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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/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.02%3A_The_Main-Group_Elements-_Bonding_and_Properties.txt |
While you are admiring this beautiful picture of faujasite, remember that the oxygen atoms have two unshared electron pairs in addition to the (Al,Si)-O-Si(or Al) bonds. Thus the oxygen atoms are sites to interact with positive site of molecules that passes by these structures. At present over 150 synthetic zeolites & zeotypes and 40 natural zeolites are known. Synthesis of zeolite is a very active field of study.
Aluminosilicates have three major minerals: Andalusite, sillimanite, and kyanite. Zeochem has been developing and manufacturing molecular sieve adsorbents since 1977. Simply put, their adsorbents are used to "screen" out impurities from a variety of applications by attracting and trapping the targeted contaminants. For example, in natural gas processing, molecular sieves are used to remove specific molecules from the gas stream to allow for more efficient downstream processing. Faujasite is a typical zeolite.
Applications of Zeolites?
As you have read above that there are many different kinds of zeolites, each with a definite structure and associate with it are unique properties. In terms of applications, we are assuming zeolites as porous aluminosilicates with large tunnels and cages for a fluid (gas and liquid) to pass through. The applications are based on the interactions between the fluid phase and the atoms or ions of the zeolites. In general terms, zeolites have many applications:
1. As selective and strong adsorbers: remove toxic material, selective concentrate a particular chemical, as Molecular Sieve. This link will be a very good to discuss zeolites. Currently, the site is under construction, but it has a very good framework. Even many deorderants are zeolite type.
2. As selective ion exchangers: for example used in water softener.
3. Superb solid acid catalysts, when the cations are protons H+. As catalysts, their environmental advantages include decreased corrosion, improved handling, decreased environmentally toxic waste and minimal undesirable byproducts.
4. As builder: a material that enhance or protecting the cleaning power of a detergent. Sodium aluminosilicate is an ion exchange builder often used in laundry detergent as a builder. A builder inactive the hardness of water by either keeping calcium ions in solution, by precipitation, or by ion exchange.
123 ppm CaCO3 = 123 g per 106 g of water.
``` 1 mol CaCO3 2 mol H+ 1 mol z-A 1926 g z-A 100
123 g CaCO3 ----------- ----------- ---------- ---------- ---
100 g CaCO3 1 mol CaCO3 12 mol H+ 1 mol z-A 80
```
= 494 g zeolite A
That 80 % of protons of the zeolite A is used means that we require a little more zeolite A than stoichiometric quantities.
DISCUSSION
Zeolites are aluminosilicates with open frames strcutures discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H+, Na K+, Ca2+, Cu2+ or Mg2+. Water molecules are also trapped in the frame work of zeolites.
In this example, we assume that when we soak the zeolite in water containing Ca2+, and Mg2+ ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
58.5 g/mol.
``` 1 mol 0.8*12 mol NaCl 100 58.5 g NaCl
10 kg z-A -------- --------------- --- -----------
2.190 kg 1 mol z-A 20 1 mol NaCl
= 12822 g NaCl = 12.8 kg NaCl
```
DISCUSSION
How much salt is required if 60% of the sodium ions are effectively used to replace all the divalent ions? | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.03%3A_The_Most_Common_Matter-_Silicates.txt |
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/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.04%3A_Boron_and_Its_Amazing_Structures.txt |
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.
Preparation and General Properties of Carbon
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 (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.
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. 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 C2− 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 $1$ ) 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)
Summary
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.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.05%3A_Carbon_Carbides_and_Carbonates.txt |
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. Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens (Group 15 elements) 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 Nitrogen
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:
$\ce{2NaN3(s) ->[\Delta] 2Na(l) + 3N2(g)} \label{Eq1}$
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.
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):
$\ce{2NO2(g) + H2O(l) -> HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms $\ce{N2O}$ (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with $\ce{CO2}$ 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 $\ce{Li3N}$ and $\ce{Ca3N2}$. These compounds consist of ionic lattices formed by $\ce{M^{n+}}$ and $\ce{N^{3−}}$ 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 $\ce{Si3N4}$ and $\ce{BN}$, 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 $\ce{O2}$ atmosphere:
$\ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)} \label{Eq3}$
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 ($\ce{HN3}$), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine ($\ce{N2H4}$) 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. $\ce{Sr(s) + N2O(g) ->[\Delta] SrO(s) + N2(g)}$
2. $\ce{NH4NO2(s) ->[\Delta] N2(g) + 2H2O(g)}$
3. $\ce{Pb(NO3)2(s) ->[\Delta] PbO2(s) + 2NO2(g)}$
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 $1$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. $\ce{NO(g) + H2O(l) ->[\Delta]}$
2. $\ce{NH4NO3(s) ->[\Delta]}$
3. $\ce{Sr(s) + N2(g) ->}$
Answer
1. $\ce{NO(g) + H2O(l) ->[\Delta] no reaction}$
2. $\ce{NH4NO3(s) ->[\Delta] N2O(g) + 2H2O(g)}$
3. $\ce{3Sr(s) + N2(g) -> Sr3N2(s)}$
Summary
Nitrogen behaves chemically like nonmetals, Nitrogen forms compounds in nine different oxidation states. 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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.06%3A_Nitrogen_and_Phosphorus-_Essential_Elements_for_Life.txt |
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen not only is important to supporting life, but also plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion stands at the heart of the phlogiston controversy (see below).
The Origin and History
Oxygen is found in the group 16 elements and is considered a chalcogen. Named from the Greek oxys + genes, "acid-former", oxygen was discovered in 1772 by Scheele and independently by Priestly in 1774. Oxygen was given its name by the French scientist, Antoine Lavoisier.
Scheele discovered oxygen through an experiment which involved burning manganese oxide. Scheele came to find that the hot manganese oxide produced a gas which he called "fire air". He also came to find that when this gas was able to come into contact with charcoal, it produced beautiful bright sparks. All of the other elements produced the same gas. Although Scheele discovered oxygen, he did not publish his work until three years after another chemist, Joseph Priestly, discovered oxygen. Joseph Priestly, an English chemist, repeated Scheele's experiment in 1774 using a slightly different setup. Priestly used a 12 in burning glass and aimed the sunlight directly towards the compound that he was testing, mercuric oxide. As a result, he was able to "discover better air" that was shown to expand a mouse's lifetime to four times as long and caused a flame to burn with higher intensity. Despite these experiments, both chemists were not able to pinpoint exactly what this element was. It was not until 1775 that Antoine Lavoisier, a French chemist, was able to recognize this unknown gas as an element.
Our atmosphere currently contains about 21% of free oxygen. Oxygen is produced in various ways. The process of photochemical dissociation in which water molecules are broken up by ultraviolet rays produces about 1-2% of our oxygen. Another process that produces oxygen is photosynthesis which is performed by plants and photosynthetic bacteria. Photosynthesis occurs through the following general reaction:
$\ce{CO2 + H2O + h\nu \rightarrow} \text{organic compounds} \ce{+ O2} \nonumber$
The Dangers of Phlogiston
Phlogiston theory is the outdated belief that a fire-like element called phlogiston is contained within combustible bodies and released during combustion. The name comes from the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). It was first stated in 1667 by Johann Joachim Becher, and then put together more formally by Georg Ernst Stahl. The theory attempted to explain burning processes such as combustion and rusting, which are now collectively known as oxidation.
Properties
• Element number: 8
• Atomic weight 15.9994
• Color: gas form- colorless, liquid- pale blue
• Melting point: 54.36K
• Boiling point: 90.2 K
• Density: .001429
• 21% of earth's atmosphere
• Third most abundant element in the universe
• Most abundant element in Earth's crust at 45.4%
• 3 Stable isotopes
• Ionization energy: 13.618 eV
• Oxygen is easily reduced and is a great oxidizing agent making it readily reactive with other elements
Magnetic Properties of Oxygen
Oxygen (O2) is paramagnetic. An oxygen molecule has six valence electrons, so the O2 molecule has 12 valence electrons with the electron configuration shown below:
As shown, there are two unpaired electrons, which causes O2 to be paramagnetic. There are also eight valence electrons in the bonding orbitals and four in antibonding orbitals, which makes the bond order 2. This accounts for the double covalent bond that is present in O2.
Video $1$: A chemical demonstration of the paramagnetism of molecular oxygen, as shown by the attraction of liquid oxygen to magnets.
As shown in Video $1$, since molecular oxygen ($O_2$) has unpaired electrons, it is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen ($N_2$) has no unpaired electrons and is not attracted to the magnet.
General Chemistry of Oxygen
Oxygen normally has an oxidation state of -2, but is capable of having oxidation states of -2, -1, -1/2, 0, +1, and +2. The oxidation states of oxides, peroxides and superoxides are as follows:
• Oxides: O-2 ,
• peroxides: O2-2 ,
• superoxide: O2-1.
Oxygen does not react with itself, nitrogen, or water under normal conditions. Oxygen does, however, dissolve in water at 20 degrees Celsius and 1 atmosphere. Oxygen also does not normally react with bases or acids. Group 1 metals (alkaline metals) are very reactive with oxygen and must be stored away from oxygen in order to prevent them from becoming oxidized. The metals at the bottom of the group are more reactive than those at the top. The reactions of a few of these metals are explored in more detail below.
Lithium: Reacts with oxygen to form white lithium oxide in the reaction below.
$\ce{4Li + O_2 \rightarrow 2Li_2O} \label{1}$
Sodium: Reacts with oxygen to form a white mixture of sodium oxide and sodium peroxide. The reactions are shown below.
• Sodium oxide: $\ce{4Na + O_2 \rightarrow 2Na_2O} \label{2}$
• Sodium peroxide: $\ce{2Na + O_2 \rightarrow Na_2O_2} \label{3}$
Potassium: Reacts with oxygen to form a mixture of potassium peroxide and potassium superoxide. The reactions are shown below.
• Potassium peroxide: $\ce{2K + O_2 \rightarrow 2K_2O_2} \label{4}$
• Potassium superoxide: $\ce{K + O_2 \rightarrow KO_2} \label{5}$
Rubidium and Cesium: Both metals react to produce superoxides through the same process as that of the potassium superoxide reaction.
The oxides of these metals form metal hydroxides when they react with water. These metal hydroxides make the solution basic or alkaline, hence the name alkaline metals.
Group 2 metals (alkaline earth metals) react with oxygen through the process of burning to form metal oxides but there are a few exceptions.
Beryllium is very difficult to burn because it has a layer of beryllium oxide on its surface which prevents further interaction with oxygen. Strontium and barium react with oxygen to form peroxides. The reaction of barium and oxygen is shown below, and the reaction with strontium would be the same.
$\ce{Ba(s) + O2 (g) \rightarrow BaO2 (s) }\label{6}$
Group 13 reacts with oxygen in order to form oxides and hydroxides that are of the form $X_2O_3$ and $X(OH)_3$. The variable X represents the various group 13 elements. As you go down the group, the oxides and hydroxides get increasingly basic.
Group 14 elements react with oxygen to form oxides. The oxides formed at the top of the group are more acidic than those at the bottom of the group. Oxygen reacts with silicon and carbon to form silicon dioxide and carbon dioxide. Carbon is also able to react with oxygen to form carbon monoxide, which is slightly acidic. Germanium, tin, and lead react with oxygen to form monoxides and dioxides that are amphoteric, which means that they react with both acids and bases.
Group 15 elements react with oxygen to form oxides. The most important are listed below.
• Nitrogen: N2O, NO, N2O3, N2O4, N2O5
• Phosphorus: P4O6, P4O8, P2O5
• Arsenic: As2O3, As2O5
• Antimony: Sb2O3, Sb2O5
• Bismuth: Bi2O3, Bi2O5
Group 16 elements react with oxygen to form various oxides. Some of the oxides are listed below.
• Sulfur: SO, SO2, SO3, S2O7
• Selenium: SeO2, SeO3
• Tellurium: TeO, TeO2, TeO3
• Polonium: PoO, PoO2, PoO3
Group 17 elements (halogens) fluorine, chlorine, bromine, and iodine react with oxygen to form oxides. Fluorine forms two oxides with oxygen: F2O and F2O2. Both fluorine oxides are called oxygen fluorides because fluorine is the more electronegative element. One of the fluorine reactions is shown below.
$\ce{O2 (g) + F2 (g) \rightarrow F2O2 (g)} \label{7}$
Group 18: Some would assume that the Noble Gases would not react with oxygen. However, xenon does react with oxygen to form $\ce{XeO_3}$ and $\ce{XeO_4}$. The ionization energy of xenon is low enough for the electronegative oxygen atom to "steal away" electrons. Unfortunately, $\ce{XeO_3}$ is HIGHLY unstable, and it has been known to spontaneously detonate in a clean, dry environment.
Transition metals react with oxygen to form metal oxides. However, gold, silver, and platinum do not react with oxygen. A few reactions involving transition metals are shown below:
$2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{(s)} \label{8}$
$4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \label{9A}$
$4Al_{(s)} + 3O_{2(g)} \rightarrow 2Al_2O_{3(s)} \label{9B}$
Reaction of Oxides
We will be discussing metal oxides of the form $X_2O$. The variable $X$ represents any metal that is able to bond to oxygen to form an oxide.
• Reaction with water: The oxides react with water to form a metal hydroxide.
$X_2O + H_2O \rightarrow 2XOH \nonumber$
• Reaction with dilute acids: The oxides react with dilute acids to form a salt and water.
$X_2O + 2HCl \rightarrow 2XCl + H_2O \nonumber$
Reactions of Peroxides
The peroxides we will be discussing are of the form $X_2O_2$. The variable $X$ represents any metal that can form a peroxide with oxygen.
Reaction with water: If the temperature of the reaction is kept constant despite the fact that the reaction is exothermic, then the reaction proceeds as follows:
$X_2O_2+ 2H_2O \rightarrow 2XOH + H_2O_2 \nonumber$
If the reaction is not carried out at a constant temperature, then the reaction of the peroxide and water will result in decomposition of the hydrogen peroxide that is produced into water and oxygen.
Reaction with dilute acid: This reaction is more exothermic than that with water. The heat produced causes the hydrogen peroxide to decompose to water and oxygen. The reaction is shown below.
$X_2O_2 + 2HCl \rightarrow 2XCl + H_2O_2 \nonumber$
$2H_2O_2 \rightarrow 2H_2O + O_2 \nonumber$
Reaction of Superoxides
The superoxides we will be talking about are of the form $XO_2$, with $X$ representing any metal that forms a superoxide when reacting with oxygen.
Reaction with water: The superoxide and water react in a very exothermic reaction that is shown below. The heat that is produced in forming the hydrogen peroxide will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2 \nonumber$
Reaction with dilute acids: The superoxide and dilute acid react in a very exothermic reaction that is shown below. The heat produced will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2 \nonumber$
Allotropes of Oxygen
There are two allotropes of oxygen; dioxygen (O2) and trioxygen (O3) which is called ozone. The reaction of converting dioxygen into ozone is very endothermic, causing it to occur rarely and only in the lower atmosphere. The reaction is shown below:
$3O_{2 (g)} \rightarrow 2O_{3 (g)} \;\;\; ΔH^o= +285 \;kJ \nonumber$
Ozone is unstable and quickly decomposes back to oxygen but is a great oxidizing agent.
Miscellaneous Reactions
Reaction with Alkanes: The most common reactions that involve alkanes occur with oxygen. Alkanes are able to burn and it is the process of oxidizing the hydrocarbons that makes them important as fuels. An example of an alkane reaction is the reaction of octane with oxygen as shown below.
C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l) ∆Ho = -5.48 X 103 kJ
Reaction with ammonia: Oxygen is able to react with ammonia to produce dinitrogen (N2) and water (H2O) through the reaction shown below.
$4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O \nonumber$
Reaction with Nitrogen Oxide: Oxygen is able to react with nitrogen oxide in order to produce nitrogen dioxide through the reaction shown below.
$NO + O_2 \rightarrow NO_2 \nonumber$
Problems
1. Is oxygen reactive with noble gases?
2. Which transition metals does oxygen not react with?
3. What is produced when an oxide reacts with water?
4. Is oxygen reactive with alkali metals? Why are the alkali metals named that way?
5. If oxygen is reactive with alkali metals, are oxides, peroxides or superoxides produced?
Solutions
1. No, noble gases are unreactive with oxygen.
2. Oxygen is mostly unreactive with gold and platinum.
3. When an oxide reacts with water, a metal hydroxide is produced.
4. Oxygen is very reactive with alkali metals. Alkali metals are given the name alkali because the oxides of these metals react with water to form a metal hydroxide that is basic or alkaline.
5. Lithium produces an oxide, sodium produces a peroxide, and potassium, cesium, and rubidium produce superoxides.
Contributors and Attributions
• Phillip Ball (UCD), Katharine Williams (UCD)
23.08: Sulfur- A Dangerous and Useful Element
Learning Objectives
• Describe the properties, preparation, and uses of sulfur
Sulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure $1$. Sulfur is a constituent of many proteins and is essential for life.
The Frasch process, illustrated in Figure $2$, is important in the mining of free sulfur from enormous underground deposits in Texas and Louisiana. Superheated water (170 °C and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is 99.5% to 99.9% pure and requires no purification for most uses.
Larger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.
Sulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is S8. However, chemists commonly use S to simplify the coefficients in chemical equations; we will follow this practice in this book.
Like oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2−). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:
$\ce{S}(s)+\ce{6HNO3}(aq)⟶\ce{2H3O+}(aq)+\ce{SO4^2-}(aq)+\ce{6NO2}(g) \nonumber$
The chemistry of sulfur with an oxidation state of 2− is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states.
Summary
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
Glossary
Frasch process
important in the mining of free sulfur from enormous underground deposits | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.07%3A_Oxygen.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.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/23%3A_Chemistry_of_the_Nonmetals/23.09%3A_Halogens-_Reactive_Chemicals_with_High_Electronegativity.txt |
Vanadium takes its name from the Scandinavian goddess Vanadis and was discovered in 1801 by Andrés Manuel del Rio. It was isolated in 1867 by Henry Roscoe as a silvery-white metal that is somewhat heavier than aluminum but lighter than iron. It has excellent corrosion resistance at room temperature.
The history of its discovery is an interesting tale. del Rio sent his brown ore samples, containing what he thought was a new element to Paris for analysis and confirmation, along with a brief explanation that was ambiguous. The complete analysis and description of his work were lost in a shipwreck so the Paris lab saw nothing but brown powder and a brief confusing note. A second sample sent to Berlin was mislabeled lead chromate when it arrived. del Rio gave up, losing confidence in his discovery. The element was rediscovered in 1867 by Nils Sefstrôm.
Vanadium has an unusually large number of stable oxidation states (+2, +3, +4, +5)each of which is characterized by a unique color in solution. The metal is used as an alloying agent for steel. It combines with nearly all non-metals in compounds.
Vanadium(V) oxide as a Catalyst
During the Contact Process for manufacturing sulfuric acid, sulfur dioxide has to be converted into sulfur trioxide, which is done by passing sulfur dioxide and oxygen over a solid vanadium(V) oxide catalyst.
$SO_2 + \dfrac{1}{2}O_2 \ce{->[V_2O_5]} SO_3 \nonumber$
This is a good example of the ability of transition metals and their compounds to act as catalysts because of their ability to change their oxidation state (oxidation number). The sulfur dioxide is oxidized to sulfur trioxide by the vanadium(V) oxide. In the process, the vanadium(V) oxide is reduced to vanadium(IV) oxide.
$SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4 \nonumber$
The vanadium(IV) oxide is then re-oxidized by the oxygen.
$V_2O_4 + \dfrac{1}{2} O_2 \rightarrow V_2O_5 \nonumber$
Although the catalyst has been temporarily changed during the reaction, at the end it is chemically the same as it started.
Vanadium's oxidation states
Vanadium has oxidation states in its compounds of +5, +4, +3 and +2. This section looks at ways of changing between them. It starts with a bit of description, and then goes on to look at the reactions in terms of standard redox potentials (standard electrode potentials).
Reducing vanadium(V) in stages to vanadium(II)
The usual source of vanadium in the +5 oxidation state is ammonium metavanadate, NH4VO3. This isn't very soluble in water and is usually first dissolved in sodium hydroxide solution. The solution can be reduced using zinc and an acid - either hydrochloric acid or sulfuric acid, usually using moderately concentrated acid. The exact vanadium ion present in the solution is very complicated, and varies with the pH of the solution. The reaction is done under acidic conditions when the main ion present is VO2+ - called the dioxovanadium(V) ion.
If you do the reaction in a small flask, it is normally stoppered with some cotton wool. This allows hydrogen (produced from a side reaction between the zinc and acid) to escape. At the same time it stops much air from entering. This prevents re-oxidation of the lower oxidation states of vanadium (particularly the +2 state) by oxygen in the air. The reaction is usually warmed so that the changes happen in a reasonable time. The reduction is shown in two stages. Some individual important colors are shown, but the process is one continuous change from start to finish.
The reduction from +5 to +4
It is important to notice that the green color you see isn't actually another oxidation state. it is just a mixture of the original yellow of the +5 state and the blue of the +4. Be very careful with the formulae of the two vanadium ions - they are very easy to confuse!
The reduction from +4 to +2. The color changes just continue.
The reason for the inverted commas around the vanadium(III) ion is that this is almost certainly a simplification. The exact nature of the complex ion will depend on which acid you use in the reduction process. The simplification is probably reasonable at this level.
Re-oxidation of the vanadium(II)
The vanadium(II) ion is very easily oxidized. If you remove the cotton wool from the flask and pour some solution into a test tube, it turns green because of its contact with oxygen in the air. It is oxidized back to vanadium(III). If it is allowed to stand for a long time, the solution eventually turns blue as the air oxidizes it back to the vanadium(IV) state - VO2+ ions. Adding nitric acid (a reasonably powerful oxidizing agent) to the original vanadium(II) solution also produces blue VO2+ ions. The vanadium(II) is again oxidized back to vanadium(IV).
Using Zinc as the reducing agent
Let's look at the first stage of the reduction - from VO2+ to VO2+. The redox potential for the vanadium half-reaction is given by:
The corresponding equilibrium for the zinc is:
The simple principle is that if you couple two of these half-reactions together, the one with the more positive E° value will move to the right; the one with the more negative (or less positive) E° moves to the left. If you mix together zinc and VO2+ ions in the presence of acid to provide the H+ ions:
That converts the two equilibria into two one-way reactions. You can write these down and combine them to give the ionic equation for the reaction if you want to.
The other stages of the reaction
Here are the E° values for all the steps of the reduction from vanadium(V) to vanadium(II):
. . . and here is the zinc value again:
Remember that for the vanadium reactions to move to the right (which is what we want), their E° values must be more positive than whatever you are reacting them with. In other words, for the reactions to work, zinc must always have the more negative value - and that's the case. Zinc can reduce the vanadium through each of these steps to give the vanadium(II) ion.
Using other reducing agents
Suppose you replaced zinc as the reducing agent by tin. How far would the set of reductions go this time?
Here are the E° values again:
. . . and here is the tin value:
For each reduction to happen, the vanadium reaction has to have the more positive E° value because we want it to go to the right. That means that the tin must have the more negative value.
• In the first vanadium equation (from +5 to +4), the tin value is more negative. That works OK.
• In the second vanadium equation (from +4 to +3), the tin value is again the more negative. That works as well.
• But in the final vanadium reaction (from +3 to +2), tin no longer has the more negative E° value. Tin won't reduce vanadium(III) to vanadium(II).
Re-oxidation of the vanadium(II)
The vanadium(II) oxidation state is easily oxidized back to vanadium(III) - or even higher.
Oxidation by hydrogen ions
You will remember that the original reduction we talked about was carried out using zinc and an acid in a flask stoppered with a piece of cotton wool to keep the air out. Air will rapidly oxidize the vanadium(II) ions - but so also will the hydrogen ions present in the solution!
The vanadium(II) solution is only stable as long as you keep the air out, and in the presence of the zinc. The zinc is necessary to keep the vanadium reduced. What happens if the zinc isn't there? Look at these E° values:
The reaction with the more negative E° value goes to the left; the reaction with the more positive (or less negative) one to the right. That means that the vanadium(II) ions will be oxidized to vanadium(III) ions, and the hydrogen ions reduced to hydrogen.
Will the oxidation go any further - for example, to the vanadium(IV) state? Have a look at the E° values and decide:
In order for the vanadium equilibrium to move to the left, it would have to have the more negative E° value. It hasn't got the more negative E° value and so the reaction does not happen.
Oxidation by nitric acid
In a similar sort of way, you can work out how far nitric acid will oxidize the vanadium(II). Here's the first step:
The vanadium reaction has the more negative E° value and so will move to the left; the nitric acid reaction moves to the right.
Nitric acid will oxidize vanadium(II) to vanadium(III). The second stage involves these E° values:
The nitric acid again has the more positive E° value and so moves to the right. The more negative (less positive) vanadium reaction moves to the left. Nitric acid will certainly oxidize vanadium(III) to vanadium(IV). Will it go all the way to vanadium(V)?
No, it won't! For the vanadium reaction to move to the left to form the dioxovanadium(V) ion, it would have to have the more negative (less positive) E° value. It hasn't got a less positive value, and so the reaction does not happen.
You can work out the effect of any other oxidizing agent on the lower oxidation states of vanadium in exactly the same way. But don't assume that because the E° values show that a reaction is possible, it will necessarily happen. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/24%3A_Metals_and_Metallurgy/24.01%3A_Vanadium-_A_Problem_and_an_Opportunity.txt |
This page looks at the various factors which influence the choice of method for extracting metals from their ores, including reduction by carbon, reduction by a reactive metal (like sodium or magnesium), and by electrolysis. Details for the extraction of aluminum, copper, iron and titanium are given in separate pages in this section.
What are "ores"?
An ore is any naturally-occurring source of a metal that you can economically extract the metal from. Aluminum, for example, is the most common metal in the Earth's crust, occurring in all sorts of minerals. However, it isn't economically worthwhile to extract it from most of these minerals. Instead, the usual ore of aluminum is bauxite - which contains from 50 - 70% of aluminum oxide.
Copper is much rarer, but fortunately can be found in high-grade ores (ones containing a high percentage of copper) in particular places. Because copper is a valuable metal, it is also worth extracting it from low-grade ores as well. Ores are commonly oxides, for example:
• bauxite (Al2O3)
• haematite (Fe2O3)
• rutile (TiO2 )
. . . or sulfides, for example:
• pyrite (FeS2 )
• chalcopyrite (CuFeS2 )
Concentrating the ore
This simply means getting rid of as much of the unwanted rocky material as possible before the ore is converted into the metal. In some cases this is done chemically. For example, pure aluminum oxide is obtained from bauxite by a process involving a reaction with sodium hydroxide solution. This is described in detail on the aluminum page in this section. Some copper ores can be converted into copper(II) sulfate solution by leaving the crushed ore in contact with dilute sulphuric acid for a long time. Copper can then be extracted from the copper(II) sulfate solution. But, in many cases, it is possible to separate the metal compound from unwanted rocky material by physical means. A common example of this involves froth flotation.
Froth flotation
The ore is first crushed and then treated with something which will bind to the particles of the metal compound that you want and make those particles hydrophobic. "Hydrophobic" literally means "water fearing". In concentrating copper ores, for example, pine oil is often used. The pine oil binds to the copper compounds, but not to the unwanted rocky material.
The treated ore is then put in a large bath of water containing a foaming agent (a soap or detergent of some kind), and air is blown through the mixture to make a lot of bubbles. Because they are water-repellent, the coated particles of the metal compound tend to be picked up by the air bubbles, float to the top of the bath, and are allowed to flow out over the sides. The rest of the rocky material stays in the bath.
Reducing the metal compound to the metal
Why is this reduction?
At its simplest, where you are starting from metal oxides, the ore is being reduced because oxygen is being removed.
However, if you are starting with a sulfide ore, for example, that's not a lot of help! It is much more helpful to use the definition of reduction in terms of addition of electrons. To a reasonable approximation, you can think of these ores as containing positive metal ions. To convert them to the metal, you need to add electrons - reduction.
Choosing a method of reduction
There are various economic factors you need to think about in choosing a method of reduction for a particular ore. These are all covered in detail on other pages in this section under the extractions of particular metals. What follows is a quick summary.
You need to consider:
• the cost of the reducing agent;
• energy costs;
• the desired purity of the metal.
There may be various environmental considerations as well - some of which will have economic costs.
Chemical Reduction
Carbon (as coke or charcoal) is cheap. It not only acts as a reducing agent, but it also acts as the fuel to provide heat for the process. However, in some cases (for example with aluminum) the temperature needed for carbon reduction is too high to be economic - so a different method has to be used. Carbon may also be left in the metal as an impurity. Sometimes this can be removed afterwards (for example, in the extraction of iron); sometimes it can't (for example in producing titanium), and a different method would have to be used in cases like this.
Other more reactive metals can be used to reduce the ore. Titanium is produced by reducing titanium(IV) chloride using a more reactive metal such as sodium or magnesium. As you will see if you read the page about titanium extraction, this is the only way of producing high purity metal.
$TiCl_4 + 4Na \rightarrow Ti + 4NaCl \nonumber$
The more reactive metal sodium releases electrons easily as it forms its ions:
$4Na \rightarrow 4Na^+ + 4e^- \nonumber$
These electrons are used to reduce the titanium(IV) chloride:
$TiCl_4 + 4e^- \rightarrow Ti + 4Cl^- \nonumber$
The downside of this is expense. You have first to extract (or to buy) the sodium or magnesium. The more reactive the metal is, the more difficult and expensive the extraction becomes. That means that you are having to use a very expensive reducing agent to extract the titanium. As you will see if you read the page about titanium extraction, there are other problems in its extraction which also add to the cost.
Reduction by electrolysis
This is a common extraction process for the more reactive metals - for example, for aluminum and metals above it in the electrochemical series. You may also come across it in other cases such as one method of extracting copper and in the purification of copper. During electrolysis, electrons are being added directly to the metal ions at the cathode (the negative electrode). The downside (particularly in the aluminum case) is the cost of the electricity. An advantage is that it can produce very pure metals. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/24%3A_Metals_and_Metallurgy/24.03%3A_Metallurgical_Processes.txt |
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 23.1).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+.
As you learned previously, 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 \(2\)). 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.
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.
Due to a small increase in successive ionization energies, most of the transition metals have multiple oxidation states separated by a single electron.
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 23.2 for the 2+ ions of the first-row transition metals.
All transition-metal cations have dn electron configurations; the ns electrons are always lost before the (n − 1)d electrons.
Table \(1\): d-Electron Configurations of the Dications of the First-Row Transition Metals
Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
d2 d3 d4 d5 d6 d7 d8 d9 d10
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.
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.
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
*The convention of using roman numerals to indicate the oxidation states of a metal is used here.
electronic structure s2d1 s2d2 s2 d3 s1d5 s2d5 s2d6 s2d7 s2d8 s1d10 s2d10
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
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 23.1, Figure 23.2, Table 23.1, Table 23.2, and Table 23.3 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).
1. Hg2+ > Fe2+ > Zr4+ > Fe3+ > Pt4+ | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/24%3A_Metals_and_Metallurgy/24.05%3A_Sources_Properties_and_Products_of_Some_of_the_3d_Transition_Metals.txt |
Why is a ruby red? The mineral corundum is a crystalline form of alumina: Al2O3. A pure crystal of corundum is colorless. However, if just 1% of the Al3+ ions are replaced with Cr3+ ions, the mineral becomes deep red in color and is known as ruby (Al2O3:Cr3+). Why does replacing Al3+ with Cr3+in the corundum structure produce a red color?
Ruby is an allochromatic mineral, which means its color arises from trace impurities. The color of an idiochromatic mineral arises from the essential components of the mineral. In some minerals the color arises from defects in the crystal structure. Such defects are called color centers.
The mineral beryl is a crystalline beryllium aluminosilicate with the chemical formula Be3Al2Si6O18. A pure crystal of beryl is colorless. However, if just 1% of the Al3+ ions are replaced with Cr3+ ions, the mineral becomes green in color and is known as emerald (Be3Al2Si6O18:Cr3+).
Why does replacing Al3+ with Cr3+ in corundum produce a red mineral (ruby) while replacing Al3+ with Cr3+ in beryl produces a green mineral (emerald)?
