the diet is too low, an organism must be able to extract the element from the environment and concentrate it. If the concentration of an essential element in the diet is too high, an organism must be able to limit its intake to avoid toxic effects. Moreover, organisms must be able to switch off the uptake process rapidly if dietary levels rise suddenly, and they must be able to store essential elements for future use.
Three distinct steps are involved in transition metal uptake. First, the metal must be “mobilized” from the environment and brought into contact with a cell in a form that can be absorbed. Second, the metal must be transported across the cell membrane into the cell. Third, the element must be transported to its point of utilization within a cell or to other cells within the organism. In our discussion, we focus on the uptake, transport, and storage of iron, which illustrates the most important points. Because iron deficiency (anemia) is the most widespread nutritional deficiency known in humans, the uptake of iron is especially well understood.
Iron complexes in biological systems. Iron(III) forms very stable octahedral complexes with hydroxamate and catecholate ligands.
The solubility of metal ions such as Fe3+, which form highly insoluble hydroxides, depends on the pH and the presence of complexing agents. In an oxygen-containing atmosphere, iron exists as Fe(III) because of the positive reduction potential of Fe3+ (Fe3+ + e → Fe2+; E° = +0.77 V). Because ferric hydroxide [Fe(OH)3] is highly insoluble (Ksp ≈ 1 × 10−39), the equilibrium concentration of Fe3+(aq) at pH 7.0 is very low, about 10−18 M. You would have to drink 2 × 1013 L of iron-saturated water per day (roughly 5 mi3) to consume the recommended daily intake of Fe for humans, which is about 1 mg/day. Animals such as humans can overcome this problem by consuming concentrated sources of iron, such as red meat, but microorganisms cannot.
Consequently, most microorganisms synthesize and secrete organic molecules called siderophores to increase the total concentration of available iron in the surrounding medium. Siderophores are generally cyclic compounds that use bidentate ligands, such as the hydroxamate and catecholate groups shown here, to bind Fe3+ in an octahedral arrangement. Typical siderophores are ferrichrome, a cyclic peptide produced by fungi, and enterobactin, a cyclic ester produced by bacteria (Figure $1$). Attaching the three iron ligands to a cyclic framework greatly increases the stability of the resulting Fe3+ complex due to the chelate effect described in Section 23.4. The formation constants for the Fe3+ complexes of ferrichrome and enterobactin are about 1032 and 1040, respectively, which are high enough to allow them to dissolve almost any Fe(III) compound.
Siderophores increase the [Fe3+] in solution, providing the bacterium that synthesized them (as well as any competitors) with a supply of iron. In addition, siderophores neutralize the positive charge on the metal ion and provide a hydrophobic “wrapping” that enables the Fe3+–siderophore complex to be recognized by a specific protein that transports it into the interior of a cell. Once it is inside a cell, the iron is reduced to Fe2+, which has a much lower affinity for the siderophore and spontaneously dissociates.
In contrast, multicellular organisms can increase the concentration of iron in their diet by lowering the pH in the gastrointestinal tract. At pH 1.0 (the approximate pH of the stomach), most Fe(III) salts dissolve to form Fe3+(aq), which is absorbed by specific proteins in the intestinal wall. A protein called transferrin forms a complex with iron(III), allowing it to be transported to other cells. Proteins that bind tightly to Fe(III) can also be used as antibacterial agents because iron is absolutely essential for bacterial growth. For example, milk, tears, and egg white all contain proteins similar to transferrin, and their high affinity for Fe3+ allows them to sequester iron, thereby preventing bacteria from growing in these nutrient-rich media.
Iron is released from transferrin by reduction to Fe2+, and then it is either used immediately (e.g., for the synthesis of hemoglobin) or stored in a very large protein called ferritin for future use (Figure $2$). Ferritin uses oxygen to oxidize Fe2+ to Fe3+, which at neutral pH precipitates in the central cavity of the protein as a polymeric mixture of Fe(OH)3 and FePO4. Because a fully loaded ferritin molecule can contain as many as 4500 Fe atoms, which corresponds to about 25% Fe by mass, ferritin is an effective way to store iron in a highly concentrated form. When iron is needed by a cell, the Fe3+ is reduced to the much more soluble Fe2+ by a reductant such as ascorbic acid (vitamin C). The structure of ferritin contains channels at the junctions of the subunits, which provide pathways for iron to enter and leave the interior of a molecule.
Metalloproteins and Metalloenzymes
A protein that contains one or more metal ions tightly bound to amino acid side chains is called a metalloprotein; some of the most common ligands provided by amino acids are shown here. A metalloprotein that catalyzes a chemical reaction is a metalloenzyme. Thus all metalloenzymes are metalloproteins, but the converse is not true. Recent estimates suggest that more than 40% of all known enzymes require at least one metal ion for activity, including almost all the enzymes responsible for the synthesis, duplication, and repair of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
Electron-Transfer Proteins
Proteins whose function is to transfer electrons from one place to another are called electron-transfer proteins. Because they do not catalyze a chemical reaction, electron-transfer proteins are not enzymes; they are biochemical reductants or oxidants consumed in an enzymatic reaction. The general reaction for an electron-transfer protein is as follows:
$M^{n+} + e^- \rightleftharpoons M^{(n−1)+} \label{23.14}$
Because many transition metals can exist in more than one oxidation state, electron-transfer proteins usually contain one or more metal ions that can undergo a redox reaction. Incorporating a metal ion into a protein has three important biological consequences:
1. The protein environment can adjust the redox potential (E0′), of the metal ion over a rather large potential range, whereas the redox potential of the simple hydrated metal ion [Mn+(aq)], is essentially fixed.
2. The protein can adjust the structure of the metal complex to ensure that electron transfer is rapid.
3. The protein environment provides specificity, ensuring that the electron is transferred to only the desired site.
Three important classes of metalloproteins transfer electrons: blue copper proteins, cytochromes, and iron–sulfur proteins, which generally transfer electrons at high (> 0.20 V), intermediate (±0 V), and low (−0.20 to −0.50 V) potentials, respectively (Table 23.12). Although these electron-transfer proteins contain different metals with different structures, they are all designed to ensure rapid electron transfer to and from the metal. Thus when the protein collides with its physiological oxidant or reductant, electron transfer can occur before the two proteins diffuse apart. For electron transfer to be rapid, the metal sites in the oxidized and reduced forms of the protein must have similar structures.
Table $1$: Some Properties of the Most Common Electron-Transfer Proteins
Protein Metal Center M/e− Transferred Reduction Potential (V)
* A sulfur bound to an organic group is represented as SR.
See Figure $\PageIndex{2b}$: for the structure of imidazole (Im).
iron–sulfur proteins* [Fe(SR)4]2− 1 Fe −0.1 to +0.1
[(RS)2FeS2Fe(SR)2]2− 2 Fe −0.2 to −0.4
[Fe3S4(SR)3]3− 3 Fe −0.1 to −0.2
[Fe4S4(SR)4]2− 4 Fe −0.3 to −0.5
cytochromes Fe-heme (low spin) 1 Fe ~0
blue copper proteins [Cu(Im)2(SR)(SR2)] 1 Cu ≥ +0.20
Blue Copper Proteins
Blue copper proteins were first isolated from bacteria in the 1950s and from plant tissues in the early 1960s. The intense blue color of these proteins is due to a strong absorption band at a wavelength of about 600 nm. Although simple Cu2+ complexes, such as [Cu(H2O)6]2+ and [Cu(NH3)4]2+, are also blue due to an absorption band at 600 nm, the intensity of the absorption band is about 100 times less than that of a blue copper protein. Moreover, the reduction potential for the Cu2+/Cu+ couple in a blue copper protein is usually +0.3 to +0.5 V, considerably more positive than that of the aqueous Cu2+/Cu+ couple (+0.15 V).
The copper center in blue copper proteins has a distorted tetrahedral structure, in which the copper is bound to four amino acid side chains (Figure $3$). Although the most common structures for four-coordinate Cu2+ and Cu+ complexes are square planar and tetrahedral, respectively, the structures of the oxidized (Cu2+) and reduced (Cu+) forms of the protein are essentially identical. Thus the protein forces the Cu2+ ion to adopt a higher-energy structure that is more suitable for Cu+, which makes the Cu2+ form easier to reduce and raises its reduction potential.
Moreover, by forcing the oxidized and reduced forms of the metal complex to have essentially the same structure, the protein ensures that electron transfer to and from the copper site is rapid because only minimal structural reorganization of the metal center is required. Kinetics studies on simple metal complexes have shown that electron-transfer reactions tend to be slow when the structures of the oxidized and reduced forms of a metal complex are very different, and fast when they are similar. You will see that other metal centers used for biological electron-transfer reactions are also set up for minimal structural reorganization after electron transfer, which ensures the rapid transfer of electrons.
Cytochromes
The cytochromes (from the Greek cytos, meaning “cell”, and chroma, meaning “color”) were first identified in the 1920s by spectroscopic studies of cell extracts. Based on the wavelength of the maximum absorption in the visible spectrum, they were classified as cytochromes a (with the longest wavelength), cytochromes b (intermediate wavelength), and cytochromes c (shortest wavelength). It quickly became apparent that there was a correlation between their spectroscopic properties and other physical properties. For examples, cytochromes c are generally small, soluble proteins with a reduction potential of about +0.25 V, whereas cytochromes b are larger, less-soluble proteins with reduction potentials of about 0 V.
All cytochromes contain iron, and the iron atom in all cytochromes is coordinated by a planar array of four nitrogen atoms provided by a cyclic tetradentate ligand called a porphyrin. The iron–porphyrin unit is called a heme group. The structures of a typical porphyrin (protoporphyrin IX) and its iron complex (protoheme) are shown here. In addition to the four nitrogen atoms of the porphyrin, the iron in a cytochrome is usually bonded to two additional ligands provided by the protein, as shown in Figure $4$.
A cytochrome. Shown here is protoporphyrin IX and its iron complex, protoheme.
In contrast to the blue copper proteins, two electron configurations are possible for both the oxidized and reduced forms of a cytochrome, and this has significant structural consequences. Thus Fe2+ is d6 and can be either high spin (with four unpaired electrons) or low spin (with no unpaired electrons). Similarly, Fe3+ is d5 and can also be high spin (with five unpaired electrons) or low spin (with one unpaired electron). In low-spin heme complexes, both the Fe2+ and the Fe3+ ions are small enough to fit into the “hole” in the center of the porphyrin; hence the iron atom lies almost exactly in the plane of the four porphyrin nitrogen atoms in both cases. Because cytochromes b and c are low spin in both their oxidized and reduced forms, the structures of the oxidized and reduced cytochromes are essentially identical. Hence minimal structural changes occur after oxidation or reduction, which makes electron transfer to or from the heme very rapid.
Electron transfer reactions occur most rapidly when minimal structural changes occur during oxidation or reduction.
Iron–Sulfur Proteins
Although all known bacteria, plants, and animals use iron–sulfur proteins to transfer electrons, the existence of these proteins was not recognized until the late 1950s. Iron–sulfur proteins transfer electrons over a wide range of reduction potentials, and their iron content can range from 1 to more than 12 Fe atoms per protein molecule. In addition, most iron–sulfur proteins contain stoichiometric amounts of sulfide (S2−).
These properties are due to the presence of four different kinds of iron–sulfur units, which contain one, two, three, or four iron atoms per Fe–S complex (Figure $5$). In all cases, the Fe2+ and Fe3+ ions are coordinated to four sulfur ligands in a tetrahedral environment. Due to tetrahedral coordination by weak-field sulfur ligands, the iron is high spin in both the Fe3+ and Fe2+ oxidation states, which results in similar structures for the oxidized and reduced forms of the Fe–S complexes. Consequently, only small structural changes occur after oxidation or reduction of the Fe–S center, which results in rapid electron transfer.
Reactions of Small Molecules
Although small molecules, such as O2, N2, and H2, do not react with organic compounds under ambient conditions, they do react with many transition-metal complexes. Consequently, virtually all organisms use metalloproteins to bind, transport, and catalyze the reactions of these molecules. Probably the best-known example is hemoglobin, which is used to transport O2 in many multicellular organisms.
Under ambient conditions, small molecules, such as O2, N2, and H2, react with transition-metal complexes but not with organic compounds.
Oxygen Transport
Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O2. At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O2 through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O2 rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O2 is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell.
Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in Table $2$. Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex.
Table $2$: Some Properties of the Three Classes of Oxygen-Transport Proteins
Protein Source M per Subunit M per O2 Bound Color (deoxy form) Color (oxy form)
hemoglobin mammals, birds, fish, reptiles, some insects 1 Fe 1 Fe red-purple red
hemerythrin marine worms 2 Fe 2 Fe colorless red
hemocyanin mollusks, crustaceans, spiders 2 Cu 2 Cu colorless blue
Myoglobin and Hemoglobin
Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (Figure 26.8.1). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O2 binds.
In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe2+ is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O2 binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; Figure 26.8.2). Because low-spin Fe2+ and Fe3+ are smaller than high-spin Fe2+, the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O2 pressure at which half of the molecules in a solution of myoglobin are bound to O2 (P1/2) is about 1 mm Hg (1.3 × 10−3 atm).
A vacant coordination site at a metal center in a protein usually indicates that a small molecule will bind to the metal ion, whereas a coordinatively saturated metal center is usually involved in electron transfer.
Hemoglobin consists of two subunits of 141 amino acids and two subunits of 146 amino acids, both similar to myoglobin; it is called a tetramer because of its four subunits. Because hemoglobin has very different O2-binding properties, however, it is not simply a “super myoglobin” that can carry four O2 molecules simultaneously (one per heme group). The shape of the O2-binding curve of myoglobin (Mb; Figure $7$) can be described mathematically by the following equilibrium:
$MbO_2 \rightleftharpoons Mb + O_ 2 \label{26.8.1a}$
$K_{diss}=\dfrac{[Mb][O_2]}{[MbO_2]} \label{26.8.1b}$
In contrast, the O2-binding curve of hemoglobin is S shaped (Figure $8$). As shown in the curves, at low oxygen pressures, the affinity of deoxyhemoglobin for O2 is substantially lower than that of myoglobin, whereas at high O2 pressures the two proteins have comparable O2 affinities. The physiological consequences of the unusual S-shaped O2-binding curve of hemoglobin are enormous. In the lungs, where O2 pressure is highest, the high oxygen affinity of deoxyhemoglobin allows it to be completely loaded with O2, giving four O2 molecules per hemoglobin. In the tissues, however, where the oxygen pressure is much lower, the decreased oxygen affinity of hemoglobin allows it to release O2, resulting in a net transfer of oxygen to myoglobin.
The S-shaped O2-binding curve of hemoglobin is due to a phenomenon called cooperativity, in which the affinity of one heme for O2 depends on whether the other hemes are already bound to O2. Cooperativity in hemoglobin requires an interaction between the four heme groups in the hemoglobin tetramer, even though they are more than 3000 pm apart, and depends on the change in structure of the heme group that occurs with oxygen binding. The structures of deoxyhemoglobin and oxyhemoglobin are slightly different, and as a result, deoxyhemoglobin has a much lower O2 affinity than myoglobin, whereas the O2 affinity of oxyhemoglobin is essentially identical to that of oxymyoglobin. Binding of the first two O2 molecules to deoxyhemoglobin causes the overall structure of the protein to change to that of oxyhemoglobin; consequently, the last two heme groups have a much higher affinity for O2 than the first two.
Oxygen is not unique in its ability to bind to a ferrous heme complex; small molecules such as CO and NO bind to deoxymyoglobin even more tightly than does O2. The interaction of the heme iron with oxygen and other diatomic molecules involves the transfer of electron density from the filled t2g orbitals of the low-spin d6 Fe2+ ion to the empty π* orbitals of the ligand. In the case of the Fe2+–O2 interaction, the transfer of electron density is so great that the Fe–O2 unit can be described as containing low-spin Fe3+ (d5) and O2. We can therefore represent the binding of O2 to deoxyhemoglobin and its release as a reversible redox reaction:
$Fe^{2+} + O_2 \rightleftharpoons \ce{Fe^{3+}–O_2^−} \label{26.8.2}$
As shown in Figure $9$, the Fe–O2 unit is bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in CO are empty and those in NO are singly occupied, these ligands interact more strongly with Fe2+ than does O2, in which the π* orbitals of the neutral ligand are doubly occupied.
Although CO has a much greater affinity for a ferrous heme than does O2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O2, which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O2 and CO provides a plausible explanation for the difference in affinities. As shown in Figure $9$, the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO.
Hemerythrin
Hemerythrin is used to transport O2 in a variety of marine invertebrates. It is an octamer (eight subunits), with each subunit containing two iron atoms and binding one molecule of O2. Deoxyhemerythrin contains two Fe2+ ions per subunit and is colorless, whereas oxyhemerythrin contains two Fe3+ ions and is bright reddish violet. These invertebrates also contain a monomeric form of hemerythrin that is located in the tissues, analogous to myoglobin. The binding of oxygen to hemerythrin and its release can be described by the following reaction, where the HO2 ligand is the hydroperoxide anion derived by the deprotonation of hydrogen peroxide (H2O2):
$\ce{2Fe^{2+} + O2 + H^{+} <=> 2Fe^{3+}–O2H} \label{23.17}$
Thus O2 binding is accompanied by the transfer of two electrons (one from each Fe2+) and a proton to O2.
Hemocyanin
Hemocyanin is used for oxygen transport in many arthropods (spiders, crabs, lobsters, and centipedes) and in mollusks (shellfish, octopi, and squid); it is responsible for the bluish-green color of their blood. The protein is a polymer of subunits that each contain two copper atoms (rather than iron), with an aggregate molecular mass of greater than 1,000,000 amu. Deoxyhemocyanin contains two Cu+ ions per subunit and is colorless, whereas oxyhemocyanin contains two Cu2+ ions and is bright blue. As with hemerythrin, the binding and release of O2 correspond to a two-electron reaction:
$\ce{2Cu^{+} + O2 <=> Cu^{2+}–O2^{2−}–Cu^{2+}} \label{23.18}$
Although hemocyanin and hemerythrin perform the same basic function as hemoglobin, these proteins are not interchangeable. In fact, hemocyanin is so foreign to humans that it is one of the major factors responsible for the common allergies to shellfish.
Myoglobin, hemoglobin, hemerythrin, and hemocyanin all use a transition-metal complex to transport oxygen.
Enzymes Involved in Oxygen Activation
Many of the enzymes involved in the biological reactions of oxygen contain metal centers with structures that are similar to those used for O2 transport. Many of these enzymes also contain metal centers that are used for electron transfer, which have structures similar to those of the electron-transfer proteins discussed previously. In this section, we briefly describe two of the most important examples: dioxygenases and methane monooxygenase.
Dioxygenases are enzymes that insert both atoms of O2 into an organic molecule. In humans, dioxygenases are responsible for cross-linking collagen in connective tissue and for synthesizing complex organic molecules called prostaglandins, which trigger inflammation and immune reactions. Iron is by far the most common metal in dioxygenases; and the target of the most commonly used drug in the world, aspirin, is an iron enzyme that synthesizes a specific prostaglandin. Aspirin inhibits this enzyme by binding to the iron atom at the active site, which prevents oxygen from binding.
Methane monooxygenase catalyzes the conversion of methane to methanol. The enzyme is a monooxygenase because only one atom of O2 is inserted into an organic molecule, while the other is reduced to water:
$\ce{CH_4 + O_2 + 2e^- + 2H^+ \rightarrow CH_3OH + H_2O} \label{23.19}$
Because methane is the major component of natural gas, there is enormous interest in using this reaction to convert methane to a liquid fuel (methanol) that is much more convenient to ship and store. Because the C–H bond in methane is one of the strongest C–H bonds known, however, an extraordinarily powerful oxidant is needed for this reaction. The active site of methane monooxygenase contains two Fe atoms that bind O2, but the details of how the bound O2 is converted to such a potent oxidant remain unclear.
Metal Ions as Lewis Acids
Reactions catalyzed by metal ions that do not change their oxidation states during the reaction are usually group transfer reactions, in which a group such as the phosphoryl group (−PO32−) is transferred. These enzymes usually use metal ions such as Zn2+, Mg2+, and Mn2+, and they range from true metalloenzymes, in which the metal ion is tightly bound, to metal-activated enzymes, which require the addition of metal ions for activity. Because tight binding is usually the result of specific metal–ligand interactions, metalloenzymes tend to be rather specific for a particular metal ion. In contrast, the binding of metal ions to metal-activated enzymes is largely electrostatic in nature; consequently, several different metal ions with similar charges and sizes can often be used to give an active enzyme.
Metalloenzymes generally contain a specific metal ion, whereas metal-activated enzymes can use any of several metal ions of similar size and charge.
A metal ion that acts as a Lewis acid can catalyze a group transfer reaction in many different ways, but we will focus on only one of these, using a zinc enzyme as an example. Carbonic anhydrase is found in red blood cells and catalyzes the reaction of CO2 with water to give carbonic acid.
$\ce{CO_2(g) + H_2O(l) \rightleftharpoons H^{+}(aq) + HCO^{-}3(aq)} \label{23.20}$
Although this reaction occurs spontaneously in the absence of a catalyst, it is too slow to absorb all the CO2 generated during respiration. Without a catalyst, tissues would explode due to the buildup of excess CO2 pressure. Carbonic anhydrase contains a single Zn2+ ion per molecule, which is coordinated by three histidine imidazole ligands and a molecule of water. Because Zn2+ is a Lewis acid, the pKa of the Zn2+–OH2 unit is about 8 versus 14 for pure water. Thus at pH 7–8, a significant fraction of the enzyme molecules contain the Zn2+–OH group, which is much more reactive than bulk water. When carbon dioxide binds in a nonpolar site next to the Zn2+–OH unit, it reacts rapidly to give a coordinated bicarbonate ion that dissociates from the enzyme:
$\ce{Zn^{2+}–OH^{-} + CO_2 \rightleftharpoons Zn^{2+}–OCO_2H^- \rightleftharpoons Zn^{2+} + HCO_3^{-}} \label{23.21}$
The active site of carbonic anhydrase.
Thus the function of zinc in carbonic anhydrase is to generate the hydroxide ion at pH 7.0, far less than the pH required in the absence of the metal ion.
Enzymes That Use Metals to Generate Organic Radicals
An organic radical is an organic species that contains one or more unpaired electrons. Chemists often consider organic radicals to be highly reactive species that produce undesirable reactions. For example, they have been implicated in some of the irreversible chemical changes that accompany aging. It is surprising, however, that organic radicals are also essential components of many important enzymes, almost all of which use a metal ion to generate the organic radical within the enzyme. These enzymes are involved in the synthesis of hemoglobin and DNA, among other important biological molecules, and they are the targets of pharmaceuticals for the treatment of diseases such as anemia, sickle-cell anemia, and cancer. In this section, we discuss one class of radical enzymes that use vitamin B12.
Vitamin B12 was discovered in the 1940s as the active agent in the cure of pernicious anemia, which does not respond to increased iron in the diet. Humans need only tiny amounts of vitamin B12, and the average blood concentration in a healthy adult is only about 3.5 × 10−8 M. The structure of vitamin B12, shown in Figure $10$, is similar to that of a heme, but it contains cobalt instead of iron, and its structure is much more complex. In fact, vitamin B12 has been called the most complex nonpolymeric biological molecule known and was the first naturally occurring organometallic compound to be isolated. When vitamin B12 (the form present in vitamin tablets) is ingested, the axial cyanide ligand is replaced by a complex organic group.
The cobalt–carbon bond in the enzyme-bound form of vitamin B12 and related compounds is unusually weak, and it is particularly susceptible to homolytic cleavage:
$\ce{CoCH_2R <=> Co^{2+⋅} + ⋅CH_2R} \label{23.22}$
Homolytic cleavage of the Co3+–CH2R bond produces two species, each of which has an unpaired electron: a d7 Co2+ derivative and an organic radical, ·CH2R, which is used by vitamin B12-dependent enzymes to catalyze a wide variety of reactions. Virtually all vitamin B12-catalyzed reactions are rearrangements in which an H atom and an adjacent substituent exchange positions:
In the conversion of ethylene glycol to acetaldehyde, the initial product is the hydrated form of acetaldehyde, which rapidly loses water:
The enzyme uses the ·CH2R radical to temporarily remove a hydrogen atom from the organic substrate, which then rearranges to give a new radical. Transferring the hydrogen atom back to the rearranged radical gives the product and regenerates the ·CH2R radical.
The metal is not involved in the actual catalytic reaction; it provides the enzyme with a convenient mechanism for generating an organic radical, which does the actual work. Many examples of similar reactions are now known that use metals other than cobalt to generate an enzyme-bound organic radical.
Nearly all vitamin B12-catalyzed reactions are rearrangements that occur via a radical reaction.
Summary
Three separate steps are required for organisms to obtain essential transition metals from their environment: mobilization of the metal, transport of the metal into the cell, and transfer of the metal to where it is needed within a cell or an organism. The process of iron uptake is best understood. To overcome the insolubility of Fe(OH)3, many bacteria use organic ligands called siderophores, which have high affinity for Fe(III) and are secreted into the surrounding medium to increase the total concentration of dissolved iron. The iron–siderophore complex is absorbed by a cell, and the iron is released by reduction to Fe(II). Mammals use the low pH of the stomach to increase the concentration of dissolved iron. Iron is absorbed in the intestine, where it forms an Fe(III) complex with a protein called transferrin that is transferred to other cells for immediate use or storage in the form of ferritin.
Proteins that contain one or more tightly bound metal ions are called metalloproteins, and metalloproteins that catalyze biochemical reactions are called metalloenzymes. Proteins that transfer electrons from one place to another are called electron-transfer proteins. Most electron-transfer proteins are metalloproteins, such as iron–sulfur proteins, cytochromes, and blue copper proteins that accept and donate electrons. The oxidized and reduced centers in all electron-transfer proteins have similar structures to ensure that electron transfer to and from the metal occurs rapidly. Metalloproteins also use the ability of transition metals to bind small molecules, such as O2, N2, and H2, to transport or catalyze the reactions of these small molecules. For example, hemoglobin, hemerythrin, and hemocyanin, which contain heme iron, nonheme iron, and copper, respectively, are used by different kinds of organisms to bind and transfer O2. Other metalloenzymes use transition-metal ions as Lewis acids to catalyze group transfer reactions. Finally, some metalloenzymes use homolytic cleavage of the cobalt–carbon bond in derivatives of vitamin B12 to generate an organic radical that can abstract a hydrogen atom and thus cause molecular rearrangements to occur.
Key Takeaway
• Organisms have developed strategies to extract transition metals from the environment and use the metals in electron-transfer reactions, reactions of small molecules, Lewis-acid catalysis, and the generation of reactive organic radicals.
Conceptual Problems
1. What are the advantages of having a metal ion at the active site of an enzyme?
2. Why does the structure of the metal center in a metalloprotein that transfers electrons show so little change after oxidation or reduction?
Structure and Reactivity
1. In enzymes, explain how metal ions are particularly suitable for generating organic radicals.
2. A common method for treating carbon-monoxide poisoning is to have the patient inhale pure oxygen. Explain why this treatment is effective. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/22%3A_The_d-Block_Elements/22.06%3A_Transition_Metals_in_Biology.txt |
Carbon is unique among the elements in its ability to catenate, to form a wide variety of compounds that contain long chains and/or rings of carbon atoms. Some of the most complex chemical structures known are those of the organic molecules found in living organisms. In spite of their size and complexity, these biological molecules obey the same chemical principles as simpler organic molecules.
• 23.1: Organic Compounds
• 23.2: Functional Groups and Classes of Organic Compounds
Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework.
• 23.3: Isomers of Organic Compounds
• 23.4: Reactivity of Organic Molecules
Electrophiles have a strong tendency to react with nucleophiles. The reactivity of a molecule is often affected by the degree of substitution of the carbon bonded to a functional group; the carbon is designated as primary, secondary, or tertiary. Identifying the transient species formed in a chemical reaction, some of which are charged, enables chemists to predict the mechanism and products of the reaction.
• 23.5: Common Classes of Organic Reactions
• 23.6: Common Classes of Organic Compounds
The general properties and reactivity of each class of organic compounds is largely determined by its functional groups. In this section, we describe the relationships between structure, physical properties, and reactivity for the major classes of organic compounds. We also show you how to apply these relationships to understand some common reactions that chemists use to synthesize organic compounds.
• 23.7: The Molecules of Life
23: Organic Compounds
Learning Objectives
• To understand the difference between organic and inorganic molecules
Organic substances have been used throughout this text to illustrate the differences between ionic and covalent bonding and to demonstrate the intimate connection between the structures of compounds and their chemical reactivity. You learned, for example, that even though NaOH and alcohols (ROH) both have OH in their formula, NaOH is an ionic compound that dissociates completely in water to produce a basic solution containing Na+ and OH ions, whereas alcohols are covalent compounds that do not dissociate in water and instead form neutral aqueous solutions. You also learned that an amine (RNH2), with its lone pairs of electrons, is a base, whereas a carboxylic acid (RCO2H), with its dissociable proton, is an acid.
The structure of a solid with a hybrid metal-organic framework. Organic and inorganic groups of the proper structure can be used to synthesize solids with very large pores (central sphere) that can accommodate a variety of small molecules. The rigid benzene rings are used as “props” to hold the metal units (carboxylate-bridged copper dimers) apart. Such solids have potential applications in hydrogen storage for use in fuel cells or automobiles.
Carbon is unique among the elements in its ability to catenate, to form a wide variety of compounds that contain long chains and/or rings of carbon atoms. Some of the most complex chemical structures known are those of the organic molecules found in living organisms. In spite of their size and complexity, these biological molecules obey the same chemical principles as simpler organic molecules. Thus we can use Lewis electron structures to understand the preferred mode of reactivity of a variety of organic compounds, relative electronegativities and bond polarities to predict how certain groups of atoms will react, and molecular orbital theory to explain why certain organic species that contain multiple bonds are especially stable or undergo particular reactions when they interact with light. In this chapter, we continue our description of organic compounds by focusing on their molecular structures and reactivity; we will also introduce some of the fundamental types of reactions and reaction mechanisms you will encounter in organic and biological chemistry. We discuss why butter is a solid and oils are liquids despite the apparent similarities in their structures, why the widely used anti-inflammatory drug ibuprofen takes longer than half an hour to relieve pain, and the identity of the major carcinogen in grilled meats and cigarette smoke. The chapter concludes with a brief introduction to the molecules of life, which will explain how the consumption of lactose can result in mental retardation and cirrhosis of the liver in some individuals, how hibernating animals survive during the winter, and how certain groups of antibiotics kill bacteria that are harmful to humans. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.01%3A_Organic_Compounds.txt |
Learning Objectives
• To know the major classes of organic compounds and identify important functional groups.
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure \(1\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
The first family listed in Figure \(1\) is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(2\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Summary
Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Conceptual Problems
1. Can two substances have the same systematic name and be different compounds?
2. Is a carbon–carbon multiple bond considered a functional group? | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.02%3A_Functional_Groups_and_Classes_of_Organic_Compounds.txt |
Learning Objectives
• To learn how organic compounds with the same molecular formula can have different three-dimensional structures.
In earlier discussions of organic compounds, we focused on differences in how the functional groups were connected to the carbon framework. Differences in connectivity resulted in different chemical compounds with different names. You learned, for example, that although 1-propanol (n-propanol) and 2-propanol (isopropanol) have the same molecular formula (C3H8O), they have different physical and chemical properties. Just as with metal complexes, compounds that have the same molecular formula but different arrangements of atoms are called isomers. (For more information on metal complexes, see Section 21.4.) In this section, we describe various types of isomers, beginning with those whose three-dimensional structures differ only as the result of rotation about a C–C bond.
Conformational Isomers
The C–C single bonds in ethane, propane, and other alkanes are formed by the overlap of an sp3 hybrid orbital on one carbon atom with an sp3 hybrid orbital on another carbon atom, forming a σ bond (Figure 23.2.1). Each sp3 hybrid orbital is cylindrically symmetrical (all cross-sections are circles), resulting in a carbon–carbon single bond that is also cylindrically symmetrical about the C–C axis. Because rotation about the carbon–carbon single bond can occur without changing the overlap of the sp3 hybrid orbitals, there is no significant electronic energy barrier to rotation. Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer)Isomers whose three-dimensional structures differ because of rotation about a σ bond..
Note the Pattern
Conformational isomers differ in their three-dimensional structure due to rotation about a σ bond.
The simplest alkane to have a conformational isomer is ethane. Differences between the conformations of ethane are depicted especially clearly in drawings called Newman projections, such as those shown in part (a) in Figure 23.2.2 In a Newman projection, the ethane molecule is viewed along the C–C axis, with the carbon that is in front shown as a vertex and the carbon that is in back shown as a circle. The three hydrogen atoms nearest the viewer are shown bonded to the front carbon, and the three hydrogen atoms farthest from the viewer are shown bonded to the circle. In one extreme, called the eclipsed conformation, the C–H bonds on adjacent carbon atoms lie in the same plane. In the other extreme, called the staggered conformation, the hydrogen atoms are positioned as far from one another as possible. Rotation about the C–C bond produces an infinite number of conformations between these two extremes, but the staggered conformation is the most stable because it minimizes electrostatic repulsion between the hydrogen atoms on adjacent carbons.
In a Newman projection, the angles between adjacent C–H bonds on the same carbon are drawn at 120°, although H–C–H angles in alkanes are actually tetrahedral angles of 109.5°, which for chains of more than three carbon atoms results in a kinked structure. (For more information on bond angles and molecular modeling, see Section 6.3) Despite this three-dimensional inaccuracy, Newman projections are useful for predicting the relative stability of conformational isomers. As shown in part (b) in Figure 23.2.2, the higher energy of the eclipsed conformation represents an energy barrier of 12.6 kJ/mol that must be overcome for rotation about the C–C bond to occur. This barrier is so low, however, that rotation about the C–C bond in ethane is very fast at room temperature and occurs several million times per second for each molecule.
Longer-chain alkanes can also be represented by Newman projections. In more complex alkanes and alkane derivatives, rotation can occur about each C–C bond in a molecule. Newman projections are therefore useful for revealing steric barriers to rotation at a particular C–C bond due to the presence of bulky substituents. Figure 23.2.3 shows a plot of potential energy versus the angle of rotation about the central C–C bond (between carbon atoms 2 and 3) of n-butane (C4H10). The structure that minimizes electrostatic repulsion is the one in which the methyl groups, corresponding to carbon atoms 1 and 4, are as far apart as possible; that is, the staggered conformation. Notice that because the substituents on C2 and C3 in n-butane are not all the same, energetically nonequivalent eclipsed and staggered conformations are possible; most molecules interconvert rapidly between these conformations by a series of simple rotations.
Example 23.2.1
Draw Newman projections showing the staggered and eclipsed conformations of 1,1,1-trichloroethane (CCl3CH3).
Given: organic molecule
Asked for: staggered and eclipsed conformations
Strategy:
A Identify the C–C bond of interest. Then draw the Newman projection by representing one carbon as a vertex and the other as a circle.
B Draw bonds to each carbon at 120° angles from one another, with one arrangement representing the staggered conformation and the other the eclipsed conformation.
C Complete the Newman projections by attaching the appropriate atoms or substituent groups to the central C atoms in each conformation.
Solution:
A There is only one C–C bond: C1 is connected to three Cl atoms and C2 to three H atoms. We draw C1 as a point and C2 as a circle.
B Now we draw bonds on each carbon at 120° angles from one another to represent the staggered conformation and the eclipsed conformation.
C We then attach the H and Cl atoms to the carbon atoms in each conformation as shown.
Exercise
Draw Newman projections to illustrate the staggered and eclipsed conformations of propane (C3H8) as viewed along the C1–C2 axis.
Answer
Structural Isomers
Unlike conformational isomers, which do not differ in connectivity, structural isomersIsomers that have the same molecular formula but differ in which atoms are bonded to one another. differ in connectivity, as illustrated here for 1-propanol and 2-propanol. (For more information on structural isomers, see Section 21.4.) Although these two alcohols have the same molecular formula (C3H8O), the position of the –OH group differs, which leads to differences in their physical and chemical properties.
In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule. Consider, for example, the following five structures represented by the formula C5H12:
Of these structures, (a) and (d) represent the same compound, as do (b) and (c). No bonds have been broken and reformed; the molecules are simply rotated about a 180° vertical axis. Only three—n-pentane (a) and (d), 2-methylbutane (b) and (c), and 2,2-dimethylpropane (e)—are structural isomers. Because no bonds are broken in going from (a) to (d) or from (b) to (c), these alternative representations are not structural isomers. The three structural isomers—either (a) or (d), either (b) or (c), and (e)—have distinct physical and chemical properties.
Note the Pattern
Structural isomers differ in their connectivity.
Example 23.2.2
Draw all the structural isomers of C6H14.
Given: organic molecule
Asked for: all structural isomers
Strategy:
A Draw the simplest structural isomer, which is often the straight-chain alkane.
B Obtain branched isomers by substituting one hydrogen along the chain with an appropriate group from the chain.
C If possible, substitute more than one hydrogen with appropriate groups to obtain isomers that are more highly branched.
Solution:
A The simplest structural isomer is the straight-chain alkane n-hexane (CH3CH2CH2CH2CH2CH3).
B Removing a methyl group from one end and reattaching it to adjacent carbons while substituting hydrogen in its place give two other structures:
C To obtain yet another structural isomer, move two methyl groups to create a molecule with two branches:
We create one more structural isomer by attaching two methyl groups to the same carbon atom:
Thus there are four structural isomers of C6H14.
Exercise
Draw all the structural isomers of C4H9Cl.
Answer
Stereoisomers
Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomersMolecules that have the same connectivity but whose component atoms have different orientations in space.. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. (For more information on stereoisomers, see Section 21.4) Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement (part (a) in Figure 6.4.4). Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.
Note the Pattern
Stereoisomers have the same connectivity but different arrangements of atoms in space.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in part (a) in Figure 23.2.4, your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in part (b) in Figure 23.2.4
Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in Figure 23.2.5 This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (part (b) in Figure 23.2.5), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”).
Example 23.2.3
Draw the cis and trans isomers of each compound.
1. 1,3-dimethylcyclopentane
2. 3,4-dichloro-3-hexene
Given: organic compounds
Asked for: cis and trans isomers
Strategy:
Draw the unsubstituted compound corresponding to the systematic name given. Then place substituents on the same side to obtain the cis isomer and on opposite sides to obtain the trans isomer.
Solution:
1. The name tells us that this compound contains a five-carbon ring with two methyl groups attached. The 1,3 notation means that the methyl groups are not adjacent in the five-membered ring:
Placing the methyl substituents on the same side of the ring gives the cis isomer, whereas placing them on opposite sides of the ring gives the trans isomer:
2. The compound 3-hexene can exist as a cis or trans isomer:
Replacing the hydrogen atoms on the third and fourth carbons by chlorine does not change the overall structures of the isomers:
Exercise
Draw the cis and trans isomers of each compound.
1. 2-butene
2. 1-methyl-3-chlorocyclopentane
Answer
Example 22.3.4
Which of these compounds exist as at least one pair of enantiomers?
Given: organic compounds
Asked for: existence of enantiomers
Strategy:
Determine whether the compound is chiral. In most cases, this means that at least one carbon is bonded to four different groups. If the compound is chiral, it exists as enantiomers.
Solution:
1. The carbons in –CH3 and –CH2– are each bound to more than one hydrogen, so they are achiral. The central carbon is bound to two identical ethyl groups, so it is also achiral. Because there are no chiral centers in this compound, it has no enantiomers.
2. All carbons are bonded to at least two H atoms except for the one bonded to Br, which is bonded to four different groups: –Br, –CH3, –H, and –CH2CH3. The compound therefore has one chiral center and, as a result, exists as enantiomers: the structure shown and its nonsuperimposable mirror image.
3. The carbons in each –CH3 are bonded to more than one hydrogen, so they are achiral. One carbon is bonded to two –CH3 groups, making it achiral also. This compound has no enantiomers.
4. One carbon is bonded to three hydrogens and one to two oxygens, so they are achiral. The central carbon is bonded to four different groups: –CH3, –H, –NH2, and –CO2H, so it is chiral, and the compound has two enantiomers.
Exercise
Which of these compounds have at least one pair of enantiomers?
Answer
(c)
Optical Activity of Enantiomers
Although enantiomers have identical densities, melting and boiling points, colors, and solubility in most solvents, they differ in their interaction with plane-polarized light, which consists of electromagnetic waves oscillating in a single plane. In contrast, normal (unpolarized) light consists of electromagnetic waves oscillating in all directions perpendicular to the axis of propagation. When normal light is passed through a substance called a polarizer, only light oscillating in one direction is transmitted. A polarizer selectively filters out light that oscillates in any but the desired plane (Figure 23.2.6).
When plane-polarized light is passed through a solution, electromagnetic radiation interacts with the solute and solvent molecules. If the solution contains an achiral compound, the plane-polarized light enters and leaves the solution unchanged because achiral molecules cause it to rotate in random directions. The solute is therefore said to be optically inactive. If the solution contains a single enantiomer of a chiral compound, however, the plane-polarized light is rotated in only one direction, and the solute is said to be optically active. A clockwise rotation is called dextrorotatory (from the Latin dextro, meaning “to the right”) and is indicated in the name of the compound by (+), whereas a counterclockwise rotation is called levorotatory (from the Latin levo, meaning “to the left”) and is designated (−). As you will soon discover, this designation is important in understanding how chiral molecules interact with one another.
Note the Pattern
Chiral molecules are optically active; achiral molecules are not.
The magnitude of the rotation of plane-polarized light is directly proportional to the number of chiral molecules in a solution; it also depends on their molecular structure, the temperature, and the wavelength of the light. Because of these variables, every chiral compound has a specific rotationThe amount (in degrees) by which the plane of polarized light is rotated when the light is passed through a solution that contains 1.0 g of a solute per 1.0 mL of solvent in a tube 10.0 cm long., which is defined as the amount (in degrees) by which the plane of polarized light is rotated when the light is passed through a solution containing 1.0 g of solute per 1.0 mL of solvent in a tube 10.0 cm long. A chiral solution that contains equal concentrations of a pair of enantiomers is called a racemic mixture. In such a solution, the optical rotations exactly cancel one another, so there is no net rotation, and the solution is optically inactive. The categories of stereoisomers are summarized in Figure 23.2.7 .
Interactions of Enantiomers with Other Chiral Molecules
In living organisms, virtually every molecule that contains a chiral center is found as a single enantiomer, not a racemic mixture. At the molecular level, our bodies are chiral and interact differently with the individual enantiomers of a particular compound. For example, the two enantiomers of carvone produce very different responses in humans: (−)-carvone is the substance responsible for the smell of spearment oil, and (+)-carvone—the major flavor component of caraway seeds—is responsible for the characteristic aroma of rye bread.
A pharmaceutical example of a chiral compound is ibuprofen, a common analgesic and anti-inflammatory agent that is the active ingredient in pain relievers such as Motrin and Advil (Figure 23.5.2). The drug is sold as a racemic mixture that takes approximately 38 minutes to achieve its full effect in relieving pain and swelling in an adult human. Because only the (+) enantiomer is active in humans, however, the same mass of medication would relieve symptoms in only about 12 minutes if it consisted of only the (+) enantiomer. Unfortunately, isolating only the (+) enantiomer would substantially increase the cost of the drug. Conversion of the (−) to (+) enantiomer in the human body accounts for the delay in feeling the full effects of the drug. A racemic mixture of another drug, the sedative thalidomide, was sold in Europe from 1956 to the early 1960s. It was prescribed to treat nausea during pregnancy, but unfortunately only the (+) enantiomer was safe for that purpose. The (−) enantiomer was discovered to be a relatively potent teratogen, a substance that causes birth defects, which caused the children of many women who had taken thalidomide to be born with missing or undeveloped limbs. As a result, thalidomide was quickly banned for this use. It is currently used to treat leprosy, however, and it has also shown promise as a treatment for AIDS (acquired immunodeficiency syndrome).
These examples dramatically illustrate the point that the biological activities of enantiomers may be very different. But how can two molecules that differ only by being nonsuperimposable mirror images cause such different responses? The biological effects of many substances—including molecules such as carvone that have a scent and drugs such as ibuprofen and thalidomide—depend on their interaction with chiral sites on specific receptor proteins. As schematically illustrated in Figure 23.2.8, only one enantiomer of a chiral substance interacts with a particular receptor, thereby initiating a response. The other enantiomer may not bind at all, or it may bind to another receptor, producing a different response.
Summary
Isomers are different compounds that have the same molecular formula. For an organic compound, rotation about a σ bond can produce different three-dimensional structures called conformational isomers (or conformers). In a Newman projection, which represents the view along a C–C axis, the eclipsed conformation has the C–H bonds on adjacent carbon atoms parallel to each other and in the same plane, representing one conformational extreme. In the staggered conformation, the opposite extreme, the hydrogen atoms are as far from one another as possible. Electrostatic repulsions are minimized in the staggered conformation. Structural isomers differ in the connectivity of the atoms. Structures that have the same connectivity but whose components differ in their orientations in space are called stereoisomers. Stereoisomers can be geometric isomers, which differ in the placement of substituents in a rigid molecule, or optical isomers, nonsuperimposable mirror images. Molecules that are nonsuperimposable mirror images are chiral molecules. A molecule and its nonsuperimposable mirror image are called enantiomers. These differ in their interaction with plane-polarized light, light that oscillates in only one direction. A compound is optically active if its solution rotates plane-polarized light in only one direction and optically inactive if its rotations cancel to produce no net rotation. A clockwise rotation is called dextrorotatory and is indicated in the compound’s name by (+), whereas a counterclockwise rotation is called levorotatory, designated by (−). The specific rotation is the amount (in degrees) by which the plane of polarized light is rotated when light is passed through a solution containing 1.0 g of solute per 1.0 mL of solvent in a tube 10.0 cm long. A solution that contains equal concentrations of each enantiomer in a pair is a racemic mixture; such solutions are optically inactive.
Key Takeaways
• Isomers can be conformational or structural.
• Stereoisomers have the same connectivity but can be optical or geometric isomers.
Conceptual Problems
1. What hybrid orbitals are used to form C–C bonds in saturated hydrocarbons? Describe the bond.
2. How are conformational isomers related? Sketch two conformational isomers of propane, looking along the C1–C2 axis.
3. Why do alkanes with more than two carbons have a kinked structure? Explain why a kinked structure is so stable.
4. Are n-pentane and 2-methylbutane conformational isomers or structural isomers? How would you separate these compounds from a mixture of the two?
5. How are structural isomers different from stereoisomers? Do stereoisomers have free rotation about all carbon–carbon bonds? Explain your answers.
6. Which of these objects is chiral?
1. a shoe
2. a laced football
3. an automobile
4. a fork
7. Which of these objects is chiral?
1. a rollerblade
2. an unmarked baseball bat
3. a bicycle
4. your arms
5. a spoon
8. Are all stereoisomers also enantiomers? Are all enantiomers stereoisomers? Explain your answers.
Answers
1. sp3; it is a σ bond that is cylindrically symmetrical (all cross sections perpendicular to the internuclear axis are circles).