Recall 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. 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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.01%3A_The_Colors_of_Rubies_and_Emeralds.txt |
Learning Objectives
• Outline the general approach for the isolation of transition metals from natural sources
• Describe typical physical and chemical properties of the transition metals
• Identify simple compound classes for transition metals and describe their chemical properties
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $2$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series.
Example $1$: Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Check Your Learning Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Uses of Lanthanides in Devices
Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 10−5% versus 0.79 × 10−5% by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.
The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure $3$). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.
As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to$470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.
The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.
Properties of the Transition Elements
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (Table P1), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as $\ce{MoO4^2-}$ and $\ce{ReO4-}$.
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $4$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Example $2$: Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (Table P1) for each oxide in the specified oxidation state:
$\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V} \nonumber$
$\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V} \nonumber$
$\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V} \nonumber$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from (Table P1).
Answer
$\ce{Co}(s)+\ce{2HCl}⟶\ce{H2}+\ce{CoCl2}(aq)$; no reaction because Pt(s) will not be oxidized by H+
Preparation of the Transition Elements
Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure $5$). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.
Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.
In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.
We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining.
1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.
2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal.
3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals.
Isolation of Iron
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $6$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g) \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $6$.
The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l) \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $7$).
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.
Isolation of Copper
The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:
$\ce{CaCO3}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)+\ce{CO2}(g) \nonumber$
$\ce{FeO}(s)+\ce{SiO2}(s)⟶\ce{FeSiO3}(l) \nonumber$
In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).
Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:
$\ce{2Cu2S}(l)+\ce{3O2}(g)⟶\ce{2Cu2O}(l)+\ce{2SO2}(g) \nonumber$
$\ce{2Cu2O}(l)+\ce{Cu2S}(l)⟶\ce{6Cu}(l)+\ce{SO2}(g) \nonumber$
The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure $8$). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).
Isolation of Silver
Silver sometimes occurs in large nuggets (Figure $9$) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores.
Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, $\ce{[Ag(CN)2]-}$, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
$\ce{2Ag2S}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{2S}(s)+\ce{4OH-}(aq) \nonumber$
$\ce{AgCl}(s)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)+\ce{Cl-}(aq) \nonumber$
The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:
$\ce{2[Ag(CN)2]-}(aq)+\ce{Zn}(s)⟶\ce{2Ag}(s)+\ce{[Zn(CN)4]^2-}(aq) \nonumber$
Example $3$: Refining Redox
One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:
$\ce{4Ag}(s)+\ce{8CN-}(aq)+\ce{O2}(g)+\ce{2H2O}(l)⟶\ce{4[Ag(CN)2]-}(aq)+\ce{4OH-}(aq) \nonumber$
Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:
$\ce{4Ag}(s)+\ce{8CN-}(aq)⟶\ce{4[Ag(CN)2]-}(aq)? \nonumber$
Solution
The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.
Exercise $3$
During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?
Answer
The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0).
Transition Metal Compounds
The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows.
Halides
Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:
$\ce{2Fe}(s)+\ce{3Cl2}(g)⟶\ce{2FeCl3}(s) \nonumber$
Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:
$\ce{Fe}(s)+\ce{2FeCl3}(s)⟶\ce{3FeCl2}(s) \nonumber$
The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.
In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:
$\ce{NiCO3}(s)+\ce{2HF}(aq)⟶\ce{NiF2}(aq)+\ce{H2O}(l)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)+\ce{2HBr}(aq)⟶\ce{CoBr2}(aq)+\ce{2H2O}(l) \nonumber$
Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:
$\ce{Cr}(s)+\ce{2HCl}(aq)⟶\ce{CrCl2}(aq)+\ce{H2}(g) \nonumber$
The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics.
The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:
$\ce{SiCl4}(l)+\ce{2H2O}(l)⟶\ce{SiO2}(s)+\ce{4HCl}(aq) \nonumber$
$\ce{TiCl4}(l)+\ce{2H2O}(l)⟶\ce{TiO2}(s)+\ce{4HCl}(aq) \nonumber$
Oxides
As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.
The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO.
Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:
$\ce{FeC2O4}(s)⟶\ce{FeO}(s)+\ce{CO}(g)+\ce{CO2}(g) \nonumber$
$\ce{Co(OH)2}(s)⟶\ce{CoO}(s)+\ce{H2O}(g) \nonumber$
With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:
$\ce{CoO}(s)+\ce{2HNO3}(aq)⟶\ce{Co(NO3)2}(aq)+\ce{H2O}(l) \nonumber$
$\ce{Sc2O3}(s)+\ce{6HCl}(aq)⟶\ce{2ScCl3}(aq)+\ce{3H2O}(l) \nonumber$
The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions $\ce{VO4^3-}$, $\ce{CrO4^2-}$, and $\ce{MnO4-}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by:
$\ce{CrO3}(s)+\ce{2Na+}(aq)+\ce{2OH-}(aq)⟶\ce{2Na+}(aq)+\ce{CrO4^2-}(aq)+\ce{H2O}(l) \nonumber$
Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively.
Hydroxides
When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:
$\ce{Co^2+}(aq)+\ce{2OH-}(aq)⟶\ce{Co(OH)2}(s) \nonumber$
In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:
$\ce{4Fe^3+}(aq)+\ce{6OH-}(aq)+\ce{nH2O}(l)⟶\ce{2Fe2O3⋅(n + 3)H2O}(s) \nonumber$
These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal.
Carbonates
Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:
$\ce{Ni^2+}(aq)+\ce{CO3^2-}⟶\ce{NiCO3}(s) \nonumber$
The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides.
Other Salts
In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.
A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:
$\ce{2Sc}(s)+\ce{6HBr}(aq)⟶\ce{2ScBr3}(aq)+\ce{3H2}(g) \nonumber$
The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:
$\ce{Ni(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2ClO4-}(aq)⟶\ce{Ni^2+}(aq)+\ce{2ClO4-}(aq)+\ce{4H2O}(l) \nonumber$
Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:
$\ce{Ba^2+}(aq)+\ce{2Cl-}(aq)+\ce{2K+}(aq)+\ce{CrO4^2-}(aq)⟶\ce{BaCrO4}(s)+\ce{2K+}(aq)+\ce{2Cl-}(aq) \nonumber$
In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.
High Temperature Superconductors
A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.
Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.
One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.
The new materials become superconducting at temperatures close to 90 K (Figure $10$), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.
Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.
Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure $11$).
Summary
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.
Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.
Glossary
actinide series
(also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103
coordination compound
stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
d-block element
one of the elements in groups 3–11 with valence electrons in d orbitals
f-block element
(also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table
first transition series
transition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29
fourth transition series
transition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111
hydrometallurgy
process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
lanthanide series
(also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71
platinum metals
group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
rare earth element
collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition series
transition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47
smelting
process of extracting a pure metal from a molten ore
steel
material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
superconductor
material that conducts electricity with no resistance
third transition series
transition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79 | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.02%3A_Properties_of_Transition_Metals.txt |
Learning Objectives
• List the defining traits of coordination compounds
• Describe the structures of complexes containing monodentate and polydentate ligands
• Use standard nomenclature rules to name coordination compounds
• Explain and provide examples of geometric and optical isomerism
• Identify several natural and technological occurrences of coordination compounds
The hemoglobin in your blood, the chlorophyll in green plants, vitamin \(B_{12}\), and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(1\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH4. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(2\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure \(3\)). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H2NCH2CH2NH2) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(4\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en)3]3+, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.
Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or “many teeth”) because it can bite into the metal center with more than one bond. The term chelate (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(4\) showed one example of a chelate and the heme complex in hemoglobin is another important example (Figure \(5\)). It contains a polydentate ligand with four donor atoms that coordinate to iron.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as NH3, Cl, and H2O, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, H2NCH2CH2NH2, and the anion of the acid glycine, \(\ce{NH2CH2CO2-}\) (Figure \(6\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The heme ligand (Figure \(5\)) is a tetradentate ligand.
The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group (e.g., Table \(1\). For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua (H2O), amine (NH3), carbonyl (CO), and nitrosyl (NO). For example, name [Pt(NH3)2Cl4] as diaminetetrachloroplatinum(IV).
3. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri- (for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris- (for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups (C5H4N).
Table \(1\): Examples of Anionic Ligands
Anionic Ligand Name
F fluoro
Cl chloro
Br bromo
I iodo
CN cyano
\(\ce{NO3-}\) nitrato
OH hydroxo
O2– oxo
\(\ce{C2O4^2-}\) oxalato
\(\ce{CO2^2-}\) carbonato
When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Tables \(2\), \(3\), and \(3\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state.
Table \(2\): Select Coordination Complexes based on total Charge
Examples in Which the Complex Is Cation
[Co(NH3)6]Cl3 hexaaminecobalt(III) chloride
[Pt(NH3)4Cl2]2+ tetraaminedichloroplatinum(IV) ion
[Ag(NH3)2]+ diaminesilver(I) ion
[Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride
[Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethylenediamine)cobalt(III) sulfate
Examples in Which the Complex Is Neutral
[Pt(NH3)2Cl4] diaminetetrachloroplatinum(IV)
[Ni(H2NCH2CH2NH2)2Cl2] dichlorobis(ethylenediamine)nickel(II)
Examples in Which the Complex Is an Anion
[PtCl6]2− hexachloroplatinate(IV) ion
Na2[SnCl6] sodium hexachlorostannate(IV)
Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Example \(1\): Coordination Numbers and Oxidation States
Determine the name of the following complexes and give the coordination number of the central metal atom.
1. Na2[PtCl6]
2. K3[Fe(C2O4)3]
3. [Co(NH3)5Cl]Cl2
Solution
1. There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
2. The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
3. In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaaminechlorocobalt(III) chloride and the coordination number is six.
Exercise \(1\)
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.
Answer
K[Ag(CN)2]; coordination number two
The Structures of Complexes
The most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (Figure \(7\)). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table \(3\) compares coordination numbers to the molecular geometry:
Table \(3\): Coordination Numbers and Molecular Geometry
Coordination Number Molecular Geometry Example
2 linear [Ag(NH3)2]+
3 trigonal planar [Cu(CN)3]2−
4 tetrahedral(d0 or d10), low oxidation states for M [Ni(CO)4]
4 square planar (d8) [NiCl4]2−
5 trigonal bipyramidal [CoCl5]2−
5 square pyramidal [VO(CN)4]2−
6 octahedral [CoCl6]3−
7 pentagonal bipyramid [ZrF7]3−
8 square antiprism [ReF8]2−
8 dodecahedron [Mo(CN)8]4−
9 and above more complicated structures [ReH9]2−
Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding d-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(8\). The chloride and nitrate anions in [Co(H2O)6]Cl2 and [Cr(en)3](NO3)3, and the potassium cations in K2[PtCl6], are outside the brackets and are not bonded to the metal ion.
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN)4]2− (Figure \(9\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH3)2Cl2], each ligand has two other ligands at 90° angles (called the cis positions) and one additional ligand at an 180° angle, in the trans position.
Isomerism in Complexes
Isomers are different chemical species that have the same chemical formula. Transition metals often form geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral [Co(NH3)4Cl2]+ ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure \(1\)). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH3)4Cl2]NO3 isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.
Example \(2\): Geometric Isomers
Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure \(9\)b. Draw the other geometric isomer and give its full name.
Solution
In the Figure \(9\)b, the two chlorine ligands occupy cis positions. The other form is shown in below. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diaminedichloroplatinum(II).
The trans isomer of [Pt(NH3)2Cl2] has each ligand directly across from an adjacent ligand.
Exercise \(2\)
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).
Answer
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(11\). These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other.
The [Co(en)2Cl2]+ ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure \(12\)).
Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH3)5SCN]2+ or [Co(NH3)5NCS]2+).
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6][Br] and [CoCl5Br][Cl].
Coordination Complexes in Nature and Technology
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(13\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.
Transition Metal Catalysts
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (Figure \(14\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.
Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(13\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.
The structure of heme (Figure \(15\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe2+; oxidation of the iron to Fe3+ prevents oxygen transport.
Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure \(16\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH2CH(SH)CH2OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(17\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.
Example \(3\): Chelation Therapy
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(18\). Identify which atoms in this molecule could act as donor atoms.
Solution
All of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal.
Exercise \(3\)
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations.1 Identify at least two biologically important metals that could be disrupted by chelation therapy.
Answer
Ca, Fe, Zn, and Cu
Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN)2] and [Au(CN)2] are used extensively in the electroplating industry.
In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diaminedichloroplatinum(II), [Pt(NH3)2(Cl)2], and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diamine (NH3)2 portion is retained with other groups, replacing the dichloro [(Cl)2] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.
Summary
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.
Footnotes
1. National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002).
Glossary
bidentate ligand
ligand that coordinates to one central metal through coordinate bonds from two different atoms
central metal
ion or atom to which one or more ligands is attached through coordinate covalent bonds
chelate
complex formed from a polydentate ligand attached to a central metal
chelating ligand
ligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configuration
configuration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compound
substance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination number
number of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination sphere
central metal atom or ion plus the attached ligands of a complex
donor atom
atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
ionization isomer
(or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
ligand
ion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomer
coordination compound that possesses a ligand that can bind to the transition metal in two different ways (CN vs. NC)
monodentate
ligand that attaches to a central metal through just one coordinate covalent bond
optical isomer
(also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
polydentate ligand
ligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
trans configuration
configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.03%3A_Coordination_Compounds.txt |
Learning Objectives
• To understand that there may be more than one way to arrange the same groups around the same atom with the same geometry (stereochemistry).
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases. For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two amines beside each other is called cis-platin.
• These two ligands are 90 degrees from each other.
• The one with the amines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
CIS/TRANS isomers have different physical properties
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
CIS/TRANS isomers have different biological properties
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
Exercise \(1\)
Draw the cis and trans isomers of the following compounds:
1. \(\ce{(NH3)2IrCl(CO)}\)
2. \(\ce{(H3P)2PtHBr}\)
3. \(\ce{(AsH3)2PtH(CO)}\)
Exercise \(2\)
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible.
Geometric Isomers
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 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.
Planar 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:
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, 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.
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 Isomers
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
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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.04%3A_Structure_and_Isomerization.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. 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. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes.
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, 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 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 (Figure $\PageIndex{1a}$). 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 Figure $\PageIndex{1b}$, 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.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), 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
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0. \nonumber$
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 Figure $\PageIndex{1a}$).
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 24.6.2, for d1–d3 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.
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 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 d8–d10 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.
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)
*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.
[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
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+: Δo = 22,900 cm−1
[Rh(NH3)6]3+: Δo = 34,100 cm−1
[Ir(NH3)6]3+: Δo = 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:
$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$
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.
The largest Δo splittings 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 d–d transition (Figure 24.6.3). 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.
Recall 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. 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.
Crystal Field Stabilization Energies
Recall 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.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Example $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
1. From the number of ligands, determine the coordination number of the compound.
2. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
3. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
4. 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.
1. 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−
Answers
1. octahedral; high spin; five
2. square planar; low spin; no unpaired electrons
Summary
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. 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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.05%3A_Bonding_in_Coordinate_Compounds.txt |
Chelating Agents for Poising (EDTA)
Biomolecules (Porphorins and Hemoglogin)
Drugs and Health (cis-Platin)
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/Map%3A_A_Molecular_Approach_(Tro)/25%3A_Transition_Metals_and_Coordination_Compounds/25.06%3A_Applications_of_Coordination_Compounds.txt |
Learning Objectives
• To identify the components of the scientific method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(1\)).
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 °C, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 °C. An example of a quantitative observation was the initial observation leading to the modern theory of the dinosaurs’ extinction: iridium concentrations in sediments dating to 66 million years ago were found to be 20–160 times higher than normal. The development of this theory is a good exemplar of the scientific method in action (see Figure \(2\) below).
After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists collected additional data that either support or refute it.
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in the dinosaur extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why.
One example of a law, the Law of Definite Proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Some solid compounds do not strictly obey the law of definite proportions. The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure \(2\) summarizes the application of the scientific method in this case.
Example \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice were added to 100 mL of water at 25 °C, the temperature of the water decreased to 15.5 °C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Given: components of the scientific method
Asked for: statement classification
Strategy: Refer to the definitions in this section to determine which category best describes each statement.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are quantitative observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
5. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas.
6. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive.
Answer a
experiment
Answer b
law
Answer c
theory
Answer d
hypothesis
Answer e
qualitative observation
Answer f
quantitative observation
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.
Summary
Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.
Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/01%3A_Chemical_Foundations/1.3%3A_The_Scientific_Method.txt |
Learning Objectives
• Recognize the SI base units and explain the system of prefixes used with them.
• Define and calculate density.
People who live in the United States measure weight in pounds, height in feet and inches, and a car’s speed in miles per hour. In contrast, chemistry and other branches of science use the International System of Units (also known as SI after Système Internationale d’Unités), which was established so that scientists around the world could communicate efficiently with each other. Many countries have also adopted SI units for everyday use as well. The United States is one of the few countries that has not.
Base SI Units
Base (or basic) units, are the fundamental units of SI. There are seven base units, which are listed in Table $1$, Chemistry uses five of the base units: the mole for amount, the kilogram for mass, the meter for length, the second for time, and the kelvin for temperature. The degree Celsius (°C) is also commonly used for temperature. The numerical relationship between kelvins and degrees Celsius is as follows:
$K = °C + 273 \label{Eq1}$
Table $1$: The Seven Base SI Units
Property Unit Abbreviation
length meter m
mass kilogram kg
time second s
amount mole mol
temperature kelvin K
electrical current ampere amp
luminous intensity candela cd
The United States uses the English (sometimes called Imperial) system of units for many quantities. Inches, feet, miles, gallons, pounds, and so forth, are all units connected with the English system of units. There have been many mistakes due to the improper conversion of units between the SI and English systems.
The size of each base unit is defined by international convention. For example, the kilogram is defined as the quantity of mass of a special metal cylinder kept in a vault in France (Figure $1$). The other base units have similar definitions and standards. The sizes of the base units are not always convenient for all measurements. For example, a meter is a rather large unit for describing the width of something as narrow as human hair. Instead of reporting the diameter of hair as 0.00012 m or as 1.2 × 10−4 m using scientific notation as discussed in section 1.4, SI also provides a series of prefixes that can be attached to the units, creating units that are larger or smaller by powers of 10.
Common prefixes and their multiplicative factors are listed in Table $2$. (Perhaps you have already noticed that the base unit kilogram is a combination of a prefix, kilo- meaning 1,000 ×, and a unit of mass, the gram.) Some prefixes create a multiple of the original unit: 1 kilogram equals 1,000 grams, and 1 megameter equals 1,000,000 meters. Other prefixes create a fraction of the original unit. Thus, 1 centimeter equals 1/100 of a meter, 1 millimeter equals 1/1,000 of a meter, 1 microgram equals 1/1,000,000 of a gram, and so forth.
Table $2$: Prefixes Used with SI Units
Prefix Abbreviation Multiplicative Factor Multiplicative Factor in Scientific Notation
giga- G 1,000,000,000 × 109 ×
mega- M 1,000,000 × 106 ×
kilo- k 1,000 × 103 ×
deca- D 10 × 101 ×
deci- d 1/10 × 10−1 ×
centi- c 1/100 × 10−2 ×
milli- m 1/1,000 × 10−3 ×
micro- µ* 1/1,000,000 × 10−6 ×
nano- n 1/1,000,000,000 × 10−9 ×
*The letter µ is the Greek lowercase letter for m and is called “mu,” which is pronounced “myoo.”
Both SI units and prefixes have abbreviations, and the combination of a prefix abbreviation with a base unit abbreviation gives the abbreviation for the modified unit. For example, kg is the abbreviation for kilogram. We will be using these abbreviations throughout this book.
The Difference Between Mass and Weight
The mass of a body is a measure of its inertial property or how much matter it contains. The weight of a body is a measure of the force exerted on it by gravity or the force needed to support it. Gravity on earth gives a body a downward acceleration of about 9.8 m/s2. In common parlance, weight is often used as a synonym for mass in weights and measures. For instance, the verb “to weigh” means “to determine the mass of” or “to have a mass of.” The incorrect use of weight in place of mass should be phased out, and the term mass used when mass is meant. The SI unit of mass is the kilogram (kg). In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. Thus, the SI unit of the quantity weight defined in this way (force) is the newton (N).
Derived SI Units
Derived units are combinations of SI base units. Units can be multiplied and divided, just as numbers can be multiplied and divided. For example, the area of a square having a side of 2 cm is 2 cm × 2 cm, or 4 cm2 (read as “four centimeters squared” or “four square centimeters”). Notice that we have squared a length unit, the centimeter, to get a derived unit for area, the square centimeter.
Volume is an important quantity that uses a derived unit. Volume is the amount of space that a given substance occupies and is defined geometrically as length × width × height. Each distance can be expressed using the meter unit, so volume has the derived unit m × m × m, or m3 (read as “meters cubed” or “cubic meters”). A cubic meter is a rather large volume, so scientists typically express volumes in terms of 1/1,000 of a cubic meter. This unit has its own name—the liter (L). A liter is a little larger than 1 US quart in volume. Below are approximate equivalents for some of the units used in chemistry.
Approximate Equivalents to Some SI Units
• 1 m ≈ 39.36 in. ≈ 3.28 ft ≈ 1.09 yd
• 1 in. ≈ 2.54 cm
• 1 km ≈ 0.62 mi
• 1 kg ≈ 2.20 lb
• 1 lb ≈ 454 g
• 1 L ≈ 1.06 qt
• 1 qt ≈ 0.946 L
As shown in Figure $3$, a liter is also 1,000 cm3. By definition, there are 1,000 mL in 1 L, so 1 milliliter and 1 cubic centimeter represent the same volume.
$1\; mL = 1\; cm^3 \label{Eq2}$
Example $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kiloliter
2. microsecond
3. decimeter
4. nanogram
Answer a
The abbreviation for a kiloliter is kL. Because kilo means “1,000 ×,” 1 kL equals 1,000 L.
Answer b
The abbreviation for microsecond is µs. Micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s.
Answer c
The abbreviation for decimeter is dm. Deci means 1/10th, so 1 dm equals 0.1 m.
Answer d
The abbreviation for nanogram is ng and equals 0.000000001 g.
Exercise $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kilometer
2. milligram
3. nanosecond
4. centiliter
Answer a
km (1,000 m)
Answer b
mg (0.001 g)
Answer c
ns (0.000000001 s)
Answer d
cL (0.01L)
Energy, another important quantity in chemistry, is the ability to perform work, such as moving a box of books from one side of a room to the other side. It has a derived unit of kg•m2/s2. (The dot between the kg and m units implies the units are multiplied together.) Because this combination is cumbersome, this collection of units is redefined as a joule (J). An older unit of energy, but likely more familiar to you, the calorie (cal), is also widely used. There are 4.184 J in 1 cal. Energy changes occur during all chemical processes and will be discussed in a later chapter.
To Your Health: Energy and Food
The food in our diet provides the energy our bodies need to function properly. The energy contained in food could be expressed in joules or calories, which are the conventional units for energy, but the food industry prefers to use the kilocalorie and refers to it as the Calorie (with a capital C). The average daily energy requirement of an adult is about 2,000–2,500 Calories, which is 2,000,000–2,500,000 calories (with a lowercase c).
If we expend the same amount of energy that our food provides, our body weight remains stable. If we ingest more Calories from food than we expend, however, our bodies store the extra energy in high-energy-density compounds, such as fat, and we gain weight. On the other hand, if we expend more energy than we ingest, we lose weight. Other factors affect our weight as well—genetic, metabolic, behavioral, environmental, cultural factors—but dietary habits are among the most important.
In 2008 the US Centers for Disease Control and Prevention issued a report stating that 73% of Americans were either overweight or obese. More alarmingly, the report also noted that 19% of children aged 6–11 and 18% of adolescents aged 12–19 were overweight—numbers that had tripled over the preceding two decades. Two major reasons for this increase are excessive calorie consumption (especially in the form of high-fat foods) and reduced physical activity. Partly because of that report, many restaurants and food companies are working to reduce the amounts of fat in foods and provide consumers with more healthy food options.
Density is defined as the mass of an object divided by its volume; it describes the amount of matter contained in a given amount of space.
$\mathrm{density=\dfrac{mass}{volume}}\label{Eq3}$
Thus, the units of density are the units of mass divided by the units of volume: g/cm3 or g/mL (for solids and liquids), g/L (for gases), kg/m3, and so forth. For example, the density of water is about 1.00 g/cm3, while the density of mercury is 13.6 g/mL. (Remember that 1 mL equals 1 cm3.) Mercury is over 13 times as dense as water, meaning that it contains over 13 times the amount of matter in the same amount of space. The density of air at room temperature is about 1.3 g/L. Table 1.6.3 shows the densities of some common substances.
Table $2$: Densities of Common Substances
Solids Liquids Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3 water 1.0 g/cm3 dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3 ethanol 0.79 g/cm3 oxygen 1.31 g/L
iron 7.9 g/cm3 acetone 0.79 g/cm3 nitrogen 1.14 g/L
copper 9.0 g/cm3 glycerin 1.26 g/cm3 carbon dioxide 1.80 g/L
lead 11.3 g/cm3 olive oil 0.92 g/cm3 helium 0.16 g/L
silver 10.5 g/cm3 gasoline 0.70–0.77 g/cm3 neon 0.83 g/L
gold 19.3 g/cm3 mercury 13.6 g/cm3 radon 9.1 g/L
Example $2$: Density of Bone
What is the density of a section of bone if a 25.3 cm3 sample has a mass of 27.8 g?
Solution
Because density is defined as the mass of an object divided by its volume, we can set up the following relationship:
\begin{align*} \mathrm{density} &=\dfrac{mass}{volume} \[4pt] &= \dfrac{27.8\:g}{25.3\:cm^3} \[4pt] &=1.10\:g/cm^3 \end{align*} \nonumber
Note that we have limited our final answer to three significant figures.
Exercise $2$: Density of Oxygen
What is the density of oxygen gas if a 15.0 L sample has a mass of 21.7 g?
Answer
1.45 g/L
Density can be used to convert between the mass and the volume of a substance. This will be discussed in the next section.
Concept Review Exercises
1. What is the difference between a base unit and a derived unit? Give two examples of each type of unit.
2. Do units follow the same mathematical rules as numbers do? Give an example to support your answer.
3. What is density?
Answers
1. Base units are the seven fundamental units of SI; derived units are constructed by making combinations of the base units; Two examples of base units: kilograms and meters (answers will vary); Two examples of derived units: grams per milliliter and joules (answers will vary).
2. yes; $\mathrm{mL\times\dfrac{g}{mL}=g}$ (answers will vary)
3. Density is defined as the mass of an object divided by its volume
Key Takeaways
• Recognize the SI base units and derived units.
• Combining prefixes with base units creates new units of larger or smaller sizes. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/01%3A_Chemical_Foundations/1.4%3A_Chemistry_in_Industry.txt |
All measurements have a degree of uncertainty regardless of precision and accuracy. This is caused by two factors, the limitation of the measuring instrument (systematic error) and the skill of the experimenter making the measurements (random error).
Introduction
The graduated buret in Figure 1 contains a certain amount of water (with yellow dye) to be measured. The amount of water is somewhere between 19 ml and 20 ml according to the marked lines. By checking to see where the bottom of the meniscus lies, referencing the ten smaller lines, the amount of water lies between 19.8 ml and 20 ml. The next step is to estimate the uncertainty between 19.8 ml and 20 ml. Making an approximate guess, the level is less than 20 ml, but greater than 19.8 ml. We then report that the measured amount is approximately 19.9 ml. The graduated cylinder itself may be distorted such that the graduation marks contain inaccuracies providing readings slightly different from the actual volume of liquid present.
Figure 1: A meniscus as seen in a burette of colored water. '20.00 mL' is the correct depth measurement. Click here for a more complete description on buret use, including proper reading. Figure used with permission from Wikipedia.
Systematic vs. Random Error
The diagram below illustrates the distinction between systematic and random errors.
Figure 2: Systematic and random errors. Figure used with permission from David DiBiase (Penn State U).
Systematic errors: When we use tools meant for measurement, we assume that they are correct and accurate, however measuring tools are not always right. In fact, they have errors that naturally occur called systematic errors. Systematic errors tend to be consistent in magnitude and/or direction. If the magnitude and direction of the error is known, accuracy can be improved by additive or proportional corrections. Additive correction involves adding or subtracting a constant adjustment factor to each measurement; proportional correction involves multiplying the measurement(s) by a constant.
Random errors: Sometimes called human error, random error is determined by the experimenter's skill or ability to perform the experiment and read scientific measurements. These errors are random since the results yielded may be too high or low. Often random error determines the precision of the experiment or limits the precision. For example, if we were to time a revolution of a steadily rotating turnable, the random error would be the reaction time. Our reaction time would vary due to a delay in starting (an underestimate of the actual result) or a delay in stopping (an overestimate of the actual result). Unlike systematic errors, random errors vary in magnitude and direction. It is possible to calculate the average of a set of measured positions, however, and that average is likely to be more accurate than most of the measurements.
1. Since Tom must rely on the machine for an absorbance reading and it provides consistently different measurements, this is an example of systematic error.
2. The majority of Claire's variation in time can likely be attributed to random error such as fatigue after multiple laps, inconsistency in swimming form, slightly off timing in starting and stopping the stop watch, or countless other small factors that alter lap times. To a much smaller extent, the stop watch itself may have errors in keeping time resulting in systematic error.
3. The researcher's percent error is about 0.62%.
4. This is known as multiplier or scale factor error.
5. This is called an offset or zero setting error.
6. Susan's percent error is -7.62%. This percent error is negative because the measured value falls below the accepted value. In problem 7, the percent error was positive because it was higher than the accepted value.
7. You would first weigh the beaker itself. After obtaining the weight, then you add the graphite in the beaker and weigh it. After obtaining this weight, you then subtract the weight of the graphite plus the beaker minus the weight of the beaker. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/01%3A_Chemical_Foundations/1.5%3A_Polymer_Chemistry.txt |
Timeline of Chemistry
• Wikipedia Link: Timeline of Chemistry
2.02 Fundamental Chemical Laws
Learning Objectives
• Explain how all matter is composed of atoms.
• Describe the modern atomic theory.
Take some aluminum foil. Cut it in half. Now there are two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can this exercise be taken, at least in theory? Can one continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century BC.
John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. The theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. This section explains the theories that Dalton used as a basis for his theory: (1) Law of Conservation of Mass, (2) Law of Definite Proportions, and (3) Law of Multiple Proportions
Law 1: The Conservation of Mass
"Nothing comes from nothing" is an important idea in ancient Greek philosophy that argues that what exists now has always existed, since no new matter can come into existence where there was none before. Antoine Lavoisier (1743-1794) restated this principle for chemistry with the law of conservation of mass, which "means that the atoms of an object cannot be created or destroyed, but can be moved around and be changed into different particles." This law says that when a chemical reaction rearranges atoms into a new product, the mass of the reactants (chemicals before the chemical reaction) is the same as the mass of the products (the new chemicals made). More simply, whatever you do, you will still have the same amount of stuff (however, certain nuclear reactions like fusion and fission can convert a small part of the mass into energy.
The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, mass is conserved. The law of conservation of mass was formulated by Lavoisier as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air.
Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, Figure $1$ shows that the burning of word does follow the law of conservation of mass. Scientists did not account for the gases that play a critical role in this reaction.
The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction.
Law 2: Definite Proportions
Joseph Proust (1754-1826) formulated the law of definite proportions (also called the Law of Constant Composition or Proust's Law). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law in action.
Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass.
The Law of Definite Proportions applies when elements are reacted together to form the same product. Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen and oxygen react to form water, the Law of Definite Proportions can not be used to compare one experiment in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another material that can be made from hydrogen and oxygen).
Example $1$: water
Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow, by condensing steam, from river, sea, pond, etc. It can be from different places: USA, UK, Australia, or anywhere. It can be made by chemical reactions like burning hydrogen in oxygen.
However, if the water is pure, it will always consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation.
Law 3: Multiple Proportions
Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here.
• Lineshows the ratio of the relative weights of the two elements in each compound. These ratios were calculated by simply taking the molar mass of each element, and multiplying by the number of atoms of that element per mole of the compound. Thus for NO2, we have (1 × 14) : (2 × 16) = 14:32. (These numbers were not known in the early days of Chemistry because atomic weights (i.e., molar masses) of most elements were not reliably known.)
• The numbers in Lineare just the mass ratios of O:N, found by dividing the corresponding ratios in line 1. But someone who depends solely on experiment would work these out by finding the mass of O that combines with unit mass (1 g) of nitrogen.
• Line is obtained by dividing the figures the previous line by the smallest O:N ratio in the line above, which is the one for N2O. Note that just as the law of multiple proportions says, the weight of oxygen that combines with unit weight of nitrogen work out to small integers.
• Of course we just as easily could have illustrated the law by considering the mass of nitrogen that combines with one gram of oxygen; it works both ways!
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers.
Example $2$: Oxides of Carbon
Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistent with the law of multiple proportions?
Solution
The Law of Multiple Proportions states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio.
Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is:
$\dfrac{0.571\; \text{g oxygen}}{0.429 \;\text{g carbon}} = 1.33\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is
$\dfrac{0.727\; \text{g oxygen}}{0.273 \;\text{g carbon}} = 2.66\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
Dividing the mass of oxygen per g of carbon of the second compound:
$\dfrac{2.66}{1.33} = 2\nonumber$
Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers.
Dalton's Atomic Theory
The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure $4$), is a fundamental concept that states that all elements are composed of atoms. Previously, an atom was defined as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of the human pinkie (about 1 cm).
Dalton’s ideas are called the modern atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word atom comes from the Greek word atomos, which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science, and the scientific method was being used to study nature. When Dalton announced a modern atomic theory, he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion.
Dalton's Theory was a powerful development as it explained the three laws of chemical combination (above) and recognized a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Six postulates are involved in Dalton's Atomic Theory:
1. All matter consists of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass, but differ from the atoms of other elements.
3. Atoms cannot be created or destroyed.
4. Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form compound atoms.
5. Atoms of same element can combine in more than one ratio to form two or more compounds.
6. The atom is the smallest unit of matter that can take part in a chemical reaction.
In light of the current state of knowledge in the field of Chemistry, Dalton’s theory had a few drawbacks. According to Dalton’s postulates,
1. The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions.
2. According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. For example, chlorine has two isotopes with mass numbers 35 and 37.
3. Dalton also claimed that atoms of different elements are different in all respects. This has been proven wrong in certain cases: argon and calcium atoms each have an same atomic mass (40 amu).
4. According to Dalton, atoms of different elements combine in simple whole number ratios to form compounds. This is not observed in complex organic compounds like sugar ($C_{12}H_{22}O_{11}$).
5. The theory fails to explain the existence of allotropes (different forms of pure elements); it does not account for differences in properties of charcoal, graphite, diamond.
Despite these drawbacks, the importance of Dalton’s theory should not be underestimated. He displayed exceptional insight into the nature of matter. and his ideas provided a framework that was later modified and expanded by other. Consequentiually, John Dalton is often considered to be the father of modern atomic theory.
Fundamental Experiments in Chemistry: Fundamental Experiments in Chemistry, YouTube(opens in new window) [youtu.be]
Summary
This article explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.01_The_Early_History_of_Chemistry.txt |
John Dalton, a British school teacher, published his theory about atoms in 1808. His findings were based on experiments and the laws of chemical combination.
2.04 Early Experiments to Characterize the Atom
Learning Objectives
• To become familiar with the components and structure of the atom.
Long before the end of the 19th century, it was well known that applying a high voltage to a gas contained at low pressure in a sealed tube (called a gas discharge tube) caused electricity to flow through the gas, which then emitted light (Figure $1$). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; this form of energy was called a cathode ray.
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure $2$). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Subsequently, the American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:
$\dfrac {mass}{charge} \times {charge} ={mass} \nonumber$
It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy. The video below shows how JJ Thompson used such a tube to measure the ratio of charge over mass of an electron
Measuring e/m For an Electron. Video from Davidson College demonstrating Thompson's e/m experiment.
Radioactivity
The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivity (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning.
Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to the lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.
These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As Figure $3$ illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. Figure $5$ shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.
The Atomic Model
Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure $6$).
In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (Figure $\PageIndex{7a}$) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (Figure $\PageIndex{7b}$). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (Figure $\PageIndex{7c}$). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
The Nuclear Atom: The Nuclear Atom, YouTube(opens in new window) [youtu.be]
Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source.
Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.”
The historical development of the different models of the atom’s structure is summarized in Figure $8$. Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons, and is therefore the nucleus of a helium atom.
Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure $9$ shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.
Summary
Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.03_Dalton%27s_Atomic_Theory.txt |
Learning Objectives
• To know the meaning of isotopes and atomic masses.
The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that
• Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element.
• Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge) is balanced by the same number of electrons outside the nucleus. The other nuclear particle is the neutron. As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom.
Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion.
Elements
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$, which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table $2$.
Table $2$: Element Symbols Based on Names No Longer in Use
Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
Figure $2$: Formalism used for identifying specific nuclide (any particular kind of nucleus)
In addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table $3$.
Table $3$: Properties of Selected Isotopes
Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
How Elements Are Represented on the Periodic Table: How Elements Are Represented on the Periodic Table, YouTube(opens in new window) [youtu.be]
Example $1$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $1$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer
$\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Summary
The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.05_The_Modern_View_of_Atomic_Structure%3A_An_Introduction.txt |
Chemical compounds can generally be classified into two broad groups: molecular compounds and ionic compounds. Molecular compounds involve atoms joined by covalent bonds and can be represented by a variety of formulas. Ionic compounds are composed of ions joined by ionic bonding, and their formulas are generally written using oxidation states.
Molecular compounds are composed of atoms that are held together by covalent bonds. These bonds are formed when electrons are shared between two atoms. The concept of chemical formulas was created to describe many characteristics of molecular compounds in a simple manner. A normal chemical formula encompasses factors such as which elements are in the molecule and how many atoms of each element there are. The number of atoms of each element is denoted by a subscript, a small number that is written to the left of the element.
$CH_3COOH \nonumber$
In the preceding formula, the subscript “3” denotes the fact that there are three hydrogen atoms present in the molecule.
Other types of formulas are used to display more detailed characteristics of molecules.
An empirical formula represents the proportions of atoms in a molecule. It gives important information about a molecule, because it displays the ratios of atoms that are present within the molecule. However, its limitations exist in the sense that it does not represent the exact number of atoms present in the molecule as the molecular formula does. In certain situations, the molecular and the empirical formula can be the same, but in other situations, the molecular formula is a multiple of the ratios of atoms indicated in the empirical formula. Since empirical formulas can be derived from molecular formulas, molecular formulas are generally more useful than empirical formulas.
Empirical vs. molecular compounds
C5H7O is a possible empirical formula, because a ratio of 5:7:1 cannot be simplified any further. In this particular case, the empirical formula could also be the molecular formula, if there are exactly 5 carbon atoms, 7 hydrogen atoms, and 1 oxygen atom per molecule. However, another possible molecular formula for this same molecule is C10H14O2, because while there are 10 carbon atoms, 14 hydrogen atoms, and 2 oxygen atoms present, the ratio 10:14:2 can be simplified to 5:7:1, giving way to the same empirical formula. Additionally, C10H14O2 is not the only possibility of a molecular formula for this molecule; any formula with the same relative proportions of these atoms that can be simplified to a 5:7:1 ratios is a possible molecular formula for this molecule. When given adequate information, the empirical formula and molecular formula can be quantitatively ascertained.
A structural formula is written to denote the details of individual atoms’ bonding. More specifically, it clarifies what types of bonds exist, between which atoms these bonds exist, and the order of the atoms’ bonding within the molecule. Covalent bonds are denoted by lines. A single line represents a single bond, two lines represent a double bond, three lines represent a triple bond, and onwards. A single covalent bond occurs when two electrons are shared between atoms, a double occurs when four electrons are shared between two atoms, etc. In this sense, the higher the number of bonds, the stronger the bond between the two atoms.
A condensed structural formula is a less graphical way of representing the same characteristics displayed by a structural formula. In this type of formula, the molecule is written as a molecular formula with the exception that it indicates where the bonding occurs.
All the representations discussed thus far have not addressed how to show a molecule’s three-dimensional structure. The two ways to illustrate a spatial structure are through the use of the ball-and-stick model as well as the space-filling model.
The ball-and-stick model uses balls to spatially represent a molecule. The balls are the atoms in a molecule and sticks are the bonds between specific atoms.
The space-filling model is also a method of spatially displaying a molecule and its characteristics. A space-filling model shows atoms’ sizes relative sizes to one another.
Ionic Compounds
Ionic compounds are composed of positive and negative ions that are joined by ionic bonds. Ionic bonds are generally formed when electrons are transferred from one atom to another, causing individual atoms to become charged particles, or ions. Ions can be referred to as either monatomic or polyatomic. Monatomic ions such as Cl- are composed of only one ion, while polyatomic ions such as NO3- are defined as polyatomic ions. A combination of these ions that forms a compound whose charge is equal to zero is known as a formula unit of an ionic compound. Ionic compounds generally tend to form crystallized salts. They generally have high boiling/melting points, and are good conductors of electricity. The formulas of ionic compounds are always written with the cation first, followed by the anion. The formula can then be completed with reference to the oxidation states of the elements present. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.06_Molecules_and_Ions.txt |
Learning Objectives
• To become familiar with the organization of the periodic table.
Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its atomic number (Z), the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the periodic table (Figure \(1\)). The rationale for the peculiar format of the periodic table is explained later. Each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons.
The elements are arranged in a periodic table, which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that facilitates the prediction of many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periods, and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groups, numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elements, listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elements, listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3.
Metals, Nonmetals, and Semimetals
The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in Figure \(1\) divides the elements into metals (in blue, below and to the left of the line) and nonmetals (in bronze, above and to the right of the line). Gold-colored lements that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetals.
The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductile; they can be hammered or pressed into thin sheets or foils because they are malleable; and most have a shiny appearance, so they are lustrous. The vast majority of the known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids.
Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals.
Example \(1\): Classifying Elements
Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal?
Given: element
Asked for: classification
Strategy:
Find selenium in the periodic table shown in Figure \(1\) and then classify the element according to its location.
Solution:
The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure \(1\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties.
Exercise \(1\)
Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal?
Answer
metal
As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of Group 1 are known as the alkali metals, Group 2 are the alkaline earth metals, Group 17 are the halogens, and Group 18 are the noble gases.
Group 1: The Alkali Metals
The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal. The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
Group 2: The Alkaline Earth Metals
The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
Group 17: The Halogens
The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).
Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
Group 18: The Noble Gases
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
The noble gases are unreactive at room temperature and pressure.
Summary
The periodic table is used as a predictive tool. It arranges of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.07_An_Introduction_to_the_Periodic_Table.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/02%3A_Atoms_Molecules_and_Ions/2.08_Naming_Simple_Compounds.txt |
Learning Objectives
• to know the meaning of isotopes and atomic masses.
Atomic and Molecular Weights
The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. $\ce{H_2O}$, for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation:
$\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1}$
not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved.
Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen:
$\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2}$
Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom:
$\dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3}$
Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. The atomic mass unit (amu) was not standardized against hydrogen, but rather, against the 12C isotope of carbon (amu = 12).
Thus, the mass of the hydrogen atom (1H) is 1.0080 amu, and the mass of an oxygen atom (16O) is 15.995 amu. Once the masses of atoms were determined, the amu could be assigned an actual value:
1 amu = 1.66054 x 10-24 grams conversely: 1 gram = 6.02214 x 1023 amu
Mass Numbers and Atomic Mass of Elements: Mass Numbers and Atomic Mass of Elements, YouTube(opens in new window) [youtu.be]
Average Atomic Mass
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 2.3.1), the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions.
Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure $1$). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.
The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom listed in Table 2.3.2 whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.
Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is
$\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}$
The masses of the other elements are determined in a similar way.
The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in Table 2.3.2 reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows:
$\rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5}$
Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation.
The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.
Example $1$: Bromine
Naturally occurring bromine consists of the two isotopes listed in the following table:
Solutions to Example 2.4.1
Isotope Exact Mass (amu) Percent Abundance (%)
79Br 78.9183 50.69
81Br 80.9163 49.31
Calculate the atomic mass of bromine.
Given: exact mass and percent abundance
Asked for: atomic mass
Strategy:
1. Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.
2. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.
3. Add together the weighted masses to obtain the atomic mass of the element.
4. Check to make sure that your answer makes sense.
Solution:
A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write
atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + …
Bromine has only two isotopes. Converting the percent abundances to mass fractions gives
$\ce{^{79}Br}: {50.69 \over 100} = 0.5069 \nonumber$
$\ce{^{81}Br}: {49.31 \over 100} = 0.4931 \nonumber$
B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:
$\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu$
$\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu$
C The sum of the weighted masses is the atomic mass of bromine is
40.00 amu + 39.90 amu = 79.90 amu
D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.
Exercise $1$
Magnesium has the three isotopes listed in the following table:
Solutions to Example 2.4.1
Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17
Use these data to calculate the atomic mass of magnesium.
Answer
24.31 amu
Finding the Averaged Atomic Weight of an Element: Finding the Averaged Atomic Weight of an Element(opens in new window) [youtu.be]
Summary
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.01_Counting_by_Weighing.txt |
The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an average of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth.
Avogadro's Constant
One mole of oxygen atoms contains $6.02214179 \times 10^{23}$ oxygen atoms. Also, one mole of nitrogen atoms contains $6.02214179 \times 10^{23}$ nitrogen atoms. The number $6.02214179 \times 10^{23}$ is called Avogadro's number ($N_A$) or Avogadro's constant, after the 19th century scientist Amedeo Avogadro.
Each carbon-12 atom weighs about $1.99265 \times 10^{-23}\; g$; therefore,
$(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber$
Applications of the Mole
The mass of a mole of substance is called the molar mass of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance).
The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH4. This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below:
$2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber$
In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of $\ce{H_2O}$.
To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×1023 of anything. Using the following relation:
$\text{1 mole} = 6.02214179 \times 10^{23}$
is analogous to saying:
$\text{1 Dozen} = \text{12 eggs}$
It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number.
Example $1$: Converting Mass to Moles
How many moles of potassium ($\ce{K}$) atoms are in 3.04 grams of pure potassium metal?
Solution
In this example, multiply the mass of $\ce{K}$ by the conversion factor (inverse molar mass of potassium):
$\dfrac{1\; mol\; K}{39.10\; grams \;K} \nonumber$
39.10 grams is the molar mass of one mole of $\ce{K}$; cancel out grams, leaving the moles of $\ce{K}$:
$3.04\; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0778\; mol\; K \nonumber$
Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of molar mass of substance/one mole of substance. One simply needs to follow the same method but in the opposite direction.
Example $2$: Converting Moles to mass
How many grams are 10.78 moles of Calcium ($\ce{Ca}$)?
Solution
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
$10.78 \cancel{\;mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\; \cancel{mol\; Ca}}\right) = 432.1\; g\; Ca \nonumber$
The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example $1$. Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of Avogadro's constant (6.02214179×1023) / one mole of substance. Verifying that the units cancel properly is a good way to make sure the correct method is used.
Example $3$: Atoms to Mass
How many atoms are in a 3.5 g sample of sodium (Na)?
Solution
$3.5\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.152\; mol\; Na \nonumber$
$0.152\; \cancel{mol\; Na} \left(\dfrac{6.02214179\times 10^{23}\; atoms\; Na}{1\;\cancel{ mol\; Na}}\right) = 9.15 \times 10^{22}\; atoms\; of\; Na \nonumber$
In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.
Using Avogadro's constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table $1$). By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. The following table provides a reference for the ways in which these various quantities can be manipulated:
Table $1$: Conversion Factors
Known Information Multiply By Result
Mass of substance (g) 1/ Molar mass (mol/g) Moles of substance
Moles of substance (mol) Avogadro's constant (atoms/mol) Atoms (or molecules)
Mass of substance (g) 1/Molar mass (mol/g) × Avogadro's constant (atoms/mol)) Atoms (or molecules)
Example $4$: Mass to Moles
How many moles are in 3.00 grams of potassium (K)?
Solution
$3.00 \; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0767\; mol\; K \nonumber$
In this example, multiply the mass of K by the conversion factor:
$\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber$
39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K.
Example $5$: Moles to Mass
How many grams is in 10.00 moles of calcium (Ca)?
Solution
This is the calculation in Example $2$ performed in reverse. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca:
$10.00\; \cancel{mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\;\cancel{ mol\; Ca}}\right) = 400.8\; grams \;of \;Ca \nonumber$
The number of atoms can also be calculated using Avogadro's Constant (6.02214179×1023) / one mole of substance.
Example $6$: Mass to Atoms
How many atoms are in a 3.0 g sample of sodium (Na)?
Solution
Convert grams to moles
$3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber$
Convert moles to atoms
$0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber$
Summary
The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to $6.02214179 \times 10^{23}$ atoms, or other elementary units such as molecules.
Problems
1. Using a periodic table, give the molar mass of the following:
1. H
2. Se
3. Ne
4. Cs
5. Fe
2. Convert to moles and find the total number of atoms.
1. 5.06 grams of oxygen
2. 2.14 grams of K
3. 0.134 kg of Li
3. Convert the following to grams
1. 4.5 mols of C
2. 7.1 mols of Al
3. 2.2 mols of Mg
4. How many moles are in the product of the reaction
1. 6 mol H + 3 mol O → ? mol H2O
2. 1 mol Cl + 1 mol Cl → ? mol Cl2
3. 5 mol Na + 4 mol Cl → ? mol NaCl
Answers
1. Question 2
1. 1.008 g/mol
2. 78.96 g/mol
3. 20.18 g/mol
4. 132.91g/mol
5. 55.85 g/mol
2. Question 2
2. 5.06g O (1mol/16.00g)= 0.316 mol of O
0.316 mols (6.022x1023 atoms/ 1mol) = 1.904x1023 atoms of O
3. 2.14g K (1mol/39.10g)= 0.055 mol of K
0.055 mols (6.022x1023 atoms/ 1mol) = 3.312x1022 atoms of K
4. 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 mols Li
19.3 (6.022x1023 atoms/ 1mol) = 1.16x1025 atoms of Li
1. Question 3
1. 4.5 mols of C (12.011g/1mol) = 54.05 g of C
2. 7.1 mols of Al (26.98g/1mol) = 191.56 g of Al
3. 2.2 mols of Mg (24.31g/1mol) = 53.48 g of MG
2. Question 4
1. 8. 6 mol H + 3 mol O → 3 mol H2O
2. 9. 1 mol Cl + 1 mol Cl → 1 mol Cl2
3. 10. 5 mol Na + 4 mol Cl → 4 mol NaCl + 1 mol Na (excess) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.02_The_Mole.txt |
Mass is a basic physical property of matter. The mass of an atom or a molecule is referred to as the atomic mass. The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems.
Introduction
In chemistry, there are many different concepts of mass. It is often assumed that atomic mass is the mass of an atom indicated in unified atomic mass units (u). However, the book Quantities, Units and Symbols in Physical Chemistry published by the IUPAC clearly states:
"Neither the name of the physical quantity, nor the symbol used to denote it, should imply a particular choice of unit."
The name "atomic mass" is used for historical reasons, and originates from the fact that chemistry was the first science to investigate the same physical objects on macroscopic and microscopic levels. In addition, the situation is rendered more complicated by the isotopic distribution. On the macroscopic level, most mass measurements of pure substances refer to a mixture of isotopes. This means that from a physical stand point, these mixtures are not pure. For example, the macroscopic mass of oxygen (O2) does not correspond to the microscopic mass of O2. The former usually implies a certain isotopic distribution, whereas the latter usually refers to the most common isotope (16O2). Note that the former is now often referred to as the "molecular weight" or "atomic weight".
Mass Concepts in Chemistry
name in chemistry physical meaning symbol units
atomic mass mass on microscopic scale m, ma Da, u, kg, g
molecular mass mass of a molecule m Da, u, kg, g
isotopic mass mass of a specific isotope Da, u, kg, g
mass of entity mass of a chemical formula m, mf Da, u, kg, g
average mass average mass of a isotopic distribution m Da, u, kg, g
molar mass average mass per mol M = m/n kg/mol or g/mol
atomic weight average mass of an element Ar = m / mu unitless
molecular weight average mass of a molecule Mr = m / mu unitless
relative atomic mass ratio of mass m and and the atomic mass constant mu Ar = m / mu unitless
atomic mass constant mu = m(12C)/12 mu = 1 Da = 1 u Da, u, kg, g
relative molecular mass ratio of mass m of a molecule and and the atomic mass constant mu Mr = m / mu unitless
relative molar mass ? ? ?
mass number nucleon number A nucleons, or unitless
integer mass nucleon number * Da m Da, u
nominal mass integer mass of molecule consisting of most abundant isotopes m Da, u
exact mass mass of molecule calculated from the mass of its isotopes (in contrast of measured ba a mass spectrometer) Da, u, kg, g
accurate mass mass (not normal mass) Da, u, kg, g
These concepts are further explained below.
Average Mass
Isotopes are atoms with the same atomic number, but different mass numbers. A different mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12, by definition). This explains how lithium can have an atomic mass of 6.941 Da. The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature; more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:
$m(E) = \sum_{n=1} m(I_n) \times p(I_n)$
where ∑ represents a n-times summation over all isotopes $I_n$ of element E, and p(I) represents the relative abundance of the isotope I.
Example 1
Find the average atomic mass of boron using the Table 1 below:
Mass and abundance of Boron isotopes
n isotope In mass m (Da) isotopic abundance p
1 10B 10.013 0.199
2 11B 11.009 0.801
Solution
The average mass of Boron is:
$m(B) = (10.013\ Da)(0.199) + (11.009\ Da)(0.801) = 1.99\ Da + 8.82\ Da = 10.81\ Da$
Relative Mass
Traditionally it was common practice in chemistry to avoid using any units when indicating atomic masses (e.g. masses on microscopic scale). Even today, it is common to hear a chemist say, "12C has exactly mass 12". However, because mass is not a dimensionless quantity, it is clear that a mass indication needs a unit. Chemists have tried to rationalize the omission of a unit; the result is the concept of relative mass, which strictly speaking is not even a mass but a ratio of two masses. Rather than using a unit, these chemists claim to indicate the ratio of the mass they want to indicate and the atomic mass constant mu which is defined analogous to the unit they want to avoid. Hence the relative atomic mass of the mass m is defined as:
$A_r = \dfrac{m}{m_u}$
The quantity is now dimensionless. As this unit is confusing and against the standards of modern metrology, the use of relative mass is discouraged.
Molecular Weight, Atomic Weight, Weight vs. Mass
Until recently, the concept of mass was not clearly distinguished from the concept of weight. In colloquial language this is still the case. Many people indicate their "weight" when they actually mean their mass. Mass is a fundamental property of objects, whereas weight is a force. Weight is the force F exerted on a mass m by a gravitational field. The exact definition of the weight is controversial. The weight of a person is different on ground than on a plane. Strictly speaking, weight even changes with location on earth.
When discussing atoms and molecules, the mass of a molecule is often referred to as the "molecular weight". There is no univerally-accepted definition of this term; however, mosts chemists agree that it means an average mass, and many consider it dimensionless. This would make "molecular weight" a synonym to "average relative mass".
Integer Mass
Because the proton and the neutron have similar mass, and the electron has a very small mass compared to the former, most molecules have a mass that is close to an integer value when measured in daltons. Therefore it is quite common to only indicate the integer mass of molecules. Integer mass is only meaningful when using dalton (or u) units.
Accurate Mass
Many mass spectrometers can determine the mass of a molecule with accuracy exceeding that of the integer mass. This measurement is therefore called the accurate mass of the molecule. Isotopes (and hence molecules) have atomic masses that are not integer masses due to a mass defect caused by binding energy in the nucleus.
Units
The atomic mass is usually measured in the units unified atomic mass unit (u), or dalton (Da). Both units are derived from the carbon-12 isotope, as 12 u is the exact atomic mass of that isotope. So 1 u is 1/12 of the mass of a carbon-12 isotope:
1 u = 1 Da = m(12C)/12
The first scientists to measure atomic mass were John Dalton (between 1803 and 1805) and Jons Jacoband Berzelius (between 1808 and 1826). Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known was Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen. Berzelius demonstrated that this is not always the case by showing that chlorine (Cl) has a mass of 35.45, which is not a whole number multiple of hydrogen's mass.
Some chemists use the atomic mass unit (amu). The amu was defined differently by physicists and by chemists:
• Physics: 1 amu = m(16O)/16
• Chemistry: 1 amu = m(O)/16
Chemists used oxygen in the naturally occurring isotopic distribution as the reference. Because the isotopic distribution in nature can change, this definition is a moving target. Therefore, both communities agreed to the compromise of using m(12C)/12 as the new unit, naming it the "unified atomic mass unit" (u). Hence, the amu is no longer in use; those who still use it do so with the definition of the u in mind. For this reason, the dalton (Da) is increasingly recommended as the accurate mass unit.
Neither u nor Da are SI units, but both are recognized by the SI.
Molar Mass
The molar mass is the mass of one mole of substance, whether the substance is an element or a compound. A mole of substance is equal to Avogadro's number (6.023×1023) of that substance. The molar mass has units of g/mol or kg/mol. When using the unit g/mol, the numerical value of the molar mass of a molecule is the same as its average mass in daltons:
• Average mass of C: 12.011 Da
• Molar mass of C: 12.011 g/mol
This allows for a smooth transition from the microscopic world, where mass is measured in daltons, to the macroscopic world, where mass is measured in kilograms.
Example 2
What is the molar mass of phenol, C6H5OH?
Average mass m = 6 × 12.011 Da + 6 × 1.008 Da + 1 × 15.999 Da = 94.113 Da
Molar mass = 94.113 g/mol = 0.094113 kg/mo
Measuring Masses in the Atomic Scale
Masses of atoms and molecules are measured by mass spectrometry. Mass spectrometry is a technique that measures the mass-to-charge ratio (m/q) of ions. It requires that all molecules and atoms to be measured be ionized. The ions are then separated in a mass analyzer according to their mass-to-charge ratio. The charge of the measured ion can then be determined, because it is a multiple of the elementary charge. The the ion's mass can be deduced. The average masses indicated in the periodic table are then calculated using the isotopic abundances, as explained above.
The masses of all isotopes have been measured with very high accuracy. Therefore, it is much simpler and more accurate to calculate the mass of a molecule of interest as a sum of its isotopes than measuring it with a commercial mass spectrometer.
Note that the same is not true on the nucleon scale. The mass of an isotope cannot be calculated accurately as the sum of its particles (given in the table below); this would ignore the mass defect caused by the binding energy of the nucleons, which is significant.
Table 2: Mass of three sub-atomic particles
Particle SI (kg) Atomic (Da) Mass Number A
Proton 1.6726×10-27 1.0073 1
Neutron 1.6749×10-27 1.0087 1
Electron 9.1094×10-31 0.00054858 0
As shown in Table 2, the mass of an electron is relatively small; it contributes less than 1/1000 to the overall mass of the atom.
Where to Find Atomic Mass
The atomic mass found on the Periodic Table (below the element's name) is the average atomic mass. For example, for Lithium:
The red arrow indicates the atomic mass of lithium. As shown in Table 2 above and mathematically explained below, the masses of a protons and neutrons are about 1u. This, however, does not explain why lithium has an atomic mass of 6.941 Da where 6 Da is expected. This is true for all elements on the periodic table. The atomic mass for lithium is actually the average atomic mass of its isotopes. This is discussed further in the next section.
One particularly useful way of writing an isotope is as follows:
Applications
Applications Include:
1. Average Molecular Mass
2. Stoichiometry
Note: One particularly important relationship is illustrated by the fact that an atomic mass unit is equal to 1.66 × 10-24 g. This is the reciprocal of Avogadro's constant, and it is no coincidence:
$\dfrac{\rm Atomic~Mass~(g)}{1 {\rm g}} \times \dfrac{1 {\rm mol}}{6.022 \times 10^{23}} = \dfrac{\rm Mass~(g)}{1 {\rm atom}}$
Because a mol can also be expressed as gram × atoms,
$1\ u = \frac{M_u\ (molar\ mass\ unit)}{N_A\ (Avogadro's\ Number)} = 1\ \frac{g}{mol\ N_A}$
1u = Mu(molar mass unit)/NA(Avogadro's Number)=1g/mol/NA
NA known as Avogadro's number (Avogadro's constant) is equal to 6.023×1023 atoms.
Atomic mass is particularly important when dealing with stoichiometry.
Practice Problems
1. What is the molecular mass of radium bicarbonate, Ra(HCO3)2?
2. List the following, from least to greatest, in terms of their number of neutrons, and then atomic mass: 14N, 42Cl, 25Na, 10Be
3. A new element, Zenium, has 3 isotopes, 59Ze, 61Ze, and 67Ze, with abundances of 62%, 27%, and 11% respectively. What is the atomic mass of Zenium?
4. An isotope with a mass number of 55 has 5 more neutrons than protons. What element is it?
5. How much mass does 3.71 moles of Fluorine have?
6. How many grams are there in 4.3 × 1022 molecules of POCl3?
7. How many moles are there in 23 grams of sodium carbonate?
Solutions
a) Molecular mass of Ra(HCO3)2
= 226 + 2(1.01 u + 12.01 u + (16.00 u)(3)) = 348 u or g/mol
b) Number of neutrons: 10Be, 14N, 25Na, 42Cl
Atomic Mass: 10Be, 14N, 25Na, 42Cl
Note: It is the same increasing order for both number of neutrons and atomic mass because more neutrons means more mass.
c) Atomic mass of Zenium:
(59 u)(0.62) + (61 u)(0.27) + (67 u)(0.11)
= 37 u + 16 u + 7.4 u
= 60.4 u or g/mol
d) Mn
e) (3.71 moles F2)(19 × 2 g/mol F2)
= (3.71 mol F2)(38 g/mol F2)
= 140 g F2
f) (4.3 × 1022 molecules POCI3)(1 mol/6.022 × 1023 molecules POCI3)(30.97 + 16.00 + 35.45 x 3 g/mol POCI3)
= (4.3 × 1022 molecules POCI3)(mol/6.022 × 1023 molecules POCI3)(153.32 g/mol POCI3)
= 11 g POCI3
g) (23 g Na2CO3)(1 mol/22.99 × 2 + 12.01 + 16.00 × 3 g Na2CO3)
= (23 g Na2CO3)(1 mol/105.99 g Na2CO3)
= (0.22 mol Na2CO3)
Contributors and Attributions
• Gunitika Dandona (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.03_Molar_Mass.txt |
`CHEMTUTOR`
Every chemist has dreamed that atoms were large enough to see and manipulate one at a time. The same chemist realizes after considering it, that if individual molecules were available for manipulation, it would take far too long to get anything done. The view from the atom is very different from the view of trillions and trillions of atoms. The mass action of the atoms that we see on our ‘macro’ view of the world is the result of the action of an incredibly large number of atoms averaged in their actions. The most usual way we count the atoms is by weighing them. The mass of material as weighed on a balance and the atomic weight of the material being weighed is the way we have of knowing how many atoms or molecules we are working with. Instead of counting eggs, we can count cartons of eggs, each carton of which has a given number, a dozen. Instead of counting B-B’s, we can count liters of B-B’s and find out how many B-B’s are in a liter. Instead of counting oats, we buy train cars of oats and know the number of oats in a full train car.
WHY DO WE NEED MOLS?
There are less than one hundred naturally occurring elements. Each element has a characteristic atomic weight. Most Periodic Charts include the atomic weight of an element in the box with the element. The atomic weight is usually not an integer because it is close to being the number of protons plus the average number of neutrons of an element. Let’s use the atomic weight as a number of grams. This will give us the same number of any atom we choose. If we weigh out 1.008 grams of hydrogen and 35.45 grams of chlorine and 24.3 grams of magnesium, we will have the same number of atoms of each one of these elements. The neat trick with this system is that we can weigh the atoms on a grand scale of number of atoms and get a count of them. This number of atoms that is the atomic weight expressed in grams is Avogadro’s number, 6.022 E 23. The name for Avogadro’s number of ANYTHING is a mole or mol. A mol of aluminum is 27.0 grams of aluminum atoms. Aluminum is an element, so the particles of aluminum are atoms. There are Avogadro’s number of aluminum atoms in 27.0 grams of it. But 1.008 grams of hydrogen is NOT a mol of hydrogen! Why not? Remember that hydrogen is one of the diatomic gases. There is really no such thing as loose hydrogen atoms. The total mass of a single hydrogen diatomic molecule (H2) is 2. 016 AMU. A mol of hydrogen gas has a mass of 2.016 grams. In that 2.016 gram mass is Avogadro’s number of H2 molecules because that is the way hydrogen comes. A mol of water is 18.016 grams because each water molecule has two hydrogen atoms and one oxygen atom. A mol of water has in it Avogadro’s number of water molecules. Another way to view the same thing is that a formula weight is the total mass of a formula in AMU expressed with units of grams per mol.