2. The sp3 hybridized orbitals form bonds at tetrahedral angles (109.5°), which forces the carbon atoms to form a zigzag chain.
3. (a), (c), and (d)
Structure and Reactivity
1. Single bonds between carbon atoms are free to rotate 360°.
1. Explain what happens to the potential energy of an n-hexane molecule as rotation occurs around the C2–C3 bond.
2. Draw Newman projections of the n-hexane conformations corresponding to the energy minima and maxima in the diagram, which shows potential energy versus degrees of rotation about the C3–C4 axis.
2. Sketch all the structural isomers of each compound.
1. C6H13Br
2. C3H6Cl2
3. Draw all the possible structural isomers of each compound.
1. C5H11Br
2. C4H8Cl2
4. Sketch all the isomers of each compound. Identify the cis- and trans-isomers.
1. monochlorobutene
2. bromochloropropene
5. Which molecules are chiral? On the structural formulas of the chiral molecules, identify any chiral centers with an asterisk.
1. 2,2-dimethylpentane
2. 1,2-dimethylpentane
3. 2,4-dimethylpentane
4. 1-chloro-2-methylpentane
6. Which molecules are chiral? On the structural formulas of the chiral molecules, identify any chiral centers with an asterisk.
1. 1,1-dichloropropane
2. 1,2-dichloropropane
3. 1,3-dichloropropane
4. 1,1,2-trichloropropane
7. Draw the structures of the enantiomers of each compound.
1. CH2CHCHClBr
2. 4-chloro-2-hexene
3. methylethylpropylamine
8. Draw the structures of the enantiomers of each compound.
1. 4-methyl-2-hexene
2. methylethylhexylamine
3. CH3CH2CH(CH3)Cl
9. Draw the structures of the enantiomers of each compound.
10. Draw the structures of the enantiomers of each compound.
Answers
1. (b) and (d);
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.03%3A_Isomers_of_Organic_Compounds.txt |
Learning Objectives
• To understand the relationship between structure and reactivity for a series of related organic compounds.
Understanding why organic molecules react as they do requires knowing something about the structure and properties of the transient species that are generated during chemical reactions. Identifying transient intermediates enables chemists to elucidate reaction mechanisms, which often allows them to control the products of a reaction. In designing the synthesis of a molecule, such as a new drug, for example, chemists must be able to understand the mechanisms of intermediate reactions to maximize the yield of the desired product and minimize the occurrence of unwanted reactions. Moreover, by recognizing the common reaction mechanisms of simple organic molecules, we can understand how more complex systems react, including the much larger molecules encountered in biochemistry.
Nearly all chemical reactions, whether organic or inorganic, proceed because atoms or groups of atoms having a positive charge or a partial positive charge interact with atoms or groups of atoms having a negative charge or a partial negative charge. Thus when a bond in a hydrocarbon is cleaved during a reaction, identifying the transient species formed, some of which are charged, allows chemists to determine the mechanism and predict the products of a reaction.
Chemists often find that the reactivity of a molecule is affected by the degree of substitution of a carbon that is bonded to a functional group. These carbons are designated as primary, secondary, or tertiary. A primary carbon is bonded to only one other carbon and a functional group, a secondary carbon is bonded to two other carbons and a functional group, and a tertiary carbon is bonded to three other carbons and a functional group.
Reactive Intermediates
Cleaving a C–H bond can generate either –C+ and H, −C· and H· or −C and H+, all of which are unstable and therefore highly reactive. The most common species formed is –C+, which is called a carbocation (part (a) in Figure 24.3.1). A carbocation has only six valence electrons and is therefore electron deficient. It is an electrophile (from “electron” and the Greek suffix phile, meaning “loving”), which is a species that needs electrons to complete its octet. (Recall that electron-deficient compounds, such as those of the group 13 elements, act as Lewis acids in inorganic reactions.) In general, when a highly electronegative atom, such as Cl, is bonded to a carbocation, it draws electrons away from the carbon and destabilizes the positive charge. In contrast, alkyl groups and other species stabilize the positive charge by increasing electron density at the carbocation. Thus a tertiary carbocation (R3C+) is more stable than a primary carbocation (RCH2+).
The reactivity of a molecule is often affected by the degree of substitution of the carbon bonded to a functional group.
Adding one electron to a carbocation produces a neutral species called a radical. An example is the methyl radical (·CH3), shown in part (b) in Figure 24.3.1. Because the carbon still has less than an octet of electrons, it is electron deficient and also behaves as an electrophile. Like carbocations, radicals can be stabilized by carbon substituents that can donate some electron density to the electron-deficient carbon center. Like carbocations, a tertiary radical (R3C·) is more stable than a primary radical (RCH2·).
Adding an electron to a radical produces a carbanion, which contains a negatively charged carbon with eight valence electrons (part (c) in Figure 24.3.1). The methyl anion (CH3) has a structure that is similar to NH3 with its lone pair of electrons, but it has a much stronger tendency to share its lone pair with another atom or molecule. A carbanion is a nucleophile (from “nucleus” and phile), an electron-rich species that has a pair of electrons available to share with another atom. Carbanions are destabilized by groups that donate electrons, so the relationship between their structure and reactivity is exactly the opposite of carbocations and radicals. That is, a tertiary carbanion (R3C) is less stable than a primary carbanion (RCH2). Carbanions are most commonly encountered in organometallic compounds such as methyllithium (CH3Li) or methylmagnesium chloride (CH3MgCl), where the more electropositive metal ion stabilizes the negative charge on the more electronegative carbon atom.
Electrophiles such as carbocations seek to gain electrons and thus have a strong tendency to react with nucleophiles, which are negatively charged species or substances with lone pairs of electrons. Reacting electrophiles with nucleophiles is a central theme in organic reactions.
Electrophiles react with nucleophiles.
Example \(1\)
Classify each species as an electrophile, a nucleophile, or neither.
1. BF3
2. CH4
3. (CH3)3C+
4. NH2
Given: molecular formulas
Asked for: mode of reactivity
Strategy:
Determine whether the compound is electron deficient, in which case it is an electrophile; electron rich, in which case it is a nucleophile; or neither.
Solution:
1. The BF3 molecule is a neutral compound that contains a group 13 element with three bonds to B. The boron atom has only six valence electrons, so it tends to accept an electron pair. The compound is therefore an electrophile.
2. The CH4 molecule has four bonds to C, which is typical of a neutral group 14 compound. The carbon atom has no lone pairs to share and no tendency to gain electrons, and each hydrogen atom forms one bond that contains two valence electrons. Thus CH4 is neither an electrophile nor a nucleophile.
3. The (CH3)3C+ cation contains a group 14 atom (carbon) with only three bonds. It therefore has only six valence electrons and seeks electrons to complete an octet. Hence (CH3)3C+, a carbocation, is an electrophile.
4. The NH2 anion contains a group 15 element with a lone pair of electrons, two bonds, and a negative charge, giving N a total of eight electrons. With its negative charge, the N atom has two lone pairs of electrons, making it a potent nucleophile.
Exercise \(1\)
Classify each compound as an electrophile, a nucleophile, or neither.
1. C6H5OH
2. AlBr3
3. (CH3)4C
Answer:
1. nucleophile
2. electrophile
3. neither
Summary
Electrophiles have a strong tendency to react with nucleophiles. The reactivity of a molecule is often affected by the degree of substitution of the carbon bonded to a functional group; the carbon is designated as primary, secondary, or tertiary. Identifying the transient species formed in a chemical reaction, some of which are charged, enables chemists to predict the mechanism and products of the reaction. One common transient species is a carbocation, a carbon with six valence electrons that is an electrophile; that is, it needs electrons to complete its octet. A radical is a transient species that is neutral but electron deficient and thus acts as an electrophile. In contrast, a carbanion has eight valence electrons and is negatively charged. It is an electron-rich species that is a nucleophile because it can share a pair of electrons. In chemical reactions, electrophiles react with nucleophiles.
Conceptual Problems
1. Arrange CH2F+, CHCl2+, CH3+, and CHF2+ in order of increasing stability. Explain your reasoning.
2. Arrange CH3CH2+, CHBr2+, CH3+, and CHBrCl+ in order of decreasing stability. Explain your reasoning.
3. Identify the electrophile and the nucleophile in each pair.
1. CH3 and Li+
2. CH3ONa and formaldehyde
3. H+ and propene
4. benzene and Cl
1. Identify the electrophile and the nucleophile in each pair.
1. CH3+ and Br
2. HC≡CNa and pentanal
3. acetone and CN
4. (CH3)2S and CH3I
Answers
1. CHF2+ < CHCl2+ < CH2F+ < CH3+; electronegative substituents destabilize the positive charge. The greater the number of electronegative substituents and the higher their electronegativity, the more unstable the carbocation.
1. CH3, nucleophile; Li+, electrophile
2. CH3O, nucleophile; formaldehyde, electrophile
3. H+, electrophile; propene, nucleophile
4. benzene, electrophile; Cl, nucleophile
Structure and Reactivity
1. Draw Lewis electron structures of the products of carbon–hydrogen cleavage reactions. What is the charge on each species?
2. Identify the electrophile and the nucleophile in each reaction; then complete each chemical equation.
1. CH3+ + Cl
2. CH3CH=CH2 + HBr →
3. (CH3)3N + BCl3 | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.04%3A_Reactivity_of_Organic_Molecules.txt |
Learning Objectives
• To become familiar with the common classes of organic reactions.
Certain patterns are encountered repeatedly in organic reactions, many reflecting the interactions of nucleophiles and electrophiles. In this section, we discuss five common types of organic reactions: substitution reactions, elimination reactions, addition reactions, radical reactions, and oxidation–reduction reactions. You have encountered many of these types of reactions previously, such as the formation of peptides by the elimination of water, the oxidation–reduction reactions that generate voltage in batteries, and chain reactions that involve organic radicals. (For more information on peptide formation, see Section 12.8. For more information on batteries, see Chapter 19 . For more information on radicals, see Section 14.6.) In this section, we expand our discussion to include some of the mechanisms behind these reactions.
Substitution
In a substitution reactionA chemical reaction in which one atom or a group of atoms in a substance is replaced by another atom or a group of atoms from another substance., one atom or a group of atoms in a substance is replaced by another atom or group of atoms from another substance. A typical substitution reaction is reacting the hydroxide ion with methyl chloride:
$CH_{3}Cl + OH^{-} \rightarrow CH_{3}OH + Cl^{-} \tag{23.4.1}$
Methyl chloride has a polar C–Cl bond, with the carbon atom having a partial positive charge. In Equation 24.1, the electronegative Cl atom is replaced by another electronegative species that is a stronger nucleophile, in this case OH. Reactions of this sort are called nucleophilic substitution reactions. For this type of reaction to occur, the nucleophilic reactant must possess a pair of electrons and have a greater affinity for the electropositive carbon atom than the original substituent.
One type of nucleophilic substitution reaction is shown in Equation 24.1. It proceeds by a mechanism in which the lone pair of electrons on the entering nucleophile (OH) attacks the partially positively charged carbon atom of the polar C–Cl bond, causing the C–Cl bond to weaken and break:
Note the Pattern
In nucleophilic substitution reactions, the nucleophile must possess a pair of electrons and have a greater affinity for the electropositive species than the original substituent.
The convention for writing such a mechanism is to draw arrows showing the direction of electron flow—that is, from the electron-rich center (the nucleophile) to the electron-poor center (the electrophile). The intermediate species, enclosed by square brackets, represents a transient arrangement of atoms that is only postulated to exist. If the atom under attack (in this case, the partially positively charged carbon atom) had –CH3 groups bonded to it rather than H atoms, the bulky methyl groups would interfere with the attack by OH, making the reaction sterically hindered. The reaction would then proceed in two discrete steps in a second type of substitution reaction: the C–Cl bond would break, forming the (CH3)3C+ carbocation (the electrophile), which would then react with hydroxide (the nucleophile) in a separate step to give the product, (CH3)3COH.
Mustard gas.
An example of a nucleophilic substitution reaction involves the chemical warfare agent known as mustard gas [(ClCH2CH2)2S], which caused about 400,000 casualties during World War I. Mustard gas is toxic because it contains a chloride that can be displaced by nucleophilic amino groups in proteins, thereby allowing the molecule to irreversibly bond to a protein. Because the other product of the reaction is HCl, mustard gas causes severe burns to mucous membranes in the respiratory tract. If mustard gas reacts with DNA (deoxyribonucleic acid), cross-linking of the DNA strands through sulfur occurs, which results in coding errors, the inhibition of replication, and disruption of other DNA functions. If mustard gas reacts with RNA (ribonucleic acid), protein synthesis is altered (see Section 22.6).
Elimination
Some reactions involve the removal, or “elimination,” of adjacent atoms from a molecule. This results in the formation of a multiple bond and the release of a small molecule, so they are called elimination reactionsA chemical reaction in which adjacent atoms are removed, or “eliminated,” from a molecule, resulting in the formation of a multiple bond and a small molecule.. They have the general form
and are similar to cleavage reactions in inorganic compounds. (For more information on cleavage reactions, see Section 7.5.) A typical example is the conversion of ethyl chloride to ethylene:
$CH_{3}CH_{2}Cl \rightarrow CH_{2}=CH_{2} +HCl \tag{23.4.2}$
Note the Pattern
Elimination reactions are similar to cleavage reactions in inorganic compounds.
Much of the approximately 26 million tons of ethylene produced per year in the United States is used to synthesize plastics, such as polyethylene. In Equation 23.4.2, the A–B molecule eliminated is HCl, whose components are eliminated as H+ from the carbon atom on the left and Cl from the carbon on the right. When an acid is produced, as occurs here, the reaction is generally carried out in the presence of a base (such as NaOH) to neutralize the acid.
Addition
A reaction in which the components of a species A–B are added to adjacent atoms across a carbon–carbon multiple bond is called an addition reactionA chemical reaction in which the components of a species A–B are added to adjacent atoms across a carbon-carbon multiple bond.. An example is the reverse of the reaction shown in Equation 24.2, reacting HCl with ethylene to give ethyl chloride:
$CH_{2}=CH_{2} +HCl \rightarrow CH_{3}CH_{2}Cl \tag{23.4.3}$
Note the Pattern
An addition reaction is the reverse of an elimination reaction.
Although a multiple bond is stronger than a single bond, the π bonds of the multiple bond are weaker than the σ bond. The high electron density located between multiply bonded carbon atoms, however, causes alkenes and alkynes to behave like nucleophiles, where nucleophilic attack occurs from the more weakly bound π electrons. Hence alkenes and alkynes are regarded as functional groups. Nucleophilic attack occurs on the Hδ+ atom of the polar HCl bond, initially producing a species with a carbon that has only three bonds, a carbocation. In a second nucleophilic attack, Cl, the electrophile in Equation 23.4.3 attacks the carbocation:
Alcohols, an important class of organic compounds, are often produced by addition reactions. Initial attack by the π bond of an alkene on a Hδ+ of H3O+ produces a carbocation. The carbocation then undergoes nucleophilic attack by a lone pair of electrons from H2O followed by elimination of H+ to form the alcohol.
Radical Reactions
Many important organic reactions involve radicals, such as the combustion of fuels. Probably the best known is reacting a saturated hydrocarbon, such as ethane, with a halogen, such as Br2. The overall reaction is as follows:
$CH_{3}CH_{3} +Br_{2} \xrightarrow[or h\nu ]{400^{o}C} CH_{3}CH_{2}Br + HBr \tag{23.4.4}$
Radical chain reactions occur in three stages: initiation, propagation, and termination. (For more information on radicals, see Section 14.6.) At high temperature or in the presence of light, the relatively weak Br–Br bond is broken in an initiation step that produces an appreciable number of Br atoms (Br·). During propagation, a bromine atom attacks ethane, producing a radical, which then reacts with another bromine molecule to produce ethyl bromide:
$\begin{matrix} \cancel{Br\cdot } + CH_{3}CH_{3} & \rightarrow \cancel{CH_{3}CH_{2}\cdot } + HBr\ \cancel{CH_{3}CH_{2}\cdot } + Br_{2} & \rightarrow CH_{3}CH_{2}Br + \cancel{Br\cdot } \ -------- & ---------\ Br_{2} + CH_{3}CH_{3} &\rightarrow CH_{3}CH_{2}Br + HBr \end{matrix} \tag{23.4.5}$
The sum of the two propagation steps corresponds to the balanced chemical equation for the overall reaction. There are three possible termination steps: the combination of (1) two bromine atoms, (2) two ethyl radicals, or (3) an ethyl and a bromine radical:
$\begin{matrix} Br\cdot + Br\cdot & \rightarrow Br_{2} \ CH_{3}CH_{2}\cdot + \cdot CH_{2}CH_{3} & \rightarrow CH_{3}CH_{2}CH_{2}CH_{3} \ CH_{3}CH_{2}\cdot + Br\cdot &\rightarrow CH_{3}CH_{2}Br \end{matrix} \tag{23.4.6}$
Because radicals are powerful nucleophiles and hence highly reactive, such reactions are not very selective. For example, the chlorination of n-butane gives a roughly 70:30 mixture of 2-chlorobutane, formed from the more stable radical by reacting a secondary carbon and 1-chlorobutane.
Note the Pattern
Because radicals are highly reactive, radical reactions are usually not very selective.
Oxidation–Reduction Reactions
Oxidation–reduction reactions, which are common in organic chemistry, can often be identified by changes in the number of oxygen atoms at a particular position in the hydrocarbon skeleton or in the number of bonds between carbon and oxygen at that position. An increase in either corresponds to an oxidation, whereas a decrease corresponds to a reduction. Conversely, an increase in the number of hydrogen atoms in a hydrocarbon is often an indication of a reduction. We can illustrate these points by considering how the oxidation state of the carbon atom changes in the series of compounds, which is shown in part (a) in Figure 23.4.3. (For a review of oxidation states and formal changes, see Section 7.5, and Section 8.10). The number of oxygen atoms or the number of bonds to oxygen changes throughout the series. Hence the conversion of methane to formic acid is an oxidation, whereas the conversion of carbon dioxide to methanol is a reduction. Also, the number of hydrogen atoms increases in going from the most oxidized to least oxidized compound. As expected, as the oxidation state of carbon increases, the carbon becomes a more potent electrophile. Thus the carbon of CO2 is a stronger electrophile (i.e., more susceptible to nucleophilic attack) than the carbon of an alkane such as methane.
Similarly, in compounds with a carbon–nitrogen bond, the number of bonds between the C and N atoms increases as the oxidation state of the carbon increases (part (b) in Figure 23.4.3). In a nitrile, which contains the –C≡N group, the carbon has the same oxidation state (+2) as in a carboxylic acid, characterized by the –CO2H group. We therefore expect the carbon of a nitrile to be a rather strong electrophile.
Example 23.4.1
Write an equation to describe each reaction. Identify the electrophile and the nucleophile in each reaction.
1. the nucleophilic substitution reaction of potassium cyanide with 1-chloropropane to give CH3CH2CH2CN (butyronitrile)
2. the electrophilic addition reaction of HBr with cis-2-butene
Given: reactants, products, and reaction mechanism
Asked for: equation and identification of electrophile and nucleophile
Strategy:
Use the mechanisms described to show how the indicated products are formed from the reactants.
Solution:
1. The CN ion of KCN is a potent nucleophile that can displace the chlorine atom of 1-chloropropane, releasing a chloride ion. Substitution results in the formation of a new C–C bond: $CN^{-}+\underset{1-chloropropane}{CH_{3}CH_{2}CH_{2}Cl}\rightarrow \underset{butylnitrile}{CH_{3}CH_{2}CH_{2}CN} + Cl^{-}$
The carbon bonded to chlorine is an electrophile because of the highly polar C–Cl bond.
2. In the electrophilic addition of a hydrogen halide to an alkene, the reaction is as follows:
The first step is nucleophilic attack of the π electrons of the double bond on the electrophilic hydrogen of the polar H–Br bond to generate the transient carbocation, followed by nucleophilic attack by the halide to give the product. Thus the alkene is the nucleophile, and the proton of the acid is the electrophile.
Exercise
Write an equation to describe each reaction. In each reaction, identify the electrophile and nucleophile.
1. the nucleophilic substitution reaction of sodium methoxide (NaOCH3) with benzyl bromide (C6H5CH2Br)
2. the acid-catalyzed electrophilic addition reaction of water with cyclopentene
Answer
1. OCH3(methoxide ion) + C6H5CH2Br (benzyl bromide) → C6H5CH2OCH3 (benzylmethyl ether) + Br ; OCH3 is the nucleophile, and C6H5CH2Br is the electrophile.
2. Cyclopentene is the nucleophile, and H3O+ is the electrophile.
Summary
There are common patterns to how organic reactions occur. In a substitution reaction, one atom or a group of atoms in a substance is replaced by another atom or a group of atoms from another substance. Bulky groups that prevent attack cause the reaction to be sterically hindered. In an elimination reaction, adjacent atoms are removed with subsequent formation of a multiple bond and a small molecule. An addition reaction is the reverse of an elimination reaction. Radical reactions are not very selective and occur in three stages: initiation, propagation, and termination. Oxidation–reduction reactions in organic chemistry are identified by the change in the number of oxygens in the hydrocarbon skeleton or the number of bonds between carbon and oxygen or carbon and nitrogen.
Key Takeaway
• The common classes of organic reactions—substitution, elimination, addition, oxidation-reduction, and radical—all involve reacting electrophiles with nucleophiles.
Conceptual Problems
1. Identify the nucleophile and the electrophile in the nucleophilic substitution reaction of 2-bromobutane with KCN.
2. Identify the nucleophile and the electrophile in the nucleophilic substitution reaction of 1-chloropentane with sodium methoxide.
3. Do you expect an elimination reaction to be favored by a strong or a weak base? Why?
4. Why do molecules with π bonds behave as nucleophiles when mixed with strong electrophiles?
Answer
1. CN is the nucleophile, and C2H5Cδ+HBrCH3 is the electrophile.
Structure and Reactivity
1. Sketch the mechanism for the nucleophilic substitution reaction of potassium cyanide with iodoethane.
2. Sketch the mechanism for the nucleophilic substitution reaction of NaSH with 1-bromopropane.
3. Sketch the mechanism for the elimination reaction of cyclohexylchloride with potassium ethoxide. Identify the electrophile and the nucleophile in this reaction.
4. What is the product of the elimination reaction of 1-bromo-2-methylpropane with sodium ethoxide?
5. Write the structure of the product expected from the electrophilic addition of HBr to cis-3-hexene.
6. Write the structure of the product expected from the electrophilic addition of 1-methylcyclopentene to HBr. Identify the electrophile and the nucleophile, and then write a mechanism for this reaction.
7. Write a synthetic scheme for making propene from propane. After synthesizing propene, how would you make 2-bromopropane?
8. Write a synthetic scheme for making ethylene from ethane. After synthesizing ethylene, how would you make iodoethane?
9. From the high-temperature reaction of Br2 with 3-methylpentane, how many monobrominated isomers would you expect to be produced? Which isomer is produced from the most stable radical?
10. For the photochemical reaction of Cl2 with 2,4-dimethylpentane, how many different monochlorinated isomers would you expect to be produced? Which isomer is produced from the most stable precursor radical?
11. How many different radicals can be formed from the photochemical reaction of Cl2 with 3,3,4-trimethylhexane?
12. How many monobrominated isomers would you expect from the photochemical reaction of Br2 with
1. isobutene?
2. 2,2,3-trimethylpentane?
13. Arrange acetone, ethane, carbon dioxide, acetaldehyde, and ethanol in order of increasing oxidation state of carbon.
14. What product(s) do you expect from the reduction of a ketone? the oxidation of an aldehyde?
15. What product(s) do you expect from the reduction of formaldehyde? the oxidation of ethanol?
Answers
1. four; 3-bromo-3-methylpentane
2. seven
3. methanol; acetaldehyde, followed by acetic acid and finally CO2
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.05%3A_Common_Classes_of_Organic_Reactions.txt |
Learning Objectives
• To understand the general properties of functional groups and the differences in their reactivity.
The general properties and reactivity of each class of organic compounds is largely determined by its functional groups. In this section, we describe the relationships between structure, physical properties, and reactivity for the major classes of organic compounds. We also show you how to apply these relationships to understand some common reactions that chemists use to synthesize organic compounds.
Alkanes, Alkenes, and Alkynes
The boiling points of alkanes increase smoothly with increasing molecular mass. They are similar to those of the corresponding alkenes and alkynes because of similarities in molecular mass between analogous structures (Table $1$). In contrast, the melting points of alkanes, alkenes, and alkynes with similar molecular masses show a much wider variation because the melting point strongly depends on how the molecules stack in the solid state. It is therefore sensitive to relatively small differences in structure, such as the location of a double bond and whether the molecule is cis or trans.
Table $1$: Boiling Points (in °C) of Alkanes, Alkenes, and Alkynes of Comparable Molecular Mass
Length of Carbon Chain
Class Two C Atoms Three C Atoms Four C Atoms
alkane −88.6 −42.1 −0.5
alkene −103.8 −47.7 −6.3
alkyne −84.7 −23.2 8.1
Because alkanes contain only C–C and C–H bonds, which are strong and not very polar (the electronegativities of C and H are similar), they are not easily attacked by nucleophiles or electrophiles. Consequently, their reactivity is limited, and often their reactions occur only under extreme conditions. For example, catalytic cracking can be used to convert straight-chain alkanes to highly branched alkanes, which are better fuels for internal combustion engines. Catalytic cracking is one example of a pyrolysis reaction (from the Greek pyros, meaning “fire,” and lysis, meaning “loosening”), in which alkanes are heated to a sufficiently high temperature to induce cleavage of the weakest bonds: the C–C single bonds. The result is a mixture of radicals derived from essentially random cleavage of the various C–C bonds in the chain. Pyrolysis of n-pentane, for example, is nonspecific and can produce these four radicals:
$\mathrm{2CH_3CH_2CH_2CH_2CH_3\xrightarrow{\Delta}CH_3\cdot+ CH_2CH_2CH_2\cdot+ CH_3CH_2\cdot+ CH_3CH_2CH_2\cdot} \tag{24.5.1}$
Recombination of these radicals (a termination step) can produce ethane, propane, butane, n-pentane, n-hexane, n-heptane, and n-octane. Radicals that are formed in the middle of a chain by cleaving a C–H bond tend to produce branched hydrocarbons. In catalytic cracking, lighter alkanes are removed from the mixture by distillation.
Radicals are also produced during the combustion of alkanes, with CO2 and H2O as the final products. Radicals are stabilized by the presence of multiple carbon substituents that can donate electron density to the electron-deficient carbon. The chemical explanation of octane ratings rests partly on the stability of radicals produced from the different hydrocarbon fuels. Recall that n-heptane, which does not burn smoothly, has an octane rating of 0, and 2,2,4-trimethylpentane (“isooctane”), which burns quite smoothly, has a rating of 100. Isooctane has a branched structure and is capable of forming tertiary radicals that are comparatively stable.
In contrast, the radicals formed during the combustion of n-heptane, whether primary or secondary, are less stable and hence more reactive, which partly explains why burning n-heptane causes premature ignition and engine knocking.
In Section 24.2, we explained that rotation about the carbon–carbon multiple bonds of alkenes and alkynes cannot occur without breaking a π bond, which therefore constitutes a large energy barrier to rotation (Figure $1$). Consequently, the cis and trans isomers of alkenes generally behave as distinct compounds with different chemical and physical properties. A four-carbon alkene has four possible isomeric forms: three structural isomers, which differ in their connectivity, plus a pair of geometric isomers from one structural isomer (2-butene). These two geometric isomers are cis-2-butene and trans-2-butene. The four isomers have significantly different physical properties.
Figure $1$: Carbon–Carbon Bonding in Alkenes and Interconversion of Cis and Trans Isomers
In butane, there is only a small energy barrier to rotation about the C2–C3 σ bond. In the formation of cis- or trans-2-butene from butane, the p orbitals on C2 and C3 overlap to form a π bond. To convert cis-2-butene to trans-2-butene or vice versa through rotation about the double bond, the π bond must be broken. Because this interconversion is energetically unfavorable, cis and trans isomers are distinct compounds that generally have different physical and chemical properties.
Alkynes in which the triple bond is located at one end of a carbon chain are called terminal alkynes and contain a hydrogen atom attached directly to a triply bonded carbon: R–C≡C–H. Terminal alkynes are unusual in that the hydrogen atom can be removed relatively easily as H+, forming an acetylide ion (R–C≡C). Acetylide ions are potent nucleophiles that are especially useful reactants for making longer carbon chains by a nucleophilic substitution reaction. As in earlier examples of such reactions, the nucleophile attacks the partially positively charged atom in a polar bond, which in the following reaction is the carbon of the Br–C bond:
Alkenes and alkynes are most often prepared by elimination reactions. A typical example is the preparation of 2-methyl-1-propene, whose derivative, 3-chloro-2-methyl-1-propene, is used as a fumigant and insecticide. The parent compound can be prepared from either 2-hydroxy-2-methylpropane or 2-bromo-2-methylpropane:
The reaction on the left proceeds by eliminating the elements of water (H+ plus OH), so it is a dehydration reaction. If an alkane contains two properly located functional groups, such as –OH or –X, both of them may be removed as H2O or HX with the formation of a carbon–carbon triple bond:
Alkenes and alkynes are most often prepared by elimination reactions.
Arenes
Most arenes that contain a single six-membered ring are volatile liquids, such as benzene and the xylenes, although some arenes with substituents on the ring are solids at room temperature. In the gas phase, the dipole moment of benzene is zero, but the presence of electronegative or electropositive substituents can result in a net dipole moment that increases intermolecular attractive forces and raises the melting and boiling points. For example, 1,4-dichlorobenzene, a compound used as an alternative to naphthalene in the production of mothballs, has a melting point of 52.7°C, which is considerably greater than the melting point of benzene (5.5°C).
Certain aromatic hydrocarbons, such as benzene and benz[a]pyrene, are potent liver toxins and carcinogens. In 1775, a British physician, Percival Pott, described the high incidence of cancer of the scrotum among small boys used as chimney sweeps and attributed it to their exposure to soot. His conclusions were correct: benz[a]pyrene, a component of chimney soot, charcoal-grilled meats, and cigarette smoke, was the first chemical carcinogen to be identified.
Although arenes are usually drawn with three C=C bonds, benzene is about 150 kJ/mol more stable than would be expected if it contained three double bonds. This increased stability is due to the delocalization of the π electron density over all the atoms of the ring. Compared with alkenes, arenes are poor nucleophiles. Consequently, they do not undergo addition reactions like alkenes; instead, they undergo a variety of electrophilic aromatic substitution reactions that involve the replacement of –H on the arene by a group –E, such as –NO2, –SO3H, a halogen, or an alkyl group, in a two-step process. The first step involves addition of the electrophile (E) to the π system of benzene, forming a carbocation. In the second step, a proton is lost from the adjacent carbon on the ring:
The carbocation formed in the first step is stabilized by resonance.
Arenes undergo substitution reactions rather than elimination because of increased stability arising from delocalization of their π electron density.
Many substituted arenes have potent biological activity. Some examples include common drugs and antibiotics such as aspirin and ibuprofen, illicit drugs such as amphetamines and peyote, the amino acid phenylalanine, and hormones such as adrenaline (Figure $2$).
Aspirin (antifever activity), ibuprofen (antifever and anti-inflammatory activity), and amphetamine (stimulant) have pharmacological effects. Phenylalanine is an amino acid. Adrenaline is a hormone that elicits the “fight or flight” response to stress. Chiral centers are indicated with an asterisk.
Alcohols and Ethers
Both alcohols and ethers can be thought of as derivatives of water in which at least one hydrogen atom has been replaced by an organic group, as shown here. Because of the electronegative oxygen atom, the individual O–H bond dipoles in alcohols cannot cancel one another, resulting in a substantial dipole moment that allows alcohols to form hydrogen bonds.
Alcohols therefore have significantly higher boiling points than alkanes or alkenes of comparable molecular mass, whereas ethers, without a polar O–H bond, have intermediate boiling points due to the presence of a small dipole moment (Table $2$). The larger the alkyl group in the molecule, however, the more “alkane-like” the alcohol is in its properties. Because of their polar nature, alcohols and ethers tend to be good solvents for a wide range of organic compounds.
Table $2$: Boiling Points of Alkanes, Ethers, and Alcohols of Comparable Molecular Mass
Name Formula Molecular Mass (amu) Boiling Point (°C)
alkane propane C3H8 44 −42.1
n-pentane C5H12 72 36.1
n-heptane C7H16 100 98.4
ether dimethylether (CH3)2O 46 −24.5.1
diethylether (CH3CH2)2O 74 34.5
di-n-propylether (CH3CH2CH2)2O 102 90.1
alcohol ethanol CH3CH2OH 46 78.3
n-butanol CH3(CH2)3OH 74 117.7
n-hexanol CH3(CH2)5OH 102 157.6
Alcohols are usually prepared by adding water across a carbon–carbon double bond or by a nucleophilic substitution reaction of an alkyl halide using hydroxide, a potent nucleophile (Figure $1$). Alcohols can also be prepared by reducing compounds that contain the carbonyl functional group (C=O; part (a) in Figure 24.5.7). Alcohols are classified as primary, secondary, or tertiary, depending on whether the –OH group is bonded to a primary, secondary, or tertiary carbon. For example, the compound 5-methyl-3-hexanol is a secondary alcohol.
Ethers, especially those with two different alkyl groups (ROR′), can be prepared by a substitution reaction in which a nucleophilic alkoxide ion (RO) attacks the partially positively charged carbon atom of the polar C–X bond of an alkyl halide (R′X):
Although both alcohols and phenols have an –OH functional group, phenols are 106–108 more acidic than alcohols. This is largely because simple alcohols have the –OH unit attached to an sp3 hybridized carbon, whereas phenols have an sp2 hybridized carbon atom bonded to the oxygen atom. The negative charge of the phenoxide ion can therefore interact with the π electrons in the ring, thereby delocalizing and stabilizing the negative charge through resonance. In contrast, the negative charge on an alkoxide ion cannot be stabilized by these types of interactions.
Alcohols undergo two major types of reactions: those involving cleavage of the O–H bond and those involving cleavage of the C–O bond. Cleavage of an O–H bond is a reaction characteristic of an acid, but alcohols are even weaker acids than water. The acidic strength of phenols, however, is about a million times greater than that of ethanol, making the pKa of phenol comparable to that of the NH4+ ion (9.89 versus 9.25, respectively):
$C_6H_5OH + H_2O \rightleftharpoons H_3O^+ + C_6H_5O^− \tag{24.5.1}$
Alcohols undergo two major types of reactions: cleavage of the O–H bond and cleavage of the C–O bond.
Cleavage of the C–O bond in alcohols occurs under acidic conditions. The –OH is first protonated, and nucleophilic substitution follows:
In the absence of a nucleophile, however, elimination can occur, producing an alkene (Figure 24.5.6). Ethers lack the –OH unit that is central to the reactivity of alcohols, so they are comparatively unreactive. Their low reactivity makes them highly suitable as solvents for carrying out organic reactions.
Aldehydes and Ketones
Aromatic aldehydes, which have intense and characteristic flavors and aromas, are the major components of such well-known flavorings as vanilla and cinnamon (Figure 24.5.3). Many ketones, such as camphor and jasmine, also have intense aromas. Ketones are found in many of the hormones responsible for sex differentiation in humans, such as progesterone and testosterone.
In compounds containing a carbonyl group, nucleophilic attack can occur at the carbon atom of the carbonyl, whereas electrophilic attack occurs at oxygen.
Aldehydes and ketones contain the carbonyl functional group, which has an appreciable dipole moment because of the polar C=O bond. The presence of the carbonyl group results in strong intermolecular interactions that cause aldehydes and ketones to have higher boiling points than alkanes or alkenes of comparable molecular mass (Table 24.5.3). As the mass of the molecule increases, the carbonyl group becomes less important to the overall properties of the compound, and the boiling points approach those of the corresponding alkanes.
Table $2$: Boiling Points of Alkanes, Aldehydes, and Ketones of Comparable Molecular Mass
Name Formula Molecular Mass (amu) Boiling Point (°C)
alkane n-butane C4H10 58 −0.5
n-pentane C5H12 72 36.1
aldehyde propionaldehyde (propanal) C3H6O 58 48.0
butyraldehyde (butanal) C4H8O 72 74.8
ketone acetone (2-propanone) C3H6O 58 56.1
methyl ethyl ketone (2-butanone) C4H8O 72 79.6
Aldehydes and ketones are typically prepared by oxidizing alcohols (part (a) in Figure 24.5.7). In their reactions, the partially positively charged carbon atom of the carbonyl group is an electrophile that is subject to nucleophilic attack. Conversely, the lone pairs of electrons on the oxygen atom of the carbonyl group allow electrophilic attack to occur. Aldehydes and ketones can therefore undergo both nucleophilic attack (at the carbon atom) and electrophilic attack (at the oxygen atom).
Nucleophilic attack occurs at the partially positively charged carbon of a carbonyl functional group. Electrophilic attack occurs at the lone pairs of electrons on the oxygen atom.
Aldehydes and ketones react with many organometallic compounds that contain stabilized carbanions. One of the most important classes of such compounds are the Grignard reagents, organomagnesium compounds with the formula RMgX (X is Cl, Br, or I) that are so strongly polarized that they can be viewed as containing R and MgX+. These reagents are named for the French chemist Victor Grignard (1871–1935), who won a Nobel Prize in Chemistry in 1912 for their development. In a Grignard reaction, the carbonyl functional group is converted to an alcohol, and the carbon chain of the carbonyl compound is lengthened by the addition of the R group from the Grignard reagent. One example is reacting cyclohexylmagnesium chloride, a Grignard reagent, with formaldehyde:
The nucleophilic carbanion of the cyclohexyl ring attacks the electrophilic carbon atom of the carbonyl group. Acidifying the solution results in protonation of the intermediate to give the alcohol. Aldehydes can also be prepared by reducing a carboxylic acid group (–CO2H) (part (a) in Figure 24.5.7), and ketones can be prepared by reacting a carboxylic acid derivative with a Grignard reagent. The former reaction requires a powerful reducing agent, such as a metal hydride.
$1$
Explain how each reaction proceeds to form the indicated product.
Given: chemical reaction
Asked for: how products are formed
Strategy:
1. Identify the functional group and classify the reaction.
2. Use the mechanisms described to propose the initial steps in the reaction.
Solution:
1. A One reactant is an alcohol that undergoes a substitution reaction.
B In the product, a bromide group is substituted for a hydroxyl group. The first step in this reaction must therefore be protonation of the –OH group of the alcohol by H+ of HBr, followed by the elimination of water to give the carbocation:
The bromide ion is a good nucleophile that can react with the carbocation to give an alkyl bromide:
1. A One reactant is a Grignard reagent, and the other contains a carbonyl functional group. Carbonyl compounds act as electrophiles, undergoing nucleophilic attack at the carbonyl carbon.
B The nucleophile is the phenyl carbanion of the Grignard reagent:
The product is benzyl alcohol.
Exercise $1$
Predict the product of each reaction.
Answer:
Carboxylic Acids
The pungent odors of many carboxylic acids are responsible for the smells we associate with sources as diverse as Swiss cheese, rancid butter, manure, goats, and sour milk. The boiling points of carboxylic acids tend to be somewhat higher than would be expected from their molecular masses because of strong hydrogen-bonding interactions between molecules. In fact, most simple carboxylic acids form dimers in the liquid and even in the vapor phase. The four lightest carboxylic acids are completely miscible with water, but as the alkyl chain lengthens, they become more “alkane-like,” so their solubility in water decreases.
Compounds that contain the carboxyl functional group are acidic because carboxylic acids can easily lose a proton: the negative charge in the carboxylate ion (RCO2) is stabilized by delocalization of the π electrons:
As a result, carboxylic acids are about 1010 times more acidic than the corresponding simple alcohols whose anions (RO) are not stabilized through resonance.
Carboxylic acids are typically prepared by oxidizing the corresponding alcohols and aldehydes (part (a) in Figure 24.5.7). They can also be prepared by reacting a Grignard reagent with CO2, followed by acidification:
$\mathrm{CO_2+ RMgCl \xrightarrow{H_3O^+} RCO_2H + Mg^{2+}+ Cl^-+ H_2O} \tag{24.5.2}$
The initial step in the reaction is nucleophilic attack by the R group of the Grignard reagent on the electrophilic carbon of CO2:
Delocalization of π bonding over three atoms (O–C–O) makes carboxylic acids and their derivatives less susceptible to nucleophilic attack than aldehydes and ketones with their single π bond. The reactions of carboxylic acids are dominated by two factors: their polar –CO2H group and their acidity. Reaction with strong bases, for example, produce carboxylate salts, such as sodium stearate:
$RCO_2H + NaOH \rightarrow RCO_2^−Na^+ + H_2O \tag{24.5.3}$
where R is CH3(CH2)16. As you learned in previously long-chain carboxylate salts are used as soaps.
Delocalization of π bonding over three atoms makes carboxylic acids and their derivatives less susceptible to nucleophilic attack as compared with aldehydes and ketones.
Carboxylic Acid Derivatives
Replacing the –OH of a carboxylic acid with groups that have different tendencies to participate in resonance with the C=O functional group produces derivatives with rather different properties. Resonance structures have significant effects on the reactivity of carboxylic acid derivatives, but their influence varies substantially, being least important for halides and most important for the nitrogen of amides. In this section, we take a brief look at the chemistry of two of the most familiar and important carboxylic acid derivatives: esters and amides.
Esters
Esters have the general formula RCO2R′, where R and R′ can be virtually any alkyl or aryl group. Esters are often prepared by reacting an alcohol (R′OH) with a carboxylic acid (RCO2H) in the presence of a catalytic amount of strong acid. The purpose of the acid (an electrophile) is to protonate the doubly bonded oxygen atom of the carboxylic acid (a nucleophile) to give a species that is more electrophilic than the parent carboxylic acid.
The nucleophilic oxygen atom of the alcohol attacks the electrophilic carbon atom of the protonated carboxylic acid to form a new C–O bond. The overall reaction can be written as follows:
Because water is eliminated, this is a dehydration reaction. If an aqueous solution of an ester and strong acid or base is heated, the reverse reaction will occur, producing the parent alcohol R′OH and either the carboxylic acid RCO2H (under strongly acidic conditions) or the carboxylate anion RCO2 (under basic conditions).
As stated earlier, esters are familiar to most of us as fragrances, such as banana and pineapple. Other esters with intense aromas function as sex attractants, or pheromones, such as the pheromone from the oriental fruit fly. Research on using synthetic insect pheromones as a safer alternative to insecticides for controlling insect populations, such as cockroaches, is a rapidly growing field in organic chemistry.
Amides
In the general structure of an amide,
the two substituents on the amide nitrogen can be hydrogen atoms, alkyl groups, aryl groups, or any combination of those species. Although amides appear to be derived from an acid and an amine, in practice they usually cannot be prepared by this synthetic route. In principle, nucleophilic attack by the lone electron pair of the amine on the carbon of the carboxylic acid could occur, but because carboxylic acids are weak acids and amines are weak bases, an acid–base reaction generally occurs instead:
$RCO_2H + R′NH_2 \rightarrow RCO_2^− + R′NH_3^+ \tag{24.5.4}$
Amides are therefore usually prepared by the nucleophilic reaction of amines with more electrophilic carboxylic acid derivatives, such as esters.
The lone pair of electrons on the nitrogen atom of an amide can participate in π bonding with the carbonyl group, thus reducing the reactivity of the amide (Figure 24.5.5) and inhibiting free rotation about the C–N bond. Amides are therefore the least reactive of the carboxylic acid derivatives. The stability of the amide bond is crucially important in biology because amide bonds form the backbones of peptides and proteins. The amide bond is also found in many other biologically active and commercially important molecules, including penicillin; urea, which is used as fertilizer; saccharin, a sugar substitute; and valium, a potent tranquilizer.
Amides are the least reactive of the carboxylic acid derivatives because amides participate in π bonding with the carbonyl group.
Amines
Amines are derivatives of ammonia in which one or more hydrogen atoms have been replaced by alkyl or aryl groups. They are therefore analogous to alcohols and ethers. Like alcohols, amines are classified as primary, secondary, or tertiary, but in this case the designation refers to the number of alkyl groups bonded to the nitrogen atom, not to the number of adjacent carbon atoms. In primary amines, the nitrogen is bonded to two hydrogen atoms and one alkyl group; in secondary amines, the nitrogen is bonded to one hydrogen and two alkyl groups; and in tertiary amines, the nitrogen is bonded to three alkyl groups. With one lone pair of electrons and C–N bonds that are less polar than C–O bonds, ammonia and simple amines have much lower boiling points than water or alcohols with similar molecular masses. Primary amines tend to have boiling points intermediate between those of the corresponding alcohol and alkane. Moreover, secondary and tertiary amines have lower boiling points than primary amines of comparable molecular mass.
Tertiary amines form cations analogous to the ammonium ion (NH4+), in which all four H atoms are replaced by alkyl groups. Such substances, called quaternary ammonium salts, can be chiral if all four substituents are different. (Amines with three different substituents are also chiral because the lone pair of electrons represents a fourth substituent.)
Alkylamines can be prepared by nucleophilic substitution reactions of alkyl halides with ammonia or other amines:
$RCl + NH_3 \rightarrow RNH_2 + HCl \tag{24.5.5}$
$RCl + R′NH_2 \rightarrow RR′NH + HCl \tag{24.5.6}$
$RCl + R′R″NH \rightarrow RR′R″N + HCl \tag{24.5.7}$
The primary amine formed in the first reaction (Equation 24.5.5) can react with more alkyl halide to generate a secondary amine (Equation 24.5.6), which in turn can react to form a tertiary amine (Equation 24.5.7). Consequently, the actual reaction mixture contains primary, secondary, and tertiary amines and even quaternary ammonium salts.
The reactions of amines are dominated by two properties: their ability to act as weak bases and their tendency to act as nucleophiles, both of which are due to the presence of the lone pair of electrons on the nitrogen atom. Amines typically behave as bases by accepting a proton from an acid to form an ammonium salt, as in the reaction of triethylamine (the ethyl group is represented as Et) with aqueous HCl (the lone pair of electrons on nitrogen is shown):
$Et_3N:(l) + HCl{(aq)} \rightarrow Et_3NH^+Cl^−_{(aq)} \tag{24.5.8}$
which gives triethylammonium chloride. Amines can react with virtually any electrophile, including the carbonyl carbon of an aldehyde, a ketone, or an ester. Aryl amines such as aniline (C6H5NH2) are much weaker bases than alkylamines because the lone pair of electrons on nitrogen interacts with the π bonds of the aromatic ring, delocalizing the lone pair through resonance (Figure 24.5.6).
Note
The reactions of amines are dominated by their ability to act as weak bases and their tendency to act as nucleophiles.
Delocalization of the lone electron pair on N over the benzene ring reduces the basicity of aryl amines, such as aniline, compared with that of alkylamines, such as cyclohexylamine. These electrostatic potential maps show that the electron density on the N of cyclohexylamine is more localized than it is in aniline, which makes cyclohexylamine a stronger base.