So Avogadro’s number is just a number, like dozen or gross or million or billion, but it is a very large number. You could consider a mol of sand grains or a mol of stars. We are more likely to speak of a mol of some chemical, for which we can find the mass of a mol of the material by adding the atomic weights of all the atoms in a formula of the chemical. The unit of atomic weight or formula weight is grams/mol.
The chemical formula of a material should tell you; (a) which elements are in the material, (b) how many atoms of each element are in the formula, (c) the total formula weight, and (d) how the elements are attached to each other. The symbols of the elements tell you which elements are in the material. The numbers to the right of each symbol tells how many atoms of that element are in the formula. The type of atoms and their arrangement in the formula will tell how the elements are attached to each other. A metal and a nonmetal or negative polyatomic ion shows an ionic compound. A pair of non-metals are bonded by covalent bonds. Some crystals have water of hydration loosely attached in the crystal. This is indicated by the dot such as in blue vitriol, Cu(SO4)·5H2O, showing five molecules of water of hydration to one formula of cupric sulfate.
The unit of the formula weight or molecular weight or atomic weight is "grams per mol," so it provides a relationship between mass in grams and mols of material.
nFw = m
'n' is the number of mols, 'Fw is the formula weight, and 'm' is the mass.
PERCENTS BY WEIGHT
All men weigh 200 pounds. All women weigh 125 pounds. What is the percent by weight of woman in married couples? A married couple is one man and one woman. The total weight is 325 pounds. The formula for percent is:
In this case the woman is the target.
125 # 325 # x 100% = 38.461% = 36.5%
Notice that the units of pound cancel to make the percent a pure number of comparison.
The weights of atoms are the atomic weights. What is the percentage of chloride in potassium chloride? The atomic weight of potassium is 39.10 g/mol. The atomic weight of chlorine is 35.45 g/mol. So the formula weight of potassium chloride is 74.55 g/mol. The chloride is the target and the potassium chloride is the total. 35.45 g/mol / x 100% = 47.55198 % or 47.6 % to three significant figures.
35.45g/mol
74.55g/mol
x 100% = 47.5520% = 47.6%
You can do that with any part of a compound. What is the percentage of sulfate in beryllium sulfate tetrahydrate?
Notice that the examples here are done to two decimal points of the atomic weights. The problems in the practice bunch at the end of this chapter are done to one decimal point of the atomic weight.
BASIC STOICHIOMETRY
Pronounce stoichiometry as “stoy-kee-ah-met-tree,” if you want to sound like you know what you are talking about, or “stoyk:,” if you want to sound like a real geek. Stoichiometry is just a five dollar idea dressed up in a fifty dollar name. You can compare the amounts of any materials in the same chemical equation using the formula weights and the coefficients of the materials in the equation. Let’s consider the equation for the Haber reaction, the combination of nitrogen gas and hydrogen gas to make ammonia.
N2 + 3 H2 2 NH3
The formula for nitrogen is N2 and the formula for hydrogen is H2. They are both diatomic gases. The formula for ammonia is NH3
The balanced equation requires one nitrogen molecule and three hydrogen molecules to make two ammonia molecules, meaning that one nitrogen molecule reacts with three hydrogen molecules to make two ammonia molecules or one MOL of nitrogen and three MOLS of hydrogen make two MOLS of ammonia. Now we are getting somewhere. The real way we measure amounts is by weight (actually, mass), so (to two significant digits) 28 grams (14 g/mol times two atoms of nitrogen per molecule) of nitrogen and 6 grams of hydrogen (1 g/mol times two atoms of hydrogen per molecule times three mols) make 34 grams of ammonia. Notice that no mass is lost or gained, since the formula weight for ammonia is 17 (one nitrogen at 14 and three hydrogens at one g/mol) and there are two mols of ammonia made. Once you have the mass proportions, any mass-mass stoichiometry can be done by good old proportionation. What is the likelihood you will get just a simple mass-mass stoich problem on your test? You should live so long. Well, you should get ONE.
Rather than thinking in terms of proportions, think in mols and mol ratios, a much more general and therefore more useful type of thinking. A mol ratio is just the ratio of one material in a chemical equation to another material in the same equation. The mol ratio uses the coefficients of the materials as they appear in the balanced chemical equation. What is the mol ratio of hydrogen to ammonia in the Haber equation? 3 mols of hydrogen to 2 mols of ammonia. Easy. In the standard stoichiometry calculations you should know, ALL ROADS LEAD TO MOLS. You can change any amount of any measurement of any material in the same equation with any other material in any measurement in the same equation. That is powerful. The setup is similar to Dimensional Analysis.
1. Start with what you know (GIVEN), expressing it as a fraction.
2. Use definitions or other information to change what you know to mols of that material.
3. Use the mol ratio to exchange mols of the material given to the mols of material you want to find.
4. Change the mols of material you are finding to whatever other measurement you need.
How many grams of ammonia can you make with 25 grams of hydrogen? (Practice your mol math rather than doing this by proportion. Check it by proportion in problems that permit it.)
You are given the mass of 25 grams of hydrogen. Start there.
25 g H2/1 Change to mols of hydrogen by the formula weight of hydrogen 1 mol of H2 = 2.0 g. (The 2.0 g goes in the denominator to cancel with the gram units in the material given.) Change mols of hydrogen to mols of ammonia by the mol ratio. 3 mols of hydrogen = 2 mols of ammonia. (The mols of hydrogen go in the denominator to cancel with the mols of hydrogen. You are now in units of mols of ammonia.) Convert the mols of ammonia to grams of ammonia by the formula weight of ammonia, 1 mol of ammonia = 17 g. (Now the mols go in the denominator to cancel with the mols of ammonia.) Cancel the units as you go.
The math on the calculator should be the last thing you do. 2 5 ÷ 2 . 0 x 2 ÷ 3 x 1 7 = and the number you get (141.66667) will be a number of grams of ammonia as the units in your calculations show. Round it to the number of significant digits your instructor requires (often three sig. figs.) and put into scientific notation if required. Most professors suggest that scientific notation be used if the answer is over one thousand or less than a thousandth. The answer is 142 grams of ammonia.
The calculator technique in the preceding paragraph illustrates a straightforward way to do the math. If you include all the numbers in order as they appear, you will have less chance of making an error. Many times students have been observed gathering all the numbers in the numerator, gathering all the numbers in the denominator, presenting a new fraction of the collected numbers, and then doing the division to find an answer. While this method is not wrong, the extra handling of the numbers has seen to produce many more errors.
See the Stoichiometry Roadmap for a way to consider this idea graphically. This example starts at "mass given" and goes throught the mol ratio to "mass find."
Notice by the chart above we may get the number of mols of material given if we change the mass by the formula weight, but in our continuous running math problem, we don’t have to stop and calculate a number of mols. Students who insist on doing so tend to get more calculator errors.
The more traditional formula for converting mols to mass would be, where Fw is the formula weight, m is the mass, and n is the number of mols: n x Fw = m. You should be able to "see" these formula relationships on the roadmap.
DENSITY TIMES MASS OF A PURE MATERIAL
Density multiplied by the volume of a pure material is equal to the mass of that material. If we know the density of a material and the volume of the pure material, with D = density and V = volume, DV = m so:
If you were given the density and volume of pure material you could calculate the volume of another material in that equation if you know it’s density. Notice that the density must be inverted to cancel the units properly if you want the volume to find. If you need to find the density, the volume must be inverted.
See the Stoichiometry Roadmap for a graphic view of this idea. Start with "Given density times volume of a pure material.
ATOMS OR MOLECULES TO MOLS
One of the hardest ideas for some students is that the individual particles of a material are a single one of a formula of that material. Copper element comes only in the form of atoms. Water only comes in the form of a molecule with one oxygen and two hydrogen atoms. A mol, then is Avogadro's number of individual particles of whatever type of pure material the substance is made. There is no such thing as a mol of mud because mud is a mixture. There is no one mud molecule.
The word "pure" also can be misunderstood. We do not mean that a material is one hundred percent the same material for us to use it, but that we are only considering the amount of that material.
The formula behind this relationship is: where n is the number of mols, A is Avogadro's number, and # is the number of individual particles of material,
A x n = #
Refer to the Stoichiometry Roadmap for a graphic view of this idea.
CONCENTRATION TIMES VOLUME OF A SOLUTION
A solution is a mixture of a fluid (often water, but not always) and another material mixed in with it. The material mixed in with it is called the solute. There is more on solutions in the chapter devoted to that. The volume of a solution, V, is measured the same way the volume of a pure liquid is measured. The concentration can be expressed in a number of ways, the most common in chemistry is the M, molar. One molar is one mol of solute in a liter of fluid. It is important to notice that the fluid is usually nothing more than a diluting agent. For most of the reactions, the fluid does not participate in any reaction.
Concentration times volume is number of mols of the solute material.
C x V = n
The "given" side of concentration times volume is easy. As with density times volume of a pure material, but the "find" side may need more work. You need one or the other of the concentration and volume before you can calculate the other. At the end of the Dimensional Analysis if you want concentration, you will be using the volume inverted. If you want the volume, you will be using the concentration inverted. This is not so difficult because the units will guide you.
Refer to the Stoichiometry Roadmap for a graphic view of this idea.
GASES
Standard temperature is zero degrees Celsius. Standard pressure is one atmosphere. A mol of ANY gas at standard temperature and pressure (STP) occupies 22.4 liters. That number is good to three significant digits. The equation would be 1 mol gas = 22.4 L @STP. The conversion factor, the Molar Volume of Gas, is 1 mol gas/22.4 L @STP or 22.4 L @STP/1 mol gas
Where n is the number of mols, V is the volume of a gas, and MVG is the molar volume of gas,
V = n x MVG
Gases not at STP will require the Ideal Gas Law Formula,
P V = n R T
where P is the pressure of the gas in atmospheres, V is the volume of the gas in liters, n is the number of mols of gas, T is the Kelvin temperature of the gas, and R is the "universal gas constant" with the measurement of 0.0821 liter-atmospheres per mol-degree. We will have to do some algebra on the PV = nRT gas equation to do the gas portion of the stoichiometry problems.
In GIVEN we only need to solve for n. n = PV/RT. If we need to find the volume, pressure, or temperature of a gas, we need to solve for the unknown and include the "mols find" as the n. More about gases later. See the Chemtutor section on Gases for math problems using the gas laws.
The earmarks of a stoichiometry problem are: There is a reaction. (A new material is made.) You know the amount of one material and you are asked to calculate the amount of another matierial in the same equation.
HOW TO USE THE "ROADMAP" FOR SOLVING CHEMISTRY PROBLEMS
1. Write all the compounds and elements in the problem correctly.
2. Write the balanced chemical equation for the problem.
3. Write the MATERIAL you have enough information about to use as GIVEN. (This has been one of the major stumbling blocks in using the roadmap.) If you know the number of moles, the mass, or the number of molecules of a material, you have all you need to start the problem. You need CONCENTRATION AND VOLUME of a solution to have the amount of solute that reacts. You need VOLUME AND DENSITY of a solid or liquid to have an amount of that. You need VOLUME, PRESSURE AND TEMPERATURE of a gas to have a complete set of information. (Notice it is useful to understand the properties of the states of matter as you do this.)
4. Write what you need to FIND and all the other pertinent information about that material. For instance, if you need to find the volume of a gas, you must also list the pressure and temperature of that gas in FIND. In this manner: FIND V, volume of gas at 79 °C and 1.8 atm.
5. Sketch out an outline of the math according to the roadmap. You know there are some points in the roadmap that you miss on the outline because they are calculated in the process, for instance if you are given a mass of one material and asked to find the density of another material with its volume, you would start at the MASS GIVEN and use the FORMULA WEIGHT to get to the MOLES GIVEN, but MOLES GIVEN does not appear in the outline because it is already calculated. You next need the MOLE RATIO to get the MOLES FIND. Again, MOLES FIND does not appear in the outline.
6. Fill in the outline with the numbers, units and materials (for instance, 15 kg Mg) and do the calculations. Be careful of numbers that need to be inverted. You can tell the coefficients that need to be inverted by the units.
STOICHIOMETRY ROADMAP
One of the really nice things about the Stoichiometry Roadmap is that once you understand it thoroughly, it can be carried around with you between your ears. Just remember that ALL ROADS LEAD TO MOLS.
MOLE AND PERCENT WORKSHEET
1. How many pennies are in a mole of pennies? How many thousand-dollar bills (k-notes!) is that mole of pennies equal to?
2. NO2 is the molecular formula for nitrous dioxide (also known as nitrogen dioxide). List the information available to you from this formula?.
3. C2H2 is the molecular formula for ethylyne (A.K.A. acetylene). (a) How many atoms are in one molecule? (b) Which atoms make up acetylene? (c) How many moles of atoms are in one molecule of acetylene? (d) How many molecules are in 5.3 moles of acetylene? (e) How many atoms are in a mole of acetylene?
4. Calculate the molar mass of a mole of the following materials: (a) Al (b) Ra (c) Co (d) CO (e) CO2 (f) HCl (g) Na2CO3 (h) Ca(NO3)2(i) (NH4)3(PO4) (j) H2O (k) Epsom salts - Mg(SO4)·7H2O (m) blue vitriol - Cu(SO4)·5H2O ?
5. Calculate the number of moles in: (a) 2.3 # of carbon (b) 0.014 g of Tin (c) a 5 Oz silver bracelet (d) a pound of table salt (e) a 350 kg cast iron engine block (f) a gal. of water (8.3 #) (g) a ton of sand (SiO2) (h) 6.2 grams of blue vitriol (i) a pound of Epsom salts ?
6. Calculate the number of atoms in: (a) 100 g of Argon (b) 1.21 kg aluminum foil (c) a 28 # lead brick (d) the E7 kg of water in an olympic swimming pool (e) 7 kg of hydrogen gas (f) a tonne of calcium nitrate ?
7. What is the percentage composition of oxygen in each of the following materials: (a) CO (b) CO2 (c) (NO3)- (d) isopropyl alcohol C3H8O (e) calcium nitrate (f) blue vitriol - Cu(SO4)·5H2O ?
8. What is the percentage composition of phosphate in each of the following materials: (a) phosphoric acid (b) sodium carbonate (c) ammonium phosphate (d) calcium phosphate ?
9. What is the percentage composition of sulfate in each of the following materials: (a) sulfuric acid (b) sodium sulfate (c) Epsom salts ( d) aluminum sulfate ?
ANSWERS TO MOL AND PERCENT PROBLEMS
1a. 6.023 E23 pennies 1b. 6.023 E18 k-Notes 2a. Covalent
2b. Elements in it (N and O) 2c. Number of atoms of each element
3a. 4 3b. C & H 3c. 6.64 E-24 3d. 3.1922 E24 3e. 2.4092 E24
4a. 27.0 4b. 226.0 4c. 58.9 4d. 28.0 4e. 44.0
4f. 36.5 4g. 106.0 4h. 164.1 4i. 149.0 4j. 18.0
4k. 246.4 4m. 249.6 5a. 86.9 5b. 1.18 E-4 5c. 1.31
5d. 7.75 5e. 6.27 E3 5f. 210 5g. 1.51 E4 5h. 0.0248
5i. 1.84 6a. 1.51 E24 6b. 2.69 E25 6c. 3.69 E25 6d. 1.00 E33
6e. 4.22E27 6f. 3.30 E28 7a. 57.1% 7b. 72.7% 7c. 77.4%
7d. 26.7% 7e. 58.5% 7f. 57.7% 8a. 96.9% 8b. 0%
8c. 63.8% 8d. 61.2% 9a. 98.0% 9b. 67.6% 9c. 39.0%
9d. 84.2%
STP GAS AND MASS STOICHIOMETRY PROBLEMS (PRELIMINARY TO GAS LAW)
All of the problems below are stoichiometry problems with at least one equation participant as a gas at STP. (a) Write and balance the chemical equation. (2) Do the math in DA style using 1 mole gas at STP = 22.4 liters as a factor. In the following problems ALL GASES ARE AT STP. Click here for a general idea of how to do the problems in this set.
1. How many moles of nitrogen gas is needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?
2. How many liters of ammonia are produced when 89.6 liters of hydrogen are used in the above reaction?
3. Ten grams of calcium carbonate was produced when carbon dioxide was added to lime water (calcium hydroxide in solution). What volume of carbon dioxide at STP was needed?
4. When 11.2 liters of hydrogen gas is made by adding zinc to sulfuric acid, what mass of zinc is needed?
5. What volume of ammonia at STP is needed to add to water to produce 11 moles of ammonia water?
6. How many grams of carbonic acid is produced when 55 liters of carbon dioxide is pressed into water?
7. magnesium hydroxide + ammonium sulfate magnesium sulfate + water + ammonia
1. How much (grams) magnesium hydroxide do you need to use in the above reaction to produce 500 liters of ammonia?
8. How much strontium bromide is needed to add to chlorine gas to produce 75 liters of bromine?
9. What mass of ammonium chlorate is needed to decompose to give off 200 liters of oxygen?
10. Your car burns mostly octane, C8H18, as a fuel. How many liters of oxygen is needed to burn a kilogram of octane?
11. copper + sulfuric acid copper II sulfate + water + sulfur dioxide
1. How many moles of copper are needed to produce 1000 L of SO2?
12. What volume of oxygen is needed to burn a pound of magnesium?
13. How many grams of sodium do you have to put into water to make 30 liters of hydrogen at STP?
14. ammonia gas and hydrogen chloride gas combine to make ammonium chloride.
1. What volume of ammonia at STP is needed to react with 47.7 liters of hydrogen chloride at STP?
15. How many liters of oxygen are needed to burn 10 liters of acetylene?
ANSWERS TO STP GAS AND MASS STOICHIOMETRY PROBLEMS
1. 0.667 mol 2. 59.7 L 3. 2.24 L 4. 32.7 g
5. 246 L 6. 152 g 7. 651 g 8. 828 g
9. 604 g 10. 2.46 kL 11. 44.6 mol 12. 210 L
13. 61.6 g 14. 47.7 L 15. 25 L
PROBLEMS ON CONCENTRATION AND DENSITY
WRITE AND BALANCE THE CHEMICAL EQUATION FOR THOSE PROBLEMS THAT NEED IT. SHOW ALL YOUR WORK. USE W5P OR DA METHOD ACCORDING TO THE ROADMAP.
1. The lead brick on my desk measures 3 by 5 by 11 cm. Lead has a density of 11.34 g/cc. How many lead atoms are in that block?
2. The lab technician at the Planter's Peanut packing factory takes a bag of peanuts, puts water into it to dissolve the salt, and dilutes the solution to one liter. She then takes ten ml of that solution and titrates it against 0.132 M silver nitrate. One bag sample takes 31.5 ml of silver nitrate to endpoint. What mass of salt was in the bag?
3. What is the concentration of sugar (C12H22O11) if twenty grams are dissolved in enough water to make 2 liters?
4. Methyl alcohol (CH3OH) has a density of 0.793 kg/l. What volume of it is needed to add to water to make five liters of 0.25 M solution?
5. Magnesium has a density of 1.741 g/cc. What volume of Mg will burn in 20 liters of oxygen at 2.1 atm and 0°C?
6. Uranium metal can be purified from uranium hexafluoride by adding calcium metal. Calcium metal has a density of 1.54 g/cc. Uranium has a density of 18.7 g/cc. What mass of uranium do you get for a Kg of Ca? What volume of uranium do you get for a cubic meter of calcium?
7. What volume of 0.27 M sodium hydroxide is needed to react with 29.5 ml of 0.55 M phosphoric acid?
8. What volume of carbon dioxide is produced at 1 atm and 87 °C when 1.6 liters of methyl alcohol burns? What volume of liquid water is produced in this reaction?
9. Seven kilograms of mercury II oxide decomposes into mercury and oxygen. Mercury has a density of 13.6 g/cc/ What volume of mercury is produced?
10. Water and calcium oxide produce calcium hydroxide. How many grams of calcium hydroxide are made if you add 275 liters of water to enough calcium oxide?
11. Gasoline (C7H16) has a density of 0.685 kg/liter. How many liters of oxygen at 37 °C and 950 mmHg are needed to burn 15 liters of gasoline?
12. Sodium hydroxide and hydrochloric acid combine to make table salt and water. 14 mL of 0.1 M sodium hydroxide is added to an excess of acid. How many moles of table salt are made? How many grams of salt is that?
13. 50 mL of 0.25 M copper II sulfate evaporates to leave CuSO4·5H2O. (That is the pentahydrate crystal of copper II sulfate.) What is the mass of this beautiful blue crystal from the solution?
14. Chlorine gas is bubbled into 100 mL of 0.25 M potassium bromide solution. This produces potassium chloride and bromine gas. The bromine (which dissolves in water) is taken from the solution and measured at 27 °C and 825 mmHg. What is the volume of bromine?
15. 95.0 mL of 0.55 M sulfuric acid is put on an excess of zinc. This produces zinc sulfate and hydrogen. How many grams of zinc sulfate are made?
16. and some dissolved sodium nitrate. (a) How many moles of silver chloride are made? (b) How many grams of silver chloride is that? (c) How many moles of sodium nitrate are made? (d) What is the concentration of sodium nitrate in the final solution?
17. How many grams of potassium permanganate, KMnO4, is needed to make 1.72 liters of 0.29 M solution?
18. By my calculations, a drop of ethyl alcohol, C2H5OH , in an olympic-sized swimming pool produces a 1.20 E-10 M solution of alcohol in water. A drop is a twentieth of a mL. How many molecules of ethyl alcohol are in a drop of the water in the pool?
19. 93.0 mL of 0.150 M magnesium hydroxide is added to 57.0 mL of 0.4 M nitric acid. (Magnesium nitrate and water are formed. What is the concentration of the magnesium nitrate after the reaction?
ANSWERS TO PROBLEMS ON CONCENTRATION AND DENSITY
1. 5.44 E24 atoms 2. 24.3 g 3. 0.0292 M 4. 0.0504 L
5. 52.3 ml(cc) Mg 6a. 1.98 kg of U 6b. 1.63 E6 mL 7. 180 mL
8a. 1.17 kL CO2 8b. 1.43 L 9. 0.477 L 10. 1.13 E 6 g
11. 23.0 kL 12a. 1.4 E-3 mols 12b. 0.0819 g 13. 3.12 g
14. 284 mL 15. 8.44 g 16a. 5.24E-3 mol 16b. 0.752 g
16c. 5.24E-3 mols 16d. 122 mmolar 17. 78.8 mg 18. 3.61E9 molecules
19. 0.152 M
PROBLEMS USING COMPLETE ROADMAP
1. How many liters of ammonia at 0 °C and 25 atm. are produced when 10 g of hydrogen is combined with nitrogen?
2. How many milliliters of hydrogen at 0 deg C and 1400 mmHg are made if magnesium reacts with 15 mL of 6 M sulfuric acid?
3. How many atoms are in 25 liters of fluorine gas at 2.85 atm and 450 °C?
4. Liquid butane (C4H10) has a density of 0.60 g/cc. It burns to make carbon dioxide at 120 °C. What volume of carbon dioxide is produced at one atm when 350 liters of liquid butane burns?
5. Isopropyl alcohol, C3H7OH , makes a good fuel for cars. What volume of oxygen at 785 mmHg and 23 °C is needed to burn 8.54 E25 molecules of isopropyl alcohol?
6. How many moles of NaCl are in a liter of a 0.15 M NaCl solution? (0.15 M NaCl is physiological saline when sterilized.)
7. How many grams of NaCl must you put into a 50 liter container to make a physiological saline solution?
8. Chlorine gas is bubbled into 100 mL of 0.25 M potassium bromide solution. This produces potassium chloride and bromine gas. The bromine dissolves completely in the water. What is the concentration of bromine?
9. 95 mL of 0.55 M sulfuric acid is put on an excess of zinc. This produces zinc sulfate and hydrogen. How many grams of zinc sulfate are made?
10. Methyl alcohol (CH3OH) has a density of 0.793 Kg/L. What volume of it is needed to add to water to make twenty-five liters of 0.15 M solution?
11. Magnesium has a density of 1.741 g/cc. What volume of Mg will burn to produce a kilogram of magnesium oxide?
12. What volume of water vapor is produced at 716 mmHg and 87°C when 2.6 liters of methyl alcohol burns?
ANSWERS TO PROBLEMS USING COMPLETE ROADMAP
1. 2.99 L 2. 1.10 E3 mL 3. 1.45 E24 atoms 4. 4.67 E5 L
5. 1.50 E4 L 6. 0.15 moles 7. 439 g 8. 0.125 M
9. 8.44 g 10. 151 mL 11. 0.346 L 12. 1.29 E5 L
Contributors
• David Wilner
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Chemistry is the experimental and theoretical study of materials on their properties at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is also a hot pursuit. Chemistry is traditionally divided into organic and inorganic chemistry. The former is the study of compounds containing at least one carbon-hydrogen bonds. By default, the chemical study of all other substances is called inorganic chemistry, a less well defined subject.
However, the boundary between organic and inorganic compounds is not always well defined. For example, oxalic acid, H2C2O4, is a compound formed in plants, and it is generally considered an organic acid, but it does not contain any C-H bond. Inorganic chemistry is also closely related to other disciplines such as materials sciences, physical chemistry, thermodynamics, earth sciences, mineralogy, crystallography, spectroscopy etc.
A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. This notation can be accredited to Swedish chemist Jons Jakob Berzeliu. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic compounds. Compounds that do not pertain to this rule are called inorganic compounds.
Molecular Geometry and Structural Formula
Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules can have the same number of atoms, and thus the same molecular formula, but can have completely different physical and chemical properties because of differences in structural formula.
Methylpropane and butane have the same molecular formula of C4H10, but are structurally different (methylpropane on the left, butane on the right).
Polymers
A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, square parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer).
Ethylene becomes the polymer polyethylene.
Molecular Formula
The molecular formula is based on the actual makeup of the compound. Although the molecular formula can sometimes be the same as the empirical formula, molecular compounds tend to be more helpful. However, they do not describe how the atoms are put together. Molecular compounds are also misleading when dealing with isomers, which have the same number and types of atoms (see above in molecular geometry and structural formula).
Ex. Molecular Formula for Ethanol: C2H6O.
Empirical Formula
An empirical formula shows the most basic form of a compound. Empirical formulas show the number of atoms of each element in a compound in the most simplified state using whole numbers. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula.
Ex. Find the empirical formula for C8H16O2.
Answer: C4H8O (divide all subscripts by 2 to get the smallest, whole number ratio).
Structural Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent.
Ex. Structural Formula for Ethanol:
Condensed Structural Formula
Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for Ethanol: CH3CH2OH (Molecular Formula for Ethanol C2H6O).
Line-Angle Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Ex. Line-Angle Formula for Ethanol:
Formulas of Inorganic Compounds
Inorganic compounds are typically not of biological origin. Inorganic compounds are made up of atoms connected using ionic bonds. These inorganic compounds can be binary compounds, binary acids, or polyatomic ions.
Binary compounds
Binary compounds are formed between two elements, either a metal paired with a nonmetal or two nonmetals paired together. When a metal is paired with a nonmetal, they form ionic compounds in which one is a negatively charged ion and the other is positvely charged. The net charge of the compound must then become neutral. Transition metals have different charges; therefore, it is important to specify what type of ion it is during the naming of the compound. When two nonmetals are paired together, the compound is a molecular compound. When writing out the formula, the element with a positive oxidation state is placed first.
Ex. Ionic Compound: BaBr2(Barium Bromide)
Ex. Molecular Compound: N2O4 (Dinitrogen Tetroxide)
Binary acids
Binary acids are binary compounds in which hydrogen bonds with a nonmetal forming an acid. However, there are exceptions such as NH3, which is a base. This is because it shows no tendency to produce a H+. Because hydrogen is positively charged, it is placed first when writing out these binary acids.
Ex. HBr (Hydrobromic Acid)
Polyatomic ions
Polyatomic ions is formed when two or more atoms are connected with covalent bonds. Cations are ions that have are postively charged, while anions are negatively charged ions. The most common polyatomic ions that exists are those of anions. The two main polyatomic cations are Ammonium and Mercury (I). Many polyatomic ions are typically paired with metals using ionic bonds to form chemical compounds.
Ex. MnO4- (Polyatomic ion); NaMnO4 (Chemical Compound)
Oxoacids
Many acids have three different elements to form ternary compounds. When one of those three elements is oxygen, the acid is known as a oxoacid. In other words, oxacids are compounds that contain hydrogen, oxgygen, and one other element.
Ex. HNO3 (Nitric Acid)
Complex Compounds
Certain compounds can appear in multiple forms yet mean the same thing. A common example is hydrates: water molecules bond to another compound or element. When this happens, a dot is shown between H2O and the other part of the compound. Because the H2O molecules are embedded within the compound, the compound is not necessarily "wet". When hydrates are heated, the water in the compound evaporates and the compound becomes anhydrous. These compounds can be used to attract water such as CoCl2. When CoCl2 is dry, CoCl2 is a blue color wherease the hexahydrate (written below) is pink in color.
Ex. CoCl2 ·6 H2O
Formulas of Organic Compounds
Organic compounds contain a combination carbon and hydrogen or carbon and hydrogen with nitrogen and a few other elements, such as phosphorous, sulfur, silicon, and the halogens. Most organic compounds are seen in biological origin, as they are found in nature.
Hydrocarbons
Hydrocarbons are compounds that consist of only carbon and hydrogen atoms. Hydrocarbons that are bonded together with only single bonds are alkanes. The simplest example is methane (shown below). When hydrocarbons have one or more double bonds, they are called alkenes. The simplest alkene is Ethene (C2H4) which contains a double bond between the two carbon atoms.
Ex. Methane on left, Ethene on right
Functional Groups
Functional groups are atoms connected to carbon chains or rings of organic molecules. Compounds that are within a functional group tend to have similar properties and characteristics. Two common functional groups are hydroxyl groups and carboxyl groups. Hydroxyl groups end in -OH and are alcohols. Carboxyl groups end in -COOH, making compounds containing -COOH carboxylic acids. Functional groups also help with nomenclature by using prefixes to help name the compounds that have similar chemical properties.
Ex. Hydroxyl Group on top; Carboxyl Group on bottom
Problems
1. Which of the following formulas are organic?
1. HClO
2. C5H10
3. CO2
2. What is the name of the following formula?
1. Classify the following formulas into their appropriate functional group
1. Acetic acid
2. Butanol
3. Oxalic acid
1. What are the empirical formulas for the following compounds?
1. C12H10O6
2. CH3CH2CH2CH2CH2CH2CH3
3. H3O
1. What is the name of the following figure and what is the molecular formula of the following figure?
Answer Key:
1. b and c. 2. Propane. 3. a. carboxyl group, b. hydroxyl group, c. carboxyl group. 4. a. C6H5O3, b. C7H16, c. H3O. 5. Methylbutane, C5H12
Contributors and Attributions
• Jean Kim (UCD), Kristina Bonnett (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.06_Determining_the_Formula_of_a_Compound.txt |
Learning Objectives
• To describe a chemical reaction.
• To calculate the quantities of compounds produced or consumed in a chemical reaction
What happens to matter when it undergoes chemical changes? The Law of conservation of mass says that "Atoms are neither created, nor destroyed, during any chemical reaction." Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements").
Chemical Equations
As shown in Figure $1$, applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. This reaction is described with a chemical equation, an expression that gives the identities and quantities of the substances in a chemical reaction.
Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H2O). The chemical equation for this reaction is written as:
$\ce{2H_2 + O_2 \rightarrow 2H_2O} \nonumber$
Chemical formulas and other symbols are used to indicate the starting materials, or reactants, which by convention are written on the left side of the equation, and the final compounds, or products, which are written on the right. An arrow points from the reactant to the products. The chemical reaction for the ammonium dichromate volcano in Figure $1$ is
$\underbrace{\ce{(NH_4)_2Cr_2O_7}}_{ reactant } \rightarrow \underbrace{\ce{Cr_2O_3 + N_2 + 4H_2O}}_{products }\label{3.1.1}$
The arrow is read as “yields” or “reacts to form.” Equation $\ref{3.1.1}$ indicates that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products). The equation for this reaction is even more informative when written as follows:
$\ce{ (NH4)2Cr2O7(s) \rightarrow Cr2O3(s) + N2(g) + 4H2O(g)} \label{3.1.2}$
Equation $\ref{3.1.2}$ is identical to Equation $\ref{3.1.1}$ except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water.
Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of Equations $\ref{3.1.1}$ and $\ref{3.1.2}$. Each side of the reaction has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms.
In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equations $\ref{3.1.1}$ and $\ref{3.1.2}$ are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms, but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.
A chemical reaction changes only the distribution of atoms, not the number of atoms.
Introduction to Chemical Reaction Equations: Introduction to Chemical Reaction Equations, YouTube(opens in new window) [youtu.be]
Balancing Simple Chemical Equations
When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane ($C_7H_{16}$), an important component of gasoline:
$\ce{C_7H_{16} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.3}$
The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water.
Equation $\ref{3.1.3}$ is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, the coefficients of the reactants and products must be adjusted to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $3$.
Balancing Combustion Reactions: Balancing Combustions Reactions, YouTube(opens in new window) [youtu.be]
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation
1. Identify the most complex substance.
2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
3. Balance polyatomic ions (if present) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Check your work by counting the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
Example $\PageIndex{1A}$: Combustion of Heptane
To demonstrate this approach, let’s use the combustion of n-heptane (Equation $\ref{3.1.3}$) as an example.
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $\ce{C_7H_{16}}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:
$\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O } \label{3.1.4}$
1. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
2. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: $\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O} \label{3.1.5}$ The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:$\ce{C_7H_{16} (l) + 11O_2 (g) \rightarrow 7CO_2 (g) + 8H_2O (g)} \label{3.1.6}$
3. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start.
Example $\PageIndex{1B}$: Combustion of Isooctane
Consider, for example, a similar reaction, the combustion of isooctane ($\ce{C8H18}$). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows:
$\ce{C_8H_{18} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.7}$
1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane.
2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + H_2O} \label{3.1.8}$
3. Balance polyatomic ions as a unit. This step does not apply to this equation.
4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + 9H_2O } \label{3.1.9}$The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: $\ce{C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O} \label{3.1.10}$Equation $\ref{3.1.10}$ is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: $\ce{2C_8H_{18} (l) + 25O_2 (g) \rightarrow 16CO_2 (g) + 18H_2O (g) }\label{3.11}$
5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.
Balancing Complex Chemical Equations: Balancing Complex Chemical Equations, YouTube(opens in new window) [youtu.be]
Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
Example $\PageIndex{1C}$: Hydroxyapatite
The reaction of the mineral hydroxyapatite ($\ce{Ca5(PO4)3(OH)}$) with phosphoric acid and water gives $\ce{Ca(H2PO4)2•H2O}$ (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.
Given: reactants and product
Asked for: balanced chemical equation
Strategy:
1. Identify the product and the reactants and then write the unbalanced chemical equation.
2. Follow the steps for balancing a chemical equation.
Solution:
A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem (recall that phosphoric acid is H3PO4). The initial (unbalanced) equation is as follows:
$\ce{ Ca5(PO4)3(OH)(s) + H_3PO4 (aq) + H_2O_{(l)} \rightarrow Ca(H_2PO_4)_2 \cdot H_2O_{(s)} } \nonumber$
1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, $\ce{Ca5(PO4)3(OH)}$, appears in the balanced chemical equation.
2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of $\ce{Ca5(PO4)3(OH)}$ contains 5 calcium atoms, so we need 5 Ca(H2PO4)2•H2O on the right side:
$\ce{Ca5(PO4)3(OH) + H3PO4 + H2O \rightarrow 5Ca(H2PO4)2 \cdot H2O} \nonumber$
3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43), shows up in three places. In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2 • H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4:
$\ce{Ca_5(PO_4)_3(OH) + 7H_3PO_4 + H_2O \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O } \nonumber$
Although OH is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately.
4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules:
$\ce{Ca_5(PO_4)_3(OH) (s) + 7H_3PO_4 (aq) + 4H_2O (l) \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O (s) } \nonumber$
The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 41 oxygen atoms on each side.
5. Check your work. Both sides of the equation contain 5 calcium atoms, 10 phosphorus atoms, 30 hydrogen atoms, and 41 oxygen atoms.
Exercise $1$: Fermentation
Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol ($CH_3CH_2OH$ and carbon dioxide $CO_2$. Write a balanced chemical reaction for the fermentation of glucose.
Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.
Answer
$C_6H_{12}O_6(s) \rightarrow 2C_2H_5OH(l) + 2CO_2(g) \nonumber$
Balancing Reactions Which Contain Polyatomics: Balancing Reactions Which Contain Polyatomics, YouTube(opens in new window) [youtu.be]
Interpreting Chemical Equations
In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it gives the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficient of that species (e.g., the 4 preceding H2O in Equation $\ref{3.1.1}$). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure $4$, the coefficients allow Equation $\ref{3.1.1}$ to be interpreted in any of the following ways:
• Two NH4+ ions and one Cr2O72 ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules.
• One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O.
• A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O.
• A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O.
These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of $H_2O$ to $N_2$ in Equation $\ref{3.1.1}$ is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:
$252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \nonumber$
yield
$152 + 28 + 72 = 252 \; g \; \text{of products.} \nonumber$
The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.
An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose ($C_6H_{12}O_6$), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.
Example $2$: Combustion of Glucose
The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:
$\ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)} \nonumber$
Construct a table showing how to interpret the information in this equation in terms of
1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.
Given: balanced chemical equation
Asked for: molecule, mole, and mass relationships
Strategy:
1. Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.
2. Use the molar masses of the reactants and products to convert from moles to grams.
3. Use Avogadro’s number to convert from moles to the number of molecules.
Solution:
This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.
1. One molecule of glucose reacts with 6 molecules of O2 to yield 6 molecules of CO2 and 6 molecules of H2O.
2. One mole of glucose reacts with 6 mol of O2 to yield 6 mol of CO2 and 6 mol of H2O.
3. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O2, 31.9988; CO2, 44.010; and H2O, 18.015.
\begin{align*} \text{mass of reactants} &= \text{mass of products} \[4pt] g \, glucose + g \, O_2 &= g \, CO_2 + g \, H_2O \end{align*} \nonumber
$1\,mol\,glucose \left ( {180.16 \, g \over 1 \, mol \, glucose } \right ) + 6 \, mol \, O_2 \left ( { 31.9988 \, g \over 1 \, mol \, O_2} \right ) \nonumber$
$= 6 \, mol \, CO_2 \left ( {44.010 \, g \over 1 \, mol \, CO_2} \right ) + 6 \, mol \, H_2O \left ( {18.015 \, g \over 1 \, mol \, H_2O} \right ) \nonumber$
$372.15 \, g = 372.15 \, g \nonumber$
C One mole of glucose contains Avogadro’s number (6.022 × 1023) of glucose molecules. Thus 6.022 × 1023 glucose molecules react with (6 × 6.022 × 1023) = 3.613 × 1024 oxygen molecules to yield (6 × 6.022 × 1023) = 3.613 × 1024 molecules each of CO2 and H2O.
In tabular form:
Solution to Example 3.1.2
$C_6H_{12}O_{6\;(s)}$ + $6O_{2\;(g)}$ $6CO_{2\;(g)}$ $6H_2O_{(l)}$
a. 1 molecule 6 molecules 6 molecules 6 molecules
b. 1 mol 6 mol 6 mol 6 mol
c. 180.16 g 191.9928 g 264.06 g 108.09 g
d. 6.022 × 1023 molecules 3.613 × 1024 molecules 3.613 × 1024 molecules 3.613 × 1024 molecule
Exercise $2$: Ammonium Nitrate Explosion
Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas.
The explosion resulted from the following reaction:
$2NH_4NO_{3\;(s)} \rightarrow 2N_{2\;(g)} + 4H_2O_{(g)} + O_{2\;(g)} \nonumber$
Construct a table showing how to interpret the information in the equation in terms of
1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
Answer:
Answer to Exercise 3.1.2
$2NH_4NO_{3\;(s)}$ $2N_{2\;(g)}$ + $4H_2O_{(g)}$ + $O_{2\;(g)}$
a. 2NH4+ ions and 2NO3 ions 2 molecules 4 molecules 1 molecule
b. 2 mol 2 mol 4 mol 1 mol
c. 160.0864 g 56.0268 g 72.0608 g 31.9988 g
d. 1.204 × 1024 formula units 1.204 × 1024 molecules 2.409 × 1024 molecules 6.022 × 1023 molecules
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]
Summary
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.07_Chemical_Equations.txt |
Learning Objectives
• To balance equations that describe reactions in solution.
• To calculate the quantities of compounds produced or consumed in a chemical reaction.
• To solve quantitative problems involving the stoichiometry of reactions in solution.
A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation. This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?
All these questions can be answered using the concepts of the mole, molar and formula masses, and solution concentrations, along with the coefficients in the appropriate balanced chemical equation.
Stoichiometry Problems
When carrying out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.
Steps in Converting between Masses of Reactant and Product
1. Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
2. From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
3. Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.
Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).
To illustrate this procedure, consider the combustion of glucose. Glucose reacts with oxygen to produce carbon dioxide and water:
$C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{3.6.1}$
Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?
The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined above can be adapted as follows:
1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:
$moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \nonumber$
2. According to the balanced chemical equation, 6 mol of CO2 is produced per mole of glucose; the mole ratio of CO2 to glucose is therefore 6:1. The number of moles of CO2 produced is thus
$moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber$
$= 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber$
$= 1.51 \, mol \, CO_2 \nonumber$
3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:
$mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \nonumber$
These operations can be summarized as follows:
$45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \nonumber$
Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (Remember that you should generally carry extra significant digits through a multistep calculation to the end to avoid this!) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.
The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example $1$, the mass of one reactant that is required to consume a given mass of another reactant.
Example $1$: The US Space Shuttle
The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).
The US space shuttle Discovery during liftoff. The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.
Given: reactants, products, and mass of one reactant
Asked for: mass of other reactant
Strategy:
1. Write the balanced chemical equation for the reaction.
2. Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.
Solution:
We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.
A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:
$H_2 (g) + O_2 (g) \rightarrow H_2O (g) \nonumber$
This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H2O and H2 gives the balanced chemical equation:
$2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \nonumber$
Thus 2 mol of H2 react with 1 mol of O2 to produce 2 mol of H2O.
1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:
$mass \, of \, O_2 = 1.00 \, tn \times { 2000 \, lb \over tn} \times {453.6 \, g \over lb} = 9.07 \times 10^5 \, g \, O_2 \nonumber$
Using the molar mass of O2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O2 contained in this mass of O2:
$mol \, O_2 = 9.07 \times 10^5 \, g \, O_2 \times {1 \, mol \, O_2 \over 32.00 \, g \, O_2} = 2.83 \times 10^4 \, mol \, O_2 \nonumber$
2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H2 needed to react with this number of moles of O2:
$mol \, H_2 = mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} \nonumber$
$= 2.83 \times 10^4 \, mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} = 5.66 \times 10^4 \, mol \, H_2 \nonumber$
3. The molar mass of H2 (2.016 g/mol) allows us to calculate the corresponding mass of H2:
$mass \, of \, H_2 = 5.66 \times 10^4 \, mol \, H_2 \times {2.016 \, g \, H_2 \over mol \, H_2} = 1.14 \times 10^5 \, g \, H_2 \nonumber$
Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:
$tons \, H_2 = 1.14 \times 10^5 \, g \, H_2 \times {1 \, lb \over 453.6 \, g} \times {1 \, tn \over 2000 \, lb} = 0.126 \, tn \, H_2 \nonumber$
The space shuttle had to be designed to carry 0.126 tn of H2 for each 1.00 tn of O2. Even though 2 mol of H2 are needed to react with each mole of O2, the molar mass of H2 is so much smaller than that of O2 that only a relatively small mass of H2 is needed compared to the mass of O2.
Exercise $1$: Roasting Cinnabar
Cinnabar, (or Cinnabarite) $HgS$ is the common ore of mercury. Because of its mercury content, cinnabar can be toxic to human beings; however, because of its red color, it has also been used since ancient times as a pigment.
Alchemists produced elemental mercury by roasting cinnabar ore in air:
$HgS (s) + O_2 (g) \rightarrow Hg (l) + SO_2 (g) \nonumber$
The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?
Answer
86.2 g
Calculating Moles from Volume
Quantitative calculations involving reactions in solution are carried out with masses, however, volumes of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. An expanded version of the flowchart for stoichiometric calculations is shown in Figure $2$. The balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used to determine the amounts of other species, as illustrated in the following examples.
The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.
Example $2$ : Extraction of Gold
Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)2] ion. Gold is then recovered by reduction with metallic zinc according to the following equation:
$Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \nonumber$
What mass of gold can be recovered from 400.0 L of a 3.30 × 10−4 M solution of [Au(CN)2]?
Given: chemical equation and molarity and volume of reactant
Asked for: mass of product
Strategy:
1. Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)2] present by multiplying the volume of the solution by its concentration.
2. From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.
Solution:
A The equation is balanced as written; proceed to the stoichiometric calculation. Figure $2$ is adapted for this particular problem as follows:
As indicated in the strategy, start by calculating the number of moles of [Au(CN)2] present in the solution from the volume and concentration of the [Au(CN)2] solution:
\begin{align} moles\: [Au(CN)_2 ]^- & = V_L M_{mol/L} \ & = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align}
B Because the coefficients of gold and the [Au(CN)2] ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)2] (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:
\begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \ &= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}
At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth$1170.
$26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\1400} {1\: \cancel{troy\: oz\: Au}} = \1170$
Exercise $2$ : Lanthanum Oxalate
What mass of solid lanthanum(III) oxalate nonahydrate [La2(C2O4)3•9H2O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl3 by adding a stoichiometric amount of sodium oxalate?
Answer
3.89 g
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]
Summary
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.08_Balancing_Chemical_Equations.txt |
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions.
Balancing
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients.
Reactants to Products
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium ($Na$), hydrogen ($H$), and chloride ($Cl$) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.
Stoichiometric Coefficients
In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:
$\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber$
we can determine that 2 moles of $HCl$ will react with 2 moles of $Na_{(s)}$ to form 2 moles of $NaCl_{(aq)}$ and 1 mole of $H_{2(g)}$. If we know how many moles of $Na$ reacted, we can use the ratio of 2 moles of $NaCl$ to 2 moles of Na to determine how many moles of $NaCl$ were produced or we can use the ratio of 1 mole of $H_2$ to 2 moles of $Na$ to convert to $NaCl$. This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.
Example 1
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
$\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber$
Solution
Start by counting the number of atoms of each element.
UNBALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
8
9
H
6
2
S
1
2
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of $H_2SO_4$ to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of $H_2O$ where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.
$\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber$
BALANCED
Element
Reactant (# of atoms)
Product (# of atoms)
Pb
1
1
O
12
12
H
8
8
S
2
2
Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.
A balanced equation ultimately has to satisfy two conditions.
1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.
Stoichiometry and Balanced Equations
In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.
Example 2
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent?
Solution
The equation for this can be written as
$\ce{I + 2S \rightarrow IS2}\nonumber$
where
• $I$ represents invitations,
• $S$ represents stamps, and
• $IS_2$ represents the sent party invitations consisting of one invitation and two stamps.
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.
Invitations Stamps Party Invitations Sent
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed.
Example 3
What is the limiting reagent in this example?
Solution
Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors.
12 I x (1IS2/1I) = 12 IS2 possible
20 S x (1IS2/2S) = 10 IS2 possible
When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions.
Types of Reactions
There are 6 basic types of reactions.
• Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2
• Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
• Decomposition: Decomposition is when complex reactants are broken down into simpler products.
• Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
• Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
• Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar Mass
Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass
Example 4
What is the molar mass of H2O?
Solution
$\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber$
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.
Example 5: Combustion of Propane
Propane ($\ce{C_3H_8}$) burns in this reaction:
$\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber$
If 200 g of propane is burned, how many g of $H_2O$ is produced?
Solution
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of $C_3H_8$ to moles of $C_3H_8$ then from moles of $C_3H_8$ to moles of $H_2O$. Then convert from moles of $H_2O$ to grams of $H_2O$.
• Step 1: 200 g $C_3H_8$ is equal to 4.54 mol $C_3H_8$.
• Step 2: Since there is a ratio of 4:1 $H_2O$ to $C_3H_8$, for every 4.54 mol $C_3H_8$ there are 18.18 mol $H_2O$.
• Step 3: Convert 18.18 mol $H_2O$ to g $H_2O$. 18.18 mol $H_2O$ is equal to 327.27 g $H_2O$.
Variation in Stoichiometric Equations
Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.
Density
Density ($\rho$) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.
Volume x (Mass/Volume) = Mass
Mass x (Volume/Mass) = Volume
Percent Mass
Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.
Example 6
A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?
Solution
10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon
0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon
Molarity
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.
Example 7
How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?
Solution
100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density, and mass percent are useful in complex examples ahead.
Determining Empirical Formulas
An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.
Example 8: Combustion of Organic Molecules
1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is the empirical formula of the organic molecule?
Solution
This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.
$\ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber$
Since all the moles of C and H in CO2 and H2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.
0.0333mol CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown
0.599g H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O) = 0.0665 mol H in unknown
Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO2 and H2O. This will give you the number of moles from both the unknown organic molecule and the O2 so you must subtract the moles of oxygen transferred from the O2.
Moles of oxygen in CO2:
0.0333mol CO2 (2mol O/1mol CO2) = 0.0666 mol O
Moles of oxygen in H2O:
0.599g H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O) = 0.0332 mol O
Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O2 reactant.
0.333mol CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2
1.466g CO2 + 0.599g H2O - 1.000g unknown organic = 1.065g O2
Moles of oxygen in O2
1.065g O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2) = 0.0666mol O
Moles of oxygen in unknown
(0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O
Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.
(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O
From this ratio, the empirical formula is calculated to be CH2O.
Determining Molecular Formulas
To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.
Example 9
In the example above, it was determined that the unknown molecule had an empirical formula of CH2O.
1. Find the molar mass of the empircal formula CH2O.
12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH2O
2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.
(120.056 g/mol) / (30.026 g/mol) = 3.9984
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.
CH2O * 4 = ?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.
molar mass of C4H8O4= 120.104 g/mol
experimentally determined mass = 120.056 g/mol
% error = | theoretical - experimental | / theoretical * 100%
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
Example 10: Complex Stoichiometry Problem
An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm3. He accidentally breaks off a 1.203 cm3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H2(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.)
Solution
Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H+(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe2+(aq).
$\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber$
Step 2: Write down all the given information
Alloy density = (3.15g alloy/ 1L alloy)
x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)
x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)
1 liter alloy = 1000cm3 alloy
alloy sample = 1.203cm3 alloy
Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.
Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.
Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.
1.203cm3 alloy(1liter alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)
Make sure all the units cancel out to give you moles of $\ce{Fe(s)}$. The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.
The balanced equation must now be used to convert moles of Fe(s) to moles of H2(g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products.
3.74 x 10-5 mol Fe (s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)
Step 5: Check units
The question asks for how many grams of H2(g) were released so the moles of H2(g) must still be converted to grams using the molar mass of H2(g). Since there are two H in each H2, its molar mass is twice that of a single H atom.
molar mass = 2(1.00794g/mol) = 2.01588g/mol
3.74 x 10-5 mol H2(g) (2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g H2(g) released
Problems
Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems.
1) Why are the following equations not considered balanced?
1. $H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}$
2. $Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}$
2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction.
3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution?
4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below.
$\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber$
5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO2 and 0.388g H2O. Knowing that all the carbon and hydrogen atoms in CO2 and H2O came from the 0.777g sample, what is the empirical formula of the organic compound?
Contributors and Attributions
• Joseph Nijmeh (UCD), Mark Tye (DVC) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.09_Stoichiometric_Calculations%3A_Amounts_of_Reactants_and_Products.txt |
Learning Objectives
• To understand the concept of limiting reactants and quantify incomplete reactions
In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess.
Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus
$1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}$
If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies.
Introduction to Limiting Reactant Problems: Introduction to Limiting Reactant Problems, YouTube(opens in new window) [youtu.be]
Now consider a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride ($\ce{TiCl4}$) and carbon dioxide.
$\ce{TiO2 (s) + Cl2 (g) \rightarrow TiCl4 (g) + CO2 (g)} \nonumber$
Titanium tetrachloride is then converted to metallic titanium by reaction with molten magnesium metal at high temperature:
$\ce{ TiCl4 (g) + 2 \, Mg (l) \rightarrow Ti (s) + 2 \, MgCl2 (l)} \label{3.7.2}$
Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about \$100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.
With 1.00 kg of titanium tetrachloride and 200 g of magnesium metal, how much titanium metal can be produced according to Equation \ref{3.7.2}?
Solving this type of problem requires that you carry out the following steps
1. Determine the number of moles of each reactant.
2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
3. Calculate the number of moles of product that can be obtained from the limiting reactant.
4. Convert the number of moles of product to mass of product.
Step 1: To determine the number of moles of reactants present, calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows: \begin{align} \text{moles} \; \ce{TiCl4} &= \dfrac{\text{mass} \, \ce{TiCl4}}{\text{molar mass} \, \ce{TiCl4}}\nonumber \[4pt] &= 1000 \, \cancel{g} \; \ce{TiCl4} \times {1 \, mol \; TiCl_4 \over 189.679 \, \cancel{g} \; \ce{TiCl4}}\nonumber \[4pt] &= 5.272 \, mol \; \ce{TiCl4} \[4pt] \text{moles }\, \ce{Mg} &= {\text{mass }\, \ce{Mg} \over \text{molar mass }\, \ce{Mg}}\nonumber \[4pt] &= 200 \, \cancel{g} \; \ce{Mg} \times {1 \; mol \, \ce{Mg} \over 24.305 \, \cancel{g} \; \ce{Mg} }\nonumber \[4pt] &= 8.23 \, \text{mol} \; \ce{Mg} \end{align}\nonumber
Step 2: There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following: ${mol \, \ce{Mg} \over mol \, \ce{TiCl4}} = {8.23 \, mol \over 5.272 \, mol } = 1.56 \nonumber$ Because the ratio of the coefficients in the balanced chemical equation is, ${ 2 \, mol \, \ce{Mg} \over 1 \, mol \, \ce{TiCl4}} = 2 \nonumber$ there is not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, there are 8.23 mol of $\ce{Mg}$, so (8.23 ÷ 2) = 4.12 mol of $\ce{TiCl4}$ are required for complete reaction. Because there are 5.272 mol of $\ce{TiCl4}$, titanium tetrachloride is present in excess. Conversely, 5.272 mol of $\ce{TiCl4}$ requires 2 × 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Therefore, magnesium is the limiting reactant.
Step 3: Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed: $mol \; \ce{Ti} = 8.23 \, mol \; \ce{Mg} = {1 \, mol \; \ce{Ti} \over 2 \, mol \; \ce{Mg}} = 4.12 \, mol \; \ce{Ti} \nonumber$ Thus only 4.12 mol of Ti can be formed.
Step 4. To calculate the mass of titanium metal that can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol): \begin{align} \text{moles }\, \ce{Ti} &= \text{mass }\, \ce{Ti} \times \text{molar mass } \, \ce{Ti}\nonumber \[6pt] &= 4.12 \, mol \; \ce{Ti} \times {47.867 \, g \; \ce{Ti} \over 1 \, mol \; \ce{Ti}}\nonumber\[6pt] &= 197 \, g \; \ce{Ti}\nonumber \end{align} \nonumber
Here is a simple and reliable way to identify the limiting reactant in any problem of this sort:
1. Calculate the number of moles of each reactant present: 5.272 mol of $\ce{TiCl4}$ and 8.23 mol of Mg.
2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation: $TiCl_4 : { 5.272 \, mol \, (actual) \over 1 \, mol \, (stoich)} = 5.272\nonumber \[6pt] Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12\nonumber$
3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant.
Density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example $1$ demonstrates.
Example $1$: Fingernail Polish Remover
Ethyl acetate ($\ce{CH3CO2C2H5}$) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol ($\ce{C2H5OH}$) with acetic acid ($\ce{CH3CO2H}$); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.
Given: reactants, products, and volumes and densities of reactants
Asked for: mass of product
Strategy:
1. Balance the chemical equation for the reaction.
2. Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.
3. Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.
4. Convert from moles of product to mass of product.
Solution:
A Always begin by writing the balanced chemical equation for the reaction:
$\ce{ C2H5OH (l) + CH3CO2H (aq) \rightarrow CH3CO2C2H5 (aq) + H2O (l)}\nonumber$
B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall that the density of a substance is the mass divided by the volume:
$\text{density} = {\text{mass} \over \text{volume} }\nonumber$
Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm3):
\begin{align*} \text{moles} \; \ce{C2H5OH} & = { \text{mass} \; \ce{C2H5OH} \over \text{molar mass } \; \ce{C2H5OH} }\nonumber \[6pt] & = {( \text{volume} \; \ce{C2H5OH} ) \times (\text{density} \, \ce{C2H5OH}) \over \text{molar mass } \; \ce{C2H5OH}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{C2H5OH} \times {0.7893 \, \cancel{g} \; \ce{C2H5OH} \over 1 \, \cancel{ml} \, \ce{C2H5OH} } \times {1 \, mol \; \ce{C2H5OH} \over 46.07 \, \cancel{g}\; \ce{C2H5OH}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \[6pt] \text{moles} \; \ce{CH3CO2H} &= {\text{mass} \; \ce{CH3CO2H} \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= { (\text{volume} \; \ce{CH3CO2H} )\times (\text{density} \; \ce{CH3CO2H}) \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{CH3CO2H} \times {1.0492 \, \cancel{g} \; \ce{CH3CO2H} \over 1 \, \cancel{ml} \; \ce{CH3CO2H}} \times {1 \, mol \; \ce{CH3CO2H} \over 60.05 \, \cancel{g} \; \ce{CH3CO2H} } \[6pt] &= 0.175 \, mol \; \ce{CH3CO2H}\nonumber \end{align*} \nonumber
C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:
\begin{align*} moles \; \text{ethyl acetate} &= mol \, \text{ethanol} \times {1 \, mol \; \text{ethyl acetate} \over 1 \, mol \; \text{ethanol}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \times {1 \, mol \, \ce{CH3CO2C2H5} \over 1 \, mol \; \ce{C2H5OH}} \[6pt] &= 0.171 \, mol \; \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:
\begin{align*} \text{ mass of ethyl acetate} &= mol \; \text{ethyl acetate} \times \text{molar mass}\; \text{ethyl acetate}\nonumber \[6pt] &= 0.171 \, mol \, \ce{CH3CO2C2H5} \times {88.11 \, g \, \ce{CH3CO2C2H5} \over 1 \, mol \, \ce{CH3CO2C2H5}}\nonumber \[6pt] &= 15.1 \, g \, \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced:
\begin{align*} \text{volume of ethyl acetate} & = 15.1 \, g \, \ce{CH3CO2C2H5} \times { 1 \, ml \; \ce{CH3CO2C2H5} \over 0.9003 \, g\; \ce{CH3CO2C2H5}} \[6pt] &= 16.8 \, ml \, \ce{CH3CO2C2H5} \end{align*} \nonumber
Exercise $1$
Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound $P_4S_{10}$. How much $P_4S_{10}$ can be prepared starting with 10.0 g of $\ce{P4}$ and 30.0 g of $S_8$?
Answer
35.9 g
Determining the Limiting Reactant and Theoretical Yield for a Reaction: Determining the Limiting Reactant and Theoretical Yield for a Reaction, YouTube(opens in new window) [youtu.be]
Limiting Reactants in Solutions
The concept of limiting reactants applies to reactions carried out in solution as well as to reactions involving pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example $2$.
Example $2$: Breathalyzer reaction
Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:
$\ce{3CH_3 CH_2 OH(aq)} + \underset{yellow-orange}{\ce{2Cr_2 O_7^{2 -}}}(aq) + \ce{16H^+ (aq)} \underset{\ce{H2SO4 (aq)}}{\xrightarrow{\hspace{10px} \ce{Ag^{+}}\hspace{10px}} } \ce{3CH3CO2H(aq)} + \underset{green}{\ce{4Cr^{3+}}}(aq) + \ce{11H2O(l)}\nonumber$
When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the $\ce{Cr2O7^{2−}}$ ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a minute. Because the $\ce{Cr2O7^{2−}}$ ion (the reactant) is yellow-orange and the Cr3+ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.
A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K2Cr2O7 in 50% H2SO4 as well as a fixed concentration of AgNO3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr6+ to Cr3+?
Given: volume and concentration of one reactant
Asked for: mass of other reactant needed for complete reaction
Strategy:
1. Calculate the number of moles of $\ce{Cr2O7^{2−}}$ ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass.
2. Find the total number of moles of $\ce{Cr2O7^{2−}}$ ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).
3. Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of $\ce{Cr2O7^{2−}}$ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass.
Solution:
A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K2Cr2O7 in 1 mL of solution, which can be used to calculate the number of moles of K2Cr2O7 contained in 1 mL:
$\dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles\nonumber$
B Because 1 mol of K2Cr2O7 produces 1 mol of $\ce{Cr2O7^{2−}}$ when it dissolves, each milliliter of solution contains 8.5 × 10−7 mol of Cr2O72. The total number of moles of Cr2O72 in a 3.0 mL Breathalyzer ampul is thus
$moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–}\nonumber$
C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of $\ce{Cr2O7^{2−}}$ ion, so the total number of moles of C2H5OH required for complete reaction is
$moles\: of\: \ce{C2H5OH} = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: \ce{Cr2O7^{2-}}} ) \left( \dfrac{3\: mol\: \ce{C2H5OH}} {2\: \cancel{mol\: \ce{Cr2O7^{2 -}}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: \ce{C2H5OH}\nonumber$
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:
$mass\: \ce{C2H5OH} = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: \ce{C2H5OH}} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: \ce{C2H5OH}}} \right) = 1 .8 \times 10 ^{-4}\: g\: \ce{C2H5OH}\nonumber$
Thus 1.8 × 10−4 g or 0.18 mg of C2H5OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.
Exercise $2$
The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction
Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess $\ce{NaOH}$ is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10−4 g of para-nitrophenol to ensure that formation of the yellow anion is complete?
Answer
4.93 × 10−5 L or 49.3 μL
In Examples $1$ and $2$, the identities of the limiting reactants are apparent: [Au(CN)2], LaCl3, ethanol, and para-nitrophenol. When the limiting reactant is not apparent, it can be determined by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation. The only difference is that the volumes and concentrations of solutions of reactants, rather than the masses of reactants, are used to calculate the number of moles of reactants, as illustrated in Example $3$.
Example $3$
When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:
$\ce{2AgNO3(aq) + K2Cr2O7(aq) \rightarrow Ag2Cr2O7(s) + 2KNO3(aq) }\nonumber$
What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M $\ce{K2Cr2O7}$ are mixed with 250 mL of 0.57 M AgNO3?
Given: balanced chemical equation and volume and concentration of each reactant
Asked for: mass of product
Strategy:
1. Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.
2. Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.
3. Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.
Solution:
A The balanced chemical equation tells us that 2 mol of AgNO3(aq) reacts with 1 mol of K2Cr2O7(aq) to form 1 mol of Ag2Cr2O7(s) (Figure 8.3.2). The first step is to calculate the number of moles of each reactant in the specified volumes:
$moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7\nonumber$
$moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3\nonumber$
B Now determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:
\begin{align*} \ce{K2Cr2O7}: \: \dfrac{0 .085\: mol} {1\: mol} &= 0.085 \[4pt] \ce{AgNO3}: \: \dfrac{0 .14\: mol} {2\: mol} &= 0 .070 \end{align*} \nonumber
Because 0.070 < 0.085, we know that $\ce{AgNO3}$ is the limiting reactant.
C Each mole of $\ce{Ag2Cr2O7}$ formed requires 2 mol of the limiting reactant ($\ce{AgNO3}$), so we can obtain only 0.14/2 = 0.070 mol of $\ce{Ag2Cr2O7}$. Finally, convert the number of moles of $\ce{Ag2Cr2O7}$ to the corresponding mass:
$mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7\nonumber$
The Ag+ and Cr2O72 ions form a red precipitate of solid $\ce{Ag2Cr2O7}$, while the $\ce{K^{+}}$ and $\ce{NO3^{−}}$ ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.)
Exercise $3$
Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:
$\ce{2NaHCO3(aq) + H2SO4(aq) \rightarrow 2CO2(g) + Na2SO4(aq) + 2H2O(l)}\nonumber$
If 13.0 mL of 3.0 M H2SO4 are added to 732 mL of 0.112 M NaHCO3, what mass of CO2 is produced?
Answer
3.4 g
Limiting Reactant Problems Using Molarities: Limiting Reactant Problems Using Molarities, YouTube(opens in new window) [youtu.be]eOXTliL-gNw (opens in new window)
Theoretical Yields
When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield, the amount obtained if the reaction occurred perfectly and the purification method were 100% efficient.
In reality, less product is always obtained than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not run to completion, resulting in a mixture of products and reactants; this last possibility is a common occurrence. Therefore, the actual yield, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:
$\text{percent yield} = {\text{actual yield } \; (g) \over \text{theoretical yield} \; (g) } \times 100\% \label{3.7.3}$
The method used to calculate the percent yield of a reaction is illustrated in Example $4$.
Example $4$: Novocain
Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction
$\underset {\text{p-amino benzoic acid}}{\ce{C7H7NO2}} + \underset {\text{2-diethylaminoethanol}}{\ce{C6H15NO}} \ce{->[\ce{H2SO4}]} \underset {\text{procaine}}{\ce{C13H20N2O2}} + \ce{H2O}\nonumber$
If this reaction were carried out with 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and 15.7 g of procaine were isolated, what is the percent yield?
The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.
Given: masses of reactants and product
Asked for: percent yield
Strategy:
1. Write the balanced chemical equation.
2. Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.
3. Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.
Solution:
A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.
B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:
$mol \; \text{p-aminobenzoic acid} = 10.0 \, g \, \times \, {1 \, mol \over 137.14 \, g } = 0.0729 \, mol \; \text{p-aminbenzoic acid}\nonumber$
$mol \; \text{2-diethylaminoethanol} = 10.0 \, g \times {1 \, mol \over 117.19 \, g} = 0.0853 \, mol \; \text{2-diethylaminoethanol}\nonumber$
The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.
$\text{theoretical yield of procaine} = 0.0729 \, mol \times {236.31 \, g \over 1 \, mol } = 17.2 \, g\nonumber$
C The actual yield was only 15.7 g of procaine, so the percent yield (via Equation \ref{3.7.3}) is
$\text{percent yield} = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \%\nonumber$
(If the product were pure and dry, this yield would indicate very good lab technique!)
Exercise $4$: Extraction of Lead
Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena ($\ce{PbS}$), which is easily converted to lead oxide ($\ce{PbO}$) in 100% yield by roasting in air via the following reaction:
$\ce{ 2PbS (s) + 3O2 \rightarrow 2PbO (s) + 2SO2 (g)}\nonumber$
The resulting $\ce{PbO}$ is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:
$\ce{PbO (s) + C(s) \rightarrow Pb (l) + CO (g)}\nonumber$
If 93.3 kg of $\ce{PbO}$ is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?
Answer
89.2%
Percent yield can range from 0% to 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.
Summary
The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained. The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/03%3A_Stoichiometry/3.10%3A_Calculations_Involving_a_Limiting_Reactant.txt |
Chemical reactions are processes that leads to the transformation of one set of chemical substances (reactants) to another set (products). Chemical reactions encompass changes that only involve the positions of electrons in the forming and breaking of chemical bonds between atoms.
04: Types of Chemical Reactions and Solution Stoichiometry
With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H2O: solid (ice), liquid (water), and gas (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including:
• Boiling Point and Freezing Point
• Surface Tension, Heat of Vaporization, and Vapor Pressure
• Viscosity and Cohesion
• Solid State
• Liquid State
• Gas State
Boiling Point and Freezing Point
If you look at the periodic table and locate tellurium (atomic number: 52), you find that the boiling points of hydrides decrease as molecule size decreases. So the hydride for tellurium: H2Te (hydrogen telluride) has a boiling point of -4°C. Moving up, the next hydride would be H2Se (hydrogen selenide) with a boiling point of -42°C. One more up and you find that H2S (hydrogen sulfide) has a boiling point at -62°C. The next hydride would be H2O (WATER!). And we all know that the boiling point of water is 100°C. So despite its small molecular weight, water has an incredibly big boiling point. This is because water requires more energy to break its hydrogen bonds before it can then begin to boil. The same concept is applied to freezing point as well, as seen in the table below. The boiling and freezing points of water enable the molecules to be very slow to boil or freeze, this is important to the ecosystems living in water. If water was very easy to freeze or boil, drastic changes in the environment and so in oceans or lakes would cause all the organisms living in water to die. This is also why sweat is able to cool our bodies.
COMPOUND BOILING POINT FREEZING POINT
Hydrogen Telluride -4°C -49°C
Hydrogen Selenide -42°C -64°C
Hydrogen Sulfide -62°C -84°C
Water 100°C 0 °C
Surface Tension, Heat of Vaporization, and Vapor Pressure
Besides mercury, water has the highest surface tension for all liquids. Water's high surface tension is due to the hydrogen bonding in water molecules. Water also has an exceptionally high heat of vaporization. Vaporization occurs when a liquid changes to a gas, which makes it an endothermic reaction. Water's heat of vaporization is 41 kJ/mol. Vapor pressure is inversely related to intermolecular forces, so those with stronger intermolecular forces have a lower vapor pressure. Water has very strong intermolecular forces, hence the low vapor pressure, but it's even lower compared to larger molecules with low vapor pressures.
• Viscosity is the property of fluid having high resistance to flow. We normally think of liquids like honey or motor oil being viscous, but when compared to other substances with like structures, water is viscous. Liquids with stronger intermolecular interactions are usually more viscous than liquids with weak intermolecular interactions.
• Cohesion is intermolecular forces between like molecules; this is why water molecules are able to hold themselves together in a drop. Water molecules are very cohesive because of the molecule's polarity. This is why you can fill a glass of water just barely above the rim without it spilling.
Solid State (Ice)
All substances, including water, become less dense when they are heated and more dense when they are cooled. So if water is cooled, it becomes more dense and forms ice. Water is one of the few substances whose solid state can float on its liquid state! Why? Water continues to become more dense until it reaches 4°C. After it reaches 4°C, it becomes LESS dense. When freezing, molecules within water begin to move around more slowly, making it easier for them to form hydrogen bonds and eventually arrange themselves into an open crystalline, hexagonal structure. Because of this open structure as the water molecules are being held further apart, the volume of water increases about 9%. So molecules are more tightly packed in water's liquid state than its solid state. This is why a can of soda can explode in the freezer.
Liquid State (Liquid Water)
It is very rare to find a compound that lacks carbon to be a liquid at standard temperatures and pressures. So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together. Each water molecule links to four others creating a tetrahedral arrangement, however they are able to move freely and slide past each other, while ice forms a solid, larger hexagonal structure.
Gas State (Steam)
As water boils, its hydrogen bonds are broken. Steam particles move very far apart and fast, so barely any hydrogen bonds have the time to form. So, less and less hydrogen bonds are present as the particles reach the critical point above steam. The lack of hydrogen bonds explains why steam causes much worse burns that water. Steam contains all the energy used to break the hydrogen bonds in water, so when steam hits your face you first absorb the energy the steam has taken up from breaking the hydrogen bonds it its liquid state. Then, in an exothermic reaction, steam is converted into liquid water and heat is released. This heat adds to the heat of boiling water as the steam condenses on your skin.
Water as the "Universal Solvent"
Because of water's polarity, it is able to dissolve or dissociate many particles. Oxygen has a slightly negative charge, while the two hydrogens have a slightly positive charge. The slightly negative particles of a compound will be attracted to water's hydrogen atoms, while the slightly positive particles will be attracted to water's oxygen molecule; this causes the compound to dissociate. Besides the explanations above, we can look to some attributes of a water molecule to provide some more reasons of water's uniqueness:
• Forgetting fluorine, oxygen is the most electronegative non-noble gas element, so while forming a bond, the electrons are pulled towards the oxygen atom rather than the hydrogen. This creates 2 polar bonds, which make the water molecule more polar than the bonds in the other hydrides in the group.
• A 104.5° bond angle creates a very strong dipole.
• Water has hydrogen bonding which probably is a vital aspect in waters strong intermolecular interaction
Why is this important for the real world?
The properties of water make it suitable for organisms to survive in during differing weather conditions. Ice freezes as it expands, which explains why ice is able to float on liquid water. During the winter when lakes begin to freeze, the surface of the water freezes and then moves down toward deeper water; this explains why people can ice skate on or fall through a frozen lake. If ice was not able to float, the lake would freeze from the bottom up killing all ecosystems living in the lake. However ice floats, so the fish are able to survive under the surface of the ice during the winter. The surface of ice above a lake also shields lakes from the cold temperature outside and insulates the water beneath it, allowing the lake under the frozen ice to stay liquid and maintain a temperature adequate for the ecosystems living in the lake to survive.
Resources
1. Cracolice, Mark S. and Edward Peters I. Basics of Introductory Chemistry. Thompson, Brooks/Cole Publishing Company. 2006
2. Petrucci, et al. General Chemistry: Principles & Modern Applications: AIE (Hardcover). Upper Saddle River: Pearson/Prentice Hall, 2007.
Contributors and Attributions
• Corinne Yee (UCD), Desiree Rozzi (UCD)
4.02 The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
• Electrolyte Solutions
An electrolyte solution is a solution that contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction.
• Enthalpy of Solution
A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution can either be positive (endothermic) or negative (exothermic). When understanding the enthalpy of solution, it is easiest to think of a hypothetical three-step process.
• Formation of Ionic Solutions
• Interionic Attractions
• Intermolecular Forces in Mixtures And Solutions
Some forces that interact within pure liquids are also present during mixtures and solutions. Forces such as Cohesive as well as Adhesive forces still apply to mixtures; however, more importantly we focus on the interaction between different molecules. Why is oil only soluble in benzene and not water? Why do only "like" molecules dissolve in "like" molecules?
• Units of Concentration
Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
Thumbnail: pixabay.com/photos/chemistry...emist-3533039/ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.01_Water_the_Universal_Solvent.txt |
Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
Relative Concentration Units
Concentrations are often expressed in terms of relative unites (e.g. percentages) with three different types of percentage concentrations commonly used:
1. Mass Percent: The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of a solution is given: $\text{Mass Percent}=\dfrac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100\% \label{1}$
2. Volume Percent: The volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given: $\text{Volume Percent}= \dfrac{\text{Volume of Solute}}{\text{Volume of Solution}} \times 100\% \label{2}$
3. Mass/Volume Percent: Another version of a percentage concentration is mass/volume percent, which measures the mass or weight of solute in grams (e.g., in grams) vs. the volume of solution (e.g., in mL). An example would be a 0.9%( w/v) $NaCl$ solution in medical saline solutions that contains 0.9 g of $NaCl$ for every 100 mL of solution (see figure below). The mass/volume percent is used to express the concentration of a solution when the mass of the solute and volume of the solution is given. Since the numerator and denominator have different units, this concentration unit is not a true relative unit (e.g. percentage), however it is often used as an easy concentration unit since volumes of solvent and solutions are easier to measure than weights. Moreover, since the density of dilute aqueous solutions are close to 1 g/mL, if the volume of a solution in measured in mL (as per definition), then this well approximates the mass of the solution in grams (making a true reletive unit (m/m)).
$\text{Mass/Volume Percent}= \dfrac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \times 100\% \label{3}$
Example $1$: Alcohol "Proof" as a Unit of Concentration
For example, In the United States, alcohol content in spirits is defined as twice the percentage of alcohol by volume (v/v) called proof. What is the concentration of alcohol in Bacardi 151 spirits that is sold at 151 proof (hence the name)?
Solution
It will have an alcohol content of 75.5% (w/w) as per definition of "proof".
When calculating these percentages, that the units of the solute and solution should be equivalent units (and weight/volume percent (w/v %) is defined in terms of grams and mililiters).
You CANNOT plug in… You CANNOT plug in…
(2 g Solute) / (1 kg Solution) (2 g Solute) / (1000 g Solution)
or (0.002 kg Solute) / (1 kg Solution)
(5 mL Solute) / (1 L Solution) (5 mL Solute) / (1000 mL Solution)
or (0.005 L Solute) / (1 L Solution)
(8 g Solute) / (1 L Solution) (8 g Solute) / (1000 mL Solution)
or (0.008 kg Solute) / (1 L Solution)
Dilute Concentrations Units
Sometimes when solutions are too dilute, their percentage concentrations are too low. So, instead of using really low percentage concentrations such as 0.00001% or 0.000000001%, we choose another way to express the concentrations. This next way of expressing concentrations is similar to cooking recipes. For example, a recipe may tell you to use 1 part sugar, 10 parts water. As you know, this allows you to use amounts such as 1 cup sugar + 10 cups water in your equation. However, instead of using the recipe's "1 part per ten" amount, chemists often use parts per million, parts per billion or parts per trillion to describe dilute concentrations.
• Parts per Million: A concentration of a solution that contained 1 g solute and 1000000 mL solution (same as 1 mg solute and 1 L solution) would create a very small percentage concentration. Because a solution like this would be so dilute, the density of the solution is well approximated by the density of the solvent; for water that is 1 g/mL (but would be different for different solvents). So, after doing the math and converting the milliliters of solution into grams of solution (assuming water is the solvent): $\dfrac{\text{1 g solute}}{\text{1000000 mL solution}} \times \dfrac{\text{1 mL}}{\text{1 g}} = \dfrac{\text{1 g solute}}{\text{1000000 g solution}}$ We get (1 g solute)/(1000000 g solution). Because both the solute and the solution are both now expressed in terms of grams, it could now be said that the solute concentration is 1 part per million (ppm). $\text{1 ppm}= \dfrac{\text{1 mg Solute}}{\text{1 L Solution}}$ The ppm unit can also be used in terms of volume/volume (v/v) instead (see example below).
• Parts per Billion: Parts per billion (ppb) is almost like ppm, except 1 ppb is 1000-fold more dilute than 1 ppm. $\text{1 ppb} = \dfrac{1\; \mu \text{g Solute}}{\text{1 L Solution}}$
• Parts per Trillion: Just like ppb, the idea behind parts per trillion (ppt) is similar to that of ppm. However, 1 ppt is 1000-fold more dilute than 1 ppb and 1000000-fold more dilute than 1 ppm. $\text{1 ppt} = \dfrac{ \text{1 ng Solute}}{\text{1 L Solution}}$
Example $2$: ppm in the Atmosphere
Here is a table with the volume percent of different gases found in air. Volume percent means that for 100 L of air, there are 78.084 L Nitrogen, 20.946 L Oxygen, 0.934 L Argon and so on; Volume percent mass is different from the composition by mass or composition by amount of moles.
Elements Name Volume Percent (v/v) ppm (v/v)
Nitrogen 78.084 780,840
Oxygen 20.946 209,460
Water Vapor 4.0% 40,000
Argon 0.934 9,340
Carbon Dioxide 0.0379 379* (but growing rapidly)
Neon 0.008 8.0
Helium 0.000524 5.24
Methane 0.00017 1.7
Krypton 0.000114 1.14
Ozone 0.000001 0.1
Dinitrogen Monoxide 0.00003 0.305
Concentration Units based on moles
• Mole Fraction: The mole fraction of a substance is the fraction of all of its molecules (or atoms) out of the total number of molecules (or atoms). It can also come in handy sometimes when dealing with the $PV=nRT$ equation. $\chi_A= \dfrac{\text{number of moles of substance A}}{\text{total number of moles in solution}}$ Also, keep in mind that the sum of each of the solution's substances' mole fractions equals 1. $\chi_A + \chi_B + \chi_C \;+\; ... \;=1$
• Mole Percent: The mole percent (of substance A) is $\chi_A$ in percent form. $\text{Mole percent (of substance A)}= \chi_A \times 100\%$
• Molarity: The molarity (M) of a solution is used to represent the amount of moles of solute per liter of the solution. $M= \dfrac{\text{Moles of Solute}}{\text{Liters of Solution}}$
• Molality: The molality (m) of a solution is used to represent the amount of moles of solute per kilogram of the solvent. $m= \dfrac{\text{Moles of Solute}}{\text{Kilograms of Solvent}}$
The molarity and molality equations differ only from their denominators. However, this is a huge difference. As you may remember, volume varies with different temperatures. At higher temperatures, volumes of liquids increase, while at lower temperatures, volumes of liquids decrease. Therefore, molarities of solutions also vary at different temperatures. This creates an advantage for using molality over molarity. Using molalities rather than molarities for lab experiments would best keep the results within a closer range. Because volume is not a part of its equation, it makes molality independent of temperature.
Example $1$
In a solution, there is 111.0 mL (110.605 g) solvent and 5.24 mL (6.0508 g) solute present in a solution. Find the mass percent, volume percent and mass/volume percent of the solute.
Solution
Mass Percent
=(Mass of Solute) / (Mass of Solution) x 100%|
=(6.0508g) / (110.605g + 6.0508g) x 100%
=(0.0518688312) x 100%
=5.186883121%
Mass Percent= 5.186%
Volume Percent
=(Volume of Solute) / (Volume of Solution) x 100%
=(5.24mL) / (111.0mL + 5.24mL) x 100%
=(0.0450791466) x 100%
=4.507914659%
Volume Percent= 4.51%
Mass/Volume Percent
=(Mass of Solute) / (Volume of Solution) x 100%
=(6.0508g) / (111.0mL + 5.24mL) x 100%
=(0.0520) x 100%
=5.205%
Mass/Volume Percent= 5.2054%
Example $2$
With the solution shown in the picture below, find the mole percent of substance C.
Solution
Moles of C= (5 C molecules) x (1mol C / 6.022x1023 C molecules) = 8.30288941x10-24mol C
Total Moles= (24 molecules) x (1mol / 6.022x1023 molecules)= 3.98538691x10-23mol total
XC= (8.30288941x10-24mol C) / (3.98538691x10-23mol) = .2083333333
Mole Percent of C
= XC x 100%
=(o.2083333333) x 100%
=20.83333333
Mole Percent of C = 20%
Example $3$
A 1.5L solution is composed of 0.25g NaCl dissolved in water. Find its molarity.
Solution
Moles of NaCl= (0.25g) / (22.99g + 35.45g) = 0.004277 mol NaCl
Molarity
=(Moles of Solute) / (Liters of Solution)
=(0.004277mol NaCl) / (1.5L)
=0.002851 M
Molarity= 0.0029M
Example $4$
0.88g NaCl is dissolved in 2.0L water. Find its molality.
Solution
Moles of NaCl= (0.88g) / (22.99g + 35.45g) = 0.01506 mol NaCl
Mass of water= (2.0L) x (1000mL / 1L) x (1g / 1mL) x (1kg / 1000g) = 2.0kg water
Molality
=(Moles of Solute) / (kg of Solvent)
=(0.01506 mol NaCl) / (2.0kg)
=0.0075290897m
Molality= 0.0075m | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.03_The_Composition_of_Solutions.txt |
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
Describing Reactions Quantitatively
Chemical reactions are constantly occurring in the world around us; everything from the rusting of an iron fence to the metabolic pathways of a human cell are all examples of chemical reactions. Chemistry is an attempt to classify and better understand these reactions.
A chemical reaction is typically represented by a chemical equation, which represents the change from reactants to products. The left hand side of the equation represents the reactants, while the right hand side represents the products. A typical chemical reaction is written with stoichiometric coefficients, which show the relative amounts of products and reactants involved in the reaction. Each compound is followed by a parenthetical note of the compound’s state of 2: (l) for liquid, (s) for solid, (g) for gas. The symbol (aq) is also commonly used in order to represent an aqueous solution, in which compounds are dissolved in water. A reaction might take the following form:
$\ce{A (aq) + B (g) \rightarrow C (s) + D (l)} \nonumber$
In the above example, $A$ and $B$, known as the reactants, reacted to form $C$ and $D$, the products.
To write an accurate chemical equation, two things must occur:
1. Each product and reactant must be written using its chemical formula, e.g., $H_2$
2. The number of atoms of each element must be equal on both sides of the equation. Coefficients are used in front of the chemical formulas in order to help balance the number of atoms, e.g.,
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
Example $1$: Balancing Reactions
Hydrogen and nitrogen react together in order to produce ammonia gas, write the chemical equation of this reaction.
Solution
Step 1: Write each product and reactant using its chemical formula.
$\ce{H_2 + N_2 \rightarrow NH_3} \nonumber$
Step 2: Ensure the number of atoms of each element are equal on both sides of the equation.
$\ce{3H_2 + N_2 \rightarrow 2NH_3} \nonumber$
In order to balance this equation, coefficients must be used. Since there are only 2 nitrogen atoms present on the left side of the equation, a coefficient of 2 must be added to $NH_3$.
Stoichiometry
The coefficient that used for balancing the equation is called the stoichiometric coefficient. The coefficients tell us the ratio of each element in a chemical equation. For example
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
means
• 2 moles of MgO is produced for every 2 moles of Mg consumed.
• 2 moles of MgO is produced for every 1 mole of O2 consumed.
When all of the reactants of a reaction are completely consumed, the reaction is in perfect stoichiometric proportions. Often times, however, a reaction is not in perfect stoichiometric proportions, leading to a situation in which the entirety of one reactant is consumed, but there is some of another reactant remaining. The reactant that is entirely consumed is called the limiting reactant, and it determines how much of the products are produced.
Example $2$: Limiting Reagent
4.00 g of hydrogen gas mixed with 20.0g of oxygen gas. How many grams of water are produced?
Solution
$n(H_2)=\dfrac{4g}{(1.008 \times2)g/mol}=1.98mol \nonumber$
So theoretically, it requires 0.99 mol of $O_2$
n(O2)=n(H2)*(1mol O2/2mol H2)=0.99 mol
m(O2)=n(O2)*(16g/mol *2) = 31.7 g O2
Because $O_2$ only has 20.0 g, less than the required mass. It is limiting.
Often, reactants do not react completely, resulting in a smaller amount of product formed than anticipated. The amount of product expected to be formed from the chemical equation is called the theoretical yield. The amount of product that is produced during a reaction is the actual yield. To determine the percent yield:
Percent yield =actual yield/theoretical yield X 100%
Chemical reactions do not only happen in the air, but also exist in solutions. In a solution, the solvent is the compound that is dissolved, and the solute is the compound that the solvent is dissolved in. The molarity of a solution is the number of moles of a solvent divided by the number of liters of solution.
$\ Molarity=\dfrac{ \text{amount of solute (mol)}}{\text{volume of solution (L)}} \nonumber$
$\ M=\dfrac{n}{V} \nonumber$
Example $3$: Concentrations
100.0 g NaCl is dissolved in 50.00 ml water. What is the molarity of the solution?
Solution
a) Find the amount of solute in moles.
100.0g/(22.99 g/mol+35.45 g/mol) =1.711 moles
b) Convert mL to L.
50.00 mL=0.05000 L
c) Find the molarity
1.711 moles/0.05000L=34.22 mol/L
Physical Changes During Chemical Reactions
Physical change is the change in physical properties. Physical changes usually occur during chemical reactions, but does not change the nature of substances. The most common physical changes during reactions are the change of color, scent and evolution of gas. However, when physical changes occur, chemical reactions may not occur.
Types of Chemical Reactions
Precipitation, or double-replacement reaction
A reaction that occurs when aqueous solutions of anions (negatively charged ions) and cations (positively charged ions) combine to form a compound that is insoluble is known as precipitation. The insoluble solid is called the precipitate, and the remaining liquid is called the supernate. See Figure2.1
Real life example: The white precipitate formed by acid rain on a marble statue:
$CaCO_3(aq)+H_2SO_4(aq) \rightarrow CaSO_4(s)+H_2O(l)+CO_2(g) \nonumber$
Example $4$: Precipitation
An example of a precipitation reaction is the reaction between silver nitrate and sodium iodide. This reaction is represented by the chemical equation :
AgNO3 (aq)+ NaI (aq) → AgI (s) + NaNO3 (aq)
Since all of the above species are in aqueous solutions, they are written as ions, in the form:
Ag+ +NO3- (aq)+ Na+ (aq) + I- (aq) → AgI (s) + Na+ (aq) + NO3- (aq)
Ions that appear on both sides of the equation are called spectator ions. These ions do not affect the reaction and are removed from both sides of the equation to reveal the net ionic equation, as written below:
Ag+ (aq) + I- (aq) → AgI (s)
In this reaction, the solid, AgI, is known as the precipitate. The formation of a precipitate is one of the many indicators that a chemical reaction has taken place.
Acid-base, or neutralization reaction
A neutralization reaction occurs when an acid and base are mixed together. An acid is a substance that produces H+ ions in solution, whereas a base is a substance that that produces OH- ions in solution. A typical acid-base reaction will produce an ionic compound called a salt and water. A typical acid-base reaction is the reaction between hydrochloric acid and sodium hydroxide. This reaction is represented by the equation:
$\ce{HCl (aq) + NaOH (aq) \rightarrow NaCl (aq)+ H_2O (l)} \nonumber$
In this reaction, $HCl$ is the acid, $NaOH$ is the base, and $NaCl$ is the salt. Real life example: Baking soda reacts with vinegar is a neutralization reaction.
Video: Vinegar and Baking Soda Reaction with Explanation
Oxidation-Reduction (Redox) Reactions
A redox reaction occurs when the oxidation number of atoms involved in the reaction are changed. Oxidation is the process by which an atom’s oxidation number is increased, and reduction is the process by which an atom’s oxidation number is decreased. If the oxidation states of any elements in a reaction change, the reaction is an oxidation-reduction reaction. An atom that undergoes oxidation is called the reducing agent, and the atom that undergoes reduction is called the oxidizing agent. An example of a redox reaction is the reaction between hydrogen gas and fluorine gas:
$H_2 (g) + F_2 (g) \rightarrow 2HF (g) \label{redox1}$
In this reaction, hydrogen is oxidized from an oxidation state of 0 to +1, and is thus the reducing agent. Fluorine is reduced from 0 to -1, and is thus the oxidizing agent.
Real life example: The cut surface of an apple turns brownish after exposed to the air for a while.
Video: Why Do Apples Turn Brown?
Combustion Reaction
A combustion reaction is a type of redox reaction during which a fuel reacts with an oxidizing agent, resulting in the release of energy as heat. Such reactions are exothermic, meaning that energy is given off during the reaction. An endothermic reaction is one which absorbs heat. A typical combustion reaction has a hydrocarbon as the fuel source, and oxygen gas as the oxidizing agent. The products in such a reaction would be $CO_2$ and $H_2O$.
$C_xH_yO_z+O_2 \rightarrow CO_2+H_2O \;\;\; \text{(unbalanced)} \nonumber$
Such a reaction would be the combustion of glucose in the following equation
$C_6H_{12}O_6 (s) + 6O_2 (g) \rightarrow 6CO_2 (g) + 6H_2O (g) \nonumber$
Real life example: explosion; burning.
Video: Combustion reactions come in many varieties. Here's a collection of various examples, all of which require oxygen, activation energy, and of course, fuel
Synthesis Reactions
A synthesis reaction occurs when one or more compounds combines to form a complex compound. The simplest equation of synthesis reaction is illustrated below.
An example of such a reaction is the reaction of silver with oxygen gas to form silver oxide:
$2Ag (s) +O_2 (g) \rightarrow 2AgO (s) \nonumber$
Real life example: Hydrogen gas is burned in air (reacts with oxygen) to form water:
$2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber$
Decomposition Reactions
A decomposition reaction is the opposite of a synthesis reaction. During a decomposition reaction, a more complex compound breaks down into multiple simpler compounds. A classic example of this type of reaction is the decomposition of hydrogen peroxide into oxygen and hydrogen gas:
$H_2O_2 (l) \rightarrow H_2 (g) + O_2 (g) \nonumber$
Single Replacement Reactions
A type of oxidation-reduction reaction in which an element in a compound is replaced by another element.
An example of such a reaction is:
$Cu (s) + AgNO_3 (aq) \rightarrow Ag(s) + Cu(NO_3)_2 (aq) \nonumber$
This is also a redox reaction.
Problems
1) C3H6O3 + O2 → CO2 (g) +H2O (g)
a) What type of reaction is this?
b) Is is exothermic or endothermic? Explain.
2) Given the oxidation-reduction reaction :
Fe (s) + CuSO4 (aq)→ FeSO4 (aq)+ Cu (s)
a) Which element is the oxidizing agent and which is the reducing agent?
b) How do the oxidation states of these species change?
3) Given the equation:
AgNO3 (aq) + KBr (aq) → AgBr (s) +KNO3 (aq)
a) What is the net ionic reaction?
b) Which species are spectator ions?
4) 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) +2 H2O (l)
a) In this reaction, which species is the acid and which is the base?
b) Which species is the salt?
c) If 2 moles of HNO3 and 1 mole of Sr(OH)2 are used, resulting in 0.85 moles of Sr(NO3)2 , what is the percent yield (with respect to moles) of Sr(NO3)2 ?
5) Identify the type of the following reactions:
a) Al(OH)3 (aq) + HCl (aq) → AlCl3 (aq) + H2O (l)
b) MnO2 + 4H+ + 2Cl- → Mn2+ + 2H2O (l) + Cl2 (g)
c) P4 (s) + Cl2 (g) → PCl3 (l)
d) Ca (s) + 2H2O (l) → Ca(OH)2 (aq) + H2 (g)
e) AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Solutions
1a) It is a combustion reaction
1b) It is exothermic, because combustion reactions give off heat
2a) Cu is the oxidizing agent and Fe is the reducing agent
2b) Fe changes from 0 to +2, and Cu changes from +2 to 0.
3a) Ag+ (aq) + Br- (aq) → AgBr (s)
3b) The spectator ions are K+ and NO3-
4a) HNO3 is the acid and Sr(OH)2 is the base
4b) Sr(NO3)2 is the salt
4c) According to the stoichiometric coefficients, the theoretical yield of Sr(NO3)2 is one mole. The actual yield was 0.85 moles. Therefore the percent yield is:
(0.85/1.0) * 100% = 85%
5a) Acid-base
5b) Oxidation-reduction
5c) Synthesis
5d) Single-replacement reaction
5e) Double replacement reaction
• Priya Muley | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.04_Types_of_Chemical_Reactions.txt |
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation. The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions.
Properties of Precipitates
Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid. The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting.
Precipitation and Double Replacement Reactions
The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as "switching partners"; that is, the two reactants each "lose" their partner and form a bond with a different partner:
A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:
$CdSO_{4(aq)} + K_2S_{(aq)} \rightarrow CdS_{(s)} + K_2SO_{4(aq)} \nonumber$
Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble. However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution.
Solubility Rules
Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule.
1. Salts formed with group 1 cations and $NH_4^+$ cations are soluble. There are some exceptions for certain $Li^+$ salts.
2. Acetates ($C_2H_3O_2^-$), nitrates ($NO_3^-$), and perchlorates ($ClO_4^-$) are soluble.
3. Bromides, chlorides, and iodides are soluble.
4. Sulfates ($SO_4^{2-}$) are soluble with the exception of sulfates formed with $Ca^{2+}$, $Sr^{2+}$, and $Ba^{2+}$.
5. Salts containing silver, lead, and mercury (I) are insoluble.
6. Carbonates ($CO_3^{2-}$), phosphates ($PO_4^{3-}$), sulfides, oxides, and hydroxides ($OH^-$) are insoluble. Sulfides formed with group 2 cations and hydroxides formed with calcium, strontium, and barium are exceptions.
If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.
Net Ionic Equations
To understand the definition of a net ionic equation, recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair:
AB(aq) + CD(aq) AD(aq) + CB(s)
The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below:
A+(aq) + B-(aq) + C+(aq) + D-(aq) A+(aq) + D-(aq) + CB(s)
In the equation above, A+and D- ions are present on both sides of the equation. These are called spectator ions because they remain unchanged throughout the reaction. Since they go through the equation unchanged, they can be eliminated to show the net ionic equation:
C+ (aq)+ B- (aq) CB (s)
The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs.
Applications and Examples
Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied.
Example 1
Complete the double replacement reaction and then reduce it to the net ionic equation.
$NaOH_{(aq)} + MgCl_{2 \;(aq)} \rightarrow \nonumber$
First, predict the products of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”).