Example $2$
Predict the products formed in each reaction and show the initial site of attack and, for part (b), the final products.
1. C6H5CH2CO2H + KOH →
Given: reactants
Asked for: products and mechanism of reaction
Strategy:
Use the strategy outlined in Example 7.
Solution:
1. The proton on the carboxylic acid functional group is acidic. Thus reacting a carboxylic acid with a strong base is an acid–base reaction, whose products are a salt—in this case, C6H5CH2CO2K+—and water.
2. The nitrogen of cyclohexylamine contains a lone pair of electrons, making it an excellent nucleophile, whereas the carbonyl carbon of ethyl acetate is a good electrophile. We therefore expect a reaction in which nucleophilic attack on the carbonyl carbon of the ester produces an amide and ethanol. The initial site of attack and the reaction products are as follows:
Exercise $2$
Predict the products of each reaction. State the initial site of attack.
1. acetic acid with 1-propanol
2. aniline (C6H5NH2) with propyl acetate [CH3C(=O)OCH2CH2CH3]
Answer:
1. Initial attack occurs with protonation of the oxygen of the carbonyl. The products are:
1. Initial attack occurs at the carbon of the carbonyl group. The products are:
Reactions like we have discussed in this section are used to synthesize a wide range of organic compounds. When chemists plan the synthesis of an organic molecule, however, they must take into consideration various factors, such as the availability and cost of reactants, the need to minimize the formation of undesired products, and the proper sequencing of reactions to maximize the yield of the target molecule and minimize the formation of undesired products. Because the synthesis of many organic molecules requires multiple steps, in designing a synthetic scheme for such molecules, chemists must often work backward from the desired product in a process called retrosynthesis. Using this process, they can identify the reaction steps needed to synthesize the desired product from the available reactants.
Summary
• The physical properties and reactivity of compounds containing the common functional groups are intimately connected to their structures.
There are strong connections among the structure, the physical properties, and the reactivity for compounds that contain the major functional groups. Hydrocarbons that are alkanes undergo catalytic cracking, which can convert straight-chain alkanes to highly branched alkanes. Catalytic cracking is one example of a pyrolysis reaction, in which the weakest bond is cleaved at high temperature, producing a mixture of radicals. The multiple bond of an alkene produces geometric isomers (cis and trans). Terminal alkynes contain a hydrogen atom directly attached to a triply bonded carbon. Removal of the hydrogen forms an acetylide ion, a potent nucleophile used to make longer carbon chains. Arenes undergo substitution rather than elimination because of enhanced stability from delocalization of their π electron density. An alcohol is often prepared by adding the elements of water across a double bond or by a substitution reaction. Alcohols undergo two major types of reactions: those involving cleavage of the O–H bond and those involving cleavage of the C–O bond. Phenols are acidic because of π interactions between the oxygen atom and the ring. Ethers are comparatively unreactive. Aldehydes and ketones are generally prepared by oxidizing alcohols. Their chemistry is characterized by nucleophilic attack at the carbon atom of the carbonyl functional group and electrophilic attack at the oxygen atom. Grignard reagents (RMgX, where X is Cl, Br, or I) convert the carbonyl functional group to an alcohol and lengthen the carbon chain. Compounds that contain the carboxyl functional group are weakly acidic because of delocalization of the π electrons, which causes them to easily lose a proton and form the carboxylate anion. Carboxylic acids are generally prepared by oxidizing alcohols and aldehydes or reacting a Grignard reagent with CO2. Carboxylic acid derivatives include esters, prepared by reacting a carboxylic acid and an alcohol, and amides, prepared by the nucleophilic reaction of amines with more electrophilic carboxylic acid derivatives, such as esters. Amides are relatively unreactive because of π bonding interactions between the lone pair on nitrogen and the carbonyl group. Amines can also be primary, secondary, or tertiary, depending on the number of alkyl groups bonded to the amine. Quaternary ammonium salts have four substituents attached to nitrogen and can be chiral. Amines are often prepared by a nucleophilic substitution reaction between a polar alkyl halide and ammonia or other amines. They are nucleophiles, but their base strength depends on their substituents.
Conceptual Problems
1. Why do branched-chain alkanes have lower melting points than straight-chain alkanes of comparable molecular mass?
2. Describe alkanes in terms of their orbital hybridization, polarity, and reactivity. What is the geometry about each carbon of a straight-chain alkane?
3. Why do alkenes form cis and trans isomers, whereas alkanes do not? Do alkynes form cis and trans isomers? Why or why not?
4. Which compounds can exist as cis and trans isomers?
1. 2,3-dimethyl-1-butene
2. 3-methyl-1-butene
3. 2-methyl-2-pentene
4. 2-pentene
1. Which compounds can exist as cis and trans isomers?
1. 3-ethyl-3-hexene
2. 1,1-dichloro-1-propene
3. 1-chloro-2-pentene
4. 3-octene
1. Which compounds have a net dipole moment?
1. o-nitrotoluene
2. p-bromonitrobenzene
3. p-dibromobenzene
1. Why is the boiling point of an alcohol so much greater than that of an alkane of comparable molecular mass? Why are low-molecular-mass alcohols reasonably good solvents for some ionic compounds, whereas alkanes are not?
2. Is an alcohol a nucleophile or an electrophile? What determines the mode of reactivity of an alcohol? How does the reactivity of an alcohol differ from that of an ionic compound containing OH, such as KOH?
3. How does the reactivity of ethers compare with that of alcohols? Why? Ethers can be cleaved under strongly acidic conditions. Explain how this can occur.
4. What functional group is common to aldehydes, ketones, carboxylic acids, and esters? This functional group can react with both nucleophiles and electrophiles. Where does nucleophilic attack on this functional group occur? Where does electrophilic attack occur?
5. What key feature of a Grignard reagent allows it to engage in a nucleophilic attack on a carbonyl carbon?
6. Do you expect carboxylic acids to be more or less water soluble than ketones of comparable molecular mass? Why?
7. Because amides are formally derived from an acid plus an amine, why can they not be prepared by the reaction of an acid with an amine? How are they generally prepared?
8. Is an amide susceptible to nucleophilic attack, electrophilic attack, or both? Specify where the attack occurs.
9. What factors determine the reactivity of amines?
Answers
1. (c) and (d)
1. The presence of a nucleophilic Cδ− resulting from a highly polar interaction with an electropositive Mg
1. Their ability to act as weak bases and their tendency to act as nucleophiles
Structure and Reactivity
1. What is the product of the reaction of 2-butyne with excess HBr?
2. What is the product of the reaction of 3-hexyne with excess HCl?
3. What elements are eliminated during the dehydrohalogenation of an alkyl halide? What products do you expect from the dehydrohalogenation of 2-chloro-1-pentene?
4. What elements are eliminated during the dehydration of an alcohol? What products do you expect from the dehydration of ethanol?
5. Predict the products of each reaction.
1. sodium phenoxide with ethyl chloride
2. 1-chloropropane with NaOH
1. Show the mechanism and predict the organic product of each reaction.
1. 2-propanol + HCl
2. cyclohexanol + H2SO4
1. A Grignard reagent can be used to generate a carboxylic acid. Show the mechanism for the first step in this reaction using CH3CH2MgBr as the Grignard reagent. What is the geometry about the carbon of the –CH2 of the intermediate species formed in this first step?
2. Draw a molecular orbital picture showing the bonding in an amide. What orbital is used for the lone pair of electrons on nitrogen?
3. What is the product of the reaction of
1. acetic acid with ammonia?
2. methyl acetate with ethylamine, followed by heat?
1. Develop a synthetic scheme to generate
1. 1,1-dichloroethane from 1,1-dibromoethane.
2. 2-bromo-1-heptene from 1-bromopentane.
Answers
1. 2,2-dibromobutane
1. C6H5OC2H5 + NaCl
2. 1-propanol + NaCl
1. CH3CO2 NH4+ (an acid-base reaction)
2. CH3CONHC2H5 + CH3OH | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.06%3A_Common_Classes_of_Organic_Compounds.txt |
Learning Objectives
• To identify the common structural units of important biological molecules.
All the functional groups described in this chapter are found in the organic molecules that are constantly synthesized and destroyed by every living organism on Earth. A detailed understanding of the reactions that occur in living organisms is the goal of biochemistry, which deals with a wide variety of organic structures and reactions. The most abundant substances found in living systems belong to four major classes: proteins, carbohydrates, lipids, and nucleic acids. Here we briefly describe the structure and some functions of these biological molecules.
Proteins
In Section 12.8, we described proteinsA biological polymer with more than 50 amino acid residues linked together by amide bonds. as biologically active polymers formed from amino acids linked together by amide bonds. In addition to an amine group and a carboxylic acid group, each amino acid contains a characteristic R group (Figure 9.7.1). In the simplest amino acid, glycine, the R group is hydrogen (–H), but in other naturally occurring amino acids, the R group may be an alkyl group or a substituted alkyl group, a carboxylic group, or an aryl group. The nature of the R group determines the particular chemical properties of each amino acid. In Figure 9.7.1, all the amino acids found in proteins except glycine are chiral compounds, which suggests that their interactions with other chiral compounds are selective. Some proteins, called enzymes, catalyze biological reactions, whereas many others have structural, contractile, or signaling functions. Because we have described proteins previously, we will not discuss them further.
The general structure of an amino acid. An amino acid is chiral except when R is an H atom.
Carbohydrates
Carbohydrates are the most abundant of the organic compounds found in nature. They constitute a substantial portion of the food we consume and provide us with the energy needed to support life. Table sugar, milk, honey, and fruits all contain low-molecular-mass carbohydrates that are easily assimilated by the human body. In contrast, the walls of plant cells and wood contain high-molecular-mass carbohydrates that we cannot digest.
Once thought to be hydrates of carbon with the general formula Cn(H2O)m, carbohydrates are actually polyhydroxy aldehydes or polyhydroxy ketones (i.e., aldehydes or ketones with several –OH groups attached to the parent hydrocarbon). The simplest carbohydrates consist of unbranched chains of three to eight carbon atoms: one carbon atom is part of a carbonyl group, and some or all of the others are bonded to hydroxyl groups. The structure of a carbohydrate can be drawn either as a hydrocarbon chain, using a Fischer projection, or as a ring, using a Haworth projection (Figure 23.6.1). The Haworth projection is named after the British chemist Sir Walter Norman Haworth, who was awarded a Nobel Prize in Chemistry in 1937 for his discovery that sugars exist mainly in their cyclic forms, as well as for his collaboration on the synthesis of vitamin C. The cyclic form is the product of nucleophilic attack by the oxygen of a hydroxyl group on the electrophilic carbon of the carbonyl group within the same molecule, producing a stable ring structure composed of five or six carbons that minimizes bond strain (Figure 23.6.1). The substituents on the right side of the carbon chain in a Fischer projection are in the “down” position in the corresponding Haworth projection. Attack by the hydroxyl group on either side of the carbonyl group leads to the formation of two cyclic forms, called anomers: an α form, with the –OH in the “down” position, and a β form, with the –OH in the “up” position.
Walter Norman Haworth (1883–1950)
At age 14, Walter Norman Haworth left school to join his father to learn linoleum design and manufacturing, but he became interested in chemistry through his use of dyes. Private tutoring enabled him to pass the entrance exam of the University of Manchester, where he received his doctorate in 1911. During World War I, Haworth organized the laboratories at St. Andrews for the production of chemicals and drugs, returning to the investigation of carbohydrates after the war.
Carbohydrates are classified according to the number of single saccharide, or sugar, units they contain (from the Latin saccharum, meaning “sugar”). The simplest are monosaccharides; a disaccharide consists of two linked monosaccharide units; a trisaccharide has three linked monosaccharide units; and so forth. Glucose is a monosaccharide, and sucrose (common table sugar) is a disaccharide. The hydrolysis of sucrose produces glucose and another monosaccharide, fructose, in a reaction catalyzed by an enzyme or by acid:
Polysaccharides hydrolyze to produce more than 10 monosaccharide units.
The common monosaccharides contain several chiral carbons and exist in several isomeric forms. One isomer of glucose, for example, is galactose, which differs from glucose in the position of the –OH bond at carbon-4:
Because carbons-2, -3, -4, and -5 of glucose are chiral, changing the position of the –OH on carbon-4 does not produce an enantiomer of glucose but a different compound, galactose, with distinct physical and chemical properties. Galactose is a hydrolysis product of lactose, a disaccharide found in milk. People who suffer from galactosemia lack the enzyme needed to convert galactose to glucose, which is then metabolized to CO2 and H2O, releasing energy. Galactose accumulates in their blood and tissues, leading to mental retardation, cataracts, and cirrhosis of the liver.
Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of biochemically important reactions. The carbonyl group, for example, can be oxidized to form a carboxylic acid or reduced to form an alcohol. The hydroxyl groups can undergo substitution reactions, resulting in derivatives of the original compound. One such derivative is Sucralose, an artificial sweetener that is six times sweeter than sucrose; it is made by replacing two of the hydroxyl groups on sucrose with chlorine. Carbohydrates can also eliminate hydroxyl groups, producing alkenes.
Note the Pattern
Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of reactions.
Two familiar polysaccharides are starch and cellulose, which both hydrolyze to produce thousands of glucose units. They differ only in the connection between glucose units and the amount of branching in the molecule (Figure 23.6.2). Starches can be coiled or branched and are hydrolyzed by the enzymes in our saliva and pancreatic juices. Animal starch, called glycogen, is stored in the liver and muscles. It consists of branched glucose units linked by bonds that produce a coiled structure. The glucose units in cellulose, in contrast, are linked to give long, unbranched chains. The chains in cellulose stack in parallel rows held together by hydrogen bonds between hydroxyl groups. This arrangement produces a rigid structure that is insoluble in water.
Cellulose is the primary structural material of plants and one of the most abundant organic substances on Earth. Because our enzymes are not able to hydrolyze the bonds between the glucose units in cellulose, we are unable to digest it. A recently marketed product containing a high percentage of cellulose was sold as a dietetic substance for rapid weight loss, but those who consumed it experienced severe intestinal discomfort because the cellulose could not be digested. The product was quickly removed from the market.
Example 23.6.1
The Fischer projection of xylose, found in many varieties of apples, is shown. Draw the ring form (Haworth projection) of xylose.
Given: Fischer projection of a sugar
Asked for: cyclic structure
Strategy:
A Identify the nucleophile and the electrophile. Indicate the point of attack, remembering that cyclic structures are most stable when they contain at least five atoms in the ring to prevent bond strain from bond angles that are too small.
B Draw the cyclic form of the structure.
Solution:
A The carbonyl carbon (C1) is a good electrophile, and each oxygen is a good nucleophile. Nucleophilic attack occurs from the –OH group on C4, producing a stable five-membered ring.
B Because of rotation about the bond between C1 and C2, ring formation gives both a and b anomers, with the following structures (H atoms have been omitted for clarity):
Exercise
Draw the cyclic form(s) of galactose, whose Fischer projection is shown in the previous discussion.
Answer
Lipids
LipidsA family of compounds that includes fats, waxes, some vitamins, and steroids and characterized by their insolubility in water. (from the Greek lipos, meaning “fat” or “lard”) are characterized by their insolubility in water. They form a family of compounds that includes fats, waxes, some vitamins, and steroids. Fatty acids, the simplest lipids, have a long hydrocarbon chain that ends with a carboxylic acid functional group. In saturated fatty acids, the hydrocarbon chains contain only C–C bonds, so they can stack in a regular array (part (a) in Figure 23.6.3). In contrast, unsaturated fatty acids have a single double bond in the hydrocarbon chain (monounsaturated) or more than one double bond (polyunsaturated). These double bonds give fatty acid chains a kinked structure, which prevents the molecules from packing tightly (part (b) in Figure 23.6.3). As a result of reduced van der Waals interactions, the melting point of an unsaturated fatty acid is lower than that of a saturated fatty acid of comparable molecule mass, thus producing an oil rather than a solid. (For more information on van der Waals interactions, see Section 11.2.) Fish oils and vegetable oils, for example, have a higher concentration of unsaturated fatty acids than does butter.
The double bonds of unsaturated fatty acids can by hydrogenated in an addition reaction that produces a saturated fatty acid:
$\underset{unsaturated\;fatty\;acid}{-CH=CH-} + H_{2}\rightarrow \underset{saturated\;fatty\;acid}{-CH_{2}CH_{2}-} \tag{23.6.1}$
They can also be oxidized to produce an aldehyde or carboxylic acid. (For more information on hydrogenation, see Section 14.8.)
Unsaturated fatty acids are the starting compounds for the biosynthesis of prostaglandins. These hormone-like substances are involved in regulating blood pressure, tissue inflammation, and contracting and relaxing smooth muscles. Drugs such as aspirin and ibuprofen inhibit the production of prostaglandins, thereby reducing inflammation.
Waxes are esters produced by the nucleophilic attack of an alcohol on the carbonyl carbon of a long-chain carboxylic acid (Figure 23.5.4). For example, the wax used in shoe polish and wax paper, which is derived from beeswax, is formed from a straight-chain alcohol with 15 carbon atoms and a fatty acid with 31 carbon atoms. Triacylglycerols are a particularly important type of ester in living systems; they are used by the body to store fats and oils. These compounds are formed from one molecule of glycerol (1,2,3-trihydroxypropane) and three fatty acid molecules. During warmer months of the year, animals that hibernate consume large quantities of plants, seeds, and nuts that have a high fat and oil content. They convert the fatty acids to triacylglycerols and store them. Hydrolysis of stored triacylglycerols during hibernation (the reverse of Figure 23.5.4) releases alcohols and carboxylic acids that the animal uses to generate energy for maintaining cellular activity, respiration, and heart rate. Derivatives of triacylglycerols with a phosphate group are major components of all cell membranes.
Steroids are lipids whose structure is composed of three cyclohexane rings and one cyclopentane ring fused together. The presence of various substituents, including double bonds, on the basic steroid ring structure produces a large family of steroid compounds with different biological activities. For example, cholesterol, a steroid found in cellular membranes, contains a double bond in one ring and four substituents: a hydroxyl group, two methyl groups, and a hydrocarbon chain.
Cholesterol is the starting point for the biosynthesis of steroid hormones, including testosterone, the primary male sex hormone, and progesterone, which helps maintain pregnancy. These cholesterol derivatives lack the long hydrocarbon side chain, and most contain one or more ketone groups.
Cholesterol is synthesized in the human body in a multistep pathway that begins with a derivative of acetic acid. We also consume cholesterol in our diets: eggs, meats, fish, and diary products all contain cholesterol, but vegetables and other plant-derived foods do not contain cholesterol. Excess cholesterol in the human body can cause gallstones, which are composed of nearly 100% cholesterol, or lipid deposits called plaque in arteries. A buildup of plaque can block a coronary artery and result in a heart attack (Figure 23.6.4).
Figure 23.6.4 Plaque in an Artery Plaque, a lipid deposit, forms from excess cholesterol in the body. This artery is nearly blocked by a thick deposit of plaque, which greatly increases the risk of a heart attack due to reduced blood flow to the heart.
Nucleic Acids
Nucleic acidsA linear polymer of nucleotides that is the basic structural component of DNA and RNA. are the basic structural components of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), the biochemical substances found in the nuclei of all cells that transmit the information needed to direct cellular growth and reproduction. Their structures are derived from cyclic nitrogen-containing compounds called pyrimidines and purines, which can engage in hydrogen bonding through the lone electron pair on nitrogen (in pyrimidine and purine) or through the hydrogen of the amine (in purine):
The same cyclic structures are found in substances such as caffeine, a purine that is a stimulant, and the antifungal agent flucytosine, a pyrimidine. (For more information on the structure of caffeine, see Section 7.2.)
When a pyrimidine or a purine is linked to a sugar by a bond called a glycosidic bond, a nucleoside is formed. Adding a phosphoric acid group to the sugar then produces a nucleotide (part (a) in Figure 23.6.5). The linkage of nucleotides forms a polymeric chain that consists of alternating sugar and phosphate groups, which is the backbone of DNA and RNA (part (b) in Figure 23.6.5).
While the function of DNA is to preserve genetic information, RNA translates the genetic information in DNA and carries that information to cellular sites where proteins are synthesized. Many antibiotics function by interfering with the synthesis of proteins in one or more kinds of bacteria. Chloramphenicol, for example, is used against infections of the eye or outer ear canal; it inhibits the formation of peptide bonds between amino acids in a protein chain. Puromycin, which is used against herpes simplex type I, interrupts extension of a peptide chain, causing the release of an incomplete protein and the subsequent death of the virus.
Mutations in the DNA of an organism may lead to the synthesis of defective proteins. Phenylketonuria (PKU), for example, is a condition caused by a defective enzyme. Left untreated, it produces severe brain damage and mental retardation. Albinism is caused by a defective enzyme that is unable to produce melanin, the pigment responsible for the color of skin and hair. Cystic fibrosis, the most common inherited disease in the United States, blocks pancreatic function and causes thick mucus secretions that make breathing difficult. An area of intense research in combating cancer involves the synthesis of drugs that stop uncontrolled cell growth by interfering with DNA replication.
Key Takeaway
The four major classes of organic compounds found in biology are proteins, carbohydrates, lipids, and nucleic acids. Their structures and reactivity are determined by the functional groups present.
Summary
Proteins are biologically active polymers formed from amino acids linked together by amide bonds. All the amino acids in proteins are chiral compounds except glycine. The most common organic compounds found in nature are the carbohydrates, polyhydroxy aldehydes or polyhydroxy ketones in unbranched chains of three to eight carbons. They are classified according to the number of sugar, or saccharide, units, and they can be drawn as a chain in a Fischer projection or in a cyclic form called a Haworth projection. The two cyclic forms in a Haworth projection are called anomers. Many sugars contain at least one chiral center. With their carbonyl and hydroxyl functional groups, carbohydrates can undergo a variety of biochemically relevant reactions. Starch and cellulose differ only in the connectivity between glucose units. Starches can be branched or unbranched, but cellulose, the structural material of plants, is unbranched, and cannot be digested by humans. Lipids are insoluble in water. The simplest lipids, fatty acids, have a long hydrocarbon chain ending in a carboxylic acid functional group. Their physical properties depend on the number of double bonds in the chain. Prostaglandins, hormone-like substances, are formed from unsaturated fatty acids, and waxes are long-chain esters of saturated fatty acids. Triacylglycerols, which the body uses to store fats and oils, consist of glycerol esterified to three fatty acid molecules. Steroids, which include cholesterol and the steroid hormones, are characterized by three cyclohexane rings and one cyclopentane ring fused together. The basic structural units of DNA and RNA are the nucleic acids, whose structures are derived from nitrogen-containing cyclic compounds called pyrimidines and purines. These structures are linked to a sugar through a glycosidic bond, forming a nucleoside. Adding a phosphoric acid group produces a nucleotide. Nucleotides link to form a polymeric chain that is the backbone of DNA and RNA.
Key Takeaway
• An understanding of the reactivity of functional groups is necessary to understanding the reactions that occur in living systems.
Conceptual Problems
1. What are the strengths and limitations of using a Haworth projection? of using a Fischer projection?
2. Nutritionists will often state that a leafy salad contains no calories. Do you agree?
3. Would you expect margarine, a polyunsaturated fat, to have a higher or lower melting point than butter, a saturated fat?
Structure and Reactivity
1. Propose a method for synthesizing the dipeptide alanylglycine (Ala-Gly), starting with the individual amino acids (Figure 9.7.1).
2. Are all the naturally occurring amino acids chiral compounds? Do you expect proteins to contain both enantiomers of alanine and other amino acids? Explain your answer.
3. The structures of cholesterol and testosterone were shown in this section. Identify the functional groups in each.
4. The structures of glucose and purine were shown in this section. Identify the functional groups in each.
Answer
1. Use a condensation reaction:
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/23%3A_Organic_Compounds/23.07%3A_The_Molecules_of_Life.txt |
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei. Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst.
We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe.
• 24.1: Introduction
• 24.2: The Components of the Nucleus
• 24.3: Nuclear Reactions
Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
• 24.4: The Interaction of Nuclear Radiation with Matter
• 24.5: Thermodynamic Stability of the Atomic Nucleus
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: \(ΔE = (Δm)c^2\).
• 24.6: Applied Nuclear Chemistry
All practical applications of nuclear power have been based on nuclear fission reactions, which nuclear power plants use to generate electricity. In nuclear power plants, nuclear reactions generate electricity. Light-water reactors use enriched uranium as a fuel. They include fuel rods, a moderator, control rods, and a powerful cooling system to absorb the heat generated in the reactor core. Fusion reactions are thermonuclear reactions because they require high temperatures for initiation.
• 24.7: The Origin of the Elements
• 24.E: Nuclear Chemistry (Exercises)
Problems and select solutions to Chapter 20.
24: Nuclear Chemistry
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei.
Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst.
We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe.
The glow caused by intense radiation. The high-energy particles ejected into the surrounding water or air by an intense radioactive source such as this nuclear reactor core produce a ghostly bluish glow.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.01%3A_Introduction.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
The Atomic Nucleus
As you learned in Chapter 1, each element can be represented by the notation <math display="inline" xml:id="av_1.0-20_m001" xmlns:xlink="http://www.w3.org/1999/xlink"><semantics><mrow><mmultiscripts/></mrow></semantics>[/itex]<mtext>X</mtext><mprescripts/><mi>Z</mi><mi>A</mi><mtext>,</mtext> where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleonsThe protons and neutrons that make up the nucleus of an atom., and an atom with a particular number of protons and neutrons is called a nuclideAn atom with a particular number of nucleons.. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
$\begin{matrix} _{Z}^{A}X: & _{8}^{16}O & _{8}^{17}O & _{8}^{18}O\ ^{A}X: & ^{16}O &^{17}O & ^{18}O\ element -A & oxygen-16 & oxygen-17 & oxygen-18 \end{matrix} \notag$
Because the number of neutrons is equal to AZ, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17.
Any nucleus that is unstable and decays spontaneously is said to be radioactiveAny nucleus that is unstable and decays spontaneously, emitting particles and electromagnetic radiation., emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopesAn isotope that emits radiation. As you learned in Chapter 14, the rate at which radioactive decay occurs is characteristic of the isotope and is generally reported as a half-life (t1/2), the amount of time required for half of the initial number of nuclei present to decay in a first-order reaction. (For more information on half-life, see Section 14.5.) An isotope’s half-life can range from fractions of a second to billions of years and, among other applications, can be used to measure the age of ancient objects. Example 1 and its corresponding exercise review the calculations involving radioactive decay rates and half-lives.
Example 24.1.1
Fort Rock Cave in Oregon is the site where archaeologists discovered several Indian sandals, the oldest ever found in Oregon. Analysis of the 14C content of the sagebrush used to make the sandals gave an average decay rate of 5.1 disintegrations per minute (dpm) per gram of carbon. The current 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g C. How long ago was the sagebrush in the sandals cut? The half-life of 14C is 5730 yr.
Given: radioisotope, current 14C/12C ratio, initial decay rate, final decay rate, and half-life
Asked for: age
Strategy:
A Use Equation 14.5.5 to calculate N0/N, the ratio of the number of atoms of 14C originally present in the sample to the number of atoms now present.
B Substitute the value for the half-life of 14C into Equation 14.5.3 to obtain the rate constant for the reaction.
C Substitute the calculated values for N0/N and the rate constant into Equation 14.5.7 to obtain the elapsed time t.
Solution:
We can use the integrated rate law for a first-order nuclear reaction (Equation 14.5.7 to calculate the amount of time that has passed since the sagebrush was cut to make the sandals:
$ln \dfrac{N}{N_{o}}=-kt$
A From Equation 14.5.5, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 and A = 5.1) to calculate N0/N:
$\dfrac{A_{o}}{A}=\dfrac{kN_{o}}{kN}=\dfrac{N_{o}}{N}=dfrac{1.5}{5.1}$
B Now we can calculate the rate constant k from the half-life of the reaction (5730 yr) using Equation 14.5.3:
$t_{1/2}=\dfrac{0.693}{k}$
Rearranging this equation to solve for k,
$k =\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730 \; yr}=1.21\times 10^{-4} \; yr^{-1}$
C Substituting the calculated values into the equation for t,
$t =\dfrac{ln\left (N/N_{o} \right )}{k}=\dfrac{ln\left (15/5.1 \right )}{1.21\times 10^{-4} \; yr^{-1}} = 8900 \; yr$
Thus the sagebrush in the sandals is about 8900 yr old.
Exercise
While trying to find a suitable way to protect his own burial chamber, the ancient Egyptian pharaoh Sneferu developed the pyramid, a burial structure that protected desert graves from thieves and exposure to wind. Analysis of the 14C content of several items in pyramids built during his reign gave an average decay rate of 8.6 dpm/g C. When were the objects in the chamber created?
Answer: about 4600 yr ago, or about 2600 BC
Nuclear Stability
As discussed in Chapter 1, the nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear forceAn extremely powerful but very short-range attractive force between nucleons that keeps the nucleus of an atom from flying apart (due to electrostatic repulsions between protons)., an extremely powerful but very short-range attractive force between nucleons (Figure 24.1.1 ). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure 24.1.2. The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., 4He, 10B, 40Ca. All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure 24.1.3, more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are 4He with 2 protons and 2 neutrons, and 208Pb with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure 24.1.2, the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Example 24.1.2
Classify each nuclide as stable or radioactive.
1. $_{15}^{30}P$
2. $_{43}^{98}Tc$
3. tin-118
4. $_{94}^{239}Pu$
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
1. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure 24.1.2, its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, $_{15}^{30}P$ is predicted to be radioactive, and it is.
2. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places $_{43}^{98}Tc$ near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that $_{43}^{98}Tc$ might be stable. However, $_{43}^{98}Tc$ has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, $_{43}^{98}Tc$ is predicted to be radioactive, and it is.
3. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus $_{50}^{118}Sn$ should be particularly stable.
4. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, $_{94}^{239}Sn$must be radioactive.
Exercise
Classify each nuclide as stable or radioactive.
1. $_{90}^{232}Th$
2. $_{20}^{40}Ca$
3. $_{8}^{15}O$
4. $_{57}^{139}La$
Answer
1. radioactive
2. stable
3. radioactive
4. stable
Superheavy Elements
In addition to the “peninsula of stability,” Figure 24.1.2 shows a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elementsAn element with an atomic number near the magic number of 126., with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature.
Key Takeaway
• Nuclei with magic numbers of neutrons or protons are especially stable, as are those nuclei that are doubly magic.
Conceptual Problems
1. What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences.
2. What do chemists mean when they say a substance is radioactive?
3. What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table?
4. In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with Z > 83 unstable?
5. What is the significance of a magic number of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons?
6. Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers.
7. Potassium has three common isotopes, 39K, 40K, and 41K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of 40K.
8. Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is 144Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is 144Sm more stable?
Answers
1. Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not.
2. Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability.
Numerical Problems
1. Write the nuclear symbol for each isotope using notation.
1. chlorine-39
2. lithium-8
3. osmium-183
4. zinc-71
2. Write the nuclear symbol for each isotope using $_{Z}^{A}X$ notation.
1. lead-212
2. helium-5
3. oxygen-19
4. plutonium-242
3. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
1. iron-57
2. $^{185}W$
3. potassium-39
4. $^{131}Xe$
4. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
1. technetium-99m
2. $^{140}La$
3. radium-227
4. $^{208}Bi$
5. Which of these nuclides do you expect to be radioactive? Explain your reasoning.
1. $^{20}ne$
2. tungsten-184
3. $^{106}Ti$
6. Which of these nuclides do you expect to be radioactive? Explain your reasoning.
1. $^{107}Ag$
2. $^{50}V$
3. lutetium-176
Answers
1. $_{17}^{39}Cl$
2. $_{3}^{8}Li$
3. $_{76}^{183}Os$
4. $_{30}^{71}Zn$
1. 26 protons; 31 neutrons; 1.19
2. 74 protons; 111 neutrons; 1.50
3. 19 protons; 20 neutrons; 1.05
4. 54 protons; 77 neutrons; 1.43
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.02%3A_The_Components_of_the_Nucleus.txt |
Learning Objectives
• To know the different kinds of radioactive decay.
• To balance a nuclear reaction.
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction, a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Classes of Radioactive Nuclei
The three general classes of radioactive nuclei are characterized by a different decay process or set of processes:
1. Neutron-rich nuclei. The nuclei on the upper left side of the band of stable nuclei have a neutron-to-proton ratio that is too high to give a stable nucleus. These nuclei decay by a process that converts a neutron to a proton, thereby decreasing the neutron-to-proton ratio.
2. Neutron-poor nuclei. Nuclei on the lower right side of the band of stable nuclei have a neutron-to-proton ratio that is too low to give a stable nucleus. These nuclei decay by processes that have the net effect of converting a proton to a neutron, thereby increasing the neutron-to-proton ratio.
3. Heavy nuclei. With very few exceptions, heavy nuclei (those with A ≥ 200) are intrinsically unstable regardless of the neutron-to-proton ratio, and all nuclei with Z > 83 are unstable. This is presumably due to the cumulative effects of electrostatic repulsions between the large number of positively charged protons, which cannot be totally overcome by the strong nuclear force, regardless of the number of neutrons present. Such nuclei tend to decay by emitting an α particle (a helium nucleus, $^4 _2 \textrm{He}$, which decreases the number of protons and neutrons in the original nucleus by 2. Because the neutron-to-proton ratio in an α particle is 1, the net result of alpha emission is an increase in the neutron-to-proton ratio.
Nuclear decay reactions always produce daughter nuclei that have a more favorable neutron-to- proton ratio and hence are more stable than the parent nucleus.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the $^A _Z \textrm{X}$ notation for nuclides to include radioactive emissions. Table $1$ lists the name and symbol for each type of emitted radiation. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.
Table $1$: Nuclear Decay Emissions and Their Symbols
Identity Symbol Charge Mass (amu)
helium nucleus $^4_2\alpha$ +2 4.001506
electron $^0_{-1}\beta$ or $\beta ^-$ −1 0.000549
photon $_0^0\gamma$
neutron $^1_0\textrm n$ 0 1.008665
proton $^1_1\textrm p$ +1 1.007276
positron $^0_{+1}\beta$ or $\beta ^+$ +1 0.000549
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol $^0_{-1}\textrm e$, which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol $^0_{-1}\beta$, which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, $^4_{2}\textrm{He}^{2+}$ refers to the nucleus of a helium atom, and $^4_{2}\alpha$ denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure $1$. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay, which results in the emission of a helium-4 nucleus as an alpha (α) particle, $^4_{2}\alpha$. The general reaction is as follows:
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\ \textrm{particle}}{^4_2 \alpha}\label{5.2.1}$
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
$^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{5.2.2}$
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta Decay
Nuclei that contain too many neutrons often undergo beta (β) decay, in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
$\underset{\textrm{unstable} \ \textrm{neutron in} \ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \ \textrm{retained} \ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \ \textrm{emitted by} \ \textrm{nucleus}}{^0_{-1} \beta}\label{5.2.3}$
The general reaction for beta decay is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{5.2.4}$
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
$^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta\label{5.2.5}$
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
$^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{5.2.6}$
The general reaction for positron emission is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+}\label{5.2.7}$
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
$^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \label{5.2.8}$
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC), in which an electron in an inner shell reacts with a proton to produce a neutron:
$^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{5.2.9}$
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
$\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray}\label{5.2.10}$
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
$^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray}\label{5.2.11}$
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
$^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray}\label{5.2.12}$
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma ($\gamma$) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
$^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \ \textrm{nuclear} \ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th}+^{0}_{0}\gamma\label{5.2.13}$
If we disregard the decay event that created the excited nucleus, then
$^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th}+^{0}_{0}\gamma\label{5.2.14}$
or more generally,
$^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X}+^{0}_{0}\gamma\label{5.2.15}$
Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a γ ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n}\label{5.2.16}$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $1$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$:
$^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows:
$^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows:
$^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus
$^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore
$^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows:
$^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta$
Exercise $1$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer
1. $^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
2. $^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
3. $^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $2$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $2$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer
1. beta decay
2. positron emission or electron capture
3. alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure $2$). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
$^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{5.2.17}$
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
$^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{5.2.18}$
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure $3$). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
$^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{5.2.19}$
Example $3$
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, $^{30}_{15}\textrm{P}$.
B The balanced nuclear equation for the reaction is as follows:
$^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n}$
Exercise $1$
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei $(^{2}_{1}\textrm{H})$ produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
Answer
neutron, $^{1}_{0}\textrm{n}$; $^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}$
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as $^{235}_{92}\textrm{U}$ and $^{239}_{94}\textrm{Pu}$ undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure $3$:
$^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{5.2.20}$
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure $4$. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure $5$, a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{5.2.21}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{5.2.22}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $2$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
Table $2$: Some Reactions Used to Synthesize Transuranium Elements
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}$
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}$
$^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}$
$^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}$
$^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}$
$^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}$
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $6$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary
In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, $^4 _2 \alpha$, and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Takeaway
Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
Key Equations
alpha decay
Equation 20.1: $^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha$
beta decay
Equation 20.4: $^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta$
positron emission
Equation 20.7: $^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta$
electron capture
Equation 20.10: $^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray}$
gamma emission
Equation 20.15: $^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma$ | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.03%3A_Nuclear_Reactions.txt |
Learning Objectives
• To know the differences between ionizing and nonionizing radiation and their effects on matter.
• To identify natural and artificial sources of radiation.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
Ionizing versus Nonionizing Radiation
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiationRadiation that is relatively low in energy. When it collides with an atom in a molecule or ion, most or all of its energy can be absorbed without causing a structural or a chemical change. is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiationRadiation of a high enough energy to transfer some as it passes through matter to one or more atoms with which it collides. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions. is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
$atom + ionizing \; radiation \rightarrow ion^{+} + e^{-} \tag{24.3.1}$
Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure 24.3.1). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle: 1 MeV/particle = 96 billion J/mol.
The Effects of Ionizing Radiation on Matter
The effects of ionizing radiation depend on four factors:
1. The type of radiation, which dictates how far it can penetrate into matter
2. The energy of the individual particles or photons
3. The number of particles or photons that strike a given area per unit time
4. The chemical nature of the substance exposed to the radiation
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in Figure 24.3.2. Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table 24.3.1
Table 24.3.1 Some Properties of Ionizing Radiation
Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air*
α particles 3–9 < 0.05 mm < 10 cm
β particles ≤ 3 < 4 mm 1 m
x-rays <10−2 < 1 cm < 3 m
γ rays 10−2–101 < 20 cm > 3 m
*Distance at which half of the radiation has been absorbed.
Wilhelm Röntgen
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
There are many different ways to measure radiation exposure, or the dose. The roentgen (R)A unit that describes the amount of energy absorbed by dry air and measures the radiation exposure or dose., which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose)A unit used to measure the effects of radiation on biological tissues; the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter.; the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
$1 \; rad=0.010 \; J/kg \;\;\;\; 1 \; Gy=1 \; J/kg \tag{24.3.2}$
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man)A unit that describes the actual amount of tissue damage caused by a given amount of radiation and equal to the number of rads multiplied by the RBE. was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
Natural Sources of Radiation
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (Figure 24.3.3). One component of background radiation is cosmic rays, high-energy particles and γ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as 14C:
$_{7}^{14}N + _{0}^{1}n\rightarrow _{6}^{14}C + _{1}^{1}p \tag{24.3.4}$
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as 232Th and 238U, as well as radioactive daughter isotopes, such as 226Ra. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the K+ ion. Naturally occurring potassium contains 0.0117% 40K, which decays by emitting both a β particle and a γ ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 40K nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of 238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
$_{86}^{222}Rn \rightarrow _{84}^{218}Po + _{2}^{4}\alpha \rightarrow _{82}^{214}Pb + _{2}^{4}\alpha \tag{24.3.5}$
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of 218Po releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The 218Po isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
Artificial Sources of Radiation
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Example 24.3.1
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
A Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
B Determine the number of decays per year for this amount of 40K.
C Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution:
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
$moles \; ^{40}K= 140 \cancel{gK}\times \dfrac{0.0117 \; mol ^{40}K}{100 \; \cancel{mol\;K}} \times \dfrac{1\;\cancel{mol\;K}}{40.0\;\cancel{gK}} = 4.10 \times 10^{-4} \;mol\;^{40}K$
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
$\dfrac{decays}{year}= 4.10 \times 10^{-4} \;mol\;^{40}K \times \dfrac{1.05 \times 10^{7} \; decays/\cancel{s}}{1 \; \cancel{mol\;^{40}K}} \times \dfrac{60\;\cancel{s}}{1\;\cancel{min}}\times \dfrac{60\;\cancel{min}}{1\;\cancel{h}} \times \dfrac{24\;\cancel{h}}{1\; \cancel{day}}\times \dfrac{365\;\cancel{day}}{1\;yr} ) C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event: \( total \; energy \; per \; year= \dfrac{1.36 \times 10^{11} \; \cancel{decays}}{yr} \times \dfrac{1.32 \; \cancel{MeV}}{\cancel{decay}}\times \dfrac{10^{6} \; \cancel{eV}}{\cancel{MeV}} \times \dfrac{1.602 \times 10^{-19} \; J}{\cancel{eV}}$
$=2.87 \times 10^{-2}\;J/yr$
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
$radiation \; dose \; per \; year= \dfrac{2.87 \times 10^{-2} \; \cancel{J}/yr}{70 \; \cancel{kg}} \times \dfrac{1 \; rad}{1 \times 10^{-2} \; \cancel{J}/\cancel{kg}}$
$=4.10 \times 10^{-2}\;rad/yr = 41\; mrad/yr$
This corresponds to almost half of the normal background radiation most people experience.
Exercise
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer: 5.7 × 103 rad/yr (which is 10 times the fatal dose)
Assessing the Impact of Radiation Exposure
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table 24.3.2. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table 24.3.2 The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) Symptoms/Effects
< 5 no observable effect
5–20 possible chromosomal damage
20–100 temporary reduction in white blood cell count
50–100 temporary sterility in men (up to a year)
100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded
> 300 permanent sterility in women
> 500 fatal to 50% within 30 days; destruction of bone marrow and intestine
> 3000 fatal within hours
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in Figure 24.3.4, but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Summary
The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure.
Key Takeaway
• Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive.
Key Equation
rad
Equation 24.3.2: 1 rad = 0.01 J/kg
Conceptual Problems
1. Why are many radioactive substances warm to the touch? Why do many radioactive substances glow?
2. Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation?
3. Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why?
4. Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters.
5. Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why?
6. List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous?
7. Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects?
8. Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases?
9. Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale.
Answers
1. Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells.
2. Ten exposures of 10 rem are less likely to cause major damage.
Numerical Problems
1. A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is 238U, which has a half-life of 4.46 × 109 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads?
2. There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 106 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why?
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.04%3A_The_Interaction_of_Nuclear_Radiation_with_Matter.txt |
Learning Objectives
• To calculate a mass-energy balance and a nuclear binding energy.
• To understand the differences between nuclear fission and fusion.
Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy.
Mass–Energy Balance
The relationship between mass (m) and energy (E) is expressed in the following equation:
$E = mc^2 \label{20.27}$
where c is the speed of light (2.998 × 108 m/s), and E and m are expressed in units of joules and kilograms, respectively. Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to Equation 20.27, every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected. As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows:
$\textrm{C(graphite)} + \frac{1}{2}\textrm O_2(\textrm g)\rightarrow \mathrm{CO_2}(\textrm g)\hspace{5mm}\Delta H^\circ=-393.5\textrm{ kJ/mol} \label{20.28}$
Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. When a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as
$\Delta{E}=(\Delta m)c^2 \label{20.29}$
we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy:
$\Delta m=\dfrac{\Delta E}{c^2} \label{20.30}$
Because 1 J = 1 (kg·m2)/s2, the change in mass is as follows:
$\Delta m=\dfrac{-393.5\textrm{ kJ/mol}}{(2.998\times10^8\textrm{ m/s})^2}=\dfrac{-3.935\times10^5(\mathrm{kg\cdot m^2})/(\mathrm{s^2\cdot mol})}{(2.998\times10^8\textrm{ m/s})^2}=-4.38\times10^{-12}\textrm{ kg/mol} \label{20.31}$
This is a mass change of about 3.6 × 10−10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible.
In contrast, for a typical nuclear reaction, such as the radioactive decay of 14C to 14N and an electron (a β particle), there is a much larger change in mass:
$^{14}\textrm C\rightarrow \,^{14}\textrm N+\,^0_{-1}\beta \label{20.32}$
We can use the experimentally measured masses of subatomic particles and common isotopes to calculate the change in mass directly. The reaction involves the conversion of a neutral 14C atom to a positively charged 14N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral 14N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral 14N atom (14.003074 amu) and the mass of a 14C atom (14.003242 amu):
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}} \&=14.003074\textrm{ amu} - 14.003242\textrm{ amu} = - 0.000168\textrm{ amu}\end{align} \label{20.33}
The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of 14C is −0.000168 g = −1.68 × 10−4 g = −1.68 × 10−7 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows:
\begin{align}\Delta E &=(\Delta m)c^2=(-1.68\times10^{-7}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \ &=-1.51\times10^{10}(\mathrm{kg\cdot m^2})/\textrm s^2=-1.51\times10^{10}\textrm{ J}=-1.51\times10^7\textrm{ kJ}\end{align} \label{20.34}
The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of 14C is a relatively low-energy nuclear reaction.
Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 103 eV), megaelectronvolts (1 MeV = 106 eV), and even gigaelectronvolts (1 GeV = 109 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of 14C is thus
$\mathrm{(-1.68\times10^{-4}\, amu) \left(\dfrac{931\, MeV}{amu}\right) = -0.156\, MeV = -156\, keV}\label{20.35}$
Example $1$
Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of 238U to 234Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom.
Given: nuclear decay reaction
Asked for: changes in mass and energy
Strategy:
A Use the mass values in Table 20.1 to calculate the change in mass for the decay reaction in atomic mass units.
B Use Equation 20.30 to calculate the change in energy in joules per mole.
C Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom.
Solution
A Using particle and isotope masses from Table 20.1, we can calculate the change in mass as follows:
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}}=(\mathrm{mass \;^{234}Th+mass\;^4_2He})-\mathrm{mass\;^{238}U} \&=(234.043601\textrm{ amu}+4.002603\textrm{ amu}) - 238.050788\textrm{ amu} = - 0.004584\textrm{ amu}\end{align}
B Thus the change in mass for 1 mol of 238U is −0.004584 g or −4.584 × 10−6 kg. The change in energy in joules per mole is as follows:
ΔE = (Δm)c2 = (−4.584 × 10−6 kg)(2.998 × 108 m/s)2 = −4.120 × 1011 J/mol
C The change in energy in electronvolts per atom is as follows:
$\Delta E = -4.584\times10^{-3}\textrm{ amu}\times\dfrac{\textrm{931 MeV}}{\textrm{amu}}\times\dfrac{1\times10^6\textrm{ eV}}{\textrm{1 MeV}}=-4.27\times10^6\textrm{ eV/atom}$
Exercise $1$
Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium (3H) to 3He and a β particle.