$2NaOH_{(aq)} + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2 \nonumber$
Second, consult the solubility rules to determine if the products are soluble. Group 1 cations ($Na^+$) and chlorides are soluble from rules 1 and 3 respectively, so $NaCl$ will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus $Mg(OH)_2$ will form a precipitate. The resulting equation is the following:
$2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Mg(OH)_{2\;(s)} \nonumber$
Third, separate the reactants into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms:
$2Na^+_{(aq)} + 2OH^-_{(aq)} + Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow Mg(OH)_{2\;(s)} + 2Na^+_{(aq)} + 2Cl^-_{(aq)} \nonumber$
Lastly, eliminate the spectator ions (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final net ionic equation is:
$Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Mg(OH)_{2(s)} \nonumber$
Example 2
Complete the double replacement reaction and then reduce it to the net ionic equation.
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow \nonumber$
Solution
The predicted products of this reaction are $CoSO_4$ and $NaCl$. From the solubility rules, $CoSO_4$ is soluble because rule 4 states that sulfates ($SO_4^{2-}$) are soluble. Similarly, we find that $NaCl$ is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows:
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl_{(aq)} \nonumber$
Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.):
Co2- (aq) + 2Cl-(aq) + 2Na+ (aq) + SO42-(aq) Co2- (aq) + SO42-(aq) + 2Na+ (aq) + 2Cl-(aq)
No precipitation reaction
This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs.
Practice Problems
Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations.
1. $Fe(NO_3)_{3\;(aq)} + NaOH_{(aq)} \rightarrow$
2. $Al_2(SO_4)_{3\;(aq)} + BaCl_{2\;(aq)} \rightarrow$
3. $HI_{(aq)} + Zn(NO_3)_{2\;(aq)} \rightarrow$
4. $CaCl_{2\;(aq)} + Na_3PO_{4\;(aq)} \rightarrow$
5. $Pb(NO_3)_{2\;(aq)} + K_2SO_{4 \;(aq)} \rightarrow$
Solutions
1. Regardless of physical state, the products of this reaction are $Fe(OH)_3$ and $NaNO_3$. The solubility rules predict that $NaNO_3$ is soluble because all nitrates are soluble (rule 2). However, $Fe(OH)_3$ is insoluble, because hydroxides are insoluble (rule 6) and $Fe$ is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows:
$Fe^{3+}_{(aq)} + NO^-_{3\;(aq)} + Na^+_{(aq)} + 3OH^-_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + Na^+_{(aq)} + NO^-_{3\;(aq)} \nonumber$
Canceling out spectator ions leaves the net ionic equation:
$Fe^{3+}_{(aq)} + OH^-_{(aq)} \rightarrow Fe(OH)_{\;3(s)} \nonumber$
2. From the double replacement reaction, the products are $AlCl_3$ and $BaSO_4$. $AlCl_3$ is soluble because it contains a chloride (rule 3); however, $BaSO_4$ is insoluble: it contains a sulfate, but the $Ba^{2+}$ ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing):
$2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3Ba^{2+}_{(aq)} + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+}_{(aq)} +6Cl^-_{(aq)} + 3BaSO_{4\;(s)} \nonumber$
Canceling out spectator ions leaves the following net ionic equation:
$Ba^{2+}_{(aq)} + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)} \nonumber$
3. From the double replacement reaction, the products $HNO_3$ and $ZnI_2$ are formed. Looking at the solubility rules, $HNO_3$ is soluble because it contains nitrate (rule 2), and $ZnI_2$ is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs.
4. The products of this double replacement reaction are $Ca_3(PO_4)_2$ and $NaCl$. Rule 1 states that $NaCl$ is soluble, and according to solubility rule 6, $Ca_3(PO_4)_2$ is insoluble. The ionic equation is:
$Ca^{2+}_{(aq)}+ Cl^-_{(aq)} + Na^+_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+_{(aq)} + Cl^-_{(aq)} \nonumber$
After canceling out spectator ions, the net ionic equation is given below:
$Ca^{2+}_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} \nonumber$
5. The first product of this reaction, $PbSO_4$, is soluble according to rule 4 because it is a sulfate. The second product, $KNO_3$, is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs.
Contributors and Attributions
• Julie Schaffer (UCD), Corinne Herman (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.05_Precipitation_Reactions.txt |
A solution consists of two or more substances dissolved in a liquid form. Not to get confused with a mixture, which is heterogeneous--multiple substances exist in varying structures-- solutions are homogenous, which means that atoms of the solute are evenly dispersed throughout the solvent (ex. water, ethanol). Think of it as comparing a cup of (dissolved) sugar water and a cup of water with lego blocks in it. The solute is the substance dissolved in the solution, and the solvent is the substance doing the dissolving.
ex.
A solution of NaCl in water A mixture of lego blocks and water
Note: All solutions are mixtures but not all mixtures are solutions.
Solvents
Water (H2O) is the most common solvent, used for dissolving many compounds or brewing coffee. Other common solvents include turpentine (a paint thinner), acetone (nail polish remover), and ethanol (used in some perfumes). Such solvents usually contain carbon and are called organic solvents. Solutions with water as the solvent are called aqueous solutions; they have special properties that are covered here.
Solutes
Different chemical compounds dissolve in solutes in varying degrees. Some compounds, such as the strong acid hydrochloric acid (HCl), dissociate completely in solution into ions. Others, like the weak base ammonia (NH3), only partly dissociate. Yet other compounds like alcohol do not dissociate at all and remain compounds. Laboratory reactions often involve acids and bases, which are covered in more detail here.
Concentration
Concentration is the measure of the amount of solute in a certain amount of solvent. Knowing the concentration of a solution is important determining the strength of an acid or base (pH), among other things. When there is so much solute present in a concentration that it no longer dissolves, the solution is saturated.
Scientists often use molarity to measure concentration.
molarity = moles/Liter
Since reaction stoichiometry relies on molar ratios, molarity is the main measurement for concentration.
A less common unit for concentration is called molality.
Molality = moles/Kg of solvent
Scientists sometimes use molality to measure concentration because liquid volumes change slightly based on the temperature and pressure. Mass, however, stays the same and can be measured accurately using a balance. Commercial concentrated products are usually expressed in mass percent; such as commercial concentrated sulfuric acid, which is 93-98% H2SO4 by mass in water (Hill, Petrucci 116).
Making a Solution
Solutions used in the laboratory are usually made from either solid solutes (often salts) or stock solutions.
To make a solution from solid solutes, first calculate how many moles of solute are in the desired solutions (using the molarity). Calculate the amount of solid you need in grams using the moles needed and the molar mass of the solute and weight out the needed amount. Transfer the solute to a container (preferably a volumetric flask, which most accurately measures volume of solution labeled on the flask) and add a small amount of solvent. Mix thoroughly to dissolve the solute. Once the solute has dissolved, add the remaining solvent to make the solution of the desired volume and mix thoroughly.
For example, to make 0.5 Liters of 0.5 molar NaCl:
1. Multiply the concentration (0.5 mols/Liters) by the volume of solution you want (0.5 Liters) to find the moles of NaCl you need.
0.5 moles/Liter * 0.5 Liters = 0.25 moles of NaCl
2. Multiply the moles of NaCl by its molar mass (58.44 g/mol) to find the grams of solute needed.
(0.25 moles NaCl)*(58.44 grams/mole) = 14.61 grams of NaCl
Making a solution of a certain concentration from a stock solution is called a dilution. When diluting a solution, keep in mind that adding a solvent to a solution changes the concentration of the solution, but not the amount of solute already present.
To dilute a solution with known concentration, first determine the number of moles of solute are in the solution by multiplying the molarity by the volume (in Liters). Then, divide by the desired molarity or volume to find the volume or concentration needed.
The equation to use is simply
M1V1 = M2V2
M1 and V1 are the concentration and volume of the original (stock) solution to dilute; M2 and V2 are the desired concentration and volume of the final solution.
Solution Stoichiometry
For reactions that take place in solutions:
1. Calculate the moles of solute reacting by multiplying the concentration (molarity) by the volume of solution (Liters)
2. Determine the Limiting Reactant, if there is one
3. Follow the stoichiometric process.
4. Convert the resulting moles of solute back to molarity by dividing by the total volume, in liters, of solution used in the reaction.
5. In the case of reactions involving ions (such as in reactions between strong acids and bases), eliminate spectator ions from the net ionic equation. Spectator ions do not react in the equations.
6. If the concentration is not given but the molar mass and volume are, use density (grams/Liter) to find the amount of solute in grams, then convert it to moles.
Problems
1. A solution is prepared by dissolving 44.6 grams of acetone (OC(CH3)2) in water to produce 1.50 Liters of solution. What is the molarity of the resulting solution?
2. A certain laboratory procedure requires 0.025 M H2SO4. How many milliliters of 1.10 M H2SO4 should be diluted in water to prepare 0.500 L of 0.025 M H2SO4?
3. A sample of saturated NaNO3 (aq) is 10.9 M at 25 degrees Celcius. How many grams of NaNO3 are in 230 mL of this solution at the same temperature?
4. A beaker of 175 mL of 0.950 M NaCl is left uncovered for a period of time. If, by the end of the time period, the volume of solution in the beaker has decreased to 137 mL (the volume loss is due to water evaporation), what is the resulting concentration of the solution?
5. A student prepares a solution by dissolving 15.0 mL ethanol (C2H5OH) in water to make a 300.0 mL solution. Calculate the concentration (molarity) of ethanol in the solution. (density = 0.789 g/mL)
1.
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Contributors and Attributions
• Stephanie Shieh (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.06_Describing_Reactions_in_Solution.txt |
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation. The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions.
Properties of Precipitates
Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid. The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting.
Precipitation and Double Replacement Reactions
The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as "switching partners"; that is, the two reactants each "lose" their partner and form a bond with a different partner:
A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:
$CdSO_{4(aq)} + K_2S_{(aq)} \rightarrow CdS_{(s)} + K_2SO_{4(aq)} \nonumber$
Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble. However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution.
Solubility Rules
Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule.
1. Salts formed with group 1 cations and $NH_4^+$ cations are soluble. There are some exceptions for certain $Li^+$ salts.
2. Acetates ($C_2H_3O_2^-$), nitrates ($NO_3^-$), and perchlorates ($ClO_4^-$) are soluble.
3. Bromides, chlorides, and iodides are soluble.
4. Sulfates ($SO_4^{2-}$) are soluble with the exception of sulfates formed with $Ca^{2+}$, $Sr^{2+}$, and $Ba^{2+}$.
5. Salts containing silver, lead, and mercury (I) are insoluble.
6. Carbonates ($CO_3^{2-}$), phosphates ($PO_4^{3-}$), sulfides, oxides, and hydroxides ($OH^-$) are insoluble. Sulfides formed with group 2 cations and hydroxides formed with calcium, strontium, and barium are exceptions.
If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.
Net Ionic Equations
To understand the definition of a net ionic equation, recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair:
AB(aq) + CD(aq) AD(aq) + CB(s)
The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below:
A+(aq) + B-(aq) + C+(aq) + D-(aq) A+(aq) + D-(aq) + CB(s)
In the equation above, A+and D- ions are present on both sides of the equation. These are called spectator ions because they remain unchanged throughout the reaction. Since they go through the equation unchanged, they can be eliminated to show the net ionic equation:
C+ (aq)+ B- (aq) CB (s)
The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs.
Applications and Examples
Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied.
Example 1
Complete the double replacement reaction and then reduce it to the net ionic equation.
$NaOH_{(aq)} + MgCl_{2 \;(aq)} \rightarrow \nonumber$
First, predict the products of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”).
$2NaOH_{(aq)} + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2 \nonumber$
Second, consult the solubility rules to determine if the products are soluble. Group 1 cations ($Na^+$) and chlorides are soluble from rules 1 and 3 respectively, so $NaCl$ will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus $Mg(OH)_2$ will form a precipitate. The resulting equation is the following:
$2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Mg(OH)_{2\;(s)} \nonumber$
Third, separate the reactants into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms:
$2Na^+_{(aq)} + 2OH^-_{(aq)} + Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow Mg(OH)_{2\;(s)} + 2Na^+_{(aq)} + 2Cl^-_{(aq)} \nonumber$
Lastly, eliminate the spectator ions (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final net ionic equation is:
$Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Mg(OH)_{2(s)} \nonumber$
Example 2
Complete the double replacement reaction and then reduce it to the net ionic equation.
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow \nonumber$
Solution
The predicted products of this reaction are $CoSO_4$ and $NaCl$. From the solubility rules, $CoSO_4$ is soluble because rule 4 states that sulfates ($SO_4^{2-}$) are soluble. Similarly, we find that $NaCl$ is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows:
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl_{(aq)} \nonumber$
Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.):
Co2- (aq) + 2Cl-(aq) + 2Na+ (aq) + SO42-(aq) Co2- (aq) + SO42-(aq) + 2Na+ (aq) + 2Cl-(aq)
No precipitation reaction
This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs.
Practice Problems
Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations.
1. $Fe(NO_3)_{3\;(aq)} + NaOH_{(aq)} \rightarrow$
2. $Al_2(SO_4)_{3\;(aq)} + BaCl_{2\;(aq)} \rightarrow$
3. $HI_{(aq)} + Zn(NO_3)_{2\;(aq)} \rightarrow$
4. $CaCl_{2\;(aq)} + Na_3PO_{4\;(aq)} \rightarrow$
5. $Pb(NO_3)_{2\;(aq)} + K_2SO_{4 \;(aq)} \rightarrow$
Solutions
1. Regardless of physical state, the products of this reaction are $Fe(OH)_3$ and $NaNO_3$. The solubility rules predict that $NaNO_3$ is soluble because all nitrates are soluble (rule 2). However, $Fe(OH)_3$ is insoluble, because hydroxides are insoluble (rule 6) and $Fe$ is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows:
$Fe^{3+}_{(aq)} + NO^-_{3\;(aq)} + Na^+_{(aq)} + 3OH^-_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + Na^+_{(aq)} + NO^-_{3\;(aq)} \nonumber$
Canceling out spectator ions leaves the net ionic equation:
$Fe^{3+}_{(aq)} + OH^-_{(aq)} \rightarrow Fe(OH)_{\;3(s)} \nonumber$
2. From the double replacement reaction, the products are $AlCl_3$ and $BaSO_4$. $AlCl_3$ is soluble because it contains a chloride (rule 3); however, $BaSO_4$ is insoluble: it contains a sulfate, but the $Ba^{2+}$ ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing):
$2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3Ba^{2+}_{(aq)} + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+}_{(aq)} +6Cl^-_{(aq)} + 3BaSO_{4\;(s)} \nonumber$
Canceling out spectator ions leaves the following net ionic equation:
$Ba^{2+}_{(aq)} + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)} \nonumber$
3. From the double replacement reaction, the products $HNO_3$ and $ZnI_2$ are formed. Looking at the solubility rules, $HNO_3$ is soluble because it contains nitrate (rule 2), and $ZnI_2$ is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs.
4. The products of this double replacement reaction are $Ca_3(PO_4)_2$ and $NaCl$. Rule 1 states that $NaCl$ is soluble, and according to solubility rule 6, $Ca_3(PO_4)_2$ is insoluble. The ionic equation is:
$Ca^{2+}_{(aq)}+ Cl^-_{(aq)} + Na^+_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+_{(aq)} + Cl^-_{(aq)} \nonumber$
After canceling out spectator ions, the net ionic equation is given below:
$Ca^{2+}_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} \nonumber$
5. The first product of this reaction, $PbSO_4$, is soluble according to rule 4 because it is a sulfate. The second product, $KNO_3$, is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs.
Contributors and Attributions
• Julie Schaffer (UCD), Corinne Herman (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.08_Stoichiometry_of_Precipitation_Reactions.txt |
Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. Simply stated, a chemical reaction is the process where reactants are transformed into products. How chemicals react is dictated by the chemical properties of the element or compound- the ways in which a compound or element undergoes changes in composition.
Describing Reactions Quantitatively
Chemical reactions are constantly occurring in the world around us; everything from the rusting of an iron fence to the metabolic pathways of a human cell are all examples of chemical reactions. Chemistry is an attempt to classify and better understand these reactions.
A chemical reaction is typically represented by a chemical equation, which represents the change from reactants to products. The left hand side of the equation represents the reactants, while the right hand side represents the products. A typical chemical reaction is written with stoichiometric coefficients, which show the relative amounts of products and reactants involved in the reaction. Each compound is followed by a parenthetical note of the compound’s state of 2: (l) for liquid, (s) for solid, (g) for gas. The symbol (aq) is also commonly used in order to represent an aqueous solution, in which compounds are dissolved in water. A reaction might take the following form:
$\ce{A (aq) + B (g) \rightarrow C (s) + D (l)} \nonumber$
In the above example, $A$ and $B$, known as the reactants, reacted to form $C$ and $D$, the products.
To write an accurate chemical equation, two things must occur:
1. Each product and reactant must be written using its chemical formula, e.g., $H_2$
2. The number of atoms of each element must be equal on both sides of the equation. Coefficients are used in front of the chemical formulas in order to help balance the number of atoms, e.g.,
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
Example $1$: Balancing Reactions
Hydrogen and nitrogen react together in order to produce ammonia gas, write the chemical equation of this reaction.
Solution
Step 1: Write each product and reactant using its chemical formula.
$\ce{H_2 + N_2 \rightarrow NH_3} \nonumber$
Step 2: Ensure the number of atoms of each element are equal on both sides of the equation.
$\ce{3H_2 + N_2 \rightarrow 2NH_3} \nonumber$
In order to balance this equation, coefficients must be used. Since there are only 2 nitrogen atoms present on the left side of the equation, a coefficient of 2 must be added to $NH_3$.
Stoichiometry
The coefficient that used for balancing the equation is called the stoichiometric coefficient. The coefficients tell us the ratio of each element in a chemical equation. For example
$\ce{2Mg + O_2 \rightarrow 2MgO} \nonumber$
means
• 2 moles of MgO is produced for every 2 moles of Mg consumed.
• 2 moles of MgO is produced for every 1 mole of O2 consumed.
When all of the reactants of a reaction are completely consumed, the reaction is in perfect stoichiometric proportions. Often times, however, a reaction is not in perfect stoichiometric proportions, leading to a situation in which the entirety of one reactant is consumed, but there is some of another reactant remaining. The reactant that is entirely consumed is called the limiting reactant, and it determines how much of the products are produced.
Example $2$: Limiting Reagent
4.00 g of hydrogen gas mixed with 20.0g of oxygen gas. How many grams of water are produced?
Solution
$n(H_2)=\dfrac{4g}{(1.008 \times2)g/mol}=1.98mol \nonumber$
So theoretically, it requires 0.99 mol of $O_2$
n(O2)=n(H2)*(1mol O2/2mol H2)=0.99 mol
m(O2)=n(O2)*(16g/mol *2) = 31.7 g O2
Because $O_2$ only has 20.0 g, less than the required mass. It is limiting.
Often, reactants do not react completely, resulting in a smaller amount of product formed than anticipated. The amount of product expected to be formed from the chemical equation is called the theoretical yield. The amount of product that is produced during a reaction is the actual yield. To determine the percent yield:
Percent yield =actual yield/theoretical yield X 100%
Chemical reactions do not only happen in the air, but also exist in solutions. In a solution, the solvent is the compound that is dissolved, and the solute is the compound that the solvent is dissolved in. The molarity of a solution is the number of moles of a solvent divided by the number of liters of solution.
$\ Molarity=\dfrac{ \text{amount of solute (mol)}}{\text{volume of solution (L)}} \nonumber$
$\ M=\dfrac{n}{V} \nonumber$
Example $3$: Concentrations
100.0 g NaCl is dissolved in 50.00 ml water. What is the molarity of the solution?
Solution
a) Find the amount of solute in moles.
100.0g/(22.99 g/mol+35.45 g/mol) =1.711 moles
b) Convert mL to L.
50.00 mL=0.05000 L
c) Find the molarity
1.711 moles/0.05000L=34.22 mol/L
Physical Changes During Chemical Reactions
Physical change is the change in physical properties. Physical changes usually occur during chemical reactions, but does not change the nature of substances. The most common physical changes during reactions are the change of color, scent and evolution of gas. However, when physical changes occur, chemical reactions may not occur.
Types of Chemical Reactions
Precipitation, or double-replacement reaction
A reaction that occurs when aqueous solutions of anions (negatively charged ions) and cations (positively charged ions) combine to form a compound that is insoluble is known as precipitation. The insoluble solid is called the precipitate, and the remaining liquid is called the supernate. See Figure2.1
Real life example: The white precipitate formed by acid rain on a marble statue:
$CaCO_3(aq)+H_2SO_4(aq) \rightarrow CaSO_4(s)+H_2O(l)+CO_2(g) \nonumber$
Example $4$: Precipitation
An example of a precipitation reaction is the reaction between silver nitrate and sodium iodide. This reaction is represented by the chemical equation :
AgNO3 (aq)+ NaI (aq) → AgI (s) + NaNO3 (aq)
Since all of the above species are in aqueous solutions, they are written as ions, in the form:
Ag+ +NO3- (aq)+ Na+ (aq) + I- (aq) → AgI (s) + Na+ (aq) + NO3- (aq)
Ions that appear on both sides of the equation are called spectator ions. These ions do not affect the reaction and are removed from both sides of the equation to reveal the net ionic equation, as written below:
Ag+ (aq) + I- (aq) → AgI (s)
In this reaction, the solid, AgI, is known as the precipitate. The formation of a precipitate is one of the many indicators that a chemical reaction has taken place.
Acid-base, or neutralization reaction
A neutralization reaction occurs when an acid and base are mixed together. An acid is a substance that produces H+ ions in solution, whereas a base is a substance that that produces OH- ions in solution. A typical acid-base reaction will produce an ionic compound called a salt and water. A typical acid-base reaction is the reaction between hydrochloric acid and sodium hydroxide. This reaction is represented by the equation:
$\ce{HCl (aq) + NaOH (aq) \rightarrow NaCl (aq)+ H_2O (l)} \nonumber$
In this reaction, $HCl$ is the acid, $NaOH$ is the base, and $NaCl$ is the salt. Real life example: Baking soda reacts with vinegar is a neutralization reaction.
Video: Vinegar and Baking Soda Reaction with Explanation
Oxidation-Reduction (Redox) Reactions
A redox reaction occurs when the oxidation number of atoms involved in the reaction are changed. Oxidation is the process by which an atom’s oxidation number is increased, and reduction is the process by which an atom’s oxidation number is decreased. If the oxidation states of any elements in a reaction change, the reaction is an oxidation-reduction reaction. An atom that undergoes oxidation is called the reducing agent, and the atom that undergoes reduction is called the oxidizing agent. An example of a redox reaction is the reaction between hydrogen gas and fluorine gas:
$H_2 (g) + F_2 (g) \rightarrow 2HF (g) \label{redox1}$
In this reaction, hydrogen is oxidized from an oxidation state of 0 to +1, and is thus the reducing agent. Fluorine is reduced from 0 to -1, and is thus the oxidizing agent.
Real life example: The cut surface of an apple turns brownish after exposed to the air for a while.
Video: Why Do Apples Turn Brown?
Combustion Reaction
A combustion reaction is a type of redox reaction during which a fuel reacts with an oxidizing agent, resulting in the release of energy as heat. Such reactions are exothermic, meaning that energy is given off during the reaction. An endothermic reaction is one which absorbs heat. A typical combustion reaction has a hydrocarbon as the fuel source, and oxygen gas as the oxidizing agent. The products in such a reaction would be $CO_2$ and $H_2O$.
$C_xH_yO_z+O_2 \rightarrow CO_2+H_2O \;\;\; \text{(unbalanced)} \nonumber$
Such a reaction would be the combustion of glucose in the following equation
$C_6H_{12}O_6 (s) + 6O_2 (g) \rightarrow 6CO_2 (g) + 6H_2O (g) \nonumber$
Real life example: explosion; burning.
Video: Combustion reactions come in many varieties. Here's a collection of various examples, all of which require oxygen, activation energy, and of course, fuel
Synthesis Reactions
A synthesis reaction occurs when one or more compounds combines to form a complex compound. The simplest equation of synthesis reaction is illustrated below.
An example of such a reaction is the reaction of silver with oxygen gas to form silver oxide:
$2Ag (s) +O_2 (g) \rightarrow 2AgO (s) \nonumber$
Real life example: Hydrogen gas is burned in air (reacts with oxygen) to form water:
$2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber$
Decomposition Reactions
A decomposition reaction is the opposite of a synthesis reaction. During a decomposition reaction, a more complex compound breaks down into multiple simpler compounds. A classic example of this type of reaction is the decomposition of hydrogen peroxide into oxygen and hydrogen gas:
$H_2O_2 (l) \rightarrow H_2 (g) + O_2 (g) \nonumber$
Single Replacement Reactions
A type of oxidation-reduction reaction in which an element in a compound is replaced by another element.
An example of such a reaction is:
$Cu (s) + AgNO_3 (aq) \rightarrow Ag(s) + Cu(NO_3)_2 (aq) \nonumber$
This is also a redox reaction.
Problems
1) C3H6O3 + O2 → CO2 (g) +H2O (g)
a) What type of reaction is this?
b) Is is exothermic or endothermic? Explain.
2) Given the oxidation-reduction reaction :
Fe (s) + CuSO4 (aq)→ FeSO4 (aq)+ Cu (s)
a) Which element is the oxidizing agent and which is the reducing agent?
b) How do the oxidation states of these species change?
3) Given the equation:
AgNO3 (aq) + KBr (aq) → AgBr (s) +KNO3 (aq)
a) What is the net ionic reaction?
b) Which species are spectator ions?
4) 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) +2 H2O (l)
a) In this reaction, which species is the acid and which is the base?
b) Which species is the salt?
c) If 2 moles of HNO3 and 1 mole of Sr(OH)2 are used, resulting in 0.85 moles of Sr(NO3)2 , what is the percent yield (with respect to moles) of Sr(NO3)2 ?
5) Identify the type of the following reactions:
a) Al(OH)3 (aq) + HCl (aq) → AlCl3 (aq) + H2O (l)
b) MnO2 + 4H+ + 2Cl- → Mn2+ + 2H2O (l) + Cl2 (g)
c) P4 (s) + Cl2 (g) → PCl3 (l)
d) Ca (s) + 2H2O (l) → Ca(OH)2 (aq) + H2 (g)
e) AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Solutions
1a) It is a combustion reaction
1b) It is exothermic, because combustion reactions give off heat
2a) Cu is the oxidizing agent and Fe is the reducing agent
2b) Fe changes from 0 to +2, and Cu changes from +2 to 0.
3a) Ag+ (aq) + Br- (aq) → AgBr (s)
3b) The spectator ions are K+ and NO3-
4a) HNO3 is the acid and Sr(OH)2 is the base
4b) Sr(NO3)2 is the salt
4c) According to the stoichiometric coefficients, the theoretical yield of Sr(NO3)2 is one mole. The actual yield was 0.85 moles. Therefore the percent yield is:
(0.85/1.0) * 100% = 85%
5a) Acid-base
5b) Oxidation-reduction
5c) Synthesis
5d) Single-replacement reaction
5e) Double replacement reaction
• Priya Muley | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.09_Acid-Base_Reactions.txt |
An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Redox reactions are common and vital to some of the basic functions of life, including photosynthesis, respiration, combustion, and corrosion or rusting.
Rules for Assigning Oxidation States
The oxidation state (OS) of an element corresponds to the number of electrons, e-, that an atom loses, gains, or appears to use when joining with other atoms in compounds. In determining the oxidation state of an atom, there are seven guidelines to follow:
1. The oxidation state of an individual atom is 0.
2. The total oxidation state of all atoms in: a neutral species is 0 and in an ion is equal to the ion charge.
3. Group 1 metals have an oxidation state of +1 and Group 2 an oxidation state of +2
4. The oxidation state of fluorine is -1 in compounds
5. Hydrogen generally has an oxidation state of +1 in compounds
6. Oxygen generally has an oxidation state of -2 in compounds
7. In binary metal compounds, Group 17 elements have an oxidation state of -1, Group 16 elements of -2, and Group 15 elements of -3.
The sum of the oxidation states is equal to zero for neutral compounds and equal to the charge for polyatomic ion species.
Example $1$: Assigning Oxidation States
Determine the Oxidation States of each element in the following reactions:
1. $\ce{Fe(s) + O2(g) -> Fe2O3(g)}$
2. $\ce{Fe^{2+}(aq)}$
3. $\ce{Ag(s) + H2S -> Ag2S(g) + H2(g)}$
Solutions
1. $\ce{Fe}$ and $\ce{O2}$ are free elements; therefore, they each have an oxidation state of 0 according to Rule #1. The product has a total oxidation state equal to 0, and following Rule #6, $\ce{O}$ has an oxidation state of -2, which means $\ce{Fe}$ has an oxidation state of +3.
2. The oxidation state of $\ce{Fe}$ ions just corresponds to its charge since it is a single element species; therefore, the oxidation state is +2.
3. $\ce{Ag}$ has an oxidation state of 0, $\ce{H}$ has an oxidation state of +1 according to Rule #5, $\ce{H2}$ has an oxidation state of 0, $\ce{S}$ has an oxidation state of -2 according to Rule #7, and hence $\ce{Ag}$ in $\ce{Ag2S}$ has an oxidation state of +1.
Example $2$: Assigning Oxidation States
Determine the oxidation states of the phosphorus atom bold element in each of the following species:
1. $\ce{Na3PO3}$
2. $\ce{H2PO4^{-}}$
Solutions
1. The oxidation numbers of $\ce{Na}$ and $\ce{O}$ are +1 and -2. Because sodium phosphite is neutral species, the sum of the oxidation numbers must be zero. Letting x be the oxidation number of phosphorus, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3.
2. Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-2), y= oxidation number of P= +5.
Example $3$: Identifying Reduced and Oxidized Elements
Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each):
1. $\ce{Zn + 2H^{+} -> Zn^{2+} + H2}$
2. $\ce{2Al + 3Cu^{2+} -> 2Al^{3+} +3Cu}$
3. $\ce{CO3^{2-} + 2H^{+} -> CO2 + H2O}$
Solutions
1. Zn is oxidized (Oxidation number: 0 → +2); H+ is reduced (Oxidation number: +1 → 0)
2. Al is oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0)
3. This is not a redox reaction because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.
An atom is oxidized if its oxidation number increases, the reducing agent, and an atom is reduced if its oxidation number decreases, the oxidizing agent. The atom that is oxidized is the reducing agent, and the atom that is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound).
Oxidation-Reduction Reactions
Redox reactions are comprised of two parts, a reduced half and an oxidized half, that always occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half loses electrons and the oxidation number increases. Simple ways to remember this include the mnemonic devices OIL RIG, meaning "oxidation is loss" and "reduction is gain." There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken up by another species in the reduction half reaction.
The two species that exchange electrons in a redox reaction are given special names:
1. The ion or molecule that accepts electrons is called the oxidizing agent - by accepting electrons it oxidizes other species.
2. The ion or molecule that donates electrons is called the reducing agent - by giving electrons it reduces the other species.
Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound, as in disproportionation reactions discussed below).
A good example of a redox reaction is the thermite reaction, in which iron atoms in ferric oxide lose (or give up) $\ce{O}$ atoms to $\ce{Al}$ atoms, producing $\ce{Al2O3}$.
$\ce{Fe2O3(s) + 2Al(s) \rightarrow Al2O3(s) + 2Fe(l)} \nonumber$
Example $4$: Identifying Oxidizing and Reducing Agents
Determine what is the oxidizing and reducing agents in the following reaction.
$\ce{Zn + 2H^{+} -> Zn^{2+} + H2} \nonumber$
Solution
The oxidation state of $\ce{H}$ changes from +1 to 0, and the oxidation state of $\ce{Zn}$ changes from 0 to +2. Hence, $\ce{Zn}$ is oxidized and acts as the reducing agent. $\ce{H^{+}}$ ion is reduced and acts as the oxidizing agent.
Combination Reactions
Combination reactions are among the simplest redox reactions and, as the name suggests, involves "combining" elements to form a chemical compound. As usual, oxidation and reduction occur together. The general equation for a combination reaction is given below:
$\ce{ A + B -> AB} \nonumber$
Example $5$: Combination Reaction
Consider the combination reaction of hydrogen and oxygen
$\ce{H2 + O2 -> H2O } \nonumber$
Solution
0 + 0 → (2)(+1) + (-2) = 0
In this reaction both H2 and O2 are free elements; following Rule #1, their oxidation states are 0. The product is H2O, which has a total oxidation state of 0. According to Rule #6, the oxidation state of oxygen is usually -2. Therefore, the oxidation state of H in H2O must be +1.