Answer
Δm = −2.0 × 10−5 amu; ΔE = −1.9 × 106 kJ/mol = −19 keV/atom
Nuclear Binding Energies
We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 (1H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium (2H) is 2.014102 amu, although its calculated mass is 2.016490 amu:
\begin{align}m_{^2\textrm H}&=m_{\textrm{neutron}}+m_{\textrm{proton}}+m_{\textrm{electron}} \&=1.008665\textrm{ amu}+1.007276\textrm{ amu}+0.000549\textrm{ amu}=2.016490\textrm{ amu} \end{align}\label{20.36}
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation $\ref{20.30}$, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The amount of energy released when a nucleus forms from its component nucleons is the nuclear binding energy (Figure 21.2.1). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus.
Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components.
Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in Figure $2$, the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., 4He, 12C, and 16O). As mentioned earlier, these are particularly stable combinations.
Because the maximum binding energy per nucleon is reached at 56Fe, all other nuclei are thermodynamically unstable with regard to the formation of 56Fe. Consequently, heavier nuclei (toward the right in $2$) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in Figure $2$) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern.
Heavier nuclei spontaneously undergo nuclear reactions that decrease their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that increase their atomic number.
Example $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 56Fe. The experimental mass of the nuclide is given in Table A4.
Given: nuclide and mass
Asked for: nuclear binding energy and binding energy per nucleon
Strategy:
A Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of 1H atoms (because its mass is the same as the mass of one electron and one proton).
B Calculate the mass defect by subtracting the experimental mass from the calculated mass.
C Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon.
Solution
A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 1H atoms, we can calculate the sum of the masses more quickly as follows:
\begin{align}\textrm{calculated mass}&=26(\textrm{mass }^1_1\textrm H)+30(\textrm{mass }^1_0 \textrm n)\ &=26(1.007825)\textrm{amu}+30(1.008665)\textrm{amu}=56.463400\textrm{ amu}\ \textrm{experimental mass} &=55.934938 \end{align}
B We subtract to find the mass defect:
\begin{align}\textrm{mass defect}&=\textrm{calculated mass}-\textrm{experimental mass} \&=56.463400\textrm{ amu}-55.934938\textrm{ amu}=0.528462\textrm{ amu}\end{align}
C The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon.
Exercise $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 238U.
Answer 1800 MeV/238U; 7.57 MeV/nucleon
Nuclear Fission and Fusion
Nuclear fission is the splitting of a heavy nucleus into two lighter ones. Fission was discovered in 1938 by the German scientists Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with neutrons in an attempt to produce new elements with Z > 92. They observed that lighter elements such as barium (Z = 56) were formed during the reaction, and they realized that such products had to originate from the neutron-induced fission of uranium-235:
$_{92}^{235}\textrm U+\,_0^1\textrm n \rightarrow \,_{56}^{141}\textrm{Ba}+\,_{36}^{92}\textrm{Kr}+3_0^1\textrm n \label{20.37}$
This hypothesis was confirmed by detecting the krypton-92 fission product. The nucleus usually divides asymmetrically rather than into two equal parts, and the fission of a given nuclide does not give the same products every time.
In a typical nuclear fission reaction, more than one neutron is released by each dividing nucleus. When these neutrons collide with and induce fission in other neighboring nuclei, a self-sustaining series of nuclear fission reactions known as a nuclear chain reaction can result (Figure $3$). For example, the fission of 235U releases two to three neutrons per fission event. If absorbed by other 235U nuclei, those neutrons induce additional fission events, and the rate of the fission reaction increases geometrically. Each series of events is called a generation. Experimentally, it is found that some minimum mass of a fissile isotope is required to sustain a nuclear chain reaction; if the mass is too low, too many neutrons are able to escape without being captured and inducing a fission reaction. The minimum mass capable of supporting sustained fission is called the critical mass. This amount depends on the purity of the material and the shape of the mass, which corresponds to the amount of surface area available from which neutrons can escape, and on the identity of the isotope. If the mass of the fissile isotope is greater than the critical mass, then under the right conditions, the resulting supercritical mass can release energy explosively. The enormous energy released from nuclear chain reactions is responsible for the massive destruction caused by the detonation of nuclear weapons such as fission bombs, but it also forms the basis of the nuclear power industry.
Nuclear fusion, in which two light nuclei combine to produce a heavier, more stable nucleus, is the opposite of nuclear fission. As in the nuclear transmutation reactions discussed in Section 20.2, the positive charge on both nuclei results in a large electrostatic energy barrier to fusion. This barrier can be overcome if one or both particles have sufficient kinetic energy to overcome the electrostatic repulsions, allowing the two nuclei to approach close enough for a fusion reaction to occur. The principle is similar to adding heat to increase the rate of a chemical reaction. As shown in the plot of nuclear binding energy per nucleon versus atomic number in Figure 21.2.3, fusion reactions are most exothermic for the lightest element. For example, in a typical fusion reaction, two deuterium atoms combine to produce helium-3, a process known as deuterium–deuterium fusion (D–D fusion):
$2_1^2\textrm H\rightarrow \,_2^3\textrm{He}+\,_0^1\textrm n \label{20.38}$
In another reaction, a deuterium atom and a tritium atom fuse to produce helium-4 (Figure $4$), a process known as deuterium–tritium fusion (D–T fusion):
$_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n \label{20.39}$
Initiating these reactions, however, requires a temperature comparable to that in the interior of the sun (approximately 1.5 × 107 K). Currently, the only method available on Earth to achieve such a temperature is the detonation of a fission bomb. For example, the so-called hydrogen bomb (or H bomb) is actually a deuterium–tritium bomb (a D–T bomb), which uses a nuclear fission reaction to create the very high temperatures needed to initiate fusion of solid lithium deuteride (6LiD), which releases neutrons that then react with 6Li, producing tritium. The deuterium-tritium reaction releases energy explosively. Example 9 and its corresponding exercise demonstrate the enormous amounts of energy produced by nuclear fission and fusion reactions. In fact, fusion reactions are the power sources for all stars, including our sun.
Example $3$
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when the neutron-induced fission of 235U produces 144Cs, 90Rb, and two neutrons:
$_{92}^{235}\textrm U+\,_0^1\textrm n\rightarrow \,_{55}^{144}\textrm{Cs}+\,_{37}^{90}\textrm{Rb}+2_0^1\textrm n$
Given: balanced nuclear reaction
Asked for: energy released in electronvolts per atom and kilojoules per mole
Strategy:
A Following the method used in Example $1$, calculate the change in mass that accompanies the reaction. Convert this value to the change in energy in electronvolts per atom.
B Calculate the change in mass per mole of 235U. Then use Equation $\ref{20.29}$ to calculate the change in energy in kilojoules per mole.
Solution
A The change in mass that accompanies the reaction is as follows:
\begin{align}\Delta m&=\mathrm{mass_{products}}-\mathrm{mass_{reactants}}=\textrm{mass}(_{55}^{144}\textrm{Cs}+\,_{37}^{90}\textrm{Rb}+\,_0^1\textrm n)-\textrm{mass }\,_{92}^{235}\textrm U \&=(143.932077\textrm{ amu}+89.914802\textrm{ amu}+1.008665\textrm{ amu})-\textrm{235.043930 amu} \&=-0.188386\textrm{ amu}\end{align}
The change in energy in electronvolts per atom is as follows:
$\Delta E=(-0.188386\textrm{ amu})(931\textrm{ MeV/amu})=-175\textrm{ MeV}$
B The change in mass per mole of $_{92}^{235}\textrm{U}$ is −0.188386 g = −1.88386 × 10−4 kg, so the change in energy in kilojoules per mole is as follows:
\begin{align}\Delta E&=(\Delta m)c^2=(-1.88386\times10^{-4}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \&=-1.693\times10^{13}\textrm{ J/mol}=-1.693\times10^{10}\textrm{ kJ/mol}\end{align}
Exercise $3$
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when deuterium and tritium fuse to give helium-4 and a neutron:
$_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n$
Answer
ΔE = −17.6 MeV/atom = −1.697 × 109 kJ/mol
Summary
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: ΔE = (Δm)c2. Large changes in energy are usually reported in kiloelectronvolts or megaelectronvolts (thousands or millions of electronvolts). With the exception of 1H, the experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect of the nucleus. The energy corresponding to the mass defect is the nuclear binding energy, the amount of energy released when a nucleus forms from its component particles. In nuclear fission, nuclei split into lighter nuclei with an accompanying release of multiple neutrons and large amounts of energy. The critical mass is the minimum mass required to support a self-sustaining nuclear chain reaction. Nuclear fusion is a process in which two light nuclei combine to produce a heavier nucleus plus a great deal of energy. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.05%3A_Thermodynamic_Stability_of_the_Atomic_Nucleus.txt |
Learning Objectives
• To understand how nuclear reactors function.
The ever-increasing energy needs of modern societies have led scientists and engineers to develop ways of harnessing the energy released by nuclear reactions. To date, all practical applications of nuclear power have been based on nuclear fission reactions. Although nuclear fusion offers many advantages in principle, technical difficulties in achieving the high energies required to initiate nuclear fusion reactions have thus far precluded using fusion for the controlled release of energy. In this section, we describe the various types of nuclear power plants that currently generate electricity from nuclear reactions, along with some possible ways to harness fusion energy in the future. In addition, we discuss some of the applications of nuclear radiation and radioisotopes, which have innumerable uses in medicine, biology, chemistry, and industry.
Nuclear Reactors
When a critical mass of a fissile isotope is achieved, the resulting flux of neutrons can lead to a self-sustaining reaction. A variety of techniques can be used to control the flow of neutrons from such a reaction, which allows nuclear fission reactions to be maintained at safe levels. Many levels of control are required, along with a fail-safe design, because otherwise the chain reaction can accelerate so rapidly that it releases enough heat to melt or vaporize the fuel and the container, a situation that can release enough radiation to contaminate the surrounding area. Uncontrolled nuclear fission reactions are relatively rare, but they have occurred at least 18 times in the past. The most recent event resulted from the damaged Fukushima Dai-ichi plant after the March 11, 2011, earthquake and tsunami that devastated Japan. The plant used fresh water for cooling nuclear fuel rods to maintain controlled, sustainable nuclear fission. When the water supply was disrupted, so much heat was generated that a partial meltdown occurred. Radioactive iodine levels in contaminated seawater from the plant were over 4300 times the regulated safety limit. To put this in perspective, drinking one liter of fresh water with this level of contamination is the equivalent to receiving double the annual dose of radiation that is typical for a person. Dismantling the plant and decontaminating the site is estimated to require 30 years at a cost of approximately $12 billion. There is compelling evidence that uncontrolled nuclear chain reactions occurred naturally in the early history of our planet, about 1.7 billion years ago in uranium deposits near Oklo in Gabon, West Africa (Figure $1$). The natural abundance of 235U 2 billion years ago was about 3%, compared with 0.72% today; in contrast, the “fossil nuclear reactor” deposits in Gabon now contain only 0.44% 235U. An unusual combination of geologic phenomena in this region apparently resulted in the formation of deposits of essentially pure uranium oxide containing 3% 235U, which coincidentally is identical to the fuel used in many modern nuclear plants. When rainwater or groundwater saturated one of these deposits, the water acted as a natural moderator that decreased the kinetic energy of the neutrons emitted by radioactive decay of 235U, allowing the neutrons to initiate a chain reaction. As a result, the entire deposit “went critical” and became an uncontrolled nuclear chain reaction, which is estimated to have produced about 100 kW of power. It is thought that these natural nuclear reactors operated only intermittently, however, because the heat released would have vaporized the water. Removing the water would have shut down the reactor until the rocks cooled enough to allow water to reenter the deposit, at which point the chain reaction would begin again. This on–off cycle is believed to have been repeated for more than 100,000 years, until so much 235U was consumed that the deposits could no longer support a chain reaction. In addition to the incident in Japan, another recent instance of an uncontrolled nuclear chain reaction occurred on April 25–26, 1986, at the Chernobyl nuclear power plant in the former Union of Soviet Socialist Republics (USSR; now in the Ukraine; Figure $2$). During testing of the reactor’s turbine generator, a series of mechanical and operational failures caused a chain reaction that quickly went out of control, destroying the reactor core and igniting a fire that destroyed much of the facility and released a large amount of radioactivity. Thirty people were killed immediately, and the high levels of radiation in a 20 mi radius forced nearly 350,000 people to be resettled or evacuated. In addition, the accident caused a disruption to the Soviet economy that is estimated to have cost almost$13 billion. It is somewhat surprising, however, that the long-term health effects on the 600,000 people affected by the accident appear to be much less severe than originally anticipated. Initially, it was predicted that the accident would result in tens of thousands of premature deaths, but an exhaustive study almost 20 yr after the event suggests that 4000 people will die prematurely from radiation exposure due to the accident. Although significant, in fact it represents only about a 3% increase in the cancer rate among the 600,000 people most affected, of whom about a quarter would be expected to eventually die of cancer even if the accident had not occurred.
If, on the other hand, the neutron flow in a reactor is carefully regulated so that only enough heat is released to boil water, then the resulting steam can be used to produce electricity. Thus a nuclear reactor is similar in many respects to the conventional power plants that burn coal or natural gas to generate electricity; the only difference is the source of the heat that converts water to steam.
Light-Water Reactors
We begin our description of nuclear power plants with light-water reactors, which are used extensively to produce electricity in countries such as Japan, Israel, South Korea, Taiwan, and France—countries that lack large reserves of fossil fuels. The essential components of a light-water reactor are depicted in Figure $3$. All existing nuclear power plants have similar components, although different designs use different fuels and operating conditions. Fuel rods containing a fissile isotope in a structurally stabilized form (such as uranium oxide pellets encased in a corrosion-resistant zirconium alloy) are suspended in a cooling bath that transfers the heat generated by the fission reaction to a secondary cooling system. The heat is used to generate steam for the production of electricity. In addition, control rods are used to absorb neutrons and thereby control the rate of the nuclear chain reaction. Control rods are made of a substance that efficiently absorbs neutrons, such as boron, cadmium, or, in nuclear submarines, hafnium. Pulling the control rods out increases the neutron flux, allowing the reactor to generate more heat, whereas inserting the rods completely stops the reaction, a process called “scramming the reactor.”
Despite this apparent simplicity, many technical hurdles must be overcome for nuclear power to be an efficient and safe source of energy. Uranium contains only 0.72% uranium-235, which is the only naturally occurring fissile isotope of uranium. Because this abundance is not enough to support a chain reaction, the uranium fuel must be at least partially enriched in 235U, to a concentration of about 3%, for it to be able to sustain a chain reaction. At this level of enrichment, a nuclear explosion is impossible; far higher levels of enrichment (greater than or equal to 90%) are required for military applications such as nuclear weapons or the nuclear reactors in submarines. Enrichment is accomplished by converting uranium oxide to UF6, which is volatile and contains discrete UF6 molecules. Because 235UF6 and 238UF6 have different masses, they have different rates of effusion and diffusion, and they can be separated using a gas diffusion process. Another difficulty is that neutrons produced by nuclear fission are too energetic to be absorbed by neighboring nuclei, and they escape from the material without inducing fission in nearby 235U nuclei. Consequently, a moderator must be used to slow the neutrons enough to allow them to be captured by other 235U nuclei. High-speed neutrons are scattered by substances such as water or graphite, which decreases their kinetic energy and increases the probability that they will react with another 235U nucleus. The moderator in a light-water reactor is the water that is used as the primary coolant. The system is highly pressurized to about 100 atm to keep the water from boiling at 100°C.
All nuclear reactors require a powerful cooling system to absorb the heat generated in the reactor core and create steam that is used to drive a turbine that generates electricity. In 1979, an accident occurred when the main water pumps used for cooling at the nuclear power plant at Three Mile Island in Pennsylvania stopped running, which prevented the steam generators from removing heat. Eventually, the zirconium casing of the fuel rods ruptured, resulting in a meltdown of about half of the reactor core. Although there was no loss of life and only a small release of radioactivity, the accident produced sweeping changes in nuclear power plant operations. The US Nuclear Regulatory Commission tightened its oversight to improve safety.
The main disadvantage of nuclear fission reactors is that the spent fuel, which contains too little of the fissile isotope for power generation, is much more radioactive than the unused fuel due to the presence of many daughter nuclei with shorter half-lives than 235U. The decay of these daughter isotopes generates so much heat that the spent fuel rods must be stored in water for as long as 5 yr before they can be handled. Even then, the radiation levels are so high that the rods must be stored for many, many more years to allow the daughter isotopes to decay to nonhazardous levels. How to store these spent fuel rods for hundreds of years is a pressing issue that has not yet been successfully resolved. As a result, some people are convinced that nuclear power is not a viable option for providing our future energy needs, although a number of other countries continue to rely on nuclear reactors for a large fraction of their energy.
Heavy-Water Reactors
Deuterium (2H) absorbs neutrons much less effectively than does hydrogen (1H), but it is about twice as effective at slowing neutrons. Consequently, a nuclear reactor that uses D2O instead of H2O as the moderator is so efficient that it can use unenriched uranium as fuel. Using a lower grade of uranium reduces operating costs and eliminates the need for plants that produce enriched uranium. Because of the expense of D2O, however, only countries like Canada, which has abundant supplies of hydroelectric power for generating D2O by electrolysis, have made a major investment in heavy-water reactors.
Breeder Reactors
A breeder reactor is a nuclear fission reactor that produces more fissionable fuel than it consumes. This does not violate the first law of thermodynamics because the fuel produced is not the same as the fuel consumed. Under heavy neutron bombardment, the nonfissile 238U isotope is converted to 239Pu, which can undergo fission:
\begin{align}_{92}^{238}\textrm U+\,_0^1\textrm n &\rightarrow \,_{92}^{239}\textrm U \rightarrow \,_{93}^{239}\textrm{Np}+\,_{-1}^0\beta \ _{93}^{239}\textrm{Np}\hspace{8mm}&\rightarrow \,_{94}^{239}\textrm{Pu}+\,_{-1}^{0}\beta \end{align} \label{20.40}
The overall reaction is thus the conversion of nonfissile 238U to fissile 239Pu, which can be chemically isolated and used to fuel a new reactor. An analogous series of reactions converts nonfissile 232Th to 233U, which can also be used as a fuel for a nuclear reactor. Typically, about 8–10 yr are required for a breeder reactor to produce twice as much fissile material as it consumes, which is enough to fuel a replacement for the original reactor plus a new reactor. The products of the fission of 239Pu, however, have substantially longer half-lives than the fission products formed in light-water reactors.
Nuclear Fusion Reactors
Although nuclear fusion reactions, are thermodynamically spontaneous, the positive charge on both nuclei results in a large electrostatic energy barrier to the reaction (remember that thermodynamic spontaneity is unrelated to the reaction rate.) Extraordinarily high temperatures (about 1.0 × 108°C) are required to overcome electrostatic repulsions and initiate a fusion reaction. Even the most feasible such reaction, deuterium–tritium fusion (D–T fusion), requires a temperature of about 4.0 × 107°C. Achieving these temperatures and controlling the materials to be fused are extraordinarily difficult problems, as is extracting the energy released by the fusion reaction, because a commercial fusion reactor would require such high temperatures to be maintained for long periods of time. Several different technologies are currently being explored, including the use of intense magnetic fields to contain ions in the form of a dense, high-energy plasma at a temperature high enough to sustain fusion (Figure $\PageIndex{4a}$). Another concept employs focused laser beams to heat and compress fuel pellets in controlled miniature fusion explosions (Figure $\PageIndex{4b}$).
Nuclear reactions such as these are called thermonuclear reactions because a great deal of thermal energy must be invested to initiate the reaction. The amount of energy released by the reaction, however, is several orders of magnitude greater than the energy needed to initiate it. In principle, a nuclear fusion reaction should thus result in a significant net production of energy. In addition, Earth’s oceans contain an essentially inexhaustible supply of both deuterium and tritium, which suggests that nuclear fusion could provide a limitless supply of energy. Unfortunately, however, the technical requirements for a successful nuclear fusion reaction are so great that net power generation by controlled fusion has yet to be achieved.
The Uses of Radioisotopes
Nuclear radiation can damage biological molecules, thereby disrupting normal functions such as cell division. Because radiation is particularly destructive to rapidly dividing cells such as tumor cells and bacteria, it has been used medically to treat cancer since 1904, when radium-226 was first used to treat a cancerous tumor. Many radioisotopes are now available for medical use, and each has specific advantages for certain applications.
In modern radiation therapy, radiation is often delivered by a source planted inside the body. For example, tiny capsules containing an isotope such as 192Ir, coated with a thin layer of chemically inert platinum, are inserted into the middle of a tumor that cannot be removed surgically. The capsules are removed when the treatment is over. In some cases, physicians take advantage of the body’s own chemistry to deliver a radioisotope to the desired location. For example, the thyroid glands in the lower front of the neck are the only organs in the body that use iodine. Consequently, radioactive iodine is taken up almost exclusively by the thyroid (Figure $\PageIndex{5a}$). Thus when radioactive isotopes of iodine (125I or 131I) are injected into the blood of a patient suffering from thyroid cancer, the thyroid glands filter the radioisotope from the blood and concentrate it in the tissue to be destroyed. In cases where a tumor is surgically inaccessible (e.g., when it is located deep in the brain), an external radiation source such as a 60Co “gun” is used to aim a tightly focused beam of γ rays at it. Unfortunately, radiation therapy damages healthy tissue in addition to the target tumor and results in severe side effects, such as nausea, hair loss, and a weakened immune system. Although radiation therapy is generally not a pleasant experience, in many cases it is the only choice.
A second major medical use of radioisotopes is medical imaging, in which a radioisotope is temporarily localized in a particular tissue or organ, where its emissions provide a “map” of the tissue or the organ. Medical imaging uses radioisotopes that cause little or no tissue damage but are easily detected. One of the most important radioisotopes for medical imaging is 99mTc. Depending on the particular chemical form in which it is administered, technetium tends to localize in bones and soft tissues, such as the heart or the kidneys, which are almost invisible in conventional x-rays (Figure $\PageIndex{5b}$). Some properties of other radioisotopes used for medical imaging are listed in Table $1$.
Table $1$: Radioisotopes Used in Medical Imaging and Treatment
Isotope Half-Life Tissue
18F 110 min brain
24Na 15 h circulatory system
32P 14 days eyes, liver, and tumors
59Fe 45 days blood and spleen
60Co 5.3 yr external radiotherapy
99mTc 6 h heart, thyroid, liver, kidney, lungs, and skeleton
125I 59.4 days thyroid, prostate, and brain
131I 8 days thyroid
133Xe 5 days lungs
201Tl 3 days heart
Because γ rays produced by isotopes such as 131I and 99mTc are emitted randomly in all directions, it is impossible to achieve high levels of resolution in images that use such isotopes. However, remarkably detailed three-dimensional images can be obtained using an imaging technique called positron emission tomography (PET). The technique uses radioisotopes that decay by positron emission, and the resulting positron is annihilated when it collides with an electron in the surrounding matter. In the annihilation process, both particles are converted to energy in the form of two γ rays that are emitted simultaneously and at 180° to each other:
$_{+1}^0\beta +\,_{-1}^0\textrm e \rightarrow 2_0^0\gamma\label{20.41}$
With PET, biological molecules that have been “tagged” with a positron-emitting isotope such as 18F or 11C can be used to probe the functions of organs such as the brain.
Another major health-related use of ionizing radiation is the irradiation of food, an effective way to kill bacteria such as Salmonella in chicken and eggs and potentially lethal strains of Escherichia coli in beef. Collectively, such organisms cause almost 3 million cases of food poisoning annually in the United States, resulting in hundreds of deaths. Figure $6$ shows how irradiation dramatically extends the storage life of foods such as strawberries. Although US health authorities have given only limited approval of this technique, the growing number of illnesses caused by antibiotic-resistant bacteria is increasing the pressure to expand the scope of food irradiation.
One of the more unusual effects of radioisotopes is in dentistry. Because dental enamels contain a mineral called feldspar (KAlSi3O8, which is also found in granite rocks), teeth contain a small amount of naturally occurring radioactive 40K. The radiation caused by the decay of 40K results in the emission of light (fluorescence), which gives the highly desired “pearly white” appearance associated with healthy teeth.
In a sign of how important nuclear medicine has become in diagnosing and treating illnesses, the medical community has become alarmed at the global shortage of 99Tc, a radioisotope used in more than 30 million procedures a year worldwide. Two reactors that produce 60% of the world’s radioactive 99Mo, which decays to 99Tc, are operating beyond their intended life expectancies. Moreover, because most of the reactors producing 99Mo use weapons-grade uranium (235U), which decays to 99Mo during fission, governments are working to phase out civilian uses of technology to produce 99Mo because of concerns that the technology can be used to produce nuclear weapons. Engineers are currently focused on how to make key medical isotopes with other alternatives that don’t require fission. One promising option is by removing a neutron from 100Mo, a stable isotope that makes up about 10% of natural molybdenum, transmuting it to 99Mo.
In addition to the medical uses of radioisotopes, radioisotopes have literally hundreds of other uses. For example, smoke detectors contain a tiny amount of 241Am, which ionizes the air in the detector so an electric current can flow through it. Smoke particles reduce the number of ionized particles and decrease the electric current, which triggers an alarm. Another application is the “go-devil” used to detect leaks in long pipelines. It is a packaged radiation detector that is inserted into a pipeline and propelled through the pipe by the flowing liquid. Sources of 60Co are attached to the pipe at regular intervals; as the detector travels along the pipeline, it sends a radio signal each time it passes a source. When a massive leak causes the go-devil to stop, the repair crews know immediately which section of the pipeline is damaged. Finally, radioactive substances are used in gauges that measure and control the thickness of sheets and films. As shown in Figure $7$, thickness gauges rely on the absorption of either β particles (by paper, plastic, and very thin metal foils) or γ rays (for thicker metal sheets); the amount of radiation absorbed can be measured accurately and is directly proportional to the thickness of the material.
Summary
All practical applications of nuclear power have been based on nuclear fission reactions, which nuclear power plants use to generate electricity. In nuclear power plants, nuclear reactions generate electricity. Light-water reactors use enriched uranium as a fuel. They include fuel rods, a moderator, control rods, and a powerful cooling system to absorb the heat generated in the reactor core. Heavy-water reactors use unenriched uranium as a fuel because they use D2O as the moderator, which scatters and slows neutrons much more effectively than H2O. A breeder reactor produces more fissionable fuel than it consumes. A nuclear fusion reactor requires very high temperatures. Fusion reactions are thermonuclear reactions because they require high temperatures for initiation. Radioisotopes are used in both radiation therapy and medical imaging.
Contributors
• Anonymous | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.06%3A_Applied_Nuclear_Chemistry.txt |
Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed.
In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
Relative Abundances of the Elements on Earth and in the Known Universe
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure 24.6.1. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure 24.6.1 illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the trends in nuclear stability discussed in Section 24.1, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table 24.6.1 for some common elements. Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. You may recall from Chapter 1 that the anomalously high iridium content of a 66-million-year-old rock layer was a key finding in the development of the asteroid-impact theory for the extinction of the dinosaurs. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Table 24.6.1 Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Synthesis of the Elements in Stars
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure 24.6.2). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
$_{1}^{1}H + \;_{1}^{1}H \rightarrow \;_{1}^{2}H + \;_{+1}^{\;\;0}\beta \tag{24.6.1}$
$_{1}^{2}H + _{1}^{1}H \rightarrow _{2}^{3}He + _{0}^{0}\gamma$
$_{2}^{3}He + _{2}^{3}He \rightarrow _{2}^{4}He + 2 _{1}^{1}H$
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great deal of energy:
$4_{1}^{1}H \rightarrow _{2}^{4}He + 2_{+1}^{\;\;0}\beta + 2_{0}^{0}\gamma \tag{24.6.2}$
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
$2_{2}^{4}He \rightarrow _{4}^{8}Be \tag{24.6.3}$
Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
$_{4}^{8}Be \overset{_{2}^{4}He}{\rightarrow} \;_{6}^{12}C \overset{_{2}^{4}He}{\rightarrow} \;_{8}^{16}O \overset{_{2}^{4}He}{\rightarrow} \;_{10}^{20}Ne \overset{_{2}^{4}He}{\rightarrow} \;_{12}^{24}Mg \tag{24.6.4}$
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
$_{6}^{12}C + \;_{6}^{12}C \rightarrow \;_{11}^{23}Na + \;_{1}^{1}H \tag{24.6.5}$
$_{6}^{12}C + \;_{8}^{16}O \rightarrow \;_{14}^{28}Si + \;_{0}^{0}\gamma \tag{24.6.6}$
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
$_{6}^{12}C \overset{_{2}^{4}He}{\rightarrow} \;_{8}^{16}O \overset{_{2}^{4}He}{\rightarrow} \;_{10}^{20}Ne \overset{_{2}^{4}He}{\rightarrow} \;_{12}^{24}Mg \overset{_{2}^{4}He}{\rightarrow} \;_{14}^{28}Si \overset{_{2}^{4}He}{\rightarrow} \;_{16}^{32}S \overset{_{2}^{4}He}{\rightarrow} \;_{18}^{36}Ar \overset{_{2}^{4}He}{\rightarrow} \;_{20}^{40}Ca \tag{24.6.7}$
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
The Formation of Heavier Elements in Supernovas
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure 24.6.2). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure 24.6.3). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
$_{26}^{56}Fe + \; 64_{0}^{1}n \rightarrow \;_{26}^{120}Fe \rightarrow \; _{50}^{120 }Sn + 24_{-1}^{\;\;0}\beta \tag{24.6.8}$
Figure 24.6.3 A Supernova A view of the remains of Supernova 1987A, located in the Large Magellanic Cloud, showing the circular halo of expanding debris produced by the explosion. Multiple neutron-capture events occur during a supernova explosion, forming both the heaviest elements and many of the less stable nuclides.
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
Example 24.6.1
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution:
1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $_{6}^{12}C + \; _{6}^{12}C \rightarrow \;_{12}^{24}Mg$
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $_{6}^{12}C + \; _{6}^{12}C \rightarrow \;_{10}^{20}Ne + \;_{2}^{4}\alpha$
Exercise
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer: 19 neutrons; $_{26}^{56}Fe + \; 19_{0}^{1}n \rightarrow \;_{26}^{75}Fe \rightarrow \; _{33}^{75 }As + 7_{-1}^{\;\;0}\beta$
Summary
By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova.
Key Takeaways
• Hydrogen and helium are the most abundant elements in the universe.
• Heavier elements are formed in the interior of stars via multiple neutron-capture events.
Conceptual Problems
1. Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements?
2. How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed?
3. Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers.
4. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.
5. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer.
6. If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation.
Answer
1. The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons.
Numerical Problems
1. Write a balanced nuclear reaction for the formation of each isotope.
1. 27Al from two 12C nuclei
2. 9Be from two 4He nuclei
2. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion.
1. 106Pd from nickel-58
2. selenium-79 from iron-56
3. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium?
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.07%3A_The_Origin_of_the_Elements.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: Principles, Patterns, and Applications" by Bruce A. Averill and Patricia Eldredge. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
20.1: Components of the Nucleus
Conceptual Problems
1. What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences.
2. What do chemists mean when they say a substance is radioactive?
3. What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table?
4. In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with Z > 83 unstable?
5. What is the significance of a magic number of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons?
6. Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers.
7. Potassium has three common isotopes, 39K, 40K, and 41K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of 40K.
8. Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is 144Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is 144Sm more stable?
Answers
5. Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not.
7. Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability.
Numerical Problems
1. Write the nuclear symbol for each isotope using $^A_Z \textrm X$ notation.
a. chlorine-39
b. lithium-8
c. osmium-183
d. zinc-71
2.Write the nuclear symbol for each isotope using $^A_Z \textrm X$ notation.
a. lead-212
b. helium-5
c. oxygen-19
d. plutonium-242
3. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
a. iron-57
b. 185W
c. potassium-39
d. 131Xe
4. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
a. technetium-99m
b. 140La
c. radium-227
d. 208Bi
5. Which of these nuclides do you expect to be radioactive? Explain your reasoning.
a. 20Ne
b. tungsten-184
c. 106Ti
6. Which of these nuclides do you expect to be radioactive? Explain your reasoning.
a. 107Ag
b. 50V
c. lutetium-176
Answers
1.
a. $^{39}_{17} \textrm{Cl}$
b. $^{8}_{3} \textrm{Li}$
c. $^{183}_{76} \textrm{Os}$
d. $^{71}_{30} \textrm{Zn}$
3.
a. 26 protons; 31 neutrons; 1.19
b. 74 protons; 111 neutrons; 1.50
c. 19 protons; 20 neutrons; 1.05
d. 54 protons; 77 neutrons; 1.43
20.2: Nuclear Reactions
Conceptual Problems
1. Describe the six classifications of nuclear decay reactions. What is the most common mode of decay for elements that have heavy nuclei? Why?
1. Complete the following table for these five nuclear reactions.
Alpha Decay Beta Decay Gamma Emission Positron Emission Electron Capture
identity of particle or radiation helium-4 nucleus
mass number of parent − mass number of daughter 4
atomic number of parent − atomic number of daughter −1
effect on neutron-to-proton ratio decreases
1. What is the most common decay process for elements in row 5 of the periodic table that contain too few neutrons for the number of protons present? Why?
2. Explain the difference between the symbols e and β. What is the difference in meaning between the symbols $^4 _2 \textrm{He}$ and $^4 _2 \alpha$?
3. What is a mass number? Which particles have a mass number of zero?
4. What are the key differences between the equations written for chemical reactions and for nuclear reactions? How are they similar?
5. Can all the kinds of nuclear decay reactions discussed be characterized by the general equation: parent → daughter + particle? Explain your answer.
6. Which types of nuclear decay reactions conserve both mass number and atomic number? In which do the parent and daughter nuclei have the same mass number but different atomic numbers? Which do not convert one element to another?
7. Describe a radioactive decay series. How many series occur naturally? Of these, which one no longer occurs in nature? Why?
8. Primordial nuclides are nuclides found on Earth that have existed in their current form since before Earth was formed. Primordial nuclides are residues from the Big Bang, from cosmogenic sources, and from ancient supernova explosions which occurred before the formation of the Solar System. Only four primordial nuclides with atomic numbers greater than lead (Z>82)) still exist (209Bi, 232Th, 235U, and 238U) - and all of them are radioactive. Except for some isotopes of thorium and uranium that have a very long half-life, the half-lives of the heavy elements are so short that these elements should have been completely converted to lighter, more stable elements long ago. Why are these elements still present in nature?
9. Why are neutrons preferred to protons when preparing new isotopes of the lighter elements?
10. Why are particle accelerators and cyclotrons needed to create the transuranium elements?
Answers
1. Both positron decay and electron capture increase the neutron-to-proton ratio; electron capture is more common for heavier elements such those of row 5.
1. The mass number is the sum of the numbers of protons and neutrons present. Particles with a mass number of zero include β particles (electrons) and positrons; gamma rays and x-rays also have a mass number of zero.
1. Unlike protons, neutrons have no charge, which minimizes the electrostatic barrier to colliding and reacting with a positively charged nucleus.
Numerical Problems
1. What type of particle is emitted in each nuclear reaction?
1. 238Pu → 234U
2. 32Si → 32P
3. 18F → 18O
4. 206Tl → 206Pb
1. What type of particle is emitted in each nuclear reaction?
1. 230Th →226Ra
2. 224Rn → 224Fr
3. 210Bi → 206Tl
4. 36Cl → 36S
1. Predict the mode of decay and write a balanced nuclear reaction for each isotope.
1. 235U
2. 254Es
3. 36S
4. 99Mo
1. Predict the mode of decay and write a balanced nuclear reaction for each isotope.
1. 13N
2. 231Pa
3. 7Be
4. 77Ge
1. Balance each nuclear reaction.
1. 208Po → α + Pb
2. 226Ra → α + Rn
3. 228Th → Ra + α + γ
4. 231Pa → Ac + α + γ
5. Ho → 166Er + β + γ
6. Ac → 226Th + β + γ
1. Complete each nuclear reaction.
1. $^{210}_{84}\textrm{Po}\rightarrow \,^{206}\textrm{Pb}$
2. $^{217}_{85}\textrm{At}\rightarrow \textrm{Bi}+\alpha$
3. $\textrm{Ra}\rightarrow ^{220}_{86}\textrm{Rn}+\alpha$
4. 208Tl → 82Pb + β
5. Np → 239Pu + β
6. Fe → 52Mn + β+ + γ
1. Write a balanced nuclear equation for each reaction.
1. β decay of 87Rb
2. β+ decay of 20Mg
3. α decay of 268Mt
1. Write a balanced nuclear equation for each reaction.
1. β decay of 45K
2. β+ decay of 41Sc
3. α decay of 146Sm
1. The decay products of several isotopes are listed here. Identify the type of radiation emitted and write a balanced nuclear equation for each.
1. 218Po → 214Pb
2. 32Si → 32P
3. an excited state of an iron-57 nucleus decaying to its ground state
4. conversion of thallium-204 to lead-204
1. The decay products of several isotopes are listed here. Identify the type of radiation emitted and write a balanced nuclear equation for each.
1. 218Po → 218At
2. 216Po → 212Pb
3. bismuth-211 converted to thallium-207
4. americium-242 converted to rhodium-107 with the emission of four neutrons
1. Predict the most likely mode of decay and write a balanced nuclear reaction for each isotope.
1. 238U
2. 208Po
3. 40S
4. molybdenum-93m
1. Predict the most likely mode of decay and write a balanced nuclear reaction for each isotope.
1. 16N
2. 224Th
3. 118In
4. 64Ge
1. For each nuclear reaction, identify the type(s) of decay and write a balanced nuclear equation.
1. 216Po → ? + At
2. ? → α + 231Pa
3. 228Th → ? + α + γ
4. 231Pa → ? + β + γ
1. For each nuclear reaction, identify the type(s) of decay and write a balanced nuclear equation.
1. 212Po → 208Pb + ?
2. 192Ir → Pt + ?
3. 241Am → 57Fe + 184? + ?
4. Ge → 77Ge + ?
1. Identify the parent isotope and write a balanced nuclear reaction for each process.
1. Lead-205 is formed via an alpha emission.
2. Titanium-46 is formed via beta and gamma emission.
3. Argon-36 is formed via a beta decay and a gamma emission.
1. Identify the parent isotope and write a balanced nuclear reaction for each process.
1. Iodine-130 is formed by ejecting an electron and a gamma ray from a nucleus.
2. Uranium-240 is formed by alpha decay.
3. Curium-247 is formed by releasing a helium dication and a gamma ray.
1. Write a balanced nuclear equation for each process.
1. Bromine undergoes a decay and produces a gas with an atomic mass of 80 amu.
2. An element emits two neutrons while decaying into two metals, each of which can be extracted and converted to chlorides with the formula MCl2. The masses of the two salts are 162.9 and 210.9 g/mol, respectively.
1. Write a balanced nuclear equation for each process.
1. An unknown element emits γ rays plus particles that are readily blocked by paper. The sample also contains a substantial quantity of tin-104.
2. An unstable element undergoes two different decay reactions: beta decay to produce a material with a mass of 222 amu and alpha decay to astatine.
1. Bombarding 249Cf with 12C produced a transuranium element with a mass of 257 amu, plus several neutral subatomic particles. Identify the element and write a nuclear reaction for this transmutation.
1. One transuranium element, 253Es, is prepared by bombarding 238U with 15 neutrons. What is the other product of this reaction? Write a balanced transmutation reaction for this conversion.
1. Complete this radioactive decay series:
$^{223}_{88}\textrm{Ra}\overset{\alpha}{\rightarrow}\textrm{Rn}\overset{\alpha}{\rightarrow}\textrm{Po}\overset{\alpha}{\rightarrow}\textrm{Pb}\overset{\beta^-}{\rightarrow}\textrm{Bi}\overset{\alpha}{\rightarrow}\textrm{Tl}\overset{\beta^-}{\rightarrow}\textrm{Pb}$
1. Complete each nuclear fission reaction.
1. $^{235}_{92}\textrm{U}+\,^{1}_{0}\textrm{n}\rightarrow \, ^{90}_{36}\textrm{Kr}\,+\,?+2^{1}_{0}\textrm{n}$
2. $?+\,^{1}_{0}\textrm{n}\rightarrow \, ^{140}\textrm{Cs}+\,^{96}\textrm{Y}+4^{1}_{0}\textrm{n}$
1. Complete each nuclear fission reaction.
1. $^{235}_{92}\textrm{U}+\,^{1}_{0}\textrm{n}\rightarrow \, ^{145}_{57}\textrm{La}\,+\,?+3^{1}_{0}\textrm{n}$
2. $?+\,^{1}_{0}\textrm{n}\rightarrow \, ^{95}_{42}\textrm{Mo}+\,^{139}_{57}\textrm{La}+2^{1}_{0}\textrm{n}+7^{0}_{-1}\beta$
1. A stable nuclide absorbs a neutron, emits an electron, and then splits into two α particles. Identify the nuclide.
1. Using 18O, how would you synthesize an element with atomic number 106 from 249Cf? Write a balanced nuclear equation for the reaction.
1. Using 10B and 252Cf, how would you synthesize an element with atomic number 103? Write a balanced nuclear equation for the reaction.
Answers
1. α decay; $^{235}_{92}\textrm{U}\rightarrow \, ^{4}_{2}\alpha+\,^{231}_{90}\textrm{Th}$
2. α decay; $^{254}_{99}\textrm{Es}\rightarrow \, ^{4}_{2}\alpha+\,^{250}_{97}\textrm{Bk}$
3. β decay; $^{36}_{16}\textrm{S}\rightarrow \, ^{0}_{-1}\beta+\,^{36}_{17}\textrm{Cl}$
4. β decay; $^{99}_{42}\textrm{Mo}\rightarrow \, ^{0}_{-1}\beta+\,^{99m}_{43}\textrm{Tc}$
1. $^{208}_{84}\textrm{Po}\rightarrow \, ^{4}_{2}\alpha+\,^{204}_{82}\textrm{Pb}$
2. $^{226}_{88}\textrm{Ra}\rightarrow \, ^{4}_{2}\alpha+\,^{222}_{86}\textrm{Rn}$
3. $^{228}_{90}\textrm{Th}\rightarrow \,^{224}_{88}\textrm{Ra}+\, ^{4}_{2}\alpha+\gamma$
4. $^{231}_{91}\textrm{Pa}\rightarrow \,^{227}_{89}\textrm{Ac}+\, ^{4}_{2}\alpha+\gamma$
5. $^{166}_{67}\textrm{Ho}\rightarrow \,^{166}_{68}\textrm{Er}+\, ^{0}_{-1}\beta+\gamma$
6. $^{226}_{89}\textrm{Ac}\rightarrow \,^{226}_{90}\textrm{Th}+\, ^{0}_{-1}\beta+\gamma$
1. $^{87}_{37}\textrm{Rb}\rightarrow \,^{87}_{38}\textrm{Sr}+\, ^{0}_{-1}\beta$
2. $^{20}_{12}\textrm{Mg}\rightarrow \,^{20}_{11}\textrm{Na}+\, ^{0}_{-1}\beta$
3. $^{268}_{109}\textrm{Mt}\rightarrow \, ^{4}_{2}\alpha+\,^{264}_{107}\textrm{Bh}$
1. α particle; $^{218}_{84}\textrm{Po}\rightarrow \, ^{4}_{2}\alpha+\,^{214}_{82}\textrm{Pb}$
2. β particle; $^{32}_{14}\textrm{Si}\rightarrow \, ^{0}_{-1}\beta+\,^{32}_{15}\textrm{P}$
3. γ ray; $^{57m}_{26}\textrm{Fe}\rightarrow \,^{57}_{26}\textrm{Fe}+\gamma$
4. β particle; $^{204}_{81}\textrm{Tl}\rightarrow \, ^{0}_{-1}\beta+\,^{204}_{82}\textrm{Pb}$
1. α emission; $^{238}_{92}\textrm{U}\rightarrow \, ^{4}_{2}\alpha+\,^{234}_{90}\textrm{Th}$
2. α emission; $^{208}_{84}\textrm{Po}\rightarrow \,^{204}_{82}\textrm{Pb}+\, ^{4}_{2}\alpha$
3. β emission; $^{40}_{16}\textrm{S}\rightarrow \, ^{0}_{-1}\beta+\,^{40}_{17}\textrm{Cl}$
4. γ emission; $^{93m}_{42}\textrm{Mo}\rightarrow \,^{93}_{42}\textrm{Mo}+\gamma$
1. β decay; $^{216}_{84}\textrm{Po}\rightarrow \, ^{0}_{-1}\beta+\,^{216}_{85}\textrm{At}$
2. α decay; $^{235}_{93}\textrm{Np}\rightarrow \, ^{4}_{2}\alpha+\,^{231}_{91}\textrm{Pa}$
3. α decay; γ emission; $^{228}_{90}\textrm{Th}\rightarrow \,^{224}_{88}\textrm{Ra} +\, ^{4}_{2}\alpha+\gamma$
4. β decay, γ emission; $^{231}_{91}\textrm{Pa}\rightarrow \,^{231}_{92}\textrm{U} +\, ^{0}_{-1}\beta+\gamma$
1. $^{80}_{35}\textrm{Br}\rightarrow \,^{80}_{36}\textrm{Kr} +\, ^{0}_{-1}\beta$
2. $^{234}_{94}\textrm{Pu}\rightarrow \,^{140}_{56}\textrm{Ba} +\, ^{92}_{38}\textrm{Sr}+\,2^{1}_{0}\textrm{n}$
1. $^{249}_{98}\textrm{Cf}+\,^{12}_{6}\textrm{C}\rightarrow \,^{257}_{104}\textrm{Rf} +4^{1}_{0}\textrm{n}$
1. $^{235}_{92}\textrm{U}+\,^{1}_{0}\textrm{n}\rightarrow \,^{145}_{57}\textrm{La} +\, ^{88}_{35}\textrm{Br}+3^{1}_{0}\textrm{n}$
2. $^{235}_{92}\textrm{U}+\,^{1}_{0}\textrm{n}\rightarrow \,^{95}_{42}\textrm{Mo} +\, ^{139}_{57}\textrm{La}+2^{1}_{0}\textrm{n}+7^{0}_{-1}\beta$
20.3: The Interaction of Nuclear Radiation with Matter
Conceptual Problems
1. Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements?
2. How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed?
3. Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers.
4. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.
5. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer.
6. If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation.
Answer
1. The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons.
Numerical Problems
1. Write a balanced nuclear reaction for the formation of each isotope.
1. 27Al from two 12C nuclei
2. 9Be from two 4He nuclei
2. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion.
1. 106Pd from nickel-58
2. selenium-79 from iron-56
3. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium?
20.4: Thermodynamic Stability of the Atomic Nucleus
Conceptual Problems
1. How do chemical reactions compare with nuclear reactions with respect to mass changes? Does either type of reaction violate the law of conservation of mass? Explain your answers.
2. Why is the amount of energy released by a nuclear reaction so much greater than the amount of energy released by a chemical reaction?
3. Explain why the mass of an atom is less than the sum of the masses of its component particles.
4. The stability of a nucleus can be described using two values. What are they, and how do they differ from each other?
5. In the days before true chemistry, ancient scholars (alchemists) attempted to find the philosopher’s stone, a material that would enable them to turn lead into gold. Is the conversion of Pb → Au energetically favorable? Explain why or why not.
6. Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome.
7. Imagine that the universe is dying, the stars have burned out, and all the elements have undergone fusion or radioactive decay. What would be the most abundant element in this future universe? Why?
8. Numerous elements can undergo fission, but only a few can be used as fuels in a reactor. What aspect of nuclear fission allows a nuclear chain reaction to occur?
9. How are transmutation reactions and fusion reactions related? Describe the main impediment to fusion reactions and suggest one or two ways to surmount this difficulty.
Numerical Problems
1. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules.
1. 238Pa → ? + β
2. 216Fr → ? + α
3. 199Bi → ? + β+
1. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules.
1. 194Tl → ? + β+
2. 171Pt → ? + α
3. 214Pb → ? + β
1. Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules per mole.
1. $_{91}^{234}\textrm{Pa}\rightarrow \,?+\,_{-1}^0\beta$
2. $_{88}^{226}\textrm{Ra}\rightarrow \,?+\,_2^4\alpha$
1. Using the information provided in Chapter 33, complete each reaction and then calculate the amount of energy released from each in kilojoules per mole.
1. $_{27}^{60}\textrm{Co}\rightarrow\,?+\,_{-1}^0\beta$ (The mass of cobalt-60 is 59.933817 amu.)
2. technicium-94 (mass = 93.909657 amu) undergoing fission to produce chromium-52 and potassium-40
1. Using the information provided in Chapter 33, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole.
1. the beta decay of bismuth-208 (mass = 207.979742 amu)
2. the formation of lead-206 by alpha decay
1. Using the information provided, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole.
1. alpha decay of oxygen-16
2. alpha decay to produce chromium-52
1. Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 87Sr if the measured mass of 87Sr is 86.908877 amu.
2. Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 60Ni.
3. The experimentally determined mass of 53Mn is 52.941290 amu. Find each of the following.
1. the calculated mass
2. the mass defect
3. the nuclear binding energy
4. the nuclear binding energy per nucleon
1. The experimentally determined mass of 29S is 28.996610 amu. Find each of the following.
1. the calculated mass
2. the mass defect
3. the nuclear binding energy
4. the nuclear binding energy per nucleon
1. Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 141Ba, 92Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
2. Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 90Sr, 143Xe, and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
3. Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction?
4. Calculate the amount of energy released by the fusion of 6Li and deuterium to give two helium-4 nuclei. Express your answer in electronvolts per atom and kilojoules per mole.
5. How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture?
Answers
1. $_{91}^{238}\textrm{Pa}\rightarrow\,_{92}^{238}\textrm{U}+\,_{-1}^0\beta$; −5.540 × 10−16 kJ
2. $_{87}^{216}\textrm{Fr}\rightarrow\,_{85}^{212}\textrm{At}+\,_{2}^4\alpha$; −1.470 × 10−15 kJ
3. $_{83}^{199}\textrm{Bi}\rightarrow\,_{82}^{199}\textrm{Pb}+\,_{+1}^0\beta$; −5.458 × 10−16 kJ
1. $_{91}^{234}\textrm{Pa}\rightarrow\,_{92}^{234}\textrm{U}+\,_{-1}^0\beta$; 2.118 × 108 kJ/mol
2. $_{88}^{226}\textrm{Ra}\rightarrow\,_{86}^{222}\textrm{Rn}+\,_{2}^4\alpha$; 4.700 × 108 kJ/mol
1. The beta decay of bismuth-208 to polonium is endothermic (ΔE = 1.400 MeV/atom, 1.352 × 108 kJ/mol).
2. The formation of lead-206 by alpha decay of polonium-210 is exothermic (ΔE = −5.405 MeV/atom, −5.218 × 108 kJ/mol).
1. 757 MeV/atom, 8.70 MeV/nucleon
1. 53.438245 amu
2. 0.496955 amu
3. 463 MeV/atom
4. 8.74 MeV/nucleon
1. −173 MeV/atom; 1.67 × 1010 kJ/mol
1. ΔE = + 9.0 × 106 kJ/mol beryllium-8; no
1. D–D fusion: ΔE = −4.03 MeV/tritium nucleus formed = −3.89 × 108 kJ/mol tritium; D–T fusion: ΔE = −17.6 MeV/tritium nucleus = −1.70 × 109 kJ/mol; D–T fusion
20.5: Applied Nuclear Chemistry
Conceptual Problems
1. In nuclear reactors, two different but interrelated factors must be controlled to prevent a mishap that could cause the release of unwanted radiation. How are these factors controlled?
2. What are the three principal components of a nuclear reactor? What is the function of each component?
3. What is meant by the term enrichment with regard to uranium for fission reactors? Why does the fuel in a conventional nuclear reactor have to be “enriched”?
4. The plot in a recent spy/action movie involved the threat of introducing stolen “weapons-grade” uranium, which consists of 93.3% 235U, into a fission reactor that normally uses a fuel containing about 3% 235U. Explain why this could be catastrophic.
5. Compare a heavy-water reactor with a light-water reactor. Why are heavy-water reactors less widely used? How do these two reactor designs compare with a breeder reactor?
6. Conventional light-water fission reactors require enriched fuel. An alternative reactor is the so-called heavy-water reactor. The components of the two different reactors are the same except that instead of using water (H2O), the moderator in a heavy-water reactor is D2O, known as “heavy water.” Because D2O is more efficient than H2O at slowing neutrons, heavy-water reactors do not require fuel enrichment to support fission. Why is D2O more effective at slowing neutrons, and why does this allow unenriched fuels to be used?
7. Isotopes emit γ rays in random directions. What difficulties do these emissions present for medical imaging? How are these difficulties overcome?
8. If you needed to measure the thickness of 1.0 mm plastic sheets, what type of radiation would you use? Would the radiation source be the same if you were measuring steel of a similar thickness? What is your rationale? Would you want an isotope with a long or short half-life for this device?
Answers
1. Neutron flow is regulated by using control rods that absorb neutrons, whereas the speed of the neutrons produced by fission is controlled by using a moderator that slows the neutrons enough to allow them to react with nearby fissile nuclei.
1. It is difficult to pinpoint the exact location of the nucleus that decayed. In contrast, the collision of a positron with an electron causes both particles to be annihilated, and in the process, two gamma rays are emitted in opposite directions, which makes it possible to identify precisely where a positron emitter is located and to create detailed images of tissues.
Numerical Problems
1. Palladium-103, which decays via electron capture and emits x-rays with an energy of 3.97 × 10−2 MeV, is often used to treat prostate cancer. Small pellets of the radioactive metal are embedded in the prostate gland. This provides a localized source of radiation to a very small area, even though the tissue absorbs only about 1% of the x-rays. Due to its short half-life, all of the palladium will decay to a stable isotope in less than a year. If a doctor embeds 50 pellets containing 2.50 mg of 103Pd in the prostate gland of a 73.9 kg patient, what is the patient’s radiation exposure over the course of a year?
2. Several medical treatments use cobalt-60m, which is formed by bombarding cobalt with neutrons to produce a highly radioactive gamma emitter that undergoes 4.23 × 1016 emissions/(s·kg) of pure cobalt-60. The energy of the gamma emission is 5.86 × 10−2 MeV. Write the balanced nuclear equation for the formation of this isotope. Is this a transmutation reaction? If a 55.3 kg patient received a 0.50 s exposure to a 0.30 kg cobalt-60 source, what would the exposure be in rads? Predict the potential side effects of this dose. | textbooks/chem/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.E%3A_Nuclear_Chemistry_%28Exercises%29.txt |
Chemistry is the study of the material world. What are different materials made of? How is their composition and structure related to their properties? How does one material become transformed into another? These are the sorts of questions that have driven the development of chemistry.
People have been using chemistry for a very long time. Medicines were obtained from plants in early societies all over the world. People made dyes for clothing and paints for houses. Metallurgy was practised in India and the Sahel, in Africa, before 1000 BC.
The Greek philosopher, Democritus, is often cited as the earliest person to formulate an idea of atoms, although similar ideas were recorded in India around the same time. Democritus thought that all things were made of atoms. Atoms were very small, he thought. They were also indivisible. Although you could cut a piece of wood in half, and cut each of those pieces in half, at some point you would reach the stage at which the wood could not be cut any longer, because you had a slice that was one atom thick.
There were an infinite variety of atoms, Democritus thought, making up an infinite variety of materials in the world. The properties of those atoms were directly responsible for the properties of materials. Water was made of water atoms, and water atoms were slippery. Iron was made of iron atoms, and iron atoms were strong and hard.
We don't quite think of materials in the same way today, but we have retained some of the roots of these ideas. For example, we do know that matter is composed of atoms, and that there are different kinds of atoms that make different materials. However, we also know that those atoms can combine with each other in different ways to make lots of different materials.
All of the materials in the world around us are made from atoms.
A more practical aspect of chemistry has its roots in the Islamic Golden Age. Practitioners such as Jabir Ibn Hayyan developed laboratory apparatus and experimental methods to recrystallize and distill compounds from natural sources. Like Democritus, these early chemists wanted to know what the world is made of, but they were also trying to make improvements in practical applications such as tanning leather, making glass or rust-proofing iron.
Exercise \(1\)
Natural products are widely used in food, medicine and industrial applications. See if you can match the following natural products to their original, natural source.
i. vanillin a. coffee beans
ii. caffeine b. willow tree bark
iii. polyisoprene c. indigo plants
iv. aspirin d. yew tree needles
v. tannin e. vanilla beans
vi. the chemotherapy agent, paclitaxel f. rubber trees
vii. blue dye for jeans g. oak tree
The translation of Arabic texts into Latin helped spur the European Renaissance. Practical observations from the Islamic period, such as the fact that matter could be converted into different forms but did not disappear, gave rise to some of the most fundamental ideas of modern chemistry.
• Conservation of mass: matter can be converted from one form to another, but it does not disappear.
As is usually true in science, new developments in chemistry built on earlier work as well as the work of contemporary colleagues, continually improving our understanding of nature in small steps. In the 1600's, Robert Boyle adopted the Islamic emphasis on experimental work. Among his experiments, he was able to isolate the hydrogen gas formed by reacting metals with acids, as other scientists were doing at that time. In the 1700's, Joseph Priestley isolated several different "airs" or gases, including oxygen. Henry Cavendish showed that hydrogen combined with oxygen to form water. Antoine Lavoisier argued that oxygen and nitrogen, the other major component of air, are elements -- materials that cannot be made from any other kind of atom. In contrast, water was not an element, because it could be made by combining atoms of oxygen and hydrogen. The free exchange of ideas allowed people to rapidly advance our understanding of the material world.
Lavoisier, in particular, was important in bringing a number of important ideas together. He clearly stated that elements were the basic unit of matter, that could not be obtained from other materials. Compounds were made by combining different elements. His careful use of a balance to weigh reactants and products of an experiment clearly confirmed the idea of conservation of mass: the total mass of products after a reaction equals the total mass of reactants. These conclusions were more sophisticated versions of earlier ideas, and Lavoisier was able to present them in a way that eventually gained wide acceptance.
• A compound is a mixture of different elements bonded together.
• Conservation of mass, revised: the total mass of products after a reaction equals the total mass of reactants.
In the 1800's one of the principle proponents of the developing atomic theory was John Dalton. He advanced the idea that all atoms of a particular element are identical (as far as he could tell at that time). An element is a fundamental atomic building block from which other materials are made. Dalton performed analyses to try to deduce the atomic weights of different elements. Taking these ideas together, he showed that a particular compound always contained the same elements in the same ratio.
• An element is a fundamental atomic building block from which other materials are made.
• A compound is a mixture of different elements bonded together in a specific ratio.
• A compound may have a specific number of atoms of one type combined with a specific number of atoms of another type.
• Because all atoms have weight, we can also think of a compound as a specific weight of one kind of element combined with a specific weight of another element.
For example, water is a compound made from hydrogen and oxygen. It is crucial for life, of course. Water is about 1/9th hydrogen by weight; the other 8/9ths are oxygen. However, a different compound, hydrogen peroxide, is a rocket fuel. Hydrogen peroxide is only about 1/19th hydrogen by weight. Those specific ratios of hydrogen to oxygen are inherent qualities of each compound.
Furthermore, Dalton found that he could make compounds through different methods. For example, he could make cupric oxide (CuO) by heating copper in air, or he could make it through various reactions involving copper and acids. It didn't matter how he made the cupric oxide; the ratio of copper to oxygen was always the same in the product.
There is one other compound containing copper and oxygen in a different ratio; it is called cuprous oxide, and it has the formula Cu2O. However, it is very different from cupric oxide. The most obvious difference is that cuprous oxide is red whereas cupric oxide is black. Once again, when elements are combined in different ratios, different materials are produced, and they have properties that differ from each other and from they elements of which they are comprised.
Exercise \(2\)
Ratios are commonly used in baking. Usually, ingredients must be combined in the correct proportions in order to make brownies or a cake.
Brownies: Cake:
1 cup sugar 1 cup sugar
2 eggs 2 eggs
1/2 cup butter 1/2 cup butter
1/2 teaspoon vanilla 1/2 teaspoon vanilla
1/2 cup flour 1-1/2 cup flour
1/3 cup cocoa 1/2 cup cocoa
1/4 teaspoon baking powder 1 teaspoon baking powder
1/4 teaspoon salt 1/4 teaspoon salt
1/4 teaspoon baking soda
1 c boiling water
1. Brownies and cake have a lot of ingredients in common. Looking at those ingredients that are found in both brownies and cake, do you see any difference in the proportions used? How do you think that affects the properties of the product?
2. The above cake recipe is just for one shallow cake pan. If you wanted a two-layer cake, what would you do with the recipe?
3. Suppose you are cleaning out your fridge and want to turn all of your eggs into brownies. If the above recipe makes sixteen 2" x 2" brownies, how many brownies could you make with a dozen eggs? How much flour would you need to accomplish your goal?
4. You want to make some brownies but you don't have any measuring cups or spoons. You notice that there is a really nice balance in the chem lab and you decide to measure your ingredients there (it's a terrible idea, by the way). You find a list of conversions, including the following:
1 cup flour = 125 g; 1 cup sugar = 200 g; 1 cup cocoa = 90 g.
How many grams of flour, sugar and cocoa would you need to use for a batch of brownies?
e) Why do you think a cup of flour does not weight the same as a cup of sugar or a cup of cocoa?
Exercise \(3\)
Mercury forms two different compounds with oxygen: mercuric oxide (HgO, which is red) and mercurous oxide (Hg2O, which is black).
1. How many atoms of mercury combine with one atom of oxygen to form one unit of mercuric oxide, HgO?
2. How many atoms of mercury combine with one atom of oxygen to form one unit of mercurous oxide, Hg2O?
3. Given the following approximate atomic weights, what is the total weight of one unit of mercuric oxide? 1 mercury atom (Hg): 200 amu; 1 oxygen atom (O) : 16 amu
4. What is the weight of one unit of mercurous oxide?
Answer a:
one atom of mercury with one atom of oxygen
Answer b:
two atoms of mercury with one atom of oxygen
Answer c:
200 amu (Hg) + 16 amu (O) = 216 amu (HgO)
Answer d:
2 x 200 amu (Hg) + 16 amu (O) = 416 amu (HgO)
Exercise \(4\)
It's pretty difficult to weigh an individual atom. Because we are working with ratios, we can always scale up and keep the ratio of atoms the same, and we will just make more of the compound we want. Instead of weighing things in atomic mass units, we usually weigh them in grams.
1 mercury atom (Hg): 200 amu; 1 mol mercury: 200 g
1 oxygen atom (O) : 16 amu; 1 mol oxygen (O): 16 g
1 hydrogen atom (H): 1 amu; 1 mol hydrogen (H): 1 g
200 amu of Hg plus 16 amu of O makes 216 amu of HgO, which is just on unit of HgO. An atom of mercury weighs 200 amu and an atom of oxygen weighs 16 amu, so one atom of mercury combined with one atom of oxygen weighs 216 amu.
A mole of mercury weighs 200 g. A mole of oxygen atoms weighs 16 g. A mole is just a scaled-up batch of atoms; it is just the atomic mass number of the atom, but weighed in grams instead of amu.
1. How much does a mole of HgO weigh?
2. How many grams of mercury would be needed in order to make one mole of Hg2O?
3. How many grams of mercury would be needed to make 0.25 moles of HgO?
4. How many grams of oxygen would be needed to make 2.08 g Hg2O?
Answer a:
200 g (Hg) + 16 g (O) = 216 g (HgO)
Answer b:
2 x 200 g (Hg) + 16 g (O) = 416 g (Hg2O)
Answer c:
If one mole is 216 g of HgO, then 0.25 mole must be a quarter of that amount, or 54 g.
0.25 mol x 216 g/mol = 54 g
Alternatively written as
0.25 mol x 216 g mol-1 = 54 g
The amount of mercury is just a fraction of that: 200/216. So the answer is (200/216) x 54 g = 50 g.
Answer d:
What fraction of a mole is 2.08 g of Hg2O, if one mole is 416 g?
2.08 g / 416 g mol-1 = 0.005 mol
Notice that when we divide g by g mol-1, the grams cancel and the mol-1 becomes mol.
There is 1 mol of O in 1 mol of HgO, so 0.005 mol of O are needed for 0.005 mol HgO.
0.005 mol x 16 g mol-1 = 0.08 g O needed
Exercise \(5\)
Priestley isolated oxygen by heating up mercuric oxide. How much oxygen could be made by heating 1 g of HgO?
By the late 1800's, enough different elements had been isolated that people began to notice patterns in their properties. If you listed the elements out by weight, elements with similar properties seemed to occur at regular intervals throughout the list. A Russian chemistry teacher, Dmitri Mendeleev, came up with a convincing way to convey this "periodicity" in a table. This is the modern periodic table (an example is shown in figure \(1\)).
Download a copy of the periodic table.
The noble gases are an example of one of the groups in the periodic table. They have a number of properties in common. As the name suggests, all of the noble gases exist in the vapour phase at normal room temperature. They are all pretty unreactive; they do not undergo chemical changes very easily. In terms of structure, they each exist as individual atoms in nature. The rest of the elements in the periodic table don't really exist as individual atoms in nature. In one way or another, they cluster together with other atoms. They may bunch up with other atoms of the same kind, forming an element, or else form groups with atoms of different kinds, forming compounds. Noble gases rarely do either of those things.
Because of their similar qualities, Mendeleev put the noble gases together in one column in the periodic table. They are helium, neon, argon, krypton, xenon, and radon.
Exercise \(6\)
1. Which of the following elements are in the same group as oxygen: silicon, sulfur, or neon?
2. Which of the following elements would have similar properties to carbon: silicon, nitrogen, sodium?
3. Which of the following elements would have similar properties to magnesium: calcium, lithium, potassium?
4. Which of the following elements would have similar properties to sodium: nitrogen, potassium, lithium?
5. Which of the following elements would have similar properties to nitrogen: oxygen, phosphorus,rubidium?
6. Which of the following elements are in the same group as fluorine: argon, iodine, or neon?
7. Which of the following elements are in the same group as gold: silver, copper, zinc?
8. Which of the following elements are in the same group as iron: ruthenium, copper, gold?
Answer a:
sulfur
Answer b:
silicon
Answer c:
calcium
Answer d:
potassium and lithium
Answer e:
phosphorous
Answer f:
iodine
Answer g:
silver and copper
Answer h:
ruthenium
In a similar way, chemists began thinking about other columns in the periodic table as belonging in groups. Sometimes, these groups were given specific names, like the noble gases. There are the halogens, meaning "salt-makers", because they were often contained in salts. There are the chalcogens, "ore-makers", found in iron oxides such as haematite and taconite, for example. There are the pnictogens, "choke-makers", so called after the discovery that air is mostly composed of nitrogen, but nitrogen is not the part of air that we breathe. Other groups are simply referred to as the self-explanatory "carbon group" or "boron group" after the first element in the column.
Alternatively, the columns of the periodic table are often numbered. At one time, the columns were separated into different groups, with specific focus on two of them: the main group and the transition metals. The main group was Group A and the transition metals constituted Group B. Each successive column across these blocks of the periodic table was then numbered using Roman numerals.
One of those periodic properties that is found in each column is the usual charge on an ion. Atoms sometimes capture extra electrons, which are negatively charged sub-atomic particles, becoming negatively charged overall. Other atoms sometimes lose their electrons, becoming positively charged as a result. To some extent, the column number corresponds to the charge on the common ions found in that column. For example, in the first column, the atoms are almost always found as ions with a +1 charge; in the second column, the atoms are almost always found as ions with a +2 charge, and so on.
That correlation breaks down after a while. Sometimes, as in the transition metals, the group number simply denotes the highest possible charge on an ion, although there may be many possible ions that are commonly observed. Eventually, as we move to the right, the correlation between group number and charge breaks down altogether. In the main group, the atoms farther to the right are not cations at all, but anions. Some, such as carbon and nitrogen, can be formally thought of as either anions or cations, depending on their environment.
In a sense, although we like to be organized and number things from left to right, the periodic table is naturally organized from the edges inward. Atoms closest to the left edge build up positive charge according to how far away from the edge they are. Atoms closest to the right edge build up negative charge according to how far away from the edge they are. Atoms in the middle vary a lot more, though.
This edge-orientation is related to the stability of the noble gases and what are called "filled-shell configurations". We will get to those, and to the physical reasons for their stability, on a later page.
Now, eventually in science, someone always decides that there is a better way of doing things, and so if you have just purchased a brand-new periodic table to display on your bedroom wall, you might just see the Arabic numerals 1-18 across the top. This system is meant to be a simpler and cleaner way of doing things. (And once again, the Lanthanides and Actinides, those two lonely rows orphaned at the bottom of the periodic table, are more-or-less ignored. That's partly because, for reasons that are a little complicated at this point, they mostly behave as if they were part of Group 3.)
So, the columns in the periodic table are designed to show us which elements are related to each other. What else is shown in the table? If we look at just one square of the periodic table, we can see what it tells us about an individual element.
In the middle of the square, there is a Roman letter or a pair of Roman letters. The first letter is always a Roman capital. This one- or two-letter designation is the atomic symbol. It is usually the first couple of letters in the name of the atom, although the name isn't always in English. For example, the symbol for mercury is Hg, for hydrargyrum. That's the Latin name, meaning, roughly, "liquid silver".
Above the symbol, or sometimes above and to the left, is the atomic weight of the atom. The atomic mass is one of the properties of an element, but it turns out that a given element can have a range of atomic weights. Consequently, the average weight is shown in the periodic table. For example, the smallest atom, hydrogen, can have a weight of either 1, 2, or 3 atomic mass units (or amu). The average weight of a hydrogen atom is 1.008 amu. That fact tells us that most of the hydrogen atoms out there have a weight of 1 amu; relatively few of them have atomic weights of 2 amu or 3 amu.
Most of the mass of an atom comes from the nucleus of the atom, where relatively heavy particles, the protons and neutrons, are found. Hydrogen atoms always contain a proton, with a mass of 1 amu, but they can also contain varying numbers of neutrons, which also have a mass of 1 amu.
(Many scientists prefer the term "mass" when talking about atoms, reserving the term "weight" to describe how heavy something feels because of the pull of gravity. Here, we will use the term "force due to gravity" if we want to discuss the latter concept.)
The number to the lower left is called the atomic number. Notice that each element in the periodic table has a number assigned from 1 to 112. The elements are numbered in order from "smallest" to "largest", roughly speaking. Element 1, hydrogen, is the smallest, and element 112, copernium, is the largest (at least, that's as far as the teams of chemists and physicists who look for new elements have gotten so far). More correctly, the atomic number simply tells us the number of protons in the nucleus of an atom of a given element. Hydrogen always has one proton; copernium always has 112 of them.
Finally, the name of the element may or may not be written on the periodic table. The other three pieces of information are always there. Sometimes additional information is given, such as specific properties of the element.
Exercise \(7\)
Supply the following information about each element.
1. The atomic number of calcium.
2. The atomic symbol of sodium.
3. The atomic mass of oxygen.
4. The name of P.
5. The name of element 6.
6. The symbol of tin.
7. The atomic mass of sulfur.
8. The name of K.
Answer a:
20
Answer b:
Na
Answer c:
16.00
Answer d:
phosphorous
Answer e:
carbon
Answer f:
Sn
Answer g:
32.07
Answer h:
potassium
The periodic table as laid out by Mendeleev had predictive value. It presented the structure-property relationships of atoms. If you knew something about one element in the table, it would lead you to believe that other elements in the same column would have similar properties. Furthermore, there were gaps in the table where there ought to be an element, but none was known. It was predicted that these elements would eventually be discovered, and they were.
However, people were not satisfied with the idea of the atom as the basic building block of the universe. People wanted to know how atoms themselves were made. Ultimately, answering this question depended on the development of quantum mechanics. We will learn a little bit more about that view of the atom next.
Exercise \(8\)
What is the approximate weight, to the nearest g, of a mol of each of the following elements?
a) neon, Ne, found in brightly-colored store signs
b) iron, Fe,used to make steel
c) copper, Cu, found in power lines
d) gold, Au, found in jewelery
e) silicon, Si, used in high-tech devices
Answer a:
Ne: 20 g/mol
Answer b:
Fe: 56 g/mol
Answer c:
Cu: 64 g/mol
Answer d:
Au: 197 g/mol
Answer e:
Si: 28 g/mol
Exercise \(9\)
What is the weight, to the nearest gram, of a mol of each of the following compounds?
a) water, H2O
b) baking soda, NaHCO3
c) sand, which is mostly silica, SiO2
d) table salt, NaCl
e) tantalum nitride, TaN, used in electronics
f) ammonia, NH3, found in some cleaning solutions
g) sodium dihydrogen phosphate, NaH2PO4, used in toothpaste
h) acetic acid, CH3CO2H, which is dissolved in water to make vinegar.
Answer a:
H2O: 18 g/mol (2 x H + 1 x O = 2 x 1 + 1 x 16 g/mol)
The molar weight of the atom was rounded to the nearest gram. In order to be very careful and avoid "error propagation", the more exact molar weight (to several decimal places) could be used and rounding could be performed after the calculation. In these cases, the result would be the same. The result could be different if very large numbers of atoms were found in the compound.
Answer b:
NaHCO3: 84 g/mol (3 x O + 1 x Na + 1 x H + 1 x C = 3 x 16 + 1 x 23 + 1 x 1 + 1 x 12 g/mol)
Answer c:
SiO2: 60 g/mol (1 x Si + 2 x O = 1 x 28 + 2 x 16 g/mol)
Answer d:
NaCl: 58 g/mol (1 x Na + 1 x Cl = 1 x 23 + 1 x 35 g/mol)
Answer e:
TaN: 195 g/mol (1 x Ta + 1 x N = 1 x 181 + 1 x 14 g/mol)
Answer f:
NH3: 17 g/mol (1 x N + 3 x H = 1 x 14 + 3 x 1 g/mol)
Answer g:
NaH2PO4: 120 g/mol (1 x Na + 2 x H + 1 x P + 4 x O = 1 x 23 + 2 x 1 + 1 x 31 + 4 x 16 g/mol)
Answer h:
CH3CO2H: 60 g/mol (2 x C + 4 x H + 2 x O = 2 x 12 + 4 x 1 + 2 x 16 g/mol) | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.01%3A_From_Democritus_to_the_19th_Century-_H.txt |
Today, we know that atoms contain protons, neutrons and electrons. The protons have significant mass and a positive charge and are found in the nucleus of the atom. The neutrons have mass but no charge and are also found in the nucleus. The electrons have negative charge and very little mass and are found outside the atom's nucleus. The weight of an atom in atomic mass units is approximately the sum of its protons and neutrons, since the electrons don't have much mass.
For example, a typical carbon atom has six protons and six neutrons, and it has an atomic weight of 12 amu. A carbon also has six electrons, but they are so small that they don't contribute to carbon's weight.
• An element is defined by the number of protons in its nucleus.
• The number of protons in an atom is equal to the number of electrons, to balance the charge.
Some carbon atoms have an extra neutron or two, so carbon may have an atomic weight of 13 or even 14 amu. However, a carbon atom can't have an extra proton; an extra proton would make it a nitrogen atom. It is the six protons that make the atom behave like carbon. Many other elements may have slightly different atomic weights, depending on how many neutrons are found in the nucleus. Different atoms of the same element with different weights are called isotopes. For example, 12C, 13C and 14C are all isotopes of carbon. They all have six protons, but different numbers of neutrons, as seen in a model of 12C and 13C, below.
• Neutrons are also in the nucleus.
• A neutron has a mass similar to a proton, but has no charge.
• Compared to protons and neutrons, the mass of an electron is very small.
Exercise $1$
An element's atomic number is just the number of protons in an atom of that element. Given the following atomic numbers and atomic weights, identify the number of protons, neutrons and electrons in an atom of the element.
1. oxygen: atomic number = 8, atomic weight = 16
2. phosphorus: atomic number = 15, atomic weight = 31
3. zinc: atomic number = 30, atomic weight = 65
4. gold: atomic number = 79, atomic weight = 197
Answer a:
O: 8 protons, 8 electrons, 8 neutrons
The atom is neutral overall, so the number of positively charged protons is equal to the number of negatively charged electrons. The atomic weight is provided almost entirely by the protons and neutrons, so the number of protons plus number of neutrons equals the atomic weight.
Answer b:
P: 15 protons, 15 electrons, 16 neutrons
Answer c:
Zn: 30 protons, 30 electrons, 35 neutrons
Answer d:
Au: 79 protons, 79 electrons, 118 neutrons
Exercise $2$
If a proton's mass is 1.67 x 10-27 kg and the mass of an electron is 9.11 x 10-31 kg, how many times heavier is a proton than an electron?
Exercise $3$
If carbon in nature is about 99% 12C and 1% 13C, then what is the average weight of a carbon atom?
Answer
Carbon: [ (99 x 12 amu) + (1 x 13 amu) ] / 100 = [1,188 + 13 amu] / 100 = 1,201 amu / 100 = 12.01 amu
Exercise $4$
Note that 14C is even rarer than 13C, because 14C is converted into 14N via radioactive decay. In that event, a high-energy electron is emitted from the 14C nucleus. Explain how that emission must convert the carbon into a nitrogen, and indicate how many protons and neutrons are found in the resulting nucleus.
Answer
The negatively charged electron's departure from the nucleus leaves behind a positive charge. A neutron is converted into a proton. The overall atomic weight remains the same, but the atom ends up with one more proton and one more electron. A 14C is converted into a 14N.
Exercise $5$
Magnesium in nature is found in three major isotopes. It is nearly 79% 24Mg, about 11% 25Mg and 12% 26Mg. What is the average weight of a magnesium atom?
Exercise $6$
Chlorine in nature is found in two major isotopes: 35Cl and 37Cl. If the average atomic weight of chlorine is about 35.5, what percentage of each isotope is found in nature?
Answer
Suppose y is the decimal fraction of 35Cl and z is the decimal fraction of 37Cl.
35.5 amu = (y x 35 amu + z x 37 amu)/100 but y + z = 1
35.5 amu = (y x 35 amu + (1-y) x 37 amu)
35.5 amu = y x 35 amu - y x 37 amu + 37 amu
(37-35) x y amu = 37 amu - 35.5 amu
2 x y amu = 1.5 amu
y = 0.75 (75% 35Cl)
z = 0.25 (25% 37Cl)
A number of developments at the beginning of the twentieth century led to our current understanding of the structure of atoms and molecules a hundred years later. At that time, some people though protons, neutrons, and electrons were lumped together in the atom. This view of the atom was called the "pudding model" of the atom.
Ernest Rutherford first proposed that an atom contains a very small, positively charged nucleus surrounded by empty space. The electrons orbited far away from the nucleus.
Rutherford was explaining the result of an experiment in which alpha particles (positively charged helium ions) were fired at a gold foil. Most of these particles passed through the foil easily, suggesting there was a lot of empty space in the material. However, some of the particles bounced directly back, having collided with the small, highly charged nuclei. The positive alpha particles were powerfully repelled by the positive nuclei, because like charges repel each other. He didn't really know much about the location of the electrons, the negatively charged particles in the atom, but believed they orbited the nucleus like planets around the sun.
Why weren't the electrons found in the nucleus? If electrons are attracted to protons, it seems like that's where they should be. Niels Bohr suggested that electrons are found only in specific, allowed orbits at different distances from the nucleus.
That conceptual leap to specific, allowed orbits marks the introduction of quantum mechanics into the understanding of the atom. Quantum mechanics is based on the idea that on a very small scale, many properties only have specific values (like 1, 2, 3...) instead of any value at all (like all the possible fractions between these integers). In other words, in the world around us, we usually view things like walking up a ramp. We can heat a pot of water just a little bit warmer, and just a little bit warmer than that, and so on. On the atomic scale, however, the world is more like walking up a set of stairs. Maybe you could heat the water to 30 oC or 40 oC, but heating to 35 oC would be impossible, because heat only comes in 10 degree packages. That is, in fact, how the quantum world really works, but on the human scale, the steps involved are so tiny that we cannot notice them.
Bohr's model was also consistent with the earlier idea of the periodic table of the elements. The idea is that electrons are found in different "shells" that are each further and further from the nucleus. Each of those shells corresponds roughly to a different row in the periodic table. Hydrogen and helium have electrons only in the first shell, and we see those two elements in the first row of the periodic table. Carbon and oxygen's outermost electrons are found in the second shell, so they show up in the second row of the periodic table.
Each row in the periodic table corresponds to an outer layer of electrons that are found further from the nucleus than the outermost electrons in the row before it. We are going to see eventually that there is a further variation on this idea, but it is still pretty much the way we see the periodic table today. Hydrogen starts the first shell, lithium the second, sodium (the Latin, natrium) the third, potassium (Latin, kalium) the fourth, and so on.
The variation we are going to see involves that dip in the middle of the periodic table. Scandium through zinc have outer electrons that are only in the third shell, not the fourth. The third and the fourth shell overlap a little bit, so that some electrons actually start to go into the fourth shell (as in potassium and calcium), then finish filling the third shell across the transition metals. The reasons for that also have to do with quantum mechanics, but we will need to learn a little more about energy and waves before we see why.
Bohr showed that electrons might be found in specific orbits around the nucleus. He also showed that electrons in these different orbits have specific amounts of energy. By doing this mathematically, he was offering an explanation to an important problem. People knew that atoms can absorb energy (they can be heated in a flame, for example) and give the energy back again in the form of light. Rather than give off light of all colors when excited, atoms only give off very specific colors. For example, heating lithium salts in a flame produces a red color, but heating sodium salts produces an orange color, whereas potassium salts produce a purple color, and so on.
These colors can be separated and studied using a prism. When people did that, they found that a given atom does not produce just one pure color of light, but several different ones. When separated by a prism, the light given off by an excited compound could be seen against a dark surface as several different, colored lines. These were called emission lines.
It had been known since the early 1800's that light had wave properties, and that light of different colors had different wavelengths. For example, red light consistes of electromagnetic waves, with a wavelength of about 700 nm, but blue light's wavelength is about 450 nm. That means a color can actually be measured numerically. Because of that fact, people can look for mathematical relationships between the emission lines observed for different atoms. Those mathematical relationships may reveal something about the atoms themselves.
Furthermore, it was known that different wavelengths of light corresponded to different amounts of energy. In one of the first developments in quantum mechanics, Max Planck in 1900 proposed that light travels in bundles called photons. Although they are particles, these photons do have wave properties. The amount of energy in a photon of light corresponds to its wavelength.
By proposing that electrons could be found only in specific orbits, specific distances away from the nucleus, Bohr was trying to explain observations from atomic spectroscopy reported by another scientist named Rydberg. Rydberg had found a mathematical relationship between the wavelengths of these emission lines. Bohr thought that, when energy was added, electrons could be excited from one energy level (or orbit) to a higher one. When the electron relaxed back to its original orbit, it gave off the energy it had gained in the form of light. The specific emission lines occur because electrons are found at very specific energy levels in an atom, so a drop from one level to another always produces the same amount of light energy. That specific amount of light energy has a specific color.
Bohr then used the mathematical relationships describing electrostatic attraction and centripetal force to show that his model of the atom was consistent with Rydberg's relationship. In fact, he could use his model to predict the emission lines of an atom.
Exercise $7$
Bohr's explanation of atomic structure built on Rydberg's observation of a numerical series in spectral emission lines. Solving a series involves finding a pattern in numbers. Find the patterns among the following sequences of numbers, and predict the next number in the sequence.
1. 1, 2, 3, 4...
2. 2, 4, 6, 8...
3. 3, 5, 7, 9...
4. 1, 4, 9, 16...
5. 2, 4, 8, 16...
6. 1, 1/2, 1/4, 1/9...
Answer a:
1, 2, 3, 4, 5... a series of whole numbers, or n.
Answer b:
2, 4, 6, 8, 10... a series of even numbers, or 2n (because every even number is two times another number).
Answer c:
3, 5, 7, 9, 11... a series of odd numbers, or 2n + 1 (because every odd number is one more than some even number).
Answer d:
1, 4, 9, 16, 25... a series of squares, or n2.
Answer e:
2, 4, 8, 16, 32... a series in which each numer is double the last number, or 2n.
Answer f:
1, 1/2, 1/4, 1/9, 1/16... a series of reciprocals of squares, or 1/n2.
Exercise $8$
Bohr's idea depended partly on the use of Coulomb's Law of electrostatic attraction. Coulomb's law is expressed mathematically as follows:
$F = \frac{kq_{1}q_{2}}{r^{2}}$
in which F is the attractive force between two charged particles, q1 and q2 are the charges on the two particles, r is the distance between the two particles and k is a constant. A large value of F means that the charges are strongly attracted to each other.
1. Suppose q1 is the charge on the nucleus of an atom and q2 is the charge on an electron. What happens to the force of attraction between an electron and the nucleus when the charge in the nucleus increases? Explain.
2. Suppose r is the distance from the electron to the nucleus. What happens to the force of attraction between an electron and nucleus when the electron gets further from the nucleus?
3. Using the ideas of Coulomb's law, compare the attraction of an electron to the nucleus in a helium atom versus a hydrogen atom.
Answer a:
When q increases, F increases. The increase is linear: if q1 doubles, F doubles. As the charge in the nucleus gets larger, the force of attraction gets larger.
Answer b:
When r increases, F decreases. The decrease is nonlinear: if r doubles, F drops by a factor of four, rather than a factor of two. As the distance from the nucleus gets longer, the attraction to the nucleus drops sharply.
Answer c:
Hydrogen and helium are both in the first row of the periodic table. To a rough approximation, the distance between nucleus and electron is similar in these two atoms. However, helium has a charge of 2+ in its nucleas, compared to the 1+ charge in the nucleus of a hydrogen atom. As a result, the attraction of an electron on helium to the nucleus would be about twice as great as the attraction of an electron on hydrogen to its nucleus.
Helium's electrons are much more tightly held than hydrogen's.
The situation is really much more complicated than that. For example, if helium's electrons are more tightly attracted to the nucleus than hydrogen's, then helium's electrons ought to be pulled closer to the nucleus than hydrogen's. That means helium's electrons are held even more tightly than we at first thought.
Another complicating factor is that helium has two electrons, whereas hydrogen has only one. An electron may be attracted to the nucleus, but electrons repel each other. That second electron in helium should offset the extra attraction to helium's more positive nucleus. That means helium's electrons may be less tightly attracted than we originally thought.
However, the effect of the second electron is much smaller than it first appears. That's because the second electron could be anywhere around the helium atom. It has a 50% chance of being further away from the first electron than that positively charged nucleus. The farther it is away, the smaller its influence. Helium's electrons are definitely more strongly attracted to the nucleus than are hydrogen's, but it is difficult to say exactly how much more without the help of some more sophisticated tools.
Exercise $9$
Max Planck described the energy of a photon using the following relationship:
$E = h \nu \) or E = \frac{hc}{\lambda} \nonumber$
In which E = energy; ν =frequency; λ = wavelength; c = speed of light; h = Planck's constant
1. What happens to the energy of light as its wavelength gets longer?
2. What happens to the energy of light as its frequency gets higher?
Answer a:
As wavelength gets longer (value of λ increases), energy decreases.
Answer b:
As frequency gets higher (value of ν increases), energy increases.
Other people were familiar with these ideas and already knew about the relationship between light and energy. Bohr's model of the atoms put all of these ideas together to successfully explain a specific atomic property:
color = wavelength = energy of light = energy between electron levels.
In other words, an excited electron can drop back to its original orbit by giving off a photon with an energy exactly the same as the difference in energy between the two orbits ("excited state" and "ground state" orbits).
However, Bohr did not explain why electrons would be found at specific energy levels in the first place. Louis de Broglie, a historian-turned-physicist, solved this problem with the idea of wave-particle duality. de Broglie put together the following ideas:
• All moving particles have wave properties.
• Electrons move around the nucleus and they have wavelengths.
• To maintain a complete standing wave along its orbit, an electron can only adopt orbits of specific circumferences. Otherwise, one end of the wave would not meet up with the other end, and it would interfere with itself.
• Orbits with specific circumferences have specific radii.
• Electrons are found at specific distances from the nucleus, but not at other distances.
One way to illustrate why an electron might have only certain allowed orbits is via the "particle in a box", a basic concept from quantum mechanics. If a particle has wave properties, then it has a wavelength. Its wavelength depends on certain conditions. By analogy, if you take a guitar string and attach it to the ends of a box, the string can only vibrate at certain frequencies. That's how guitarists can change the note played on a guitar string. By pressing one end of the string against a fret on the guitar neck, the length of the string is changed, and so is its allowed wavelength, so it makes a different sound.
The string can't move at the two points where it is held. That means the wave has to form in such a way that it returns to the same position at both ends. Because of that, certain wavelengths won't work, because the wave won't be able to return to that correct position at the far end.
Furthermore, the allowed wavelengths of a guitar string also depend on the thickness of the string. As a result, there are two conditions that control the tone that is played: which of the six guitar strings is plucked, and where the string is held against the frets.
The same thing is true with very small particles that have wave properties. These particles can have only certain wavelengths that fit their surroundings. An electron has some property, analogous to the thickness of a guitar string, that limits its possible wavelengths. Given those limits, there are only certain orbits allowed the electron. If its orbit does not have the right circumference, the electron will not be able to form a complete wave along that orbit.
These ideas ushered in a revolution in science. Quantum mechanics is a very powerful tool. It can be used to accurately predict how molecules will behave. Unfortunately, the mathematics involved in quantum mechanics are one or two math courses beyond what most introductory chemistry students are familiar with. Even so, a qualitative feel for some of the consequences of quantum mechanics is important enough that we should explore it.
Exercise $10$
Complete the table with the appropriate information.
Table for Exercise 1.2.10
Element Symbol Atomic Number Mass Number Number of Protons Number of Neutrons Number of Electrons Charge
H 1 1
H 1 0
H 1 2
H 2 1
H 3 1
4 9 2 +2
6 12 6 6
12 13 12 0
43 55 43
20 40 +2
Si 14 28 0
19 28 +4
Fe 26 30 23
35 44 -1
K 22 17 21
15 15 0
13 27 +3
S 16 16 0
Pd 106 46 +1
24 28 21
50 68 50
Hg 80 120 79
79 118 78
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.02%3A_Old_Quantum_Mechanics-_Basic_Developme.txt |
Bohr's idea that electrons are found in different orbitals or energy levels was an important step in understanding the structure of an atom. Louis de Broglie's particle-wave relationship was also a crucial development.
Werner Heisenberg and Erwin Schrödinger were able to take these ideas and develop modern quantum mechanics. The main significance of quantum mechanics is the ability to very accurately predict physical properties using basic mathematical principles. Heisenberg and Schrödinger used different kinds of mathematics to explain atomic structure, but they ultimately reached similar results.
The key approach by Schrödinger was to recognize the factors influencing the energy of an electron in an atom. He saw that there would be contributions from Coulomb's law, because of attractive forces between the electron and the nucleus as well as repulsive factors between different electrons. In addition, Schrödinger recognized that there would be a kinetic energy component related to the wavelength of the electron. Schrödinger combined these factors into "the Schrödinger wave equation."
Solving the wave equation is very useful. A solution to the equation, which is called a wave function, can indicate the energy of an electron in an atom. The solution can also be combined with other mathematical relations that will give specific predictions of different properties of atoms and molecules.
• The Schrödinger wave equation is an approach to describing an electron.
• The solution to the wave equation is called a wavefunction.
If you aren't familiar with how waves behave, you might start by looking here.
Exercise \(1\)
1. What happens to the energy of an electron as its wavelength gets shorter?
2. What happens to the energy of an electron as its wavelength gets longer?
3. If an electron is confined in a small space, what will happen to the electron's wavelength?
4. If an electron is allowed to expand into a large volume, what will happen to the electron's wavelength?
Answer a:
The energy of the electron gets higher.
Answer b:
The energy of the electron gets lower.
Answer c:
The wavelength gets shorter.
Answer d:
The wavelength gets longer.
Apart from showing us the energy of the electron, the wavefunction carries other information. One additional item is the idea of phase. This idea is illustrated with a sine wave (although a wavefunction of an electron is a more complex, three-dimensional wave).
One of the consequences of wave behavior is the presence of nodes along a wave. A node is a place where the wave changes phase; in a sine wave, the wave goes from a "trough" or valley or negative phase to a "peak" or hill or positive phase. A node is also place where the amplitude of the wave is zero. For an electron, the amplitude of the wave is related to the chance of an electron being found at that position (you'll see more about that below). We wouldn't find an electron at a node, but it could be in one of the other positions along the wave.
Other important consequences of wave properties in electrons include interference behavior. If two electrons are sent through parallel slits, a diffraction pattern results, like ripples from two different stones overlapping in a pond. That's because the waveforms of the two electrons interfere with each other. When the two waves overlap, they interact with each other in different ways, depending on whether they are in phase or out of phase.
If the waves are in phase with each other, constructive interference occurs, and a taller wave results. The peak of one wave adds onto the peak of another, and the trough of one wave falls into the trough of another. However, if the two waves are out of phase with each other, the peak of one falls into the trough of each other, and the wave disappears. That phenomenon is called destructive interference.
One of the funny consequences of quantum mechanics is that you don't need two different electrons to cause an interference pattern. A single electron can be sent through a pair of parallel slits and result in an interference pattern. The electron passes through both slits at the same time. Sometimes an electron does not behave like a solid particle at all.
• Wave properties of electrons lead to interference patterns.
The wave properties of electrons also become important when considering bonding between atoms. As atoms are brought together to share electrons, constructuve and destructive interference between electrons leads to different consequences in different situations. We will look at these factors in a later chapter.
We mentioned that mathematical operations are sometimes performed on the wavefunction in order to predict other properties. For example, if you want to get an idea of where an electron is, you have to square the wavefunction. You take the wavefunction and multiply it by itself.
We lose some information when we perform math on the wavefunction. We no longer have "amplitude"; instead we have a probability. The curve we see is a statistical distribution that tells us where the electron is most likely to be found (the highest point on the curve) as well as many other places where the electron might also be found (the rest of the curve). We no longer have phase information, either; when the negative portion of the wave was multiplied by itself, it gave a positive number.
• The wavefunction can be used to narrow down the probable location of an electron.
Exercise \(2\)
What is the probability of finding an electron at a node?
Answer
The probability of finding an electron at a node is zero.
It is useful to keep the idea of particle-wave duality in mind, for a number of reasons. When we think of an electron as a particle, like a tennis ball, it is easy to expect that it will behave like a real tennis ball. After all, we know where the tennis ball is; you can see it, right there. But if an electron is a wave, like a wave on the ocean, it is not really restricted to one location. First of all, it's spread out (maybe over many miles, in the case of some waves). Secondly, it does not sit still long enough for you to decide it's exactly here or there. That's a little more like an electron. Also, if you had two holes in a wall, you could only throw a tennis ball through one hole at once. If a wave splashed against the wall, though, it could pass through both holes at once.
So, instead of thinking of an electron as a ball, maybe we should think about it as a cup of water. There's a stiff wind blowing, so there are waves on the surface of the water. Also, there's no gravity, so we can take the cup away and the water will hold together there, with its waves. Of course, the electron is no more liquid than it is a solid, but we often need familiar analogies to remind us of these different aspects of something we can't really see.
When Schrödinger worked out solutions for his equation, he found evidence of quantization again. Bohr had found that the distance between the electron and nucleus was quantized. However, Schrödinger found four different electronic properties that were quantized. One of these properties of an electron, called the principle quantum number, corresponds roughly to Bohr's quantized distance from the nucleus.
Two of the other properties have to do with the electron's direction from the nucleus. As it happens, a specific electron might be found in any direction from the nucleus. Alternatively, it might only be found in one direction or another.
Thinking in just two dimensions, suppose you have three tennis balls: a yellow one, a red one and a blue one. Imagine you are the nucleus of an atom and the tennis balls are electrons. In Bohr's model, the tennis balls might be found different distances away from you. Maybe one is five feet away, one is ten feet away, and one is twelve feet away. They cannot be found three feet away from you, or nine feet away or eleven feet away. They have to be found at those exact distances. However, they might be found in any direction: in front of you, behind you, to the left or right. They could all be lying in the same direction or all in different directions.
What Schrödinger's math seemed to be saying was that there was an additional restriction on where electrons could be found. In addition to a specific distance, the electron might be found only in certain directions from the nucleus. That direction relied on those other two quantized properties, called the magnetic and azimuthal quantum numbers. For one set of values, the electron could be found in any direction at all. For another set of values, though, the electron was restricted to lie along one particular axis.
• According to Schrödinger's solution, an electron in an atom is described by a group of "quantum numbers".
• The quantum numbers limit an electron to a general distance from the nucleus (like the shell in the Bohr model).
• Other quantum numbers limit the electron to certain regions of space around the atom.
In this model, maybe the yellow tennis ball has to be five feet way, but can be in any direction. But maybe the blue tennis ball has different quantum numbers. It is ten feet away and it can only be directly in front of you or directly behind you, but cannot be to your left or right. The red tennis ball, on the other hand, can only be found to your left or right, but never in front of you or behind you.
There is also a fourth quantum number, and it is called spin. Spin can't be described very well because it does not have an analogous big-world property such as distance or direction. However, it is an inherent property of an electron that can have either of two values. These values are sometimes called alpha and beta, or sometimes +1/2 and -1/2.
Despite the abstract nature of spin, it does have a real physical quality associated with it: if you place an electron in a magnetic field, it will interact differently with the magnetic field, depending on the value of its spin. Hence, spin is closely associated with magnetic properties of materials.
One of the basic rules of quantum mechanics is that no two electrons on the same atom can have the same two quantum numbers. You'll need some extra tennis balls to see how that works. Suppose you have two red balls, two yellow and two blue. The first yellow ball is found five feet away from you, in any direction. The second yellow tennis ball is also found five feet away from you and in any direction. So far, the yellow tennis balls have the same properties, so they must have different spin. Of course, you can't tell that they have different spins unless you put them in a very strong magnetic field, but you know they are different because you have faith that Schrödinger did his math correctly.
The two blue balls are ten feet away from you and they can only be found in front of you or behind you. It does not matter which. They could both be in front of you or both behind you or one in either direction. They also have two different spins.
Finally, the two red balls are also ten feet away from you, but they are found to your right or left (or both). They don't have to be a different distance away from you than the blue ones because they already have a different directional property, so we can repeat the distance property. The red balls also have two different spins, but you are having trouble telling which is which.
• Spin is another quantum mechanical property of electrons.
• Two electrons in the same atom with all other quantum numbers the same must have different spin.
• Spin has no real "big world" analogy. However, two electrons with different spin interact differently with a strong magnetic field.
One more thing. The quantum world is a little more subtle than that. The quantum world also involves concepts such as the uncertainty principle. The uncertainty principle says it is difficult to find an electron's exact location. Instead, we work with probability. Instead of being five feet away exactly, we only know that the yellow ball has a very good chance of being five feet away. There is a slight chance it is four and a half feet away, and a very small chance it is only four feet away.
Similarly, the red tennis ball may not be exactly in front of you, although it is probably no more than a few feet to the left or right. The blue tennis balls may also be a few feet off their axis someplace. We can't predict exactly where they will be.
Exercise \(3\)
Let's do an exercise in pointillism. You will need a pencil, a ruler, and pens of four different colors. Gel pens work better than ballpoint pens. In a pinch, you could use crayons, colored pencils or colored markers.
1. Draw a line in pencil 6 cm long. Mark the midpoint of the line.
2. Draw a second line 6 cm long and perpendicular to the first line, so that the midpoints of the two lines intersect. You should have a nice X or cross with arms 3 cm long.
3. Lightly draw in dashed lines, 10 cm long, halfway in between the prvious lines, so that the midpoints of all the lines intersect. You should have a new X or cross with arms 5 cm long, superimposed on the old cross but rotated by 45 degrees.
4. Choose one of the solid lines to work on. Place two marks along that line, 1 cm away from the midpoint in each direction. Using the first pen, start putting dots on the paper, centered on one of those 1 cm points. Keep adding dots as you move away from that 1 cm point in all directions, but there should be fewer and fewer dots as you go out. Expand your dots the same distance in all directions. Try not to cross the dashed lines, however. Aim for about a hundred dots (or 50 if you aren't very patient or if you are using a marker). You should end with a roughly circular mass of dots, centered 1 cm from the midpoint; the mass should get fainter along its edges and heavier and darker towards its center.
5. Now do the same thing, using the same pen, at the other 1 cm point along the same line.
6. Now do the same thing, with a second pen, 1 cm away from the midpoint in both directions along the second solid line.
7. Now do the same thing, with a third pen, 3 cm away from the midpoint in both directions along the first solid line.
8. Now do the same thing, with a fourth pen, 3 cm away from the midpoint in both directions along the second solid line. You should end up with four pairs of colored spots. Each spot should have higher value (heavier and darker) in the center and should fade to lower value away from its center.
Answer
Your drawing is beautiful. You should put it on your fridge, or else send it to your mother.
What you have produced is a simplified, two-dimensional electron probability map. Each color corresponds to a different "orbital", described by a set of principle, azimuthal and magnetic quantum numbers. Suppose your four colors describe the probability of finding a 2px, a 2py, a 3px and a 3py electron. With each dot, you sucessfully found the electron. The collection of dots tells you where a particular electron is most likely to be found; the highest value corresponds to the highest probability of locating an electron.
Compare and contrast your sets of dots with each other.
Exercise \(4\)
Construct another pointillist electron probability map, showing a two-dimensional slice through a 2s and 3s orbital in two different colors. Start with the same cross drawn in pencil, but this time when you put down dots with the first pen, centerd 1 cm from the midpoint of the cross, don't restrict them along one line. Let them be any direction from the midpoint. Use a hundred dots. Add a second set of dots with the second pen, centerd 3 cm from the midpoint of the cross, and in any direction from the midpoint. Compare and contrast the two sets of dots with each other, and with your previous map.
Why do electrons sort themselves out this way? One of the factors affecting electrons is electrostatic repulsion. They repel each other. To minimize that repulsion, the first pair of electrons might be a certain distance from the nucleus. A second pair may be a little farther away, so they're not too close to the first two. Once we reach a certain distance from the nucleus, electrons may be spread out enough that more than one pair could be found at that distance. Still, it is best to keep them a little farther away from each other. One pair may lie along one axis, and another pair may lie along a perpendicular axis. Of course, there is some repulsion between two electrons within the same pair, but the repulsion is actually decreased a little by having opposite spins (this is called spin pairing). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.03%3A_New_Quantum_Mechanics.txt |
On the very, very small scale of the universe, things don't behave exactly as we are used to. Electrons are one example. They are not solid objects. They are waves. Or are they? On the nanoscale, things can be both particles and waves. An individual particle like an electron has mass, but it is spread out, not located in one position, and it has a wavelength.
The recognition of the wave properties of the electron provided a breakthrough in understanding its properties. Schrödinger's wave equation allowed scientists to make predictions about the electronic structure of atoms. Where are the electrons found? What is their energy?
The wavefunction is a mathematical expression that describes the electron. It can be plotted, like a graph, although it is a graph in three dimensions instead of two. The three-dimensional plot of the wavefunction is sometimes called an orbital. Often, chemists find it useful to look at pictures of orbitals in order to gain some sense of where electrons may be and how they may behave. Orbitals are something like the wave form of the electron.
In another sense, orbitals are related to probability maps -- the wavefunction squared reveals the probability of an electron being located at a particular position in space. It might be helpful to picture them that way. Suppose you could take a series of pictures of an electron and superimpose all of those pictures, like time-lapse photography. The result might look something like the drawing below, in which every dot represents where the electron showed up in one of our pictures. It's impossible to predict exactly where the electron would show up in the next picture, but it would be a pretty good guess that it would show up somewhere in the same rough circle as it did all of the other times.
Of course, in three-dimensional space, we would be looking at a sphere instead of a circle. Maybe our picture is just a thin slice through the middle of that sphere. In the very center of the sphere, we would find the nucleus of the atom. So this electron is found within a certain distance of the nucleus, but it can be found in any direction. We call an electron that behaves this way an s electron.
The drawing above is labelled 1s, specifically. An atom can have many s electrons, but this is the one that is closest to the nucleus. It is the first s electron we would encounter if we started at the nucleus and moved outward through the atom.
Other electrons have different distributions about the atom. They are more likely to be located in different places. In solving the Schrödinger wave equation, it turns out that there are four variables that must be restricted to certain values if the equation is to have a sensible solution. The values of these variables determine where the electron is likely to be found around the atom, and roughly how much energy it will have.
The four quantum numbers are given the symbols n, m, l, and s. If you need to, you might remember that via the word aNiMaLS. Each quantum number tells you something different about the electron that it describes. The first one, n, tells you about distance. This number has to be an integer, so it could be 1, 2, 3, 4,... How far is the electron from the nucleus? If n = 1, the electron is as close to the nucleus as it could be; if n = 4, it is further away.
We sometimes think of n as describing the "shell" of the atom. Think of the old Bohr model of the atom, with electrons being found at different distances from the nucleus. It's like the atom is a set of Russian nested dolls, with one shell outside another. The first shell (n = 1) is close to the nucleus, the second (n = 2) is farther out, and so on.
The second number, m, tells you about spatial distribution, or shape. Maybe we know how close the electron is to the nucleus, but can it be found anywhere within a certain distance, or only in certain directions? This number has to be an integer, and it has to be less than n. If n = 1, m = 0. If m = 0, there is no "shape" preference, and the orbital is spherical. These numbers describe our 1s orbital, above.
The third number, l, can be a positive or a negative integer. It can be anywhere from +m to -m. If n = 1, and m = 0, then l = 0, too. The number l roughly correlates to orientation: in what direction is the orbital oriented? If there is only one possible value of l, then there must be only one possible orientation. That's true for something spherical; it does not matter which direction you look at it, because it will always look the same.
The final number is s. That's the spin quantum number and it can have only two values, regardless of the values of the other quantum numbers. It is either +1/2 or -1/2. Spin does not have a very tangible explanation in the macroscopic world, but if electrons have opposite spin values then they can occupy the same orbital. Thus, one pair of electrons can be described by each orbital.
Every electron in an atom must have a unique set of quantum number. If n, m, and l are the same for two electrons, then s must be different.
What happens when n = 2? In that case, the electron could be farther from the nucleus. The orbital will be bigger. In this case, m could either equal 0 again, or it could equal 1. If m = 0, like in the case of the 1s orbital, the orbital will be spherical. It will be a 2s orbital. Of course, l = 0, too, because l can only range from +m to -m. This orbital will be spherical, too, because m = 0. It will look just like a 1s orbital, only bigger.
However, if we could look at a thin slice through the orbital, we would see a difference between a 1s and a 2s orbital. The 1s orbital is uniform throughout, but the 2s orbital has a phase change. Remember, it is a wave, and it has peaks and troughs. This is a three-dimensional, spherical wave, but it has a trough nested inside and the peak is on the outside (or vice versa).
If we go further from the nucleus, we may encounter a 3s orbital and a 4s orbital. These, too, would look like bigger 1s orbitals from the outside. However, a thin slice through the middle would reveal the phase behavior of the orbotal, with peaks and troughs alternating from the center to the edge.
But let's step back for a moment. If n = 2, then m = 0 or 1, because m can be any integer less than n. We looked at the case in which m = 0; that's an s orbital. What if m = 1? The quantum number m describes the number of nodal planes that slice through the center of an orbital. If m = 1, we would take the regular s orbital and chop it through the middle (or loop a rope around the middle and tie it tight) to get two different halves. These halves would be out of phase with each other. This type of orbital is called a p orbital. It has two lobes. On lobe is on one side of the nucleus and the other lobe is on the other side of the nucleus.
If m = 1, then l = 1, 0, or -1. That means this p orbital could be oriented in any of three directions. Since we are really talking about three different electrons, and they would repel each other, then they will get as far as possible from each other. The easiest way to do that, if you can imagine x, y, and z axes meeting at the nucleus, is to have one p orbital pointing along the x axis, one along the y axis, and one along the z axis. These three p orbitals are called 2px, 2py, and 2pz.
Just like with the s orbital, we could also have p orbitals farther from the nucleus. We can't have one closer to the nucleus; there isn't enough room closer in to the nucleus, and the quantum rules prevent a p orbital at energy level n = 1. A 3p orbital would look just like a 2p orbital, and could lie in three different directions. However, if we could take a thin slice through its middle and pull it out, we could see the same internal phase changes that we saw in the 2s, 3s, and 4s orbitals.
So we can have one spherical s orbital at energy level 1, another spherical s orbital at energy level two, and three propellor-shaped or dumbell-shaped p orbitals at energy level 2. What about energy level 3? If n = 3, then m = 0, 1, or 2. That means there is a third shape possible at this level. To see the shape, we would take a p orbital and pinch it in half again, but in a direction perpendicular to the one we chose the first time. This would result in a cloverleaf shape. This is called a d orbital. There would be five different orientations of a d orbital, because l = 2, 1, 0, -1, or -2.
The d orbitals can point their lobes between the axes: the dxy has lobes in the xy plane but between the x and y axes; the dyz is in the yz plane with lobes between the y and z axes; the dxz is in the xz plane and its lobes point between the x and z axes. The two remaining d orbitals point along the axees. The dx2-y2 orbital points the lobes of its cloverleaf along the x and y axes. The dz2 orbital is not a clover shape like the others; it looks like a p orbital emerging from a donut. The two lobes extend along the z axis, and the donut (usually called the toroid) sits in the xy plane.
We could go one step further, although you are not likely to encounter the next level (many students only need to work with s and p orbitals). If n = 4, we could have s, p, or d orbitals for the values of m that we have seen before, but if m = 3, then we will have to chop the d orbital shape into additional lobes again, producing a funky f orbital. There are actually several ways this happens, producing several different shapes; only one is illustrated below. An f orbital can have seven possible orientations.
More commonly, the overall shapes of these orbitals are depicted in "cartoon" form. For example, an s orbital is usually just shown as a ball or sphere.
The three p orbitals are shown here. Each one is a dumbell shape pointing in a different direction along Cartesian coordinates (x, y, z).
There are five d orbitals. Four of them are cloverleaf shapes, but the fifth contains a toroid (or donut shape).
The seven f orbitals are shown below. Four of them look like orange segments, lying on one plane. Two of them look like the cloverleaf d orbitals that have split into two layers. The last one is like the dz2 orbital, only it has two toroids instead of one.
We can summarize quantum numbers in a couple of ways. Remember, quantum numbers are just variables in the Schrödinger wave equation that have to take certain values in order for the equation to have a sensible solution. These values are usually integers, and they depend on each other. There is something different about the quantum numbers that describe each electron in an atom; in part, that helps to keep the electrons away from each other.
• Quantum numbers are constrained to specific values.
• Each electron in an atom has a unique set of quantum numbers.
number n m l s
rules integer less than n -m to +m 1/2 or -1/2
For example, the 1s electrons on a helium atom have to be different from each other somehow. They have the same values of n, m, and l, so s is different.
number n m l s
rules integer less than n -m to +m 1/2 or -1/2
1st set (1s) 1 0 0 1/2
1 0 0 -1/2
As we proceed to the second "shell" of electrons around the atom, m can vary, and so can l. That means we get different shapes and orientations of the allowed waves for the electron. We'll ignore s in the following table; in each case, there are two possibilities, s = 1/2 and s = -1/2; two electrons can be described by each orbital.
number n m l orbital description
rules integer less than n -m to +m
1st set (1s) 1 0 0 spherical
2nd set (2s) 2 0 0 spherical
3rd set (2p) 2 1 -1 dumbbell (orthogonal directions)
2 1 0
2 1 -1
When we get to the 3rd shell, there are even more variations.
number n m l orbital description
rules integer less than n -m to +m
4th set (3s) 3 0 0 spherical
5th set (3p) 3 1 -1 dumbbell
3 1 0
3 1 1
6th set (3d) 3 2 -2 cloverleaf
3 2 -1
3 2 0
3 2 1
3 2 2
Overall, the quantum number n indicates the shell in which an electron is found. That is, it indicates distance from the nucleus.
In addition, n can also indicate the number of spherical nodes in an orbital. As we saw in the progression of s orbitals, each type of orbital (s, p, d, f) becomes "nested" when we move to higher levels; a 2s orbital is like a 1s orbital nested inside a bigger, spherical shell, with a spherical node in between the two parts. A 3s orbital contains one more spherical node (two total), and so on.
Quantum number m indicates the shape of the orbital. Is the orbital divided into different lobes? Described in a different way, it indicates how many nodes or nodal planes there are between these lobes. An s orbital is completley shpherical, with no lobes. A p orbital has a nodal plane cutting it in half, forming two lobes. A d orbital has two nodal planes, usually resulting in four lobes, and so on.
If the orbital does have a non-spherical shape, l corresponds to different directions in which it can be oriented. Of course, s is the spin, and two electrons in the same orbital have to have opposite spins.
Exercise \(1\)
Provide labels (e.g. 1s, 3p, etc) for the following orbital descriptions.
1. a spherical orbital in the second shell of the atom.
2. a dumbbell-shaped orbital in the third shell of the atom.
3. a cloverleaf-shaped orbital in the third shell of the atom.
4. an orange-slice shaped orbital in the fifth shell of the atom.
5. an orbital that looks like one ball nested inside a second ball netsed inside a third ball
Answer a:
2s
Answer b:
3p
Answer c:
3d
Answer d:
5f
Answer e:
4s
Exercise \(2\)
Draw cartoons of the following orbitals.
1. a 4p orbital.
2. a 4d orbital
3. a 5d orbital
4. a 6f orbital
Answer
Exercise \(3\)
As we go to higher shells in the atom, we start to see additional spherical nodes in the orbitals. Make a sketch of each orbital below, and superimpose a repeating sine wave over the drawing such that the nodes in the sine wave correspond to the nodes in the orbital.
1. a 2p orbital
2. a 3p orbital
3. a 4p orbital
Answer
From one end of the orbital to another, the 2p orbital covers a full sine wave, a 3p orbital covers two full sine waves, and a 4p orbital covers three full sine waves.
What happens to the number of sine wavelengths as we go from one shell to the next? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.04%3A_Quantum_Numbers.txt |
The quantum mechanical model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
How electrons fill in their positions around an atom is called the Aufbau Process (German: "building-up" process). The Aufbau Process is all about keeping electrons at their lowest possible energy.
A corollary of Coulomb's law is that the energy of an electron is affected by attractive and repulsive forces. The closer an electron to the nucleus, the lower its energy. The closer an electron is to another electron, the higher its energy.
Of course, a basic principle of thermodynamics is that a system will proceed to the lowest energy possible. That means, if an atom has only one electron, the electron will have quantum numbers that place it at the lowest possible energy. It will be as close as possible to the positive nucleus.
If an atom has a second electron, it will also be as close as possible to the nucleus. It could have the same quantum numbers as the first electron, except for spin. There is a trade-off, of course, because those two electrons will be close enough to repel each other. However, if it is a choice between that and taking a position much farther from the nucleus, the second electron will go ahead and pair up. These two electrons are sometimes described as being "in the same orbital"; their first three quantum numbers are the same, so that are probably found somewhere in the same region of space. This first orbital, which has no directional restrictions, is called the 1s orbital.
There is only room for one orbital at this distance from the nucleus. A third electron has to occupy another orbital farther away, the 2s orbital. Again, this is a spherical orbital: the electron can be found in any direction. The 2 in 2s means the principle quantum number is two (corresponding to the second general energy level). The s is a code for other quantum numbers; it means the electron can be found in any direction, just like the 1s electrons.
The second energy level is large enough to accommodate additional orbitals, but they are a little farther from the nucleus. These are called the 2p orbitals. They are regions of space along the x, y and z axes. There are three orbitals of this type, and they are just called px, py and pz to remind us that they are orthogonal to each other.
Expressed in a different way, an electron with principle quantum number 2 can have four different combinations of its other quantum numbers. These combinations are denoted 2s, 2px, 2py and 2pz. The three 2p combinations are a little higher in energy than the 2s orbital.
Why is the 2p level higher than the 2s level? That's related to the wave behavior we saw before. With a node in the middle, the 2p orbital behaves as a higher-frequency wave than the 2s orbital. A higher-frequency wave has higher energy. So, a p orbital, with one node, is always higher in energy than an s orbital, with no node.
Once again, we have several energy levels available for an electron, but they will surround the atom in a way that lowers energy. A second electron remains in the lower-energy 1s orbital.
A third electron will go into the 2s orbital. It's the lowest in energy. What about a fourth? Does it go into the 2s or a 2p? Once again the pairing energy is not quite as big as the energy jump up to the 2p orbital. The fourth electron pairs up in the 2s orbital.
A fifth electron goes into one of the 2p orbitals. It does not matter which one. We will say it is the px, arbitrarily. A sixth electron again could either pair up in the px, or it could go into the py. But the py level is really the same as the px, just in a different direction. The energy is the same. That means a sixth electron will go into the py rather than pair up in the px, where it would experience extra repulsion.
Note that the p orbitals are often drawn a little differently. For example, p orbitals are usually drawn in a way that shows that they have phase. Either the two lobes are colored differently to show that they are out of phase with each other, or they are shown with one lobe shaded and the other left blank.
This pattern of "filling" electrons is generally followed for all of the elements. The tally of how many electrons are found in each orbital is called the electron configuration. For example, hydrogen has only one electron. Its ground state configuration (that means, assuming the electron hasn't been excited to another orbital via addition of energy) is 1s1.
On the other hand, an atom with six electrons, such as the element carbon, has the configuration 1s22s22px12py1. There is one electron in each of two p orbitals in order to avoid repulsion, which would happen if they were in the same one.
Exercise \(1\)
Write electron configurations for the following elements.
1. oxygen, O
2. sulfur, S
3. silicon, Si
4. nitrogen, N
5. argon, Ar
6. neon, Ne
Answer a:
O: 1s22s22px22py12pz1
Answer b:
S: 1s22s22p63s23px23px23py13pz1
Answer c:
Si: 1s22s22p63s23px23px13py1
Answer d:
N: 1s22s22px12py12pz1
Answer e:
Ar: 1s22s22p63s23p6
Answer f:
Ne: 1s22s22p6
Electron configuration and the periodic table
You may already know that electron configuration is the reason the periodic table works the way it does. Mendeleev and others noticed certain elements had very similar properties, and that's because they have very similar electron configurations. Lithium has configuration 1s22s1 and its alkali sister, sodium, has configuration 1s22s22px22py22pz23s1 . In both cases, the last electron added is an unpaired s electron. The last electron, or last few electrons, added to an atom generally play a strong role in how the atom behaves. This "frontier" electron is the one at the outer limits of the atom. If the atom is to interact with anything, the frontier electron will encounter the thing first. In contrast, the "core" electrons closer to the nucleus are more protected from the outside.
• The order of electrons in an atom, from lowest to highest energy, is:
• 1s
• 2s
• 2p
There are some shortcuts we take with electron configurations. We tend to abbreviate "filled shells" (meaning all the possible orbitals with a given principle quantum number are filled with electrons) and filled "sub-shells" (like 2s or 2p). First of all, in the case of p orbitals, if all the p orbitals are filled, we might just write 2p6 instead of 2px22py22pz2, because there is only one way all the orbitals could be filled. However, we wouldn't necessarily write 2p2 instead of 2px12py1, because we may wish to make clear that the configuration does not involve two electrons in one p orbital at that point, as in 2px22py0.
Also, we dispense with orbital labels entirely to ignore core electrons in a filled shell. For example, instead of writing 1s22s1 for lithium, we can write [He]2s1. Instead of writing 1s22s22px22py22pz23s1 for sodium, we write [Ne]3s1. The [He] means all the electrons found in a helium atom, which is a noble gas. The [Ne] means all the electrons found in a neon atom. A noble gas is an unreactive element with a filled shell: helium, neon, argon, krypton, xenon or radon.
[He] = 1s2 so [He]2s1 = 1s22s1
[Ne] = 1s22s22p6 so [Ne]3s1 = 1s22s22p63s1
The electrons beyond the noble gas shell are called the valence electrons.
Exercise \(2\)
Write abbreviated electron configurations for the following elements.
1. chlorine, Cl
2. calcium, Ca
3. aluminum, Al
4. phosphorus, P
Answer a:
Cl: [Ne]3s23px23py23pz1
Answer b:
Ca: [Ar]4s2
Answer c:
Al: [Ne]3s23px1
Answer d:
P: [Ne]3s23px13py13pz1
Principle quantum number 3 actually allows a third set of orbitals. These are called the d orbitals. The d orbitals are a little like p orbitals, but they are two-dimensional rather than one-dimensional. A d electron, for example, might extend along the x axis and the y axis, but not in between the axes.
The d orbitals have five allowed orientations. They can be found along the x and y axes (called dxy), along the x and y axes (called dxz), or along the y and z axes (called dyz). Alternatively, they might be found in between the axes instead, rotated 45 degrees away from one of the other d orbitals. One of these, called the dx2-y2 orbital, is found between the x and y axes. In the same way, you could imagine an orbital between the x and z and between the y and z axes, but that would make six different orientations. Quantum mechanical rules don't allow that. As a result, two of the possible combinations collapse into a mathematical sum, making just one orbital. We call this one the dz2 orbital.
If we look at the third row in the periodic table, we see those three sub-shells (the s, then the p, then the d). Again, the 3s sub-shell fills first. With fewer nodes, this orbital is lower in energy than either the 3p or the 3d. The 3d orbital has more nodes than the 3p, so it is even higher in energy than the 3p. In fact, it is even a little higher in energy than the 4s orbital, the first sub-shell of the next shell.
So, the 3d orbitals are higher in energy than the 3p orbitals. The 3d level is very similar in energy to the 4s level. For that reason, calcium's last electrons go into a 4s orbital, not a 3d orbital. Calcium behaves much like magnesium as a result.
• The order of electrons in an atom, from lowest to highest energy, is:
• 1s
• 2s
• 2p
• 3s
• 3p
• 4s
• 3d
• 4p
• As a consequence, all of the alkali metals (Li, Na, K, Rb, Cs, Fr) have similar properties, because they have similar electron configurations [Noble]s1
• All of the alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) have similar properties, because they have similar electron configurations [Noble]s2
Exercise \(3\)
Write abbreviated electron configurations for the following elements.
1. iron, Fe
2. nickel, Ni
3. mercury, Hg
4. lead, Pb
5. arsenic, As
6. titanium, Ti
Answer a:
iron, Fe: [Ar]4s23d6
Answer b:
nickel, Ni: [Ar]4s23d8
Answer c:
mercury, Hg: [Xe]6s24f145d10
Answer d:
lead, Pb: [Xe]6s24f145d106p2
Answer e:
arsenic, As: [Ar]4s23d104p3
Answer f:
titanium, Ti: [Ar]4s23d2
Complications in transition metals
Some subtle variations are encountered across the transition metals. Filling the d orbitals is not as straightforward as the s and p orbitals. For example, in the first row of the transition metals, all but two elements have configurations [Ar]4s23dx. However, chromium has configuration [Ar]4s13d5, and copper has configuration [Ar]4s13d10. Similarly, two elements in the third row of the transition metals do not have the configuration [Xe]6s25dx. Platinum has configuration [Xe]6s15d9, and gold has configuration [Xe]6s15d10. Things are even worse in the second row of transition metals, in which half of the elements do not follow the "correct" order of filling. Niobium is [Kr]5s14d4, molybdenum is [Kr]5s14d5, ruthenium is [Kr]5s14d7, silver is [Kr]5s14d10, and palladium does not have any s electrons at all in its outer shell: it is [Kr]4d10.
Sc = [Ar]4s23d1 and Ti = [Ar]4s23d2 and V = [Ar]4s23d3 but Cr = [Ar]4s13d5
What's going on? The main reason things are complicated here is that the 4s and 3d levels are quite close to each other in energy (as are 5s and 4d, and 6s and 5d). As a result, slight changes are causing the electron configuration to vary from one element to the next. Pairing energy is certainly a culprit; that's the amount of energy it costs to put two electrons in the same orbital. If the electron configuration is [Ar]4s23dx, then two electrons are always being forced to occupy the same space, the 4s orbital. That costs energy, because electrons repel each other. Pairing energy changes from one element to another, but by the time we reach chromium, pairing energy is evidently high enough (or the difference in energy between the 4s and 3d levels is low enough) that the energy is lower if the electrons just spread themsleves out.
So, a balance has to be struck between pairing energy and orbital energy. Both are changing as we move from one element to the next. Sometimes the pairing energy of the s orbital is small compared to the energy jump to the d orbital, so two electrons go into an s orbital. Sometimes the pairing energy of the s orbital is large compared to the energy jump to the d orbital, so the electron goes in a d orbital. Of course, the pairing energy of the d orbitals also plays a role in some cases, and it also varies from one element to another.
Well, what are you supposed to do with that information? Usually, you are expected to know the most general rule (such as filling like [Ar]4s23dx). Twenty-one out of thirty transition metals have two s electrons and some number of d electrons. Sometimes, you are expected to know the most common exceptions; those are chromium and copper (there's a lot more chromium and copper in the world than there is niobium), and they are relatively easy to remember because one has a half-filled d shell and the other has a completely filled d shell.
• Most transition metals have two s electrons and some d electrons.
• Copper and chromium have only one s electron; the other one is "promoted" into a d orbital.
• You can keep track those two exceptions if you remember that copper has one electron in each d orbital and copper has an electron pair in each d orbital ("half-filled d" and "filled d" is a good rule to remember).
One more important thing to know is that these cases describe only the transition metals in their elemental state. They do not apply to compounds, in which the transition metal is found bound with atoms of different types to form salts or other materials. The atoms in a chunk of silver metal have the electronic configuration [Kr]5s14d10, but the atoms in a silver ion, which have one less electron, have configuration [Kr]4d10. The missing electron is lost from the s orbital, not from the d. In general, compounds and ions of the transition metals do not have s electrons in the valence shells. For example, the atoms in pure molybdenum metal may have configuration [Kr]5s14d5, but the molybdenum atoms in compounds, if they still have all their electrons, have configuration [Kr]4d6.
• Most often, transition metals in ions and compounds have only d electrons in their valence shell.
Why would compounds be different? The difference is easiest to see in the case of ions, in which the metal loses one or more electrons. Because it not longer has the same amount of electrons as protons, it becomes positively charged (it has more positive protons than it has negative electrons). The positive charge causes the electrons to become more attracted to the nucleus; that atom contracts or shrinks.
Different orbitals may have different responses to this change in the metal atom. Because the d orbital is one level below the s orbital of similar energy (it is a 3d orbital, for example, compared to a 4s orbital), it is closer to the nucleus. It experiences that positive charge more strongly and contracts more than the s orbital. As a result, its energy is lowered even more than the s orbital's energy by this stronger electrostatic interaction with the nucleus.
As a result of the charge when a transition metal atom becoms an ion, the 3d level falls below the 4s level. Remember, these two orbitals are very close in energy to begin with, so small changes can reverse their order. Similarly, the 3d level generally falls below the 4s level anytime a transition metal joins other atoms to become part of a compound.
The way the 3d electrons fall in energy with increasing charge is one of the factors making the electron configurations of transition metals complicated.
• In ions and compounds, the d orbital is lower in energy than the s orbital of the next level, not the other way around.
Exercise \(4\)
Draw an orbital filling diagram (arrows and energy levels) for:
Reminder: The filling diagrams for transition metals and transition metal ions/complexes differ.
A. Ni
B. Co+2
C. Fe+3
D. Ti+4
E. V
F. Co
G. Mn+2
H. Zn+2
I. Zn
Answer
Finishing Touches
Earlier, we saw that electrons tend to go into unfilled orbitals before pairing up in the same one, provided other orbitals are available at the same energy level.
This idea is part of Hund's rule. Hund's rule says, in part, that if you have two electrons, and there are two orbitals available at the same energy level, then one electron will go into each orbital. It's partly about avoiding electron-electron repulsion that would occur if you put two electrons into the same orbital -- that is, into the exact same region of space.
It's partly something else though, and that is a quantum mechanical bias toward high multiplicity. Multiplicity refers to the number of unpaired electrons there are in an atom or molecule. By paired, we mean two electrons that have opposite spin. Remember, spin is a fundamental quantum mechanical property of an electron. It can only have two values, and the numerical values seem arbitrary but it's important to know that the two possible options are opposite numbers: they can be either +1/2 or -1/2. Unpaired electrons would be those that don't have a partner somewhere with an opposite spin value.
To illustrate that idea, consider the following drawing. It shows three different ways a set of five electrons might fill in a group of five orbital (maybe the 3d level; suppose these are the valence electrons on a vanadium atom). Often, electrons in orbital diagrams are indicated by arrows, with the direction of the arrow indicating the spin. An up arrow means spin = 1/2; a down arrow means spin = -1/2.
In two cases, some of the electrons are paired; they have a partner somewhere with opposite spin. The multiplicity is basically a tally of how many unpaired electrons are left over; you get the multiplicity by adding up the spin value of all the electrons. One case has all the electrons unpaired. All of the electrons have the same spin. Multiplicity is maximized in this case. Hund's rule says this case has the lowest energy.
What about if two electrons occupy the same orbital? We could still maximize multiplicity by keeping their spins "parallel"; that is, they could both have spin = 1/2 or both spin = -1/2. That does not happen, though. Remember the quantum rule that no two electrons on the same atom can be described by the same set of quantum numbers. In other words, each electron on the atom must have a unique identity. This rule has a name, too: the Pauli exclusion principle.
• Electrons always occupy the lowest energy orbital available.
• Multiplicity is maximised; electrons are given the same spins when possible.
• However, when they are found in the same orbital, two electrons must have opposite spin. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.05%3A_The_Aufbau_Process.txt |
The aufbau process is a set of rules that allows us to predict the electronic configuration of an atom if we know how many electrons there are in the atom. If the periodic table is used as a tool, this process is pretty easy.
For atoms found in the first two columns of the periodic table (figure \(1\)), the configuration is a closed shell of core electrons, plus s electrons in a new shell. For example, potassium has a configuration [Ar]4s1. These atoms are often called the alkali and alkaline earth elements. Alkali elements, from the first column, have a configuration ending in s1; alkaline earth elements, from the second column, have configurations ending in s2. Together, these elements are often called the s-block elements, because their valence electrons are s electrons. Remember, the valence electrons are the ones beyond the noble gas core. In the case of potassium, they are the ones beyond [Ar].
The first two and the last six columns of the periodic table are called the main group elements. Alternatively, they are sometimes called the s-block and p-block elements, respectively. For example, phosphorus has a configuration, [Ne]4s24px1py1pz1, or simply [Ne]4s24p3.
The middle block of the periodic table consists of the transition metals or the d-block elements. For example, scandium has configuration [Ne]4s23d1.
The final two rows of the periodic table are the lanthanides and actinides. Collectively, they are called the f-block elements. Samarium, for example, is [Xe]6s24f6. These elements could really be inserted at the left-hand side of the d-block in the appropriate rows. Notice that lanthanum, element 57, is followed by hafnium, element 72, in the table. The element that really occurs next is element 58, cerium, and it is shown in the lanthanide row down below. The f-block elements are usually shown below in order to save space.
Really, the periodic table should look like this:
• The periodic table is divided into columns of atoms with similar electron configurations.
• Atoms with similar electron configurations have similar properties.
Chemical reactions depend on the movement of electrons. In a reaction, one atom may accept electrons from another atom. One atom may donate electrons to another atoms. The valence electrons are the outermost electrons in an atom; they are closest to the surface of an atom. That fact makes the valence electrons more likely to interact with other atoms. The valence are also the highest-energy electrons in an atom, and most likely to participate in a reaction.
For these reasons, atoms with similar electron configurations generally behave in similar ways. The repeating properties in each row of the periodic table, as observed by Mendeleev and others, reflect the repeating electron configurations in subsequent rows. The periodic table organizes atoms with similar configurations and properties together in columns.
Exercise \(1\)
For the following elements, suggest two other elements that would have similar properties.
a) zinc, Zn b) calcium, Ca c) oxygen, O d) chlorine, Cl e) chromium, Cr
Answer a:
Zn: Cd, Hg
Answer b:
Ca: Mg, Ba
Answer c:
O: S, Se
Answer d:
Cl: F, Br
Answer e:
Cr: Mo, W
Exercise \(2\)
Make a diagram showing the energy levels of different orbitals, arranged by principal quantum number.
"Periodic trends" refer to the way in which physical properties of atoms change across the periodic table. One of the most commonly used periodic trends in chemistry is electronegativity. Electronegativity is closely connected to the basic idea of chemical reactions: the transfer of an electron from one neutral atom to another. It refers to how strongly an atom attracts electrons from other atoms.
• Electronegativity is a measure of an atom's ability to draw electrons towards itself, or the ability of the nucleus to hold electrons tightly.
There are many scales of electronegativity, based on different physical measurements. Usually, electronegativity is set to an approximately 4-point scale. Atoms with electronegativity of around 4 draw electrons very strongly toward themselves. Atoms with electronegativity of 1 (or lower) only weakly draw electrons toward themselves.
The following data use the Allen scale of electronegativity. The Allen scale uses spectroscopic measurements to estimate the energy of valence electrons in an atom. From these values, the relative attraction of the atom for its valence electrons is placed on a 4 point scale (approximately).
Table \(1\): The Allen electronegativity values of the second-row elements.
Some electronegativity scales do not have values for the noble gases, because they are based on experimental measurements of compounds, and noble gases do not commonly form compounds with other elements. Instead, they exist as single atoms. The Allen scale just depends on the ability of an atom to interact with light, which is something even noble gases can do. As a result, noble gases are also given electronegativity values on this scale. However, on many scales, fluorine would be the most electronegative atom here. As a result, fluorine is usually thought of as the most electronegative element.
Often it is useful to plot data on a graph. That way, we can get a better look at the relationship. For example, a quick glance at Figure AT5.2. shows that there is a smooth increase in electronegativity as we move across a row in the periodic table.
Exercise \(3\)
Take a look at the graph in figure \(3\). Can you explain why the electronegativity increases as atomic number increases?.
Answer
The atomic number is the number of protons in the nucleus. For two atoms in the same row of the periodic table, the outermost electrons are roughly the same distance away from the nucelus. The more positive protons there are in the nucleus, the more tighly held are the electrons.
Exercise \(4\)
Suppose you need an electron. You have a boron atom and an oxygen atom. You try to take an electron away from one. Use Figure \(3\). to predict which atom will give up the electron more easily.
Answer
Take it from the boron. The oxygen atom is holding its electrons much more tightly.
Exercise \(5\)
Suppose you have an electron. You are able to send it into a vessel that contains a carbon atom and a fluorine atom. Use Figure \(3\). to predict which atom is more likely to take the electron.
Answer
According to the drawing, the neon would take the electron, because of all the atoms depicted in the graph, neon attracts electrons most strongly. However, there is a complication. Although neon strongly attracts its own electrons, it can't accommodate an extra electron as easily as could fluorine, the next-best candidate. In "Lewis" terms, neon has a "full octet". In quantum terms, an additional electron would have a higher principle quantum number and be placed in the next "shell", farther from the nucleus. With spin-pairing, fluorine can accept another electron into its valence shell.
Exercise \(6\)
A covalent chemical bond is a pair of electrons shared between two atoms. Suppose you have a carbon-oxygen bond. Will the electrons be shared evenly between the two atoms, or will one atom pull the electrons more tightly towards itself? Use Figure \(3\). to make your prediction.
Answer
The oxygen would pull the electrons in the bond more tightly to itself.
What is happening as we move across a row in the periodic table? Why does electronegativity increase?
Keep in mind that the only difference from one element to the next is the number of protons in the nucleus. The number of protons is called the atomic number. If you know the number of protons you have, then you know what atoms you have. Electronegativity may have something to do with the number of protons in the nucleus. In fact, it should. The more protons there are in the nucleus, the more strongly electrons should be attracted to it. Each additional proton should add more electrostatic attraction for an electron. Fluorine, with nine protons, should attract electrons much more strongly than lithium, which has only three protons.
• Moving across a row of the periodic table, as protons are added to the nucleus, electrons are held more tightly.
• Electronegativity increases across a row in the periodic table.
It seems like that effect should be offset by the increasing number of electrons in the atom. Each time a proton is added, so is an electron. That electron should repel other electrons in the atom, cancelling out the effect of more protons in the nucleus.
However, the structure of the atom minimizes electron-electron repulsion a little bit. Remember that all the protons are in one place, the nucleus. All the electrons are a relatively great distance from the nucleus, in many different directions. Chances are, an additional electron is much farther away; it may be twice as far away as an additional proton in the nucleus. It may be all the way on the other side of the atom. Because electrons are spread out in the atom, and the distances between them is pretty large, the repulsive effect is a little smaller than the attractive effect of additional protons.
We can see that this trend is generally true across the periodic table, with a few exceptions here and there.
What happens as we move down a column in the periodic table? Table \(2\). shows the Allen electronegativities of the alkali metals. These elements are also called the Group 1 elements or the Group IA elements. The "Group 1" designation is used because they are the first column or group in the periodic table. The data are also presented in Figure \(5\).
Table \(2\): The Allen electronegativity values of the alkali elements.
There is a different trend here. In this case, lithium (atomic number 3) has more protons than hydrogen (atomic number 1). However, hydrogen is a lot more electronegative than lithium. Francium, with 87 protons in its nucleus, is the least electronegative alkali element.
Exercise \(7\)
Take a look at the graph in Figure \(5\). Can you explain why the electronegativity decreases as atomic number increases, going down this column?
Answer
Moving from one row to the next in the periodic table signifies that the outermost electron is in a shell farther from the nucleus. Those outermost electrons are less tightly held if they are farther from the nucleus.
Exercise \(8\)
An ionic chemical bond is a pair of ions attracted by their opposite charges. A cation is a positively charged ion; it may be an atom that has lost an electron. An anion is a negatively charged ion; it may be an atom that has gained an extra electron. Ions can form by moving an electron from one atom to another.
1. Suppose you have an ionic cesium-fluorine bond. Which ion is the cesium and which is the fluorine? Use Figure \(3\) and Figure \(5\). to make your prediction. Cesium is Cs.
2. Suppose you have an ionic sodium-oxygen bond. Which ion is the sodium and which is the oxygen? Use Figure \(3\) and Figure \(5\). to make your prediction. Sodium is Na (from the Latin, natrium).
3. Suppose you have an ionic potassium-hydrogen bond. Which ion is the potassium and which is the hydrogen? Use Figure \(3\) and Figure \(5\). to make your prediction. Potassium is K (from the Latin, kalium).
Answer a:
Cs+ F-
Answer b:
Na+ O-
Answer c:
K+ H-
Periodic Trends and Atomic Radius
The biggest difference between two atoms in the same group (column) in the periodic table is the principal quantum number. Remember, that corresponds to the "valence shell". Think of electrons as forming layers around the nucleus. Electrons with principal quantum number one form a first layer. Those with principal quantum number 2 form a second layer, and so on. Each layer is further away from the nucleus. Remember, electrostatic attraction gets weaker as charges get further away from each other. As electrons get further from the nucleus, they are less tightly held.
• Moving down a column in the periodic table, valence electrons are held less tightly because they get further from the nucleus.
• Electronegativity decreases as we move down a column in the periodic table.
We can see this general size trend in the following periodic table. This table presents covalent radii, which are related to the sizes of the atoms (although not exactly the same; data on atomic radii are not available for all atoms, however).
We can clearly see the expanding radii of atoms if we look at Group 1, the first column; these elements are called the alkali metals. Hydrogen, at the top, is very small. Lithium is much bigger. Sodium is much bigger than lithium, however, and potassium is much bigger than sodium. And so on: francium is bigger than cesium, which is bigger than rubidium, which is bigger than potassium.
Each time an electron is added to an orbital that is significantly farther from the nucleus, of course it is going to result in a bigger atom. Remember, the atom is mostly empty space, and its size is described by the outermost reaches of its electrons. So when we go to the next principal quantum number -- that is, to the next row in the periodic table, from the first row to the second row, for example -- the next electron is much further away from the nucleus. It has to be that way, because electrons repel each other. They can't all be equally close to the nucleus, because there would be too much repulsion. Instead, they form these layers, and when the first layer is so full that there would be too much repulsion if anothe relectron were added, we start the next layer.
Of course, the very first layer is very, very small. There just isn't that much room so close to the nucleus. For the first row, only two electrons are allowed. Then they have to start the next layer. For the second row, eight electrons are allowed; that's the origin of something called the "octet rule" (think "octopus") for common compounds, which you'll see later on. Eventually we get to eighteen electrons in a shell, then thirty two, as the shells get bigger and bigger like layers of an onion, or like nested Russian dolls.
There is another important trend if you look carefully. As you move from left to right across the periodic table, from one group to the next, the atoms get bigger. That does not make any sense, does it? If we are adding more electrons, why would the atom get smaller?
The key thing is, not only are we adding more electrons, but we are also adding more protons in the nucleus. The new electrons we are adding are all roughly equidistant from the nucleus; they are all equally close to the protons. So as the charge on the nucleus gets bigger, those electrons are all more strongly attracted to the center. The atom shrinks.
Eventually, we get to the point at which we couldn't possibly add more electrons; the radius has shrunk so much that repulsion would become too great if we added one more electron. Then we just start another row. Just before that point, however, we hit a sweet spot: the point at which the attraction between the nucleus and the outermost electrons is so strong, and the electrons are held so tightly, that the atom becomes very, very stable. This last column in the table contains the noble gases, which are particularly stable and unreactive.
Exercise \(9\)
Why does electronegativity fall so sharply between hydrogen and lithium, and much more subtly between lithium and sodium?
Answer
The attraction for an electron falls off with 1/r2. As the value of r gets larger and larger, the quantity 1/r2 will begin to approach a limit (of zero). As a result, the difference between two successive values of 1/r2 in a series gets smaller and smaller.
Exercise \(10\)
Which atom, in the following pairs, is more electronegative?
1. magnesium, Mg, or calcium, Ca
2. lead, Pb, or tin, Sn
3. silver, Ag, or antimony, Sb
4. gallium, Ga, or arsenic, As
5. tungsten, W, or copper, Cu
6. thallium, Tl, or sulfur, S
Answer a:
Mg > Ca
Answer b:
Sn > Pb
Answer c:
Sb > Ag
Answer d:
As > Ga
Answer e:
Cu > W
Answer f:
S > Tl
Exercise \(11\)
Electron ionization is the energy that must be added in order to pull an electron away from an atom.
1. Why do you think energy has to be used to pull an electron away from an atom? What is holding the electron there?
2. Explain the general trend in ionization energies seen in the following table (a larger value means more energy must be added to remove a first electron from the atom).
Answer a:
The electron is held by its attraction to the nucleus.
Answer b:
As the number of protons in the nucleus increases, the electron becomes more tightly held, and harder to remove.
The relationship is similar to the one seen for electronegativity, and not coincidentally. Electronegativity can be calculated in a number of ways, but one of those ways uses the ionization energy as a factor.
Electronegativity is a calculated value, whereas ionization energy is an experimentally determined one. This difference brings up an important philosophical distinction. To a beginning student, an experimental value seems faulty, whereas a calculated value sounds good. To an experienced chemist, an experimental value is verifiable; it is real. In contrast, a calculated value is enhanced in prestige only if it can be shown to agree with experiment.
Table \(3\): The ionization energies of the second row elements.
Exercise \(12\)
Sometimes a plot of the data can be revealing. Ionization energies do not follow a smooth trend. Explain why it is a little easier to remove an electron from boron and oxygen than expected. (Electron configurations may be helpful here.)
Exercise \(13\)
Explain the trend in the following data on ionization energy.
Table \(4\): The ionization energies of the alkali elements.
Exercise \(14\)
Electron affinity is the energy released when a free electron is picked up by an atom.
1. Why would energy be released when a free electron is taken by an atom?
2. Explain the general trend in the following electron affinity data.
Exercise \(15\)
1. Explain a general trend in the following electron affinity data.
2. There are several exceptions to the general trend. Why do beryllium and neon have such low electron affinities (almost zero)?
3. Nitrogen also has an electron affinity that is close to zero. Why?
Exercise \(16\)
Usually, elements become bigger as we go down a column in the periodic table. However, in a phenomenon called "the lanthanide contraction", some elements are actually smaller than the ones in the row above them. Specifically, osmium, iridium, platinum, gold, and mercury are smaller than their relatives, ruthenium, rhodium, palladium, silver, and cadmium, respectively.
Use the periodic table in Figure \(2\). to offer a possible explanation for this phenomenon.
Answer
Perhaps the simplest explanation is this: The addition of 14 extra protons between lanthanum and hafnium, compared to between yttrium and zirconium, results in a considerable contraction of the atom. The electrons are pulled inward by the added positive charge in the nucleus. Thus, elements in the third row of the transition metal block are not as large as might otherwise be expected, and in some cases are even smaller than their precedent elements.
Note that this is not the only explanation. Relativistic effects are also believed to play a role in the behaviour of these massive elements. Einstein's theory of relativity states that objects get heavier the faster they move. The velocity of an electron can be shown to increase with the charge in the nucleus of the atom. Thus, atoms that have very high atomic numbers have very, very fast electrons, and consequently very heavy ones. These heavy electrons sink toward the heavy nucleus, and the atom shrinks further. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.06%3A_The_Periodic_Table_and_Periodic_Trends.txt |
Exercise 1.1.3
1. one atom of mercury with one atom of oxygen.
2. two atoms of mercury with one atom of oxygen.
3. 200 amu (Hg) + 16 amu (O) = 216 amu (HgO)
4. 2 x 200 amu (Hg) + 16 amu (O) = 416 amu (HgO)
Exercise 1.1.4
1. 200 g (Hg) + 16 g (O) = 216 g (HgO)
2. 2 x 200 g (Hg) + 16 g (O) = 416 g (Hg2O)
3. If one mole is 216 g of HgO, then 0.25 mole must be a quarter of that amount, or 54 g.
$0.25 mol \times 216 \frac{g}{mol} = 54 g \nonumber$
Alternatively written as
$0.25 mol \times 216 g \: mol^{-1} = 54 g \nonumber$
The amount of mercury is just a fraction of that: 200/216. So the answer is $\frac{200}{216} \times 54 g = 50 g$.
d) What fraction of a mole is 2.08 g of Hg2O, if one mole is 416 g?
$\frac{2.08g}{416 g \: mol^{-1}} = 0.005 mol \nonumber$
Notice that when we divide g by g mol-1, the grams cancel and the mol-1 becomes mol.
There is 1 mol of O in 1 mol of HgO, so 0.005 mol of O are needed for 0.005 mol HgO.
$0.005 mol \times 16 g \: mol^{-1} = 0.08 g$ O needed
Exercise 1.1.6
a) sulfur b) silicon c) calcium d) potassium and lithium
e) phosphorus f) iodine g) silver and copper h) ruthenium
Exercise 1.1.7
a) 20 b) Na c) 16.00 d) phosphorus
e) carbon f) Sn g) 32.07 h) potassium
Exercise 1.1.8
a) Ne: 20 g/mol b) Fe: 56 g/mol c) Cu: 64 g/mol d) Au: 197 g/mol e) Si: 28 g/mol
Exercise 1.1.9
1. H2O: $18 \frac{g}{mol} (2 \times H + 1 \times O = 2 \times 1 + 1 \times 16 \frac{g}{mol})$
2. NaHCO3 : $84 \frac{g}{mol} (3 \times O + 1 \times Na + 1 \times H + 1 \times C = 3 \times 16 + 1 \times 23 + 1 \times 1 + 1 \times 12 \frac{g}{mol})$
3. SiO2 : $60 \frac{g}{mol} (1 \times Si + 2 \times O = 1 \times 28 + 2 \times 16 \frac{g}{mol})$
4. NaCl : $58 \frac{g}{mol} (1 \times Na + 1 \times Cl = 1 \times 23 + 1 \times 35 \frac{g}{mol})$
5. TaN : $195 \frac{g}{mol} (1 \times Ta + 1 \times N = 1 \times 181 + 1 \times 14 \frac{g}{mol})$
6. NH3 : $17 \frac{g}{mol} (1 \times N + 3 \times H = 1 \times 14 + 3 \times 1 \frac{g}{mol})$
7. NaH2PO4 : $120 \frac{g}{mol} (1 \times Na + 2 \times H + 1 \times P = 4 \times O = 1 \times 23 + 2 \times 1 + 1 \times 31 + 4 \times 16 \frac{g}{mol})$
8. CH3CO2H : $60 \frac{g}{mol} (2 \times C + 4 \times H + 2 \times O = 2 \times 12 + 4 \times 1 + 2 \times 16 \frac{g}{mol})$
Tthe molar weight of the atom was rounded to the nearest gram. In order to be very careful and avoid "error propagation", the more exact molar weight (to several decimal places) could be used and rounding could be performed after the calculation. In these cases, the result would be the same. The result could be different if very large numbers of atoms were found in the compound.
Exercise 1.2.1
a) O: 8 protons, 8 electrons, 8 neutrons
The atom is neutral overall, so the number of positively charged protons is equal to the number of negatively charged electrons. The atomic weight is provided almost entirely by the protons and neutrons, so the number of protons plus number of neutrons equals the atomic weight.
b) P: 15 protons, 15 electrons, 16 neutrons
c) Zn: 30 protons, 30 electrons, 35 neutrons
d) Au: 79 protons, 79 electrons, 118 neutrons
Exercise 1.2.3
Carbon: $\frac{(99 \times 12amu) + (1 \times 13 amu)}{100} = \frac{1188 + 13amu}{100} = \frac{1201amu}{100} = 12.01 amu$
Exercise 1.2.4
The negatively charged electron's departure from the nucleus leaves behind a positive charge. A neutron is converted into a proton. The overall atomic weight remains the same, but the atom ends up with one more proton and one more electron. A 14C is converted into a 14N.
Exercise 1.2.6
Suppose y is the decimal fraction of 35Cl and z is the decimal fraction of 37Cl.
$35.5 amu = \frac{(y \times 35 amu + z \times 37 amu)}{100}$ but y + z = 1
$35.5 amu = (y \times 35 amu + (1-y) \times 37 amu) \nonumber$
$35.5 amu = y \times 35 amu - y \times 37 amu + 37 amu \nonumber$
$(37-35) \times y amu = 37 amu - 35.5 amu \nonumber$
$2 \times y amu = 1.5 amu \nonumber$
$y = 0.75 (75 \% ^{35}Cl) \nonumber$
$z = 0.25 (25 \% ^{37}Cl \nonumber$
Exercise 1.2.7
1. 1, 2, 3, 4, 5... a series of whole numbers, or n.
2. 2, 4, 6, 8, 10... a series of even numbers, or 2n (because every even number is two times another number).
3. 3, 5, 7, 9, 11... a series of odd numbers, or 2n + 1 (because every odd number is one more than some even number).
4. 1, 4, 9, 16, 25... a series of squares, or n2.
5. 2, 4, 8, 16, 32... a series in which each numer is double the last number, or 2n.
6. 1, 1/2, 1/4, 1/9, 1/16... a series of reciprocals of squares, or 1/n2.
Exercise 1.2.8
1. When q increases, F increases. The increase is linear: if q1 doubles, F doubles. As the charge in the nucleus gets larger, the force of attraction gets larger.
2. When r increases, F decreases. The decrease is nonlinear: if r doubles, F drops by a factor of four, rather than a factor of two. As the distance from the nucleus gets longer, the attraction to the nucleus drops sharply.
3. Hydrogen and helium are both in the first row of the periodic table. To a rough approximation, the distance between nucleus and electron is similar in these two atoms. However, helium has a charge of 2+ in its nucleas, compared to the 1+ charge in the nucleus of a hydrogen atom. As a result, the attraction of an electron on helium to the nucleus would be about twice as great as the attraction of an electron on hydrogen to its nucleus.
Helium's electrons are much more tightly held than hydrogen's.
The situation is really much more complicated than that. For example, if helium's electrons are more tightly attracted to the nucleus than hydrogen's, then helium's electrons ought to be pulled closer to the nucleus than hydrogen's. That means helium's electrons are held even more tightly than we at first thought.
Another complicating factor is that helium has two electrons, whereas hydrogen has only one. An electron may be attracted to the nucleus, but electrons repel each other. That second electron in helium should offset the extra attraction to helium's more positive nucleus. That means helium's electrons may be less tightly attracted than we originally thought.
However, the effect of the second electron is much smaller than it first appears. That's because the second electron could be anywhere around the helium atom. It has a 50% chance of being further away from the first electron than that positively charged nucleus. The farther it is away, the smaller its influence. Helium's electrons are definitely more strongly attracted to the nucleus than are hydrogen's, but it is difficult to say exactly how much more without the help of some more sophisticated tools.
Exercise 1.2.9
1. As wavelength gets longer (value of λ increases), energy decreases.
2. As frequency gets higher (value of ν increases), energy increases.
Exercise 1.2.10
Periodic of Elements in a Table
Element Symbol Atomic Number Mass Number Number of Protons Number of Neutrons Number of Electrons Charge
H 1 1 1 0 1 0
H 1 1 1 0 0 +1
H 1 1 1 0 2 -1
H 1 2 1 1 1 0
H 1 3 1 2 1 0
Be 4 9 4 5 2 +2
C 6 12 6 6 6 0
Mg 12 25 12 13 12 0
Tc 43 98 43 55 43 0
Ca 20 40 20 20 18 +2
Si 14 28 14 14 14 0
K 19 47 19 28 15 +4
Fe 26 56 26 30 23 +3
Br 35 79 35 44 36 -1
Ti 22 39 22 17 21 +1
P 15 30 15 15 15 0
Al 13 27 13 14 10 +3
S 16 32 16 16 16 0
Pd 46 106 46 60 45 +1
Cr 24 52 24 28 21 +3
Sn 50 118 50 68 50 0
Hg 80 200 80 120 79 +1
Au 79 197 79 118 78 +1
Exercise 1.3.1
1. The energy of the electron gets higher.
2. The energy of the electron gets lower.
3. The wavelength gets shorter.
4. The wavelength gets longer.
Exercise 1.3.2
The probability of finding an electron at a node is zero.
Exercise 1.3.3
Your drawing is beautiful. You should put it on your fridge, or else send it to your mother.
Exercise 1.4.1
a) 2s b) 3p c) 3d e) 5f e) 4s
Exercise 1.4.3
From one end of the orbital to another, the 2p orbital covers a full sine wave, a 3p orbital covers two full sine waves, and a 4p orbital covers three full sine waves.
Exercise 1.5.1
1. O: 1s22s22px22py12pz1
2. S: 1s22s22p63s23px23px23py13pz1
3. Si: 1s22s22p63s23px23px13py1
4. N: 1s22s22px12py12pz1
5. Ar: 1s22s22p63s23p6
6. Ne: 1s22s22p6
Exercise 1.5.2
1. Cl: [Ne]3s23px23py23pz1
2. Ca: [Ar]4s2
3. Al: [Ne]3s23px1
4. P: [Ne]3s23px13py13pz1
Exercise 1.5.3
1. iron, Fe: [Ar]4s23d6
2. nickel, Ni: [Ar]4s23d8
3. mercury, Hg: [Xe]6s24f145d10
4. lead, Pb: [Xe]6s24f145d106p2
5. arsenic, As: [Ar]4s23d104p3
6. titanium, Ti: [Ar]4s23d2
1. Zn: Cd, Hg
2. Ca: Mg, Ba
3. O: S, Se
4. Cl: F, Br
5. Cr: Mo, W
Exercise 1.6.3
The atomic number is the number of protons in the nucleus. For two atoms in the same row of the periodic table, the outermost electrons are roughly the same distance away from the nucelus. The more positive protons there are in the nucleus, the more tighly held are the electrons.
Exercise 1.6.4
Take it from the boron. The oxygen atom is holding its electrons much more tightly.
Exercise 1.6.5
According to the drawing, the neon would take the electron, because of all the atoms depicted in the graph, neon attracts electrons most strongly. However, there is a complication. Although neon strongly attracts its own electrons, it can't accommodate an extra electron as easily as could fluorine, the next-best candidate. In "Lewis" terms, neon has a "full octet". In quantum terms, an additional electron would have a higher principle quantum number and be placed in the next "shell", farther from the nucleus. With spin-pairing, fluorine can accept another electron into its valence shell.
Exercise 1.6.6
The oxygen would pull the electrons in the bond more tightly to itself.
Exercise 1.6.7
Moving from one row to the next in the periodic table signifies that the outermost electron is in a shell farther from the nucleus. Those outermost electrons are less tightly held if they are farther from the nucleus.
1. Cs+ F-
2. Na+ O-
3. K+ H-
Exercise 1.6.9
The attraction for an electron falls off with 1/r2. As the value of r gets larger and larger, the quantity 1/r2 will begin to approach a limit (of zero). As a result, the difference between two successive values of 1/r2 in a series gets smaller and smaller.
1. Mg > Ca
2. Sn > Pb
3. Sb > Ag
4. As > Ga
5. Cu > W
6. S > Tl
Exercise 1.6.11
1. The electron is held by its attraction to the nucleus.
2. As the number of protons in the nucleus increases, the electron becomes more tightly held, and harder to remove.
The relationship is similar to the one seen for electronegativity, and not coincidentally. Electronegativity can be calculated in a number of ways, but one of those ways uses the ionization energy as a factor.
Electronegativity is a calculated value, whereas ionization energy is an experimentally determined one. This difference brings up an important philosophical distinction. To a beginning student, an experimental value seems faulty, whereas a calculated value sounds good. To an experienced chemist, an experimental value is verifiable; it is real. In contrast, a calculated value is enhanced in prestige only if it can be shown to agree with experiment.
Exercise 1.6.12
In order to escape the atom, an electron must gain energy. As shown in problem AT5.2., the 2p energy level is higher than the 2s energy level. That means a 2p electron already has more energy than a 2s electron. The 2p electron will not need as much additional energy in order to escape from the atom. As a result, boron's ionization energy is a little lower than beryllium's.
The additional protons in the nucleus of carbon and nitrogen more than make up for that effect.
In the case of oxygen, the next electron is just added to a 2p level, but in this case it must be paired with another electron in the same region of space (in the same "orbital"). The repulsion between these electrons, or "pairing energy", slightly destabilizes the oxygen, so less energy will be needed to remove an electron.
Once again, the continued addition of extra protons eventually compensates for this pairing effect.
Exercise 1.6.13
As with electronegativity, ionization energy decreases as the distance to the nucleus increases because of the 1/r2 relationship in Coulomb's Law.
Exercise 1.6.14
1. Energy is released because of the attraction of the free electron to a nucleus. The electron moves to lower energy as it becomes stabilized by its interaction with the nucleus.
2. The electron can get much closer to the hydrogen nucleus thanto the lithium nucleus, and so on. More energy will be released owing to the strong interaction between the electron and the hydrogen nucleus compared to the interaction between the electron and the lithium (or sodium or potassium...) nucleus.
Exercise 1.6.15
1. The data zig-zags, but broadly speaking there is an increase in electron affinity as the number of protons in the nucleus increases.
2. Beryllium and neon have zero electron affinity because the next electron in each case would be added to a higher energy level. In the case of beryllium, the next electron would go into the 2p energy level. An electron added to neon would go into the 3s energy level.
3. Although the next electron added to nitrogen would be added to a 2p level, it would have to be paired in the same region of space as an electron that was already there. The pairing energy in this case must be enough to offset the attraction of the electron to the nucleus.
Exercise 1.6.16
Perhaps the simplest explanation is this: The addition of 14 extra protons between lanthanum and hafnium, compared to between yttrium and zirconium, results in a considerable contraction of the atom. The electrons are pulled inward by the added positive charge in the nucleus. Thus, elements in the third row of the transition metal block are not as large as might otherwise be expected, and in some cases are even smaller than their precedent elements.
Note that this is not the only explanation. Relativistic effects are also believed to play a role in the behavior of these massive elements. Einstein's theory of relativity states that objects get heavier the faster they move. The velocity of an electron can be shown to increase with the charge in the nucleus of the atom. Thus, atoms that have very high atomic numbers have very, very fast electrons, and consequently very heavy ones. These heavy electrons sink toward the heavy nucleus, and the atom shrinks further. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/01%3A_Introduction_to_Atoms/1.07%3A_Atoms-_Solutions_to_Selected_Problems.txt |
Most of the elements found on earth are metals. A look at the periodic table shows that these elements occupy the entire left-hand stretch of the table, from the main group, through the transition metals, lanthanides, actinides, alkali and alkaline earth elements.
One element included here, hydrogen, is rarely classified as a metal. On earth, hydrogen is a gas, and it is usually classified as a non-metal, like oxygen and nitrogen. However, at very low temperatures and very high pressures, hydrogen is a solid, and under the right conditions it is expected to behave more like a metal. It is thought that gas giants, such as Jupiter and Saturn in our solar system, may have metallic hydrogen cores.
Metallic elements are not generally found as single atoms. Instead, the atoms in an element such as iron cluster together to make a larger structure. The materials formed in this way have some similar properties. Metals are shiny. Metals are malleable; they can be bent and formed into different shapes (at least when heated). Metals are good conductors of electricity.
• In metals, large groups of atoms cluster together.
• Metal atoms form large, extended arrays, with the atoms repeating in specific patterns throughout the solid.
The properties of metals are really important. The fact that metals are malleable allows them to be formed into sheets that can be used to make cars, airplanes, railway lines, cargo containers and ships, as well as more delicate items such as jewelry and surgical tools. A related property, ductility, allows metals to be stretched into long, thin wires. Together with the conductivity of metals, this property allows transmission lines to carry electricity from generating stations to people like you. Sometimes, the source of electricity is hundreds of miles away; electricity used to power a laptop in New York may come from places like La Grande Baleine or James Bay, in northwestern Quebec.
Some of these properties can be understood by thinking about the structure of metallic elements. A great deal of our structural understanding of metals and other materials comes from x-ray diffraction studies. A very focused beam of x-rays can be sent into a material, where they will bounce off the atoms and scatter in different directions. The outcome sounds chaotic, but if the solid is highly organized, the x-rays behave in very predictable ways. The result is an x-ray diffraction pattern. A diffraction pattern is a little like the pattern of ripples on a pond when a stone is thrown into calm water. The pattern can be studied and decoded mathematically to find the locations of the atoms within the material.
• X-ray diffraction can reveal the atomic-level structures of highly ordered materials such as metals.
What does x-ray diffraction tell us? Evidently, a chunk of metal is not just a mass of atoms stuck together randomly. Instead, the atoms arrange themselves in neat layers in very specific ways. These layers of atoms sit on top of each other to form a three-dimensional solid.
One of the properties that results from this organized arrangement of atoms is the malleability of metals. If you take a nice, soft metal such as copper, after annealing it in a fire or oven, it can be bent and shaped easily. With copper, this can be done even after the metal has cooled to room temperature. When you bend the copper, you are actually causing layers of atoms to slide over each other, until you stop bending and they come to rest in a new location.
If you have ever done this, you'll know that the more you work with the copper, the harder it is to bend. That's because while you are sliding layers of atoms back and forth, occasionally an atom (or an entire row of atoms) slips out of place. It is no longer part of a smooth layer, and so other atoms can't slide past it as easily. This situation is called a defect. Once there are enough defects in the metal, it is impossible to bend the material anymore.
• Metal atoms are found in organized layers.
• Because these layers can roll over each other, metals can be worked into different shapes.
Exercise \(1\)
An alloy is a mixture of two metals. Steel is an alloy of iron with any of a number of other elements, such as chromium or vanadium. Alloys are often harder than metals composed of a pure element. Show how alloying introduces a defect into the metal, and how that makes the metal stronger.
Answer
The second metal atom is a different size than the principle metal atom. It will not quite fit into the array of atoms. Consequently, the atoms will not be able to slide past each other as easily.
What is it that holds these metal atoms together? To answer that question, it's important to realize where metallic materials are found in the periodic table. The bottom left part of the periodic table is where the least electronegative elements are found. In fact, all of these elements lose electrons easily, and they are frequently found as cations in naturally-occurring compounds. For example, hematite is a common iron ore, containing iron cations (Fe3+) and oxygen anions (O2-). The formula of this compound is Fe2O3, meaning there is always a ratio of three oxygen anions for every two iron cations in hematite.
Exercise \(2\)
Show how the ratio of elements in hematite leads to a charge-balanced (overall neutral) compound.
Answer
2 Fe: (3+) x 2 = (6+)
3 O: (2-) x 3 = (6-)
Fe2O3: neutral
Many of the atoms in a metallic material are present as cations. But where did their lost electrons go? Well, those electrons are still in the material, moving between the iron atoms and cations. In a piece of iron, the attraction between the iron cations and the freely moving electrons helps hold the metal together. This way of thinking about metals is sometimes called the "electron sea" model of bonding.
• Bonding in metals is often described through the "electron sea model".
• Metal ions are surrounded by delocalized electrons.
• Delocalized electrons are not restricted to one atom or another; they are distributed across several atoms in the solid.
Why do metals conduct electricity? Electricity is the movement of electrons through a material. But the conduction of current through a metal probably takes place through a series of events. If an electron is introduced at one end of the material, it will probably be attracted by a metal cation. It may even be captured by that metal. Sometimes, we describe this electron as moving into a "hole"; a hole, in conductivity terms, is just a positive charge that captures an electron. But remember that metals are still pretty electropositive (the opposite of electronegative), and that metal atom is likely to lose another electron. This may not be the same electron as the one you put in; it is probably another one. That electron may in turn be captured by another hole on another metal. That metal may lose another electron, and so on. Electrons will hop and skip from one metal to another throughout the material. An electric current results because these hopping electrons in the metal move away from the electrons that are being supplied at one end. They move towards the other end, instead.
The shininess of metals is also attributed to the electron-sea aspect of metallic bonding. Collisions between incoming photons and the "free" electrons at the surface of a metal cause the photons to bounce off the surface. The reddish color of copper results from a limit on the wavelengths of visible light that bounce off the metal.
• The electron sea model is used to explain several unique properties of metals.
So far, we have looked at the electron sea model purely in terms of electrostatics: the negatively charged electron is attracted to the positively charged nucleus. However, we already saw in the discussion of quantum mechanics in the atom that kinetic energy and the wavelength of the electron is also an important factor in chemistry.
Exercise \(3\)
Suppose the following boxes are half-filled with water. Show the longest wavelength possible in each of the boxes.
Answer
The longer the box, the longer the possible wavelength.
In the electron sea model of metallic bonding, the electrons can be delocalized. They are no longer confined to a single atom, but can be spread out over multiple atoms. As a result, the wavelength of the electron increases. Because wavelength is inversely proportional to energy, as an electron's wavelength increases, its energy goes down. As a result, spreading electrons out over a group of metals results in a decrease in energy because of the wave properties of electrons.
• Delocalization is an important concept in chemistry.
• Delocalization is related to the "particle-in-a-box" concept: the longer the container, the longer the possible wavelength.
• By spreading out over a larger group of atoms, an electron can adopt a longer wavelength.
• An electron with a longer wavelength has a lower energy.
Some of the properties of metals can be understood, given a basic outline of the structure of a metal. We are going to look in more detail at exactly how metal atoms arrange themselves into solids, before we look at some of the ways metals form compounds with other elements. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/02%3A_Metals/2.01%3A_Introduction_to_Metals.txt |
Metals are composed of atoms in ordered layers. These atoms form a three-dimensional, crystalline structure. That means that the individual units within the solid -- the atoms, in this case -- are organized in a regularly repeating pattern, like the stars on the American flag.
Three-dimensional things can be difficult to think about. It's much easier to start in two dimensions. On this page, we're going to look at a couple of ways that metal atoms could arrange themselves in a single layer. We'll build up from there later on. In the meantime, we'll try to develop familiarity with a number of terms that are used to describe crystalline structures.
Simple Squares
We live in a Cartesian society. Our houses and buildings are mostly square, our rooms are square, our streets are straight and meet at right angles, unless we live in a meandering place on the water, like Boston or Venice. It's probably easiest to imagine layers of atoms that form nice squares.
Suppose a group of atoms forms a nice, straight line. Another group forms a second line, and each atom in the second line sits directly behind an atom in the first line. A third row forms in the same way. The resulting layer has neat rows and columns of atoms. The pattern of atoms regularly repeats as you look from right to left, and as you look from front to back.
We will call this type of layer a simple square layer.
• In a square layer, atoms are arranged in rows and columns.
Hexagonal Layers
Some metals atoms may be arranged in simple square layers. Others adopt a slightly different arrangement.
Start with the same, neat line of atoms in the first row. This time, when the second row of atoms forms, we won't put them directly behind the atoms in the first row. We'll put them just in between the atoms in front. Each atom in the second row sits behind two atoms in the first row, as if it were peeking between them, like kids in a kindergarten class photo (before they all squirm around).
The third row will also be slightly offset from the row in front of it. If you look carefully, you can see that puts each of the third row atoms directly behind an atom in the first row, but an extra row farther away. The fourth row will be slightly offset from the third, and so on.
We still have a crystalline structure. There is a regularly repeating pattern as we look from left to right. There is also a repeating pattern as we move from front to back, but now that pattern repeats every other row, instead of every row.
This type of layer is called a hexagonal layer. Sometimes, it is called a close-packed layer.
• In a hexagonal layer, the rows are offset from each other by half an atom.
• Each atom in a hexagonal layer is surrounded by six other atoms.
Exercise $1$
1. Look at the simple square layer. Can you find repeating patterns in other directions (rather than left-right or front-back)? If so, in what direction?
2. Look at the hexagonal layer. Can you find repeating patterns in other directions (rather than left-right or front-back)? If so, in what direction?
Answer a:
The pattern also repeats diagonally.
Answer b:
The pattern also repeats diagonally.
Exercise $2$
We usually think of atoms as little spheres, or circles if we are working in just two dimensions. How might other shapes form an organized layer? Draw an organized layer for each of the following shapes, using about a dozen units in each case.
1. squares.
2. triangles (equilateral: all sides the same length).
3. hexagons.
4. pentagons.
In each case, is there only one way to pack the shapes in a regular way? Can you find repeating patterns in the layer? If so, in what direction?
Are there any shapes that seemed more difficult to arrange in a repeating layer?
Answer
The squares can form a number of regularly repeating patterns. The rows can exactly repeat (an aaa pattern), or they can be shifted slightly, so that every other row exactly repeats the first row (an ababab pattern). They could even be shifted so that every third row is an exact repeat of the first (abcabcabc).
The pentagons do not form a repeating layer in two dimensions.
Holes
Unlike some other geometric shapes, circles do not pack together with no space in between them. The spaces in between atoms within a layer are called holes. Note that the use of the word "holes" here has nothing to do with holes in conductivity-speak. On this page, holes are just spaces between atoms. In conductivity, holes are places where there is positive charge that could capture an electron.
Sometimes, the spaces between the atoms are described in other ways, such as "interstitial sites".
Later on, these interstitial sites or holes will be important in building compounds from mixtures of elements. You will see that the empty spaces in an array of atoms leave room for a second kind of atom to pack into the same structure.
Exercise $3$
Holes generally have particular shapes. The shapes vary, depending on how the atoms are arranged. For example, how would you describe the shape of the holes in the following layers:
1. simple square packing.
2. hexagonal or close packing.
Answer a:
In the simple square packing, the holes are roughly diamond shaped.
Answer b:
In the hexagonal pattern, the holes are roughly triangular.
You can come up with an answer by looking at a drawing, or you can try packing a layer of dimes or pennies or quarters (as long as they are all the same) and look at the spaces between them
Unit Cells
The simplest repeating unit in a pattern is sometimes called the unit cell. In a layer, the unit cell can be traced again and again over the layer, so that every part of the layer is accounted for (but depending where you start tracing, you might run out of room at the edges).
Because each layer is composed of individual atoms, we might call a single atom the unit cell. However, we also use the unit cell to illustrate the spatial relationship between the atoms in the layer. For that reason, we often shift the unit cell to show multiple atoms.
For example, in a row of atoms, a unit cell might be a square. The ends of the square would cut through two neighbouring atoms. The unit cell contains two half-atoms, for a total of one atom. The square could be repeated in a straight line to reproduce the row of atoms.
• The unit cell is the smallest repeating unit in the structure.
• The unit cell shows the pattern that is found in the structure.
In a layer of atoms, the unit cell might once again be thought of as a single atom. However, we can more clearly show the relationships between these atoms by choosing a slightly different cell. The unit cell in a simple square layer is just a square that contains the corners of four neighbouring atoms. The unit cell in a hexagonal layer is a rhombus that contains the corners of four neighbouring atoms.
Exercise $4$
In the following cases, show that the given unit cells repeat throughout the structure.
1. simple square layer: a square unit cell
2. hexagonal layer: a rhombic unit cell
Note: in the hexagonal layer, you can find an additional way to do this if you rotate copies of the unit cell to get unit cells facing in three different directions.
Exercise $5$
In the following cases, show what fractions of atoms are found in the unit cell. Add the fractions to find the combined total of atoms in the unit cell.
1. simple square layer: a square unit cell
2. hexagonal layer: a rhombic unit cell
Packing Efficiency
Packing efficiency is one of the questions people wonder about in crystal structures. It deals with how tightly the atoms are packed together in a structure. Often, the more tightly packed the structure, the better (but it does not always happen that way).
Why might tighter packing of metal atoms be favored? It has to do with the fact that the metal atoms are interacting with each other. The closer they are together, the more strongly they can interact.
There are a couple of ways we can think about that interaction. In the simplest idea of the electron-sea model, metal atoms have become metal cations, and are surrounded by electrons. The closer the metal atoms can get to the electrons, the more strongly they will interact. In general, we think of the metal atoms as sharing their electrons; the closer the metal atoms are to each other, the more efficiently the electrons can be shared.
Let's just work with that idea of electrostatic attraction. Imagine a group of atoms in a simple square layer. All of the atoms are touching their neighbors on either side. Let's think about the distance between a nucleus and an individual electron in the electron sea. We know that the nucleus is in the center of the atom. We don't know exactly where a free electron is, but let's assume it is in the center of the nearest hole. Remember, the closer the electron is to the nucleus, the lower the energy.
For comparison, maybe there is another set of atoms, also in a simple square layer. Suppose they are well-separated from each other; maybe they are far enough apart that you could fit an extra atom between each pair if you wanted to. If the free electron is in the same place -- the middle of the nearest hole -- you can see that it is much farther from the nucleus in this case. The force of attraction is much lower in this case, and the overall energy is not as low.
• Most metals pack very efficiently together to form a solid.
• Efficient packing leads to stronger bonding interactions.
That first case, with atoms packed more tightly together, may be preferable, because of the stronger interaction between the metal nucleus and the free electron. For reasons like this, understanding the packing efficiency in a crystal can be very important.
Exercise $6$
Go to your piggy bank and break out a bunch of pennies, nickels, quarters or dimes. You will need at least four of each, but nine of them is a good number. Pack one type of coin into the following layers:
1. simple square
2. hexagonal or close packed
Answer a:
Using quarters, the distance is about 5 mm.
Answer b:
Using quarters, the distance is about 2.5 mm.
A free electron would be closer to the atoms in a hexagonal close packed layer. There would be stronger electron-ion attraction in that case.
In each case, measure the distance from the edge of one coin to the center of the hole.
In which case is the free electron closer to the atom? Does the packing system (the type of layer) have an effect on how tightly bonded that atoms are?
Exercise $7$
Go to your piggy bank and break out a bunch of quarters AND a bunch of dimes. You will need at least four of each, but nine of them is a good number. Pack each type of coin into the following layers:
1. simple square
2. hexagonal or close packed
Answer a:
Using dimes, the distance is about 3.5 mm.
Answer b:
Using dimes, the distance is about 1.5 mm.
A free electron would be closer to the atoms if the atoms were smaller. There would be stronger electron-ion attraction in that case.
In each case, measure the distance from the edge of one coin to the center of the hole.
In which case is the free electron closer to the atom? Does the size of the atom have an effect on how tightly bonded that atoms are?
Exercise $8$
When a metal melts, the atoms go from a tightly-bound, regular array of atoms to a loosely-bound set of mobile atoms.
Compare the melting points of the following metals, given in degrees Celsius.
potassium: 64 cesium: 29 lithium 181 sodium: 98 francium: 27
1. Can you determine any trends among these values?
2. Can you provide a physical explanation for your observation? (You may have found a trend, but can you explain it?)
Answer a:
melting points (and force of attraction between atoms): Fr < Cs < K < Na < Li
Answer b:
This trend mirrors the sizes of the atoms. Lithium is the smallest and francium the largest. The electron / ion attraction is greatest in Li and weakest in Fr.
Exercise $9$
Packing efficiency is often determined more rigorously in terms of the percentage of a unit cell that is actually occupied by atoms. What is the total area of a unit cell in the following layers?
1. simple square layer.
2. hexagonal layer.
Answer a:
The total area of this square is
$Area = w^{2} \nonumber$
in which w = width of the unit cell. The width of the unit cell is
$w = 2r \nonumber$
in which r = radius of titanium atom.
$Area = 4r^{2} = 4 (2.00 \times 10^{-10} m)^{2} = 4(4.00 \times 10^{-20} m^{2}) = 1.60 \times 10^{-19} m^{2} \nonumber$
Answer b:
The total area of this rhombus is
$Area = s^{2} sin \theta \nonumber$
in which s = one side of unit cell and θ = an angle of the unit cell (either 60° or 120°). But
$s = 2r \nonumber$
in which r = radius of titanium atom.
$Area = (2 \times 2.00 \times 10^{-10} m)^{2} sin (60) = 1.60 \times 10^{-19} (0.87) m^{2} = 1.39 \times 10^{-19} m^{2} \nonumber$
You can assume the atoms are titanium. Titanium has an atomic radius of 2.00 Angstroms (or 2.00 x 10-10 m).
Exercise $10$
How many titanium atoms are there in a unit cell in the following layers?
1. simple square layer.
2. hexagonal layer.
Answer a:
$4 \times \frac{1}{4} = 1 \: atom \nonumber$
Answer b:
$2 \times \frac{1}{6} + 2 \times \frac{1}{3} = \frac{2}{6} + \frac{4}{6} = 1 \: atom \nonumber$
You may need to add up fractions of titanium atoms to arrive at the answer. The answer may or may not be a whole number.
Exercise $11$
What is the area occupied by a titanium atom?
(If you think the question is ambiguous, assume we're concerned about the area of a cross-section of the atom at its widest part. It is a spherical atom, but we are dealing with its projection in two dimensions, i.e. its shadow when the sun is directly overhead.)
Answer
The cross-sectional (two dimensional) area of a titanium atom is the area of a circle
$Area = \pi r^{2} \nonumber$
$Area = 3.1415 \times (2.00 \times 10^{-10} m)^{2} = 3.1415 \times 4.00 \times 10^{-20} m^{2} = 1.257 \times 10^{-19} m^{2} = 1.26 \times 10^{-19} m^{2} \nonumber$
Exercise $12$
In the following cases, what percentage of the unit cell is filled with titanium atoms?
1. simple square layer.
2. hexagonal layer.
Answer a:
$Efficiency = \frac{occupied}{total} \times 100 \% = \frac{1.26 \times 10^{-19} m^{2}}{1.39 \times 10^{-19} m^{2}} \times 100 \% = 91 \% \nonumber$
Answer b:
$Efficiency = \frac{occupied}{total} \times 100 \% = \frac{1.26 \times 10^{-19} m^{2}}{1.39 \times 10^{-19} m^{2}} \times 100 \% = 91 \% \nonumber$
A greater percentage of space is occupied in hexagonal close packing.
The percentage of unit cell that is occupied by atoms is called "the packing efficiency".
Coordination Number
There are a number of terms used in describing the relationships between atoms and the atoms to which they are directly bonded. For example, we use the term "coordination number" to describe the number of other atoms that a specific atom is touching in a regular layer.
Exercise $13$
What is the coordination number of an atom in
1. simple square layer
2. hexagonal layer
Answer a:
coordination number = 4
Answer b:
coordination number = 6
Exercise $14$
What is the coordination number of an atom in the following layers:
Coordination Geometry
Take a look at all of the atoms surrounding one specific atom in a layer. If you were to draw a line from the center of one of these atoms to the next, and go all the way around until you got back to the beginning, you would get a specific shape. That shape is described as the coordination geometry of the central atom. It's closely related to the coordination number. In two dimensions, two atoms with the same coordination number always have the same coordination geometry. Occasionally in three dimensions, two atoms with the same coordination number can have different coordination geometries; the atoms that they are touching may be arranged in slightly different ways.
Exercise $15$
What is the coordination geometry of an atom in
1. simple square layer
2. hexagonal layer
Exercise $16$
What is the coordination geometry of an atom in the following layers:
Lattice
Sometimes, the arrangement of atoms in a crystal is referred to as "an array" or "a lattice". At this point, an array or lattice is just the two-dimensional network of atoms in a layer. Later, we will see three-dimensional arrays of atoms. Metallic solids are usually three-dimensional, not two-dimensional. However, as we start to build up to a third dimension, you will see that most of the ideas on this page have analogies in 3D. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/02%3A_Metals/2.02%3A_A_Layer_of_Metal_Atoms.txt |
What sort of structures compose the world around us? What sort of shapes do we see in the building blocks of the material world? The ancient Greeks placed special significance on certain perfect shapes, or Platonic solids. These shapes were considered to be perfect because they were so regular, with equally sized sides arranged at identical angles to each other, so that the solid would look the same when viewed from different angles. For example, a cube has four square sides, all connected at right angles to each other. These Platonic solids were sometimes associated with the classic elements as perceived at the time: earth, air, fire and water.
As it happens, the Greeks were not that far from the truth. Certain solids do appear frequently in the structures of nature, including the classic Platonic solids. These shapes just happen to describe the natural way that a group of atoms might arrange themselves when they come together to build a larger structure. Very often, we find an atom at the center of a Platonic solid, with other atoms arranged around it to form the corners of the solid.
In addition to the cube shown above, the tetrahedron and octahedron are extremely common Platonic solids found in natural structures. A couple of non-Platonic solids that we sometimes see are the square pyramid (think of the ones in Egypt) and the trigonal bipyramid; however, we won't encounter those in nature for a few more chapters.
Exercise $1$
For each of the following solids, identify the number of (i) corners and (ii) faces.
a) cube b) tetrahedron c) square pyramid d) trigonal bipyramid e) octahedron
Answer a:
(i) 8
(ii) 6
Answer b:
(i) 4
(ii) 4
Answer c:
(i) 5
(ii) 5
Answer d:
(i) 5
(ii) 6
Answer e:
(i) 6
(ii) 8
We will see cubic, octahedral and tetrahedral shapes cropping up as we look at the structures of extended solids such as metals and salts. In the solid state, metals are very neat, orderly structures. Metal atoms form a three-dimensional, crystalline structure. We will start thinking about those structures by seeing how they can be built up from simpler, two-dimensional layers.
Building up from Squares: Simple Cubic Packing
Take a simple square layer of atoms. Suppose another layer forms on top of that one, with atoms arranged in the same way as the first layer. Each atom in the second layer sits directly on top of an atom in the first layer.
This is a very neat, ordered, simple structure. All of the atoms are lined up in straight rows, aisles and columns. The structure looks exactly the same if viewed along the x-axis, along the y-axis, or along the z-axis.
The simplest repeating unit in three dimensions is usually thought of as a box or cube with atoms at each of its eight corners. Alternatively, we might think of the repeating unit as a cube with one atom in its center. However, it's useful to see the relationships between multiple atoms in the unit cell. For that reason, we usually picture the unit cell as the cube with an atom at each corner. Because each of those corners is shared between several unit cells, it is better to think of each unit cell as only owning a fraction of each corner atom.
• A simple cubic unit cell is based on a square layer.
• Squares of atoms are repeated in the x, y and z directions.
In the following sections, unit cells will be shown with atoms separated even further from each other because the relationships between the atoms are even more complex. In the case of a simple cubic packing system, the unit cell would be shown as follows:
Exercise $2$
Take a look at the simple cubic unit cell.
1. What fraction of an atom is found in each corner of the simple cubic unit cell?
2. How many corners are there in the cube?
3. What is the total number of atoms in the simple cubic unit cell?
Answer a:
1/8 atom at each corner
Answer b:
8 corners in a cube
Answer c:
$8 \times \frac{1}{8} = 1$ atom per simple cube
Building up from Squares: Body-Centered Cubic Packing
That isn't the only way a three-dimensional solid could be built up from a simple square layer. Remember from the discussion of layers that atoms can sit a little more tightly together by staggering, so that one atom does not sit directly behind another. Instead, an atom in the second row sits behind a pair of atoms in the first row, and right in between them as viewed from the front. This sort of staggering happens easily in three dimensions, too.
In fact, if you take four oranges and make a square with them, and then decide to place a fifth orange on top of that layer, where is the obvious place to put this last orange? You would put it right in the middle of the first four, perched in the little valley on top of them.
If you have a whole produce box full of oranges, you can make a bigger first layer of oranges, and start placing additional oranges in all of the little square valleys between the oranges of the first layer. Soon you will find that you have built a second, simple square layer. It is just like the first layer, but offset by half a unit cell to the side and half a unit cell to the back.
You can keep adding more layers until you run out of oranges or the whole thing falls down and rolls away. If you do so, you might notice that all the oranges in the third layer sit directly above the oranges in the first layer, and all the oranges in the fourth layer sit directly above the ones in the second layer. That's because each new layer is offset by half a unit cell, so every two layers, that shift adds up to one unit cell, and the oranges line up again. This pattern is sometimes described as an ABABAB pattern; that means the odd layers line up exactly with each other, and the even layers line up exactly with each other.
• Body-centered cubic packing allows the second layer of atoms to get a little closer to the atoms in the first layer.
The unit cell in this system, like the simple cubic system, is a cube with an eighth of an atom (or orange) at each corner. However, the corner atoms are not found in adjacent layers, but in alternating layers (such as the first and the third). An atom from the middle layer is also included in the cell; the entire atom is right in the center of the unit cell. This packing system is called body-centered cubic.
• The unit cell of a body-centered cubic system is a lot like the simple cubic unit cell.
• The body-centered cubic unit cell has an extra atom in its center.
One important note: in a body-centered cubic system (and the other systems to follow), the atoms in the first layer are not quite touching each other. By backing off slightly from each other within a layer, there are closer interactions from one layer to another. The extra room allowed by spreading an individual layer out a little bit allows subsequent layers of atoms to nestle more deeply in those valleys. Stronger bonding interactions result from those close connections between layers.
By the way, you would have a hard time building up a simple cubic structure with oranges, unless you glued them together or used slightly flattened oranges, like clementines. There is more structural support in the body-centered cubic structure, because the oranges resting in valleys are distributing their weight to and getting support from four oranges beneath them, not just one.
Exercise $3$
Take a look at the body-centered cubic unit cell.
1. What fraction of an atom is found in each corner of the body-centered cubic unit cell?
2. How many corners are there in the cube?
3. Are there atoms anywhere else in the cube?
4. What is the total number of atoms in the body-centered cubic unit cell?
Answer a:
1/8 atom at each corner
Answer b:
8 corners in a cube
Answer c:
1 atom in middle of cube
Answer d:
$1 + 8 \times \frac{1}{8} = 2$ atoms per body-centered cube
Building up from Hexagonal Layers: Hexagonal Close Packing
Some metal atoms form hexagonal layers. Instead of being in close contact with four neighbouring atoms in a layer, an atom in a hexagonal layer is in close contact with six neighbouring atoms.
Do you still have your oranges? Suppose you start with a hexagonal layer of them. When another layer of oranges is added on top of a hexagonal layer, the new oranges don't sit right on top of the ones in the first layer. Instead, they sit in the little valleys between the oranges of the first layer. This is a little like the second layer of body-centered cubic packing. The difference is, in body centered cubic packing, all of the valleys between the oranges of the first layer are occupied by oranges from the second layer. In hexagonal packing, each orange in the first layer has six small valleys around it. You can't easily fit six oranges into all of those valleys at the same time; there isn't enough room. Instead, only every other valley is occupied by an orange, forming the second layer.
In the drawing, you might notice that the atoms in one layer are separated slightly from each other. That's also to the second layer of atoms to fit more deeply into the valleys between the atoms in the first layer. Some close interaction between the atoms within a layer is given up in order to get better interactions between the atoms in different layers. It's a trade-off that leads to slightly stronger interactions overall.
Again, just like the systems based on simple squares, the second layer that forms looks a lot like the first. It is another hexagonal layer. It is shifted away from the first layer by half an atom to the side, and half an atom to the front or back.
• Hexagonal layers are not usually stacked with atoms directly on top of each other.
• Hexagonal layers get a little closer to each other by shifting so that atoms above can sit in valleys below.
• There isn't room for every valley to be occupied by an atom. Every other valley is left open.
Once we begin to add a third layer, we have a choice. Just like with the second layer, we can't add a new orange into every valley. We will have to add them into every other valley. We can either add them directly over the oranges in the first layer, or directly over the valleys that were left empty when we filled in the second layer.
Let's put them directly over the oranges in the first layer. When we add the oranges for the fourth layer, we'll put them directly over the ones in the second layer. This system of packing is called hexagonal close-packing.
• In hexagonal close packing, the third layer of atoms is directly above the first layer.
• This system is a little like body-cenetered cubic, but with a hexagonal base instead of a square one.
Before we consider the unit cell in this system (which is a little more complicated than in the cubic packing systems), let's take a look at these layers of atoms, with a little bit greater separation.
Because the atoms in the top layer are directly over the atoms in the bottoms layer, we can think of these two layers as being identical. We can show that by drawing them in the same color.
In a hexagonal single layer, the unit cell was a rhombus. We could use that to build the unit cell for a hexagonal close packed system. We would just add walls into the cell to build the simplest repeating unit.
Exercise $4$
The pattern of layers in the simple cubic packing system was AAAAAA; each layer was placed so that each atom was directly above an atom in the layer below it. The pattern of layers in the body centered cubic arrangement was ABABAB; each atom in a layer was directly above one that was two layers down. Which pattern fits hexagonal close packing?
Answer
Hexagonal close packed pattern is ABAB.
Exercise $5$
Take a look at the hexagonal close-packed unit cell.
1. What fraction of an atom is found in each corner of the hexagonal close-packed cubic unit cell?
2. Adding up all those fractions, how many corner atoms are within the cell?
3. Are there atoms anywhere else in the cell?
4. What is the total number of atoms in the hexagonal close-packed unit cell?
Answer a:
At both the top and bottom layer, two atoms are 1/3 within the cell and two atoms are 1/6 within the cell.
Answer b:
That makes $2 \times (2 \times \frac{1}{3} + 2 \times \frac{1}{6}) = \frac{4}{3} + \frac{4}{6} = \frac{4}{3} + \frac{2}{3} = 1$ atom total in corners
Answer c:
1 atom inside the cell.
Answer d:
2 atoms total.
Exercise $6$
In the following variation on hexagonal close packing, the two unit cells appear to be slightly different.
1. What is the relationship between the two unit cells in the above structure?
2. How would you describe the relationships between the layers in the above structure (i.e. ABCABC etc)?
Answer
The pattern in this variation is ABACA.
Building up from Hexagonal Layers: Cubic Close Packing (or Face-Centered Cubic Packing)
Start with the same hexagonal layer of oranges. Once again, when another layer is added on top of that hexagonal layer, the new oranges don't sit right on top of the ones in the first layer. Instead, they sit in the little valleys between the oranges of the first layer. Remember, in hexagonal packing, each orange in the first layer has six small valleys around it. You can't easily fit six oranges into all of those valleys at the same time; there isn't enough room. Instead, only every other valley is occupied by an orange, forming the second layer.
Again, just like in the other systems, the second layer looks a lot like the first. It is another hexagonal layer. It is shifted away from the first layer.
Once we begin to add a third layer, we have a choice. Just like with the second layer, we can't add a new orange into every valley. We will have to add them into every other valley. We can either add them directly over the oranges in the first layer, or directly over the valleys that were left empty when we filled in the second layer.
This time, let's put them directly over the holes in the first layer that were not filled before. When we add the oranges for the fourth layer, we'll put them directly over the oranges in the first layer. The oranges in the fifth layer will sit above the ones in the second layer, and the ones in the sixth layer will sit over those in the third layer.
• In cubic close-packing, the layers are actually hexagonal (also called "close-packed").
• In the third layer, the atoms do not lie directly above atoms in the first layer.
• Instead, the atoms in the third layer sit above the empty valleys from the first layer.
This system of packing is called cubic close-packing or face centered cubic packing.
Exercise $7$
The pattern of layers in the simple cubic packing system was AAAAAA. The pattern of layers in the body centered cubic arrangement was ABABAB. How would you describe the pattern in cubic close-packing?
Answer
ABCABC.
Exercise $8$
The unit cell in the hexagonal close-packed system was based on a unit cell from a single layer. Propose a similar unit cell for a cubic close-packed system.
1. What two-dimensional unit cell forms the base of this three-dimensional unit cell?
2. How many layers does the cubic close-packed unit cell cut through?
3. Draw the cubic close-packed unit cell.
Answer a:
The rhombus-shaped hexagonal unit cell.
Answer b:
4 layers.
Hey, wait a minute. Why do they call it cubic packing if the unit cell is based on a rhombus instead of a square? Maybe there is another way of looking at this structure. Maybe there is an alternative unit cell.
There is a cubic unit cell, but it is harder to see. The cube is tipped on one of its corners. We will strip away some of the extra atoms in the hexagonal layers so we can get a better look.
There is an atom in the lowest layer that forms one corner of the cube. The opposite corner of the cube is an atom in the fourth layer. That atoms sits directly above the one in the first layer, the bottom corner of the cube. In between, three faces of the cube are tilted downwards. There is an atom in the middle of each of these faces; together with atoms at the three corners in between them, these form a second layer in the structure. There are also three faces tilted upwards. There is an atom in the center of each of these faces; together with atoms at the three corners in between them, these form a third layer in the structure.
Note that this cube has an atom in the center of each face; hence the name, face-centered cubic.
• A cubic close-packed takes the name "cubic" from the presence of a cubic unit cell.
• This cubic unit cell is oriented diagonally to the close-packed layers.
• This cubic unit cell has an atom in the ceneter of each face; it is also called "face-centered cubic".
Exercise $9$
Take a look at your new cubic close-packed unit cell.
1. What fraction of an atom is found in each corner of the cubic close-packed cubic unit cell?
2. How many corners are there in the cell?
3. Are there atoms anywhere else in the cell?
4. What is the total number of atoms in the cubic close-packed unit cell?
Answer a:
1/8 atom at each corner
Answer b:
8 corners in a cube
Answer c:
1/2 atom on each face of cube
Answer d:
$6 \times \frac{1}{2} + 8 \times {1}{8} = 4$ atoms per face-centered cube
The type of packing adopted by a metal actually has some influence on the physical properties of the metal. For example, the metals that are closely associated with metal-working during antiquity, such as lead, gold, silver, and copper, generally adopt cubic close-packed structures. These metals have high values of "physical ductility" -- an index of how easily they can be stretched into wire. Titanium, cobalt, and zinc are hexagonal close-packed and have moderate values of physical ductility. Metals that adopt body-centered cubic packing, such as iron, tungsten, and chromium, usually have lower physical ductility.
This trend isn't a strict rule; some body-centered metals are more ductile than some cubic close-packed metals. Nevertheless, metal packing seems to be one factor in how easily metals can be stretched, bent, and worked. The reason it plays a role is because of the presence of "slip planes" in metal structure. Remember the idea that rows of metal atoms can slide across each other? It turns out that cubic close-packed metals form rows in lots of different directions, so their atoms can slip past each other pretty easily. In contrast, that central atom in a body-centered cubic cell acts like a speed bump for the layer of atoms above and below it, slowing down their movement.
Exercise $10$
We saw previously that the number of near neighbors of a metal atom is called the coordination number. What is the coordination number in each of the following cases?
1. an atom at the corner of a simple cubic cell. Remember, it is surrounded by other unit cells, too.
2. an atom in the center of a body-centered cubic cell.
3. an atom in the corner of a body-centered cubic cell.
4. an atom in the corner of a face-centered cubic cell.
5. an atom on the face of a face-centered cubic cell.
Answer a:
coordination number = 6 (3 within the cube, along the three adjacent edges, plus three on adjacent cubes).
In the picture, consider the green atom coordinated by its nearest neighbors, the blue atoms. One unit cell is highlighted in red.
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown for comparison. The three other nearest atoms are in cells below, behind and to the right of the one shown.
Answer b:
coordination number = 8 (corners of the cube).
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown below for simplicity. All of the coordinated atoms are found within the cell.
Answer c:
The closest atoms are in the centers of the adjacent cells; the other corners are further away; coordination number = 8.
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown below, for simplicity. Because the atom is in a corner, there are eight other cells arranged around this one: one on the right, two behind, and then four more below.
Answer d:
The closest atoms are on the faces adjacent to the corner position; these faces are found along three planes, with four faces surrounding the corner on each plane; coordination umber = 12.
Here is the coordination environment, without the atoms that are further from the green one.
A single unit cell is shown below. Because the atom is in a corner, there are seven other cells arranged around this one: one on the right, two behind, and then four more below.
Answer e:
The closest atoms are at the corners around the face (4), as well as the atoms in the middle of each adjacent face (8); coordination number = 12.
Here is the coordination environment without the other atoms.
A single unit cell is shown below. There is another cell directly below it.
Exercise $11$
Coordination geometry is related to the coordination number. What is the coordination geometry in each of the following cases?
1. an atom at the corner of a simple cubic cell. Remember, it is surrounded by other unit cells, too.
2. an atom in the center of a body-centered cubic cell.
3. an atom in the corner of a body-centered cubic cell.
Answer a:
Octahedral. An octahedron has six vertices and eight faces.
Answer b:
Cubic. A cube has eight vertices and six faces.
Answer c:
This is also a cube, translated from the last one (i.e. the cube is just shifted to a different position in the lattice).
Exercise $12$
We learned earlier about the holes or interstitial spaces between atoms in a layer. There are also holes between atoms in a three dimensional structure. For example, how would you describe the shape of the holes in the following cases:
1. the hole in the middle of a simple cubic cell.
2. a hole between the central atom and the face of a body-centered cubic cell.
3. the hole right in the middle of the face of a body-centered cubic cell.
4. the hole between the atoms forming a valley in one hexagonal layer and the atom sitting in the valley.
5. the hole between the atoms forming an empty valley in one hexagonal layer and the atoms in the layer above.
Answer a:
A cubic hole.
Answer b:
A square pyramidal hole. The atom in the hole is above a square of atoms, with an additional atom right above it.
Answer c:
An octahedral hole. The atom occupying the hole has been pushed all the way to the face of the cube. It is right in the middle of a square of atoms. One of the vertices is in the center of the next cube.
Answer d:
A tetrahedral hole. Note that the layers in the cubic closest-packed system run diagonally through the cubic unit cell. The atom in the hole sits above a trio of atoms on three adjacent faces, and is topped by the atom in the corner of the cube. Because a tetrahedron is so symmetric, the arrangement can be described in a number of other ways, choosing any three surrounding atoms as the base and the fourth atom as the cap of the tetrahedron.
Answer e:
An octahedron. Again, the hexagonal layers run diagonally through the unit cell.
Exercise $13$
Packing efficiency is often determined in terms of the percentage of the volume of a unit cell that is actually occupied by atoms.
In the following cases, calculate the volume of the entire unit cell.
1. a simple cubic unit cell.
2. a body-centered cubic unit cell.
3. a face-centered cubic unit cell.
4. a hexagonal close-packed unit cell (remember, this cell is a rhombic prism, not a cube).
You can assume the atoms in the cell are titanium. Titanium has an atomic radius of 2.00 Angstroms (or 2.00 x 10-10 m).
Exercise $14$
How many atoms are there in a unit cell in the following layers?
1. a simple cubic unit cell.
2. a body-centered cubic unit cell.
3. a face-centered cubic unit cell.
4. a hexagonal close-packed unit cell (remember, this cell is a rhombic prism, not a cube).
You may need to add up fractions of titanium atoms to arrive at the answer. The answer may or may not be a whole number.
Exercise $15$
What is the volume occupied by a titanium atom?
Exercise $16$
In the following cases, what percentage of the unit cell would be filled with titanium atoms?
1. a simple cubic unit cell.
2. a body-centered cubic unit cell.
3. a face-centered cubic unit cell.
4. a hexagonal close-packed unit cell.
The percentage of unit cell that is occupied by atoms is called "the packing efficiency".
Exercise $17$
Use the appropriate template to draw one unit cell for each of the following metals.
a) iron, Fe, is BCC b) zirconium, Zr, is HCP c) copper, Cu, is FCC
d) manganese, Mn, is BCC e) platinum, Pt, is CCP f) osmium, Os, is HCP
Answer a:
Answer b:
Answer c:
Answer d:
Answer e:
Answer f:
Want to see some additional information? Try these other sites that provide three-dimensional pictures of how atoms pack together.
Kings College Crystal Packing Lab
Naval Research Lab site
Bodner site at Purdue
Solid State site at Liverpool
Davidson College Crystals
Visualization of solid state structures: unit cells, etc.
Oxford University Solid Structures
Visualization of solid state structures: unit cells, etc. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/02%3A_Metals/2.03%3A_Building_Metal_Atoms_in_Three_Dimensions.txt |
Exercise $1$
Maraging steel is an alloy, Fe3Ni, that forms a face-centered cubic unit cell. It is used for high-strength applications such as in the forging of fencing swords (epées).
1. Draw a cube.
2. Add circles to denote atoms in a face-centered cubic array.
3. Calculate the number of atoms in the unit cell (always show work).
4. Shade in selected atoms in the unit cell to give a 3:1 ratio of iron:nickel in the unit cell.
5. Add a legend with shaded and non-shaded circles so that it is clear which circle represents which atom.
6. Face centered cubic can also be thought of as forming hexagonal layers. The layers alternate to give which pattern? ABABAB or ABCABCABC
7. In the cubic diagram you made, put a letter beside each atom to indicate which of those layers (A or B or C) it is found in.
8. Why does the addition of nickel to the steel make the material more difficult to bend or break?
Answer a), b
Answer c
There are eight atoms in the corner; each is shared by eight neighbouring cubes. There are six atoms on the faces; each is shared by two neighbouring cubes.
# atoms = $8 (\frac{1}{8}) + 6 (\frac{1}{2}) = 1 + 3 = 4$ atoms
Answer d), e), g
Answer h
Nickel is a little smaller than iron. A row of atoms rolling along in the layer above would fall into the "pothole" caused by the smaller nickel atom and get stuck.
2.05: Metals- Solutions to Selected Problems
Exercise 2.1.1
The second metal atom is a different size than the principle metal atom. It will not quite fit into the array of atoms. Consequently, the atoms will not be able to slide past each other as easily.
Exercise 2.1.2
$2Fe: \: (3^{+}) \times 2 = (6^{+}) \nonumber$
$3 O: \: (2^{-}) \times 3 = (6^{-}) \nonumber$
Fe2O3: neutral
Exercise 2.1.3
The longer the box, the longer the possible wavelength.
Exercise 2.2.1
The pattern also repeats diagonally.
Exercise 2.2.2
The squares can form a number of regularly repeating patterns. The rows can exactly repeat (an aaa pattern), or they can be shifted slightly, so that every other row exactly repeats the first row (an ababab pattern). They could even be shifted so that every third row is an exact repeat of the first (abcabcabc).
The pentagons do not form a repeating layer in two dimensions.
Exercise 2.2.3
In the simple square packing, the holes are roughly diamond shaped.
In the hexagonal pattern, the holes are roughly triangular.
Exercise 2.2.6
1. Using quarters, the distance is about 5 mm.
2. Using quarters, the distance is about 2.5 mm.
A free electron would be closer to the atoms in a hexagonal close packed layer. There would be stronger electron-ion attraction in that case.
Exercise 2.2.7
1. Using dimes, the distance is about 3.5 mm.
2. Using dimes, the distance is about 1.5 mm.
A free electron would be closer to the atoms if the atoms were smaller. There would be stronger electron-ion attraction in that case.
Exercise 2.2.8
1. melting points (and force of attraction between atoms): Fr < Cs < K < Na < Li
2. This trend mirrors the sizes of the atoms. Lithium is the smallest and francium the largest. The electron / ion attraction is greatest in Li and weakest in Fr.
Exercise 2.2.9
a) The total area of this square is
$Area \:= w^{2} \nonumber$
in which w = width of the unit cell. The width of the unit cell is
$w = 2r \nonumber$
in which r = radius of titanium atom.
$Area \: = 4r^{2} = 4(2.00 \times 10^{-10} m)^{2} = 4(4.00 \times 10^{-20} m^{2}) = 1.60 \times 10^{-19} m^{2} \nonumber$
b) The total area of this rhombus is
$Area \: = s^{2} sin \theta \nonumber$
in which s = one side of unit cell and θ = an angle of the unit cell (either 60° or 120°). But
$s = 2r \nonumber$
in which r = radius of titanium atom.
$Area \: = (2 \times 2.00 \times 10^{-10} m)^{2} \: sin(60) = 1.60 \times 10^{-19} \: (0.87) m^{2} = 1.39 \times 10^{-19} m^{2} \nonumber$
Exercise 2.2.10
1. $4 \times \frac{1}{4} = \: 1 \: atom$
2. $2 \times \frac{1}{6} + 2 \times \frac{1}{3} = \frac{2}{6} + \frac{4}{6} = \: 1 \: atom$
Exercise 2.2.11
The cross-sectional (two dimensional) area of a titanium atom is the area of a circle
$Area \: = \pi r^{2} \nonumber$
$Area \: = 3.1415 \times (2.00 \times 10^{-10} m)^{2} = 3.1415 \times 4.00 \times 10^{-20} m^{2} = 1.257 \times 10^{-19} m^{2} = 1.26 \times 10^{-19} m^{2} \nonumber$
Exercise 2.2.12
1. $Efficiency \: = \frac{occupied}{total} \times 100 \% = (\frac{1.26 \times 10^{-19} m^{2}}{1.60 \times 10^{-19} m^{2}}) \times 100 \% = 79 \%$
2. $Efficiency \: = (\frac{occupied}{total}) \times 100 \% = (\frac{1.26 \times 10^{-19} m^{2}}{1.39 \times 10^{-19} m^{2}}) \times 100 \% = 91 \%$
A greater percentage of space is occupied in hexagonal close packing.
Exercise 2.2.13
1. coordination number = 4
2. coordination number = 6
Exercise 2.2.14
1. Assuming the pattern extends linearly, coordination number = 2. Otherwise, coordination number = 2 for interior atoms and coordination number = 1 for terminal atoms.
2. Assuming the pattern extends in two dimensions, coordination number = 3. Otherwise, coordination number = 3 for interior atoms, coordination number = 2 for side edge atoms and coordination number = 1 for corner atoms.
Exercise 2.3.0
1. (i) 8; (ii) 6
2. (i) 4; (ii) 4
3. (i) 5; (ii) 5
4. (i) 5; (ii) 6
5. (i) 6; (ii) 8
Exercise 2.3.1
1. 1/8 atom at each corner
2. 8 corners in a cube
3. $8 \times \frac{1}{8} = 1$ atom per simple cube
Exercise 2.3.2
1. 1/8 atom at each corner
2. 8 corners in a cube
3. 1 atom in middle of cube
4. $1 + 8 \times \frac{1}{81 + 8 x 1/8 = 2 atoms per body-centered cube Exercise 2.3.3 Hexagonal close packed pattern is ABAB. Exercise 2.3.4 1. At both the top and bottom layer, two atoms are 1/3 within the cell and two atoms are 1/6 within the cell. 2. That makes \(2 \times (2 \times \frac{1}{3} + 2 \times \frac{1}{6}) = \frac{4}{3} + \frac{4}{6} = \frac{4}{3} + \frac{2}{3} = 1$ atom total in corners
3. 1 atom inside the cell.
4. 2 atoms total.
Exercise 2.3.5
The pattern in this variation is ABACA.
ABCABC.
Exercise 2.3.7
1. The rhombus-shaped hexagonal unit cell.
2. 4 layers.
Exercise 2.3.8
1. 1/8 atom at each corner
2. 8 corners in a cube
3. 1/2 atom on each face of cube
4. $6 \times \frac{1}{2} + 8 \times \frac{1}{8} = 4$ atoms per face-centered cube
Exercise 2.3.9
a) coordination number = 6 (3 within the cube, along the three adjacent edges, plus three on adjacent cubes).
In the picture, consider the green atom coordinated by its nearest neighbors, the blue atoms. One unit cell is highlighted in red.
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown for comparison. The three other nearest atoms are in cells below, behind and to the right of the one shown.
b) coordination number = 8 (corners of the cube).
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown below for simplicity. All of the coordinated atoms are found within the cell.
c) the closest atoms are in the centers of the adjacent cells; the other corners are further away; coordination number = 8.
Here is the coordination environment without the surrounding atoms.
A single unit cell is shown below, for simplicity. Because the atom is in a corner, there are eight other cells arranged around this one: one on the right, two behind, and then four more below.
d) The closest atoms are on the faces adjacent to the corner position; these faces are found along three planes, with four faces surrounding the corner on each plane; coordination number = 12.
Here is the coordination environment, without the atoms that are further from the green one.
A single unit cell is shown below. Because the atom is in a corner, there are seven other cells arranged around this one: one on the right, two behind, and then four more below.
e) The closest atoms are at the corners around the face (4), as well as the atoms in the middle of each adjacent face (8); coordination number = 12.
Here is the coordination environment without the other atoms.
A single unit cell is shown below. There is another cell directly below it.
Exercise 2.3.10
a) Octahedral. An octahedron has six vertices and eight faces.
b) Cubic. A cube has eight vertices and six faces.
c) This is also a cube, translated from the last one (i.e. the cube is just shifted to a different position in the lattice).
Exercise 2.3.11
a) A cubic hole.
b) A square pyramidal hole. The atom in the hole is above a square of atoms, with an additional atom right above it.
c) An octahedral hole. The atom occupying the hole has been pushed all the way to the face of the cube. It is right in the middle of a square of atoms. One of the vertices is in the center of the next cube.
d) A tetrahedral hole. Note that the layers in the cubic closest-packed system run diagonally through the cubic unit cell. The atom in the hole sits above a trio of atoms on three adjacent faces, and is topped by the atom in the corner of the cube. Because a tetrahedron is so symmetric, the arrangement can be described in a number of other ways, choosing any three surrounding atoms as the base and the fourth atom as the cap of the tetrahedron.
e) An octahedron. Again, the hexagonal layers run diagonally through the unit cell.
Exercise 2.4.1
c) There are eight atoms in the corner; each is shared by eight neighbouring cubes. There are six atoms on the faces; each is shared by two neighbouring cubes.
$\# \: atoms = 8 (\frac{1}{8})+ 6 (\frac{1}{2}) = 1 + 3 = 4 \: atoms \nonumber$
h) Nickel is a little smaller than iron. A row of atoms rolling along in the layer above would fall into the "pothole" caused by the smaller nickel atom and get stuck. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/02%3A_Metals/2.04%3A_Application_Problems_with_Metals.txt |
Periodic trends tell us that some atoms gain electrons easily to obtain a stable configuration; these atoms have a high electron affinity. These atoms are principally found in nature as anions (ANN-eye-ons). Anions are atoms that have gained extra electrons, and have overall negative charge. Examples include the halogens, such as chlorine and bromine.
• Atoms in the upper right section of the periodic table can easily form anions.
One of the most commonly-used periodic properties is electronegativity. A look at the periodic table shows the high electronegativity of atoms in the upper right part of the table. The relatively high positive charge in the nuclei of these atoms makes their attraction for electrons very strong.
Another thing to remember is that atoms towards the right hand side of the periodic table are likely to gain enough electrons to attain the same electron configuration as a noble gas.
The noble gases are exceptional in the periodic table because they do not easily undergo reactions with other elements. In a chemical reaction, electrons are somehow redistributed between atoms. The noble gases have particularly stable electron configurations, however, and do not gain or lose electrons easily. This is true even though they have high electronegativities; in the case of noble gases, high electronegativity might be thought of as a measure of the stability of the electrons in these elements. However, they have a "closed shell" configuration, meaning that any additional electrons would have to go into a much higher energy level in the next "shell". Other atoms can gain a similarly stable electron configuration by gaining or losing electrons. For example, if oxygen gains two electrons, it attains the same electron configuration as neon. Oxygen is frequently found as an O2- anion in many compounds.
Periodic trends also tell us that some atoms lose electrons easily to obtain a stable configuration; these atoms have a low ionization potential. These atoms are principally found in nature as cations (CAT-eye-ons). Cations are atoms that have lost electrons, so that they have an overall positive charge. Examples include the alkalis, such as sodium and potassium.
• Atoms in the vast expanses to the lower left part of the periodic table easily become cations.
Just as some elements attain a noble gas configuration by gaining some extra electrons, other elements can more easily gain that configuration by losing some electrons. For a sodium atom, it is simply closer to get to the same configuration as neon than it is to get to the same configuration as argon. Sodium is frequently found as Na+ in many compounds.
In general, atoms that have an overall charge are called ions (EYE-ons). In an ion, the number of protons in the nucleus is not exactly balanced by the number of electrons, so there is either an overall positive or an overall negative charge on the atom.
What happens to these atoms as they gain or lose an electron? Of course, having an overall charge on an atom introduces an additional electrostatic problem. Two atoms might not repel each other, but two cations would. In the electron sea model of metals, metal cations exist in a block of metal. Those like charges should push the metal cations away from each other, but most metals don't spontaneously fly apart into tiny particles. The cations are partly held together by the presence of delocalized electrons in the block of metal. The attraction of the delocalized electrons for the metal cations offsets any repulsion between the metal cations.
For that reason, we will see that cations are generally found in the presence of anions, and vice versa, so that repulsive forces between like-ions are balanced by attractive forces with oppositely-charged ions.
Exercise \(1\)
Given the following pairs of atoms, which one is most likely to give up an electron?
1. tantalum or vanadium
2. mercury or zinc
3. silicon or sulfur
4. tungsten or copper
Answer a:
Ta (tantalum) is lower in the periodic table than V (vanadium)
Answer b:
Hg (mercury) is lower in the periodic table than Zn (zinc)
Answer c:
Si (silicon) is to the left in the periodic table compared to S (sulfur)
Answer d:
W (tungsten) is to the left and lower in the periodic table compared to Cu (copper)
Exercise \(2\)
Provide the electron configuration for the following atoms or ions (without using the nobel gas abbreviation for the outermost shell). If the atom or ion has the same electron configuration as a noble gas, state which noble gas it resembles.
1. oxygen, O
2. oxide, O2-
3. calcium, Ca
4. calcium ion, Ca2+
5. aluminum, Al
6. aluminum ion, Al3+
7. nitrogen, N
8. nitride, N3-
9. chlorine, Cl
10. chloride, Cl-
Answer a:
[He]2s22px22py12pz1
Answer b:
[He]2s22px22py22pz2
Answer c:
[Ar]4s2
Answer d:
[Ar]
Answer e:
[Ne]3s22px1
Answer f:
[Ne]
Answer g:
[He]2s22px12py12pz1
Answer h:
[He]2s22px22py22pz2
Answer i:
[Ne]2s22px22py22pz1
Answer j:
[Ne]2s22px22py22pz2
Notice the change in the name of chloride: frequently, when an element becomes an anion, the end of its name changes to "ide". A chlorine atom that has an extra electron is called a chloride ion. An oxygen atom with two extra electrons (to get to the same electron configuration as neon) is called oxide. A nitrogen atom with three extra electrons (to get to the same electron configuration as neon) is called nitride.
There are other changes that occur when an atom is ionized (when an atom loses or gains an electron, becoming an ion). Cations tend to be relatively small. The higher than normal ratio of positive to negative charge causes the remaining electrons to contract towards the nucleus.
• A cation is always smaller than a neutral atom of the same element.
For similar reasons, anions tend to be relatively large. The low ratio of positive to negative charge causes the remaining electrons to repel each other more and expand away from the nucleus.
• An anion is always much larger than a neutral atom of the same element.
As a result, many of the elements in the periodic table are very different in size depending on whether they form cations or anions.
In describing the size of an atom or ion, typically the radius is reported rather than the diameter. Note that different units are sometimes used to report the measured sizes of atoms or ions. The most common unit is an Angstrom, Å. An Angstrom is 10-10 m; it is based on the metric system although it is not, strictly speaking, a metric unit. Alternatively, sometimes atomic and ionic radii are reported in nanometers, nm (10-9 m) or picometers, pm (10-12 m). The reason for the prevalence of the Angstrom is that this unit is about the size of an atom; most atoms have radii between 0.75 and 2 Å. An atom with a 1.0 Å radius has a radius of 100 pm, or 0.10 nm.
Exercise \(3\)
Given the following atomic radii, select the correct radius for the given ion.
1. Li : 0.134 nm. Li+ : 0.076 or 0.176 nm?
2. O : 0.073 nm. O2- : 0.040 or 0.140 nm?
3. Cs : 0.235 nm. Cs+ : 0.167 or 0.267 nm?
4. H : 0.037 nm. H- : 0.024 or 0.144 nm?
Answer a:
0.076 nm; smaller
Answer b:
0.140 nm; larger
Answer c:
0.167 nm; smaller
Answer d:
0.144; larger
Exercise \(4\)
Given the following ionic radii, select the correct radius for the given atom.
1. Pb2+ : 0.119 nm. Pb : 0.054 or 0.154 nm?
2. F- : 0.133 nm. F : 0.071 or 0.171 nm?
3. Pd2+ : 0.086 nm. Pd : 0.057 or 0.128 nm?
4. I- : 0.220 nm. I : 0.133 or 0.235 nm?
Answer a:
0.154; larger
Answer b:
0.071; smaller
Answer c:
0.128; larger
Answer d:
0.133; smaller | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.01%3A_Ionic_Atoms.txt |
Compounds are mixtures of elements bonded together to make one material. Compounds can contain two or more elements. Those elements have to be found in a specific ratio. Two materials that contain the same two elements bonded together, but in different ratios, are two different compounds. Two materials that contain two different pairs of elements bonded together, but in the same ratio, would still be different compounds.
In nature, elements are found bound up in compounds more often than not. One common way for elements to be bound together is to form a salt. Salts are found very frequently in the earth's crust. A salt contains both anions and cations. Cations are not generally found alone. Anions are not generally found alone. They need counterions to balance out their charges.
Imagine two atoms get together to form a salt. A sodium atom comes together with a fluorine atom. The sodium atom lets one of its electrons go, and the fluorine atoms snatches it up. Now there is a sodium cation and a fluoride anion. The two ions are held together by the attractive forces between their opposite charges. They have formed a salt.
• Anions and cations are always found together.
• The charges on the cations must balance the charges on the anions.
For example, sodium chloride is a very common ionic compound. It is the main component of table salt, used in cooking. Sodium chloride contains sodium ions, each with a +1 charge, and chloride ions, each with a -1 charge. Overall, the compound has no charge, because the positive sodiums balance out the charge on the negative chlorides, and vice versa.
\[\ce{Na^{+} + Cl^{-} = NaCl \: (no \: overall \: charge)}\]
In a similar way, other ionic compounds form so that there is no overall charge on the compound. Magnesium oxide forms in a ratio so that the positive charges on the magnesium balance the negative charges on the oxygen.
\[\ce{Mg^{2+} + O^{2-} = MgO \: (no \: overall \: charge)}\]
However, sodium oxide would need to form in a different ratio in order to keep the positive and negative charges balanced. Instead of forming in a one-to-one ratio, there would need to be double the number of sodium ions as oxygen ions in order to have the charges cancel each other exactly.
\[\ce{2 Na^{+} + O^{2-} = Na2O \: (no \: overall \: charge)}\]
The two positive charges on the two sodium cations balance out the two negative charges on the oxide anion.
Exercise \(1\)
What would be the ratio of elements if each of the following ions formed a salt with chloride ions, Cl-?
a) K+ b) Fe3+ c) Mo6+ d) Zr4+
Answer a:
1:1 K:Cl or KCl
Answer b:
1:3 Fe:Cl or FeCl3
Answer c:
1:6 Mo:Cl or MoCl6
Answer d:
1:4 Zr:Cl or ZrCl4
Exercise \(2\)
What would be the ratio of elements if each of the following ions formed a salt with oxide ions, O2-?
a) Li+ b) Fe3+ c) Cr6+ d) Ti4+
Answer a:
Li:O 2:1 or Li2O
Answer b:
Fe:O 2:3 or Fe2O3
Answer c:
Cr:O 1:3 or CrO3
Answer d:
Ti:O 1:2 or TiO2
Exercise \(3\)
What would be the ratio of elements if each of the following ions formed a salt with nitride ions, N3-?
a) Li+ b) Ta3+ c) W6+ d) Co2+
Answer a:
Li:N 3:1 or Li3N
Answer b:
Ta:N 1:1 or TaN
Answer c:
W:N 1:2 or WN2
Answer d:
Co:N 3:2 or Co3N2
Exercise \(4\)
Some compounds are more complicated. In the following polyoxoanions, the metal and oxide charges do not quite cancel. Given the charge on the metal, and assuming oxide has a 2- charge, what would be the overall charge on the polyoxoanion? Suggest the right ratio of counterion(s) that could balance the charge in each case (e.g. 3 Li+ ions, etc).
a) WO4n- (W6+) b) V4O12n- (V5+) c) Mo4O14n- (Mo6+) d) Cr(OH)6Mo6O18n- (OH-, Cr4+, Mo6+)
Answer a:
Li4WO4 because tungstate anion would be 4-
Answer b:
Li4V4O12 because the tetravanadate ion would be 4-
Answer c:
Li2Mo4O14 because the tetramolybdate ion would be 2-
Answer d:
Li2Cr(OH)6Mo6O18 because the complex polyoxoanion would be 2-
Ionic compounds have no overall charge. Instead, they consist of anions and cations found together to balance charge. The number of positive charges in an ionic compound equal the number of negative charges. Overall, the charge is balanced out.
Why does the charge need to be balanced out? Imagine a bottle of sodium cations. There are a few problems with the cations in that bottle. First of all, the cations will all repel each other. You'd better stand back when you take the lid off. The repulsion between the sodium atoms will make them erupt out of the bottle like a volcano.
By the way, what happened to all of the electrons when somebody made that bottle of sodium ions? Did they also make a jar of electrons with the leftovers? Did the two bottles sit on their shelves in a factory in Milwaukee until someone in Cleveland ordered a bottle of sodium ions and somebody in Tampa ordered a bottle of electrons?
The problem with that story lies in Coulomb's law, the mathematical relationship that talks about the force of attraction between two charges. A transformation of Coulomb's law deals with the energy needed to separate opposite charges. It would take a lot of energy to separate a whole bottle of electrons from a whole bottle of sodium cations, and even more energy to send one bottle to Cleveland and another to Tampa. You would need a power station to generate a bazillion gigawatts of energy to get the job done.
On the other hand, people really can generate cations or anions in instruments such as mass spectrometers. They can even keep those ions for a while in "ion traps". They would only do this with extremely small numbers of ions, though (nowhere near a visible amount), and it does take a generous amount of power. Once they switch the mass spectrometer off, the ions won't stick around for long. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.02%3A_Counterions.txt |
The strong electrostatic forces of attraction of a cation for its surrounding anions, and the strong electrostatic forces of attraction of an anion for its surrounding cations, keep the compound together.
Most of the time, we think of ionic compounds as solids. In the solid state, the ions are very close together. The forces of attraction between cations and anions are very high. In the liquid (melted) state, the ions would move around independently, and would be able to move a little farther apart from each other. In order to accomplish this independent movement, a great deal of energy would have to be supplied to get over the strong forces of attraction in the solid state. It takes a lot of heat to supply enough energy to convert them into liquid.
• Ionic compounds often have high melting points.
In the gas phase, ions would be very far from each other and would move very freely. They would no longer attract each other very strongly because of the distances between ions. In order to convert an ionic compound into vapour, an enormous amount of energy would have to be supplied to overcome the attractions in the solid or liquid phase. Often, so much energy is supplied that they undergo decomposition into different compounds rather than boiling.
• Ionic compounds have extremely high boiling points.
The changes of state in ionic compounds are governed by simple electrostatic forces between the ions. These electrostatic forces are governed by Coulomb's Law, in which the force of attraction depends on the amount of charge and the distance between the ions. That means that there are sometimes predictable variations in the properties of ionic compounds.
For example, among the potassium halides, the melting point is lowest for the iodide (681 oC) and highest for the fluoride (858 oC). The reason for that comes from the distance dependence in Coulomb's law.
Iodide is a bigger ion than bromide, chloride or fluoride. That means the distance between atoms is greater in potassium iodide than in potassium fluoride. In other words, the average distance between the positive potassium nucleus and the negative electrons surrounding the anion is greater for iodide than for chloride or fluoride. Note that this average distance really amounts to the distance between the two nuclei.
The greater the distance between ions, the lower the forces of attraction. The ions in potassium iodide can move around more easily than the ions in potassium fluoride. Potassium iodide has a lower melting point than potassium fluoride.
• Compounds containing smaller ions often have higher melting points than similar compounds that contain larger ions.
Exercise \(1\)
Select the compound that would have the lowest melting point in each of the following pairs.
a) KCl or LiCl b) NaF or NaBr c) CaO or BeO d) LiF or KBr
Answer a:
KCl would have the lowest melting point. It would be easier to melt than LiCl because the smaller Li+ ion would more strongly attract the counterion, owing to the smaller distance separating the opposite charges.
Answer b:
NaBr would have a lower melting point than NaF.
Answer c:
CaO would have a lower melting point than BeO.
Answer d:
KBr would have a lower melting point than LiF.
There is another factor, too: the force of attraction between ions also depends on the magnitude of the charges involved. The greater the size of the charge on an ion, the greater the force of attraction for its counterion.
For example, calcium fluoride, CaF2, has a lower melting point (1418 oC) than calcium oxide, CaO (2572 oC).
This difference is probably not due to differences in the distance between the charges. Fluoride and oxide are almost the same size, and if anything fluoride is a little smaller. Based on interionic distance alone, fluoride could have a slightly higher melting point than calcium oxide.
However, each fluoride has a 1- charge, but the oxide has a 2- charge. As a result of this greater charge, the force of attraction between an oxide and a calcium ion is stronger than the force of attraction between a fluoride and a calcium ion. It is more difficult to get the calcium and oxide ions to move away from each other, and the melting point is higher than for calcium fluoride.
• Compounds containing more highly charged ions often have higher melting points than similar compounds that contain lower charged ions.
Exercise \(2\)
Select which of the following compounds would have the lowest melting point. Assume that the cations and the anions in each structure are all +1 or -1, respectively.
Exercise \(3\)
Select the compound that would have the highest melting point in each of the following pairs.
a) KCl or CaCl2
b) NaF or Na2O
c) CaO or NaF
Answer a:
CaCl2
Answer b:
Na2O
Answer c:
CaO
Exercise \(4\)
The melting point of NaCl is 801°C. Estimate (guess!) the melting point of the following compounds. (More important than the actual number you come up with is whether it is larger or smaller than the melting point of NaCl and by a lot or a little.)
a) KCl
b) NaF
c) Na2O
d) MgS
Answer a:
A little lower since K+ is a little larger than Na+ (actual= 770°C)
Answer b:
A little higher since F- is a little smaller than Cl- (actual =930°C)
Answer c:
A lot higher since the anion charge is 2x higher (actual =1275°C (sublimes but doesn’t melt))
Answer d:
A whole lot higher since cation and anion charge are 2x higher (actual = 2000+ °C (decomposes above this temp)) | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.03%3A_Physical_Properties.txt |
One of the other general properties of ionic compounds is that they often have some solubility in water. The oceans, of course, are saltwater. In a mixture, two or more materials are mixed together but they remain essentially separate, like sand and water. You can still easily tell the difference between the sand and the water, because even if you shake them up they will separate again on their own.
In a mixture, two or more materials are mixed together but they remain essentially separate, like sand and water. You can still easily tell the difference between the sand and the water, because even if you shake them up they will separate again on their own.
In a suspension, one or more materials is mixed into a liquid, and the mixture becomes somewhat homogeneous. Instead of having easily identifiable layers, the liquid looks the same throughout. However, suspensions are generally cloudy liquids. Milk is a suspension. It contains water, fats and proteins. They may settle out into separate layers eventually, but it takes time.
In a solution, one or more materials is mixed into a liquid, and the mixture becomes a completely homogeneous liquid. Solutions are transparent, not cloudy. They may be colored or colorless, but you can always see through them. Saltwater is a solution.
You can't see chunks of salt in the solution because the salt particles are too small for you to see. The salt is separated into individual ions, surrounded by water molecules.
Of course, if you put some salt in water, it might not dissolve right away. You might have to stir it for a while.
Eventually more of the salt would dissolve in the water.
However, at some point, the system might come to "equilibrium": the water has dissolved all of the salt that it can, so the rest of the salt stays solid. This equilibrium may be "dynamic": different ions may become dissolved in the water or may be deposited from solution into the solid state. However, the overall ratio of dissolved ions to water stays the same.
Exercise \(1\)
Let's take a look at the idea that a given amount of water is only able to dissolve a specific amount of salt.
1. In the diagram above, how many water molecules are there?
2. How many units of salt (an anion and a cation) are dissolved?
3. If there were only a dozen water molecules present, how many units of salt would dissolve?
4. If a hundred water molecules were present, how many units of salt would dissolve?
Answer a:
25 waters.
Answer b:
2 units (2 anions and 2 cations).
Answer c:
Half the water might only dissolve half the salt: 1 unit.
Answer d:
Four times the water may dissolve four times the salt: 4 units.
Why do salts dissolve in water? Water is a molecular compound; the atoms are directly attached to each other, rather than being ions that are attracted to each other. Because of electronegativity differences, the oxygen in water has a partial negative charge and the hydrogens have partial positive charges. Ionic compounds can dissolve in polar liquids like water because the ions are attracted to either the positive or negative part of the molecule.
Note that there is a sort of tug-of-war involved when things dissolve in water. The water is pulling individual ions away from the solid. The solid is pulling individual ions back out of the water. There exists an equilibrium at some point, based on how strongly the water attracts the ions, versus how strong the ionic solid attracts the ions.
You might expect to be able to predict vaying degrees of solubility in water for different ionic compounds. You would just use the principles of Coulomb's law that we used in melting points. The smaller the ions, the closer together they would be, and the harder it would be for the water molecules to pull the ions away from each other.
Exercise \(2\)
Predict which of the following pairs should be more soluble in water, based on what you know about Coulombic attraction between ions.
1. LiF or NaF
2. NaK or KF
3. BeO or LiF
Answer a:
LiF should be more soluble.
Answer b:
KF should be more soluble (but keep reading).
Answer c:
LiF should be more soluble.
Exercise \(3\)
Although lithium fluoride and magnesium oxide contain cations and anions of roughly the same size, lithium fluoride is much more soluble in water (2.7 g/L) than magnesium oxide (0.087 g/L) at room temperature. Propose a reason why.
Answer
The MgO contain more highly charged ions (Mg2+ and O2-) than LiF (Li+ and F-) and so it is more difficult to separate the ions from their solid state.
However, the trends we saw in melting points in ionic compounds become more complicated when it comes to solubility. The water solubility of alkali chlorides does not follow a simple trend (Table \(1\)).
Table \(1\) Water solubility among alkali chlorides.
Table \(1\) Water solubility among alkali chlorides.
Compound Water Solubility in g/100 mL at 20oC
LiCl 83
NaCl 359
KCl 344
Lithium chloride is certainly the least water-soluble of the three compounds. That makes sense, since the lithium ions are small and the attraction for the chloride would be stronger over that shorter distance. However, we would expect potassium chloride to be the most soluble by far, and it is hardly different from sodium chloride.
Exercise \(4\)
Propose an explanation for why the water solubility of the alkali chlorides does not simply increase as the cation gets larger.
Answer
The ions interact with the water via electrostatic interactions, too. The same distance factor that allows small ions to attract each other more strongly also allows small ions to interact more strongly with the water.
If we change the halides, we see similar trends.
Table \(2\) Water solubility among lithium halides.
Table \(2\) Water solubility among lithium halides.
Compound Water Solubility in g/100 mL at 20oC
LiCl 83
LiBr 166
LiI 150
Once again, it isn't surprising that the lithium chloride is the least soluble, but the most soluble seems to be the lithium bromide, not the lithium iodide.
This sort of behavior, in which we start to see a trend but it then reverses, often means there is more than one factor at work. In this case, there are a couple of other factors, some of which are more complicated. One of them simply involves the fact that there are two interactions going on here. We are not just overcoming the attraction of the ionic solid for individual ions, like when something melted. In this case, there is also the attraction of the water for the ion to think about. That attraction should also be governed by Coulomb's Law. At some stage, there must be a tipping point, when the factors that increase attraction between the ions also increase the attraction between the ion and the water. One or the other of these factors may become the dominant player under different circumstances.
• Several interactions are involved in dissolution.
• Cation - anion attraction is just one of these interactions.
• Cation - water and anion - water interactions are important, too.
• Water - water interactions also play a role.
Remember that an additional factor that influences lattice energy is the amount of charge. If we look at solubility of some alkaline earth chlorides and compare them to alkali halides, we can see the difference charge makes. Alkali metals are in the first column of the periodic table and their ions have a +1 charge. Alkaline earth metals are in the second column of the periodic table and their ions have a +2 charge.
Table \(3\) Water solubility among alkali and alkaline earth chlorides.
Table \(3\) Water solubility among alkali and alkaline earth chlorides.
Compound Water Solubility in g/100 mL at 20oC
LiCl 83
BeCl2 15
MgCl2 54
We expect the higher charges in beryllium and magnesium chloride to result in higher lattice energies. If the ions in beryllium chloride become harder to separate, then it will be more difficult to dissolve this compound in water. Instead, the ions will remain tightly stuck together.
Exercise \(5\)
Let's review some basic points about ionic solids.
1. Define lattice energy.
2. What two properties affect the lattice energy of an ionic compound?
3. Does a stronger or weaker lattice energy result in a stronger ionic bond?
4. How will a strong lattice energy affect melting and boiling points of a crystal lattice?
5. How will a strong lattice energy affect solubility of a crystal lattice?
Answer a:
The lattice energy of an ionic solid is a measure of the strength of bonds in that ionic compound.
Answer b:
Charge of ion (directly related to lattice energy); Radius of each ion (inversely related to lattice energy)
Answer c:
Stronger lattice energy results in a stronger bond. The stronger the bond, the more energy required to separate ions.
Answer d:
Stronger lattice energy results higher mp or bp.
Answer e:
Stronger lattice energy results in less soluble crystal lattice.
Exercise \(6\)
In each pair, determine which compound will have a higher lattice energy.
1. NaCl or NaBr
2. KF or CaF2
3. MgO or Na2O
4. KF or CsCl
5. RbBr or CaCl2 | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.04%3A_Solubility.txt |
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