Decomposition Reactions
A decomposition reaction is the reverse of a combination reaction, the breakdown of a chemical compound into individual elements:
$\ce{AB -> A + B} \nonumber$
Example $6$: Decomposition Reaction
Consider the following reaction:
$\ce{H2O -> H2 + O2}\nonumber$
This follows the definition of the decomposition reaction, where water is "decomposed" into hydrogen and oxygen.
(2)(+1) + (-2) = 0 → 0 + 0
As in the previous example the $\ce{H2O}$ has a total oxidation state of 0; thus, according to Rule #6 the oxidation state of oxygen is usually -2, so the oxidation state of hydrogen in $\ce{H2O}$ must be +1.
Note that the autoionization reaction of water is not a redox nor decomposition reaction since the oxidation states do not change for any element:
$\ce{H2O -> H^{+} + OH^{-}}\nonumber$
Single Replacement Reactions
A single replacement reaction involves the "replacing" of an element in the reactants with another element in the products:
$\ce{A + BC -> AB + C} \nonumber$
Example $7$: Single Replacement Reaction
Equation:
$\ce{Cl_2 + Na\underline{Br} \rightarrow Na\underline{Cl} + Br_2 } \nonumber$
Calculation:
(0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = 0) + 0
In this equation, $\ce{Br}$ is replaced with $\ce{Cl}$, and the $\ce{Cl}$ atoms in $\ce{Cl2}$ are reduced, while the $\ce{Br}$ ion in $\ce{NaBr}$ is oxidized.
Double Replacement Reactions
A double replacement reaction is similar to a single replacement reaction, but involves "replacing" two elements in the reactants, with two in the products:
$\ce{AB + CD -> AD + CB} \nonumber$
An example of a double replacement reaction is the reaction of magnesium sulfate with sodium oxalate
$\ce{MgSO4(aq) + Na2C2O4(aq) -> MgC2O4(s) + Na2SO4(aq)} \nonumber$
Combustion Reactions
Combustion is the formal terms for "burning" and typically involves a substance reacts with oxygen to transfer energy to the surroundings as light and heat. Hence, combustion reactions are almost always exothermic. For example, internal combustion engines rely on the combustion of organic hydrocarbons $\ce{C_{x}H_{y}}$ to generate $\ce{CO2}$ and $\ce{H2O}$:
$\ce{C_{x}H_{y} + O2 -> CO2 + H2O}\nonumber$
Although combustion reactions typically involve redox reactions with a chemical being oxidized by oxygen, many chemicals can "burn" in other environments. For example, both titanium and magnesium metals can burn in nitrogen as well:
$\ce{ 2Ti(s) + N2(g) -> 2TiN(s)} \nonumber$
$\ce{ 3 Mg(s) + N2(g) -> Mg3N2(s)} \nonumber$
Moreover, chemicals can be oxidized by other chemicals than oxygen, such as $\ce{Cl2}$ or $\ce{F2}$; these processes are also considered combustion reactions.
Example $8$: Identifying Combustion Reactions
Which of the following are combustion reactions?
1. $\ce{2H2O → 2H2 + O2}$
2. $\ce{4Fe + 3O2 → 2Fe2O3}$
3. $\ce{2AgNO3 + H2S → Ag2S + 2NHO3}$
4. $\ce{2Al + N2 → 2AlN4}$
Solution
Both reaction b and reaction d are combustion reactions, although with different oxidizing agents. Reaction b is the conventional combustion reaction using $\ce{O2}$ and reaction uses $\ce{N2}$ instead.
Disproportionation Reactions
In disproportionation reactions, a single substance can be both oxidized and reduced. These are known as disproportionation reactions, with the following general equation:
$\ce{2A -> A^{+n} + A^{-n}} \nonumber$
where $n$ is the number of electrons transferred. Disproportionation reactions do not need begin with neutral molecules, and can involve more than two species with differing oxidation states (but rarely).
Example $9$: Disproportionation Reaction
Disproportionation reactions have some practical significance in everyday life, including the reaction of hydrogen peroxide, $\ce{H2O2}$ poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. The reaction is as follows:
$\ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \nonumber$
Dicussion
On the reactant side, $\ce{H}$ has an oxidation state of +1 and $\ce{O}$ has an oxidation state of -1, which changes to -2 for the product $\ce{H2O}$ (oxygen is reduced), and 0 in the product $\ce{O2}$ (oxygen is oxidized).
Exercise $9$
Which element undergoes a bifurcation of oxidation states in this disproportionation reaction:
$\ce{HNO2 -> HNO3 + NO + H2O} \nonumber$
Answer
The $\ce{N}$ atom undergoes disproportionation. You can confirm that by identifying the oxidation states of each atom in each species.
Contributors and Attributions
• Christopher Spohrer (UCD), Christina Breitenbuecher (UCD), Luvleen Brar (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.10_Oxidation-Reduction_Reactions.txt |
Oxidation-Reduction Reactions, or redox reactions, are reactions in which one reactant is oxidized and one reactant is reduced simultaneously. This module demonstrates how to balance various redox equations.
Identifying Redox Reactions
The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction. This requires that one and typically more species changing oxidation states during the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components. These are often separated into independent two hypothetical half-reactions to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:
$\ce{ Cu (s) + 2 Ag^+ (aq) \rightarrow Cu^{2+} (aq) + 2 Ag (s)} \nonumber$
The first step in determining whether the reaction is a redox reaction is to split the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:
$\ce{ Cu (s) \rightarrow Cu^{2+}(aq)} \nonumber$
The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in $\ce{Cu}$ to +2 in $\ce{Cu^{2+}}$. Now consider the silver atoms
$\ce{ 2 Ag^+ (aq) \rightarrow 2 Ag (s)} \nonumber$
In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an pure element. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.
Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.
Balancing Redox Reactions
Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions.
One method used to balance redox reactions is called the Half-Equation Method. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction.
Half-Equation Method to Balance redox Reactions in Acidic Aqueous Solutions
Each reaction is balanced by adjusting coefficients and adding $\ce{H2O}$, $\ce{H^{+}}$, and $\ce{e^{-}}$ in this order:
1. Balance elements in the equation other than $\ce{O}$ and $\ce{H}$.
2. Balance the oxygen atoms by adding the appropriate number of water ($\ce{H2O}$) molecules to the opposite side of the equation.
3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding $\ce{H^{+}}$ ions to the opposite side of the equation.
4. Add up the charges on each side. Make them equal by adding enough electrons ($\ce{e^{-}}$) to the more positive side. (Rule of thumb: $\ce{e^{-}}$ and $\ce{H^{+}}$ are almost always on the same side.)
5. The $\ce{e^{-}}$ on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.
The equation can now be checked to make sure that it is balanced.
Half-Equation Method to Balance redox Reactions in Basic Aqueous Solutions
If the reaction is being balanced in a basic solution, the above steps are modified with the the addition of one step between #3 and #4:
3b Add the appropriate number of $\ce{OH^{-}}$ to neutralize all $\ce{H^{+}}$ and to convert into water molecules.
The equation can now be checked to make sure that it is balanced.
Neutral Conditions
The first step to balance any redox reaction is to separate the reaction into half-reactions. The substance being reduced will have electrons as reactants, and the oxidized substance will have electrons as products. (Usually all reactions are written as reduction reactions in half-reaction tables. To switch to oxidation, the whole equation is reversed and the voltage is multiplied by -1.) Sometimes it is necessary to determine which half-reaction will be oxidized and which will be reduced. In this case, whichever half-reaction has a higher reduction potential will by reduced and the other oxidized.
Example $1$: Balancing in a Neutral Solution
Balance the following reaction
$\ce{Cu^+(aq) + Fe(s) \rightarrow Fe^{3+} (aq) + Cu (s)} \nonumber$
Solution
Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:
$\ce{Cu^+ (aq) + e^{-} \rightarrow Cu(s)} \nonumber$
and
$\ce{Fe^{3+} (aq) + 3e^{-} \rightarrow Fe(s)} \nonumber$
The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:
$\ce{Cu^+ (aq) + e^{-} \rightarrow Cu(s)} \nonumber$
and
$\ce{Fe (s) \rightarrow Fe^{3+}(aq) + 3e^{-}} \nonumber$
Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire $Cu^+(aq) + e^{-} \rightarrow Cu(s)$ half-reaction by 3 and leaving the other half reaction as it is. This gives:
$\ce{3Cu^+(aq) + 3e^{-} \rightarrow 3Cu(s)} \nonumber$
and
$\ce{Fe(s) \rightarrow Fe^{3+}(aq) + 3e^{-}} \nonumber$
Step 3: Adding the equations give:
$\ce{3Cu^+(aq) + 3e^{-} + Fe(s) \rightarrow 3Cu(s) + Fe^{3+}(aq) + 3e^{-}} \nonumber$
The electrons cancel out and the balanced equation is left.
$\ce{3Cu^{+}(aq) + Fe(s) \rightarrow 3Cu(s) + Fe^{3+}(aq)} \nonumber$
Acidic Conditions
Acidic conditions usually implies a solution with an excess of $\ce{H^{+}}$ concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions. However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add $\ce{H2O}$ molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons ($\ce{H^{+}}$). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.
Example $2$: Balancing in a Acid Solution
Balance the following redox reaction in acidic conditions.
$\ce{Cr_2O_7^{2-} (aq) + HNO_2 (aq) \rightarrow Cr^{3+}(aq) + NO_3^{-}(aq) } \nonumber$
Solution
Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.
$\ce{Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq) } \nonumber$
$\ce{HNO_2 (aq) \rightarrow NO_3^{-}(aq)} \nonumber$
Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:
$\ce{Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq)} \nonumber$
and
$\ce{HNO_2(aq) \rightarrow NO_3^{-}(aq)} \nonumber$
Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 $\ce{H2O}$ molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:
$\ce{Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber$
and
$\ce{HNO_2(aq) + H_2O(l) \rightarrow NO_3^{-}(aq) } \nonumber$
Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.
$\ce{14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber$
and
$\ce{HNO_2 (aq) + H2O (l) \rightarrow 3H^+(aq) + NO_3^{-}(aq)} \nonumber$
Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:
$\ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)} \nonumber$
For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:
$\ce{HNO_2(aq) + H_2O(l) \rightarrow 3H^+(aq) + NO_3^{-}(aq) + 2e^{-}} \nonumber$
Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:
$\ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l).} \nonumber$
and
\begin{align*} 3 \times \big[ \ce{HNO2 (aq) + H2O(l)} &\rightarrow \ce{3H^{+}(aq) + NO3^{-} (aq) + 2e^{-}} \big] \[4pt] \ce{3HNO2 (aq) + 3H_2O (l)} &\rightarrow \ce{9H^{+}(aq) + 3NO_3^{-}(aq) + 6e^{-}} \end{align*} \nonumber
Step 7: Add the reactions and cancel out common terms.
\begin{align*} \ce{3HNO_2 (aq) + 3H_2O (l)} &\rightarrow \ce{9H^+(aq) + 3NO_3^{-}(aq) + 6e^{-} } \[4pt] \ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq)} &\rightarrow \ce{2Cr^{3+}(aq) + 7H_2O(l)} \[4pt] \hline \ce{3HNO_2 (aq)} + \cancel{\ce{3H_2O (l)}} + \cancel{6e^{-}} + \ce{14H^+(aq) + Cr_2O_7^{2-} (aq)} &\rightarrow \ce{9H^+(aq) + 3NO_3^{-}(aq)} + \cancel{6e^{-}} + \ce{2Cr^{3+}(aq)} + \cancelto{4}{7}\ce{H_2O(l)} \end{align*}
The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:
$\ce{3HNO_2(aq) + 5H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 3NO_3^{-}(aq) + 2Cr^{3+}(aq) + 4H_2O(l)} \nonumber$
Basic Conditions
Bases dissolve into $\ce{OH^{-}}$ ions in solution; hence, balancing redox reactions in basic conditions requires $\ce{OH^{-}}$. Follow the same steps as for acidic conditions. The only difference is adding hydroxide ions to each side of the net reaction to balance any $\ce{H^{+}}$. $\ce{OH^{-}}$ and $\ce{H^{+}}$ ions on the same side of a reaction should be added together to form water. Again, any common terms can be canceled out.
Example $1$: Balancing in Basic Solution
Balance the following redox reaction in basic conditions.
$\ce{ Ag(s) + Zn^{2+}(aq) \rightarrow Ag_2O(aq) + Zn(s)} \nonumber$
Solution
Go through all the same steps as if it was in acidic conditions.
Step 1: Separate the half-reactions.
$\ce{Ag (s) \rightarrow Ag_2O (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 2: Balance elements other than O and H.
$\ce{ 2Ag (s) \rightarrow Ag_2O (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 3: Add H2O to balance oxygen.
$\ce{H_2O(l) + 2Ag(s) \rightarrow Ag_2O(aq)} \nonumber$
$\ce{Zn^{2+}(aq) \rightarrow Zn(s)} \nonumber$
Step 4: Balance hydrogen with protons.
$\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 5: Balance the charge with e-.
$\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq) + 2e^{-}} \nonumber$
$\ce{Zn^{2+} (aq) + 2e^{-} \rightarrow Zn (s)} \nonumber$
Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.
Step 7: Add the reactions and cancel the electrons.
$\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq). } \nonumber$
Step 8: Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side.
$\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) + 2OH^{-}(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq) + 2OH^{-}(aq).} \nonumber$
Step 9: Combine OH- ions and H+ ions that are present on the same side to form water.
$\ce{\cancel{H2O(l)} + 2Ag(s) + Zn^{2+}(aq) + 2OH^{-}(aq) -> Zn(s) + Ag_2O(aq) + \cancel{2}H_2O(l)} \nonumber$
Step 10: Cancel common terms.
$\ce{2Ag(s) + Zn^{2+}(aq) + 2OH^{-} (aq) \rightarrow Zn(s) + Ag_2O(aq) + H_2O(l)} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/04%3A_Types_of_Chemical_Reactions_and_Solution_Stoichiometry/4.11%3A_Balancing_Redox_Equations.txt |
Learning Objectives
• To describe the characteristics of a gas.
The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. Figure \(1\) compares the three states of matter and illustrates the differences at the molecular level.
The state of a given substance depends strongly on conditions. For example, H2O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor (the term vapor refers to the gaseous form of a substance that is a liquid or a solid under normal conditions so nitrogen (N2) and oxygen (O2) are referred to as gases, but gaseous water in the atmosphere is called water vapor) is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in Table \(1\).
Table \(1\): Properties of Water at 1.0 atm
Temperature State Density (g/cm3)
≤0°C solid (ice) 0.9167 (at 0.0°C)
0°C–100°C liquid (water) 0.9997 (at 4.0°C)
≥100°C vapor (steam) 0.005476 (at 127°C)
The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do not depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure.
Figure \(2\) shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H2, N2, O2, F2, and Cl2). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O3), which is also a gas. In contrast, bromine (as Br2) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions.
All of the gaseous elements (other than the monatomic noble gases) are molecules. Within the same group (1, 15, 16 and 17), the lightest elements are gases. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms.
Defining Gas Pressure: https://youtu.be/_CRn3cFs2CI
Summary
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.01_Pressure.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{6.2.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{6.2.2}$
or
$V \propto \dfrac{1}{P} \label{6.2.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equation 6.2.1 and Equation 6.2.3. Dividing both sides of Equation 6.2.1 by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure.
Boyle’s Law: https://youtu.be/lu86VSupPO4
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{6.2.4}$
or
$V \propto T \label{6.2.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
Charles’s Law: https://youtu.be/NBf510ZnlR0
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{6.2.6}$
or
$V \propto.n \text{@ constant T and P} \label{6.2.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
Note
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount but decreases with increasing pressure.
Avogadro’s Law: https://youtu.be/dRY3Trl4T24
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.02_The_Gas_Laws_of_Boyle_Charles_and_Avogadro.txt |
Learning Objectives
• To use the ideal gas law to describe the behavior of a gas.
In this module, the relationship between Pressure, Temperature, Volume, and Amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as V) and several other variables ($P$, $T$, and $n$) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions are as follows:
• Boyle’s law
$V \propto \dfrac{1}{P} \;\; \text{@ constant n and T}$
• Charles’s law
$V \propto T \;\; \text{@ constant n and P}$
• Avogadro’s law
$V \propto n \;\; \text{@ constant T and P}$
Combining these three expressions gives
$V \propto \dfrac{nT}{P} \tag{6.3.1}$
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
$V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \tag{6.3.2}$
By convention, the proportionality constant in Equation 6.3.1 is called the gas constant, which is represented by the letter $R$. Inserting R into Equation 6.3.2 gives
$V = \dfrac{Rnt}{P} = \dfrac{nRT}{P} \tag{6.3.3}$
Clearing the fractions by multiplying both sides of Equation 6.3.4 by $P$ gives
$PV = nRT \tag{6.3.4}$
This equation is known as the ideal gas law.
An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Note
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
$R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \tag{6.3.5}$
Because the product PV has the units of energy, R can also have units of J/(K•mol):
$R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\tag{6.3.6}$
Standard Conditions of Temperature and Pressure
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and $\rm1\; bar = 100 \;kPa = 10^5\;Pa$ pressure, referred to as standard temperature and pressure (STP).
$\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa$
Please note that STP was defined differently in the part. The old definition was based on a standard pressure of 1 atm.
We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 6.3.4:
$V=\dfrac{nRT}{P}\tag{6.3.7}$
Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm, approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table $1$: Molar Volumes of Selected Gases at 0°C and 1 atm
Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example $1$
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation 6.3.4) for n, we obtain
$\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm$
B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
$T=273+30=303{\rm K}$
Substituting these values into the expression we derived for n, we obtain
$n=\dfrac{PV}{RT}=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K}=1.23\times10^3\;mol$
Exercise $1$
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer: 1.5 atm
In Example $1$, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example $5$.
The Ideal Gas Law: https://youtu.be/rHGs23368mE
General Gas Equation
When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:
$\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f)\P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array}$
Both equations can be rearranged to give:
$R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f}$
The two equations are equal to each other since each is equal to the same constant $R$. Therefore, we have:
$\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\tag{6.3.8}$
The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.
Example $2$
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example $1$?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
1. Use the results from Example $1$ for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case $P$ and $n$.
3. Solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
Initial (August) Final (January)
$T_i=30$°C = 303 K $T_f=$−10°C = 263 K
$P_i=$0.980 atm $P_f=$0.980 atm
$n_i=$1.23 × 103 mol $n_f=$1.23 × 103 mol
$V_i=31150$ L $V_f=?$
B Both $n$ and $P$ are the same in both cases ($n_i=n_f,P_i=P_f$). Therefore, Equation can be simplified to:
$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}$
This is the relationship first noted by Charles.
C Solving the equation for $V_f$, we get:
$V_f=V_i\times\dfrac{T_f}{T_i}=\rm31150\;L\times\dfrac{263\;K}{303\;K}=2.70\times10^4\;L$
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise $2$
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer: 0.52 L
Example $1$ illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example $1$ can be applied in any such case, as we demonstrate in Example $2$ (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example $3$
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example $5$.
Solution:
Prepare a table to determine which parameters change and which are held constant:
Initial Final
$V_i=0.406\;\rm L$ $V_f=0.406\;\rm L$
$n_i=0.025\;\rm mol$ $n_f=0.025\;\rm mol$
$T_i=\rm25\;^\circ C=298\;K$ $T_i=\rm750\;^\circ C=1023\;K$
$P_i=1.5\;\rm atm$ $P_f=?$
Both $V$ and $n$ are the same in both cases ($V_i=V_f,n_i=n_f$). Therefore, Equation can be simplified to:
$P_iT_i=P_fT_f$
By solving the equation for $P_f$, we get:
$P_f=P_i\times\dfrac{T_i}{T_f}=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K}=5.1\;atm$
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise $3$
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer: 23.4 atm
In Example $1$ and Example $2$, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.
Example $4$
We saw in Example $1$ that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example $5$.
Solution:
Begin by setting up a table of the two sets of conditions:
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm750-30\;^\circ C=243\;K$
$n_i=\rm1.2\times10^3\;mol$ $n_i=\rm1.2\times10^3\;mol$
$V_i=\rm31150\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, Equation 6.3.8 is simplified to:
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}$
By solving the equation for $V_f$, we get:
$V_f=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i}=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K}=5.96\times10^4\;L$
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise $4$
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer: 4.07 × 103
Second Type of Ideal Gas Law Problems: https://youtu.be/WQDJOqddPI0
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
$\dfrac{n}{V}=\dfrac{P}{RT}\tag{6.3.9}$
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ($m$, in grams) divided by its molar mass ($M$, in grams per mole):
$n=\dfrac{m}{M}\tag{6.3.10}$
Substituting this expression for $n$ into Equation 6.3.9 gives
$\dfrac{m}{MV}=\dfrac{P}{RT}\tag{6.3.11}$
Because $m/V$ is the density $d$ of a substance, we can replace $m/V$ by $d$ and rearrange to give
$\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\tag{6.3.12}$
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example $5$
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation 6.3.12 to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
$M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol$
Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
$P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm$
B Substituting these values into Equation 6.3.12 gives
$\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L$
Exercise $5$
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer: radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation 6.3.12 is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example $6$.
Example $6$
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
1. Solve Equation 6.3.12 for the molar mass of the gas and then calculate the density of the gas from the information given.
2. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
3. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation 6.3.12 for the molar mass gives
$M=\dfrac{mRT}{PV}=\dfrac{dRT}{P}$
Density is the mass of the gas divided by its volume:
$\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.17\rm L}=1.84 \rm g/L$
B We must convert the other quantities to the appropriate units before inserting them into the equation:
$T=18+273=291 K$
$P=727\rm mmHg\times\dfrac{1\rm atm}{760\rm mmHg}=0.957\rm atm$
The molar mass of the unknown gas is thus
$\rho=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol$
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
$M({\rm NO})=14 + 16=30 \rm\; g/mol$
$M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol$
$M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol$
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise $6$
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer: 44 g/mol; $CO_2$
Density and the Molar Mass of Gases: https://youtu.be/gnkGBsvUFVk
Summary
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Ideal gas equation: $PV = nRT$,
where $R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}$
General gas equation: $\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}$
Density of a gas: $\rho=\dfrac{MP}{RT}$
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.03_The_Ideal_Gas_Law.txt |
What you Should Know
Before beginning this section, you should know and understand:
Chemical reactions between gaseous materials are quite similar to reactions between solids and liquids, except the Ideal Gas Law ($PV=nRT$) can now be included in the calculations. If a chemical reaction is reversible (such as the decay and formation of dinitrogen tetraoxide), then Dalton's Law of Partial Pressure may be used to determine the moles of reactants and products at which the reaction ceases (and subsequently, the temperature, pressure and volume of each gas can be determined as well).
Non-Reversible Reactions
A non-reversible reaction uses the reactants to form the products. The reaction goes in one direction; that is, using the product to recreate the reactants has greatly different requirements. One of the most common forms of non-reversible reactions is combustion (once an organic molecule has been converted to water and hydrogen gas, it is extremely difficult to reform). Other non-reversible reactions produce a state change, such as Hydrogen Peroxide (the gaseous material produces water, a liquid). To understand how the Ideal Gas Law applies to reactions, we shall use a nonreversible reaction as an example.
Example $1$: the decomposition of Hydrogen Peroxide to water and oxygen gas
If 4.000 grams of hydrogen peroxide is placed within a sealed 250 milliliter container at 500 K. What is the pressure of the oxygen gas produced in atmospheres?
$\ce{2H_2O_2 \rightarrow 2H_2O + O_2} \nonumber$
Solution
First, we need to determine the moles of $\ce{O_2}$ produced, just like any other stoichiometric problem.
$(4g\; \cancel{\ce{H_2O_2}}) \times \left(\dfrac{1\; mol\; \cancel{\ce{H_2O_2}}}{34.016\;g\; \cancel{ \ce{H_2O_2}}} \right) \left(\dfrac{1\; mol\; \ce{O_2}}{2\;mol\; \cancel{\ce{H_2O_2}}} \right) = 0.0588 \;mol\; \ce{O_2} \nonumber$
With the moles of oxygen determined, we can now use the Ideal Gas Law to determine the pressure.
$PV=nRT \nonumber$
The volume (250 mL = 0.25 L) and temperature (500 K) are already given to us, and R (0.0820574 Latm mol-1K-1) is a constant.
\begin{align*} P &=\dfrac{nRT}{V} \[4pt] &= \dfrac{(0.0588\; mol\; O_2) \times (0.0820\; L \;atm \;mol^{-1}\;K^{-1}) \times (500 \;K)}{0.25\;L} \[4pt] &= 9.65\; atm \end{align*}
By using the Ideal Gas Law for unit conversions, properties such as the pressure, volume, moles, and temperature of a gas involved in a reaction can be determined. However, a different approach is needed to solve reversible reactions.
For further clarification, when solving equations with gases, we must remember that gases behave differently under different conditions. For example, if we have a certain temperature or pressure, this can change the number of moles produced or the volume. This is unlike regular solids where we only had to account for the mass of the solids and solve for the mass of the product by stoichiometry. In order to solve for the temperature, pressure, or volume of a gas using chemical reactions, we often need to have information on two out of three of these variables. So we need either the temperature and volume, temperature and pressure, or pressure and volume. The mass we can find using stoichiometric conversions we have learned before.
The reason why gases require additional information is because gases behave as ideal gases and ideal gases behave differently under different conditions. To account for these conditions, we use the ideal gas equation PV=nRT where P is the pressure measured in atmosphere(atm), V is the volume measured in liters (L), n is the number of moles, R is the gas constant with a value of .08206 L atm mol-1 K-1, and T is the temperature measured in kelvin (K).
Example $2$
Suppose we have the following combustion reaction (below). If we are given 2 moles of ethane at STP, how many liters of CO2 are produced?
$\ce{2C2H6(s) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$
Solution
Step 1
First use stoichiometry to solve for the number of moles of CO2 produced.
$(2\, mol\, \ce{C2H6} )\left(\dfrac{4 \,mol \ce{CO2}}{2\, mol\, \ce{ C2H6}}\right) = 4 mol \, \ce{CO2} \nonumber$
So 4 moles of Carbon Dioxide are produced if we react 2 moles of ethane gas.
Step 2
Now we simply need to manipulate the ideal gas equation to solve for the variable of interest. In this case we are solving for the number of liters.
Since we are told ethane is at STP, we know that the temperature is 273 K and the pressure is 1 atm.
So the variables we have are:
• V = ?
• T = 273K
• P = 1 atm
• n = 4 moles CO2
• R (gas constant) = 0.08206 L atm mol-1 K-1
Isolating the variable of interest from $PV=nRT$, we get
\begin{align*} V &=\dfrac{nRT}{P} \[4pt] &= \dfrac{4\, mol \times 0.08206 \,L \,atm \,mol^{-1} K^{-1} \times 273\,K}{1\, atm} \[4pt] &= 89.61\, L \end{align*}
So we have a volume of 89.61 liters.
Reversible Reactions in Gases
A reversible reaction is a chemical reaction in which reactants produces a product, which then decays back to the reactants. This continues until the products and reactants are in equilibrium. In other words, the final state of the gas includes both the reactants and the products. For example, Reactant A combines with Reactant B to form Product AB, which then breaks apart into A and B, until an equilibrium of the three is reached. In a reaction between gases, determining gas properties such as partial pressure and moles can be quite difficult. For this example, we consider the theral decomposition of Dinitrogen Tetraoxide into Nitrogen Dioxide.
Example $3$
For this example, we shall use Dinitrogen Tetraoxide, which decomposes to form Nitrogen Dioxide.
$\ce{N_2O_4 <=> 2NO_2} \nonumber$
2 atm of dinitrogen tetraoxide is added to a 500 mL container at 273 K. After several minutes, the total pressure of N2O4 and 2NO2 at equilibrium is found to be 3.2 atm. Find the partial pressures of both gases.
Solution
The simplest way of solving this problem is to begin with an ICE table.
$\ce{N2O4}$
$\ce{2NO2}$ Description of Each Letter
Initial 2atm 0atm The initial amounts of reactants and products
Change -X +2X The unknown change, X, each one multiplied by their stoichiometric factor
Equilibrium 2-Xatm 2x atm The initial + the change
With this data, a simple equation can be derived to determine the value of X.
$P_{total} = (2-X) + 2X = 3.2\;atm$
$X = 1.2\; atm$
$P_{NO_2}= 2x = 2.4\; atm$
$P_{N_20_4} = 2-x = 0.8\; atm$
Law of Combining Volumes
This law of combining volumes was first discovered by the famous scientist Gay-Lussac who noticed this relationship. He determined that if certain gases that are products and reactions in a chemical reaction are measured at the same conditions, temperature and pressure, then the volume of gas consumed/produced is equal to the ratio between the gases or the ratio of the coefficients.
Example 5.4.4
If ozone, hydrogen, and oxygen were all measured at 35oC and at 753 mmHg, then how many liters of ozone was consumed if you had 5 liters of oxygen gas?
$\ce{O3(g) + H2O(l) -> H2(g) + 2O2(g)} \nonumber$
Solution
Step 1
Identify what we are looking for and if any relationships can be spotted. In this case, we can see that there are three gases all at the same temperature and pressure, which follows Gay-Lussac’s Law of Combining Volumes. We can now proceed to use his law.
Step 2
We simply change the coefficients to volumetric ratios. So for every 1 Liter of Ozone gas we have, we produce 1 Liter of H2gas and 2 Liter of $O_2$ gas.
We are given 5 liters of Oxygen gas and want to solve for the amount of liters of ozone consumed. We simply use the 2:1 stoichiometry of the reaction.
$5 L O_2 \left(\dfrac{1\; L\; O_3}{2\; L\; O_2}\right) = 2.5\; L\; O_3$
Problems
1. A 450 mL container of oxygen gas is at STP. Hydrogen gas is pumped into the container, producing water. What is the least amount of mL of Hydrogen gas needed in order to react the oxygen to completion? $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
2. This reaction occurred at 427 Kelvin, with 37 g of $CH_4$ and an excess of oxygen. The carbon dioxide produced was captured in a 30L sealed container. What is the pressure of the carbon dioxide within the container? $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Solution
1. $900\; mL\; H_2$: To solve this question you simply use Gay-Lussac’s law of combining volume because both gases are at the same temperature at pressure. $450\; mL\; O_2 \left(\dfrac{2\; mL\; H_2}{1\; mL\; O_2}\right) = 900\; mL\; H_2$
2. $2.7\; atm$: For a question like this, you want to first determine the number of moles of the compound in interest first using stoichiometry and then using the ideal gas law to solve for the variable of interest. $37 \;g \;CH_4 \left(\dfrac{1\; mol\; CH_4}{16\; g\; CH_4}\right)\left(\dfrac{1\; mol\; CO_2}{1\; mol\; CH_4}\right)= 2.3 \;mol\; CO_2$
Now use the ideal gas law to solve for the pressure of CO2.
• PV=nRT
• P = ?
• V = 30L
• n= 2.3 mol CO2
• T = 427 K
• P = nRT/V
P = (2.3 mol * .08206 L atm mol-1 K-1 * 427 K)/(30 L)
P=2.7 atm | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.04_Gas_Stoichiometry.txt |
Learning Objectives
• To determine the contribution of each component gas to the total pressure of a mixture of gases
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture.
Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
$P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \label{6.6.1}$
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures. We can write it mathematically as
$P_{tot}= P_1+P_2+P_3+P_4 \; ... = \sum_{i=1}^n{P_i} \label{6.6.2}$
where $P_{tot}$ is the total pressure and the other terms are the partial pressures of the individual gases (up to $n$ component gases).
For a mixture of two ideal gases, $A$ and $B$, we can write an expression for the total pressure:
$P_{tot}=P_A+P_B=n_A\bigg(\dfrac{RT}{V}\bigg) + n_B\bigg(\dfrac{RT}{V}\bigg)=(n_A+n_B)\bigg(\dfrac{RT}{V}\bigg) \label{6.6.3}$
More generally, for a mixture of $n$ component gases, the total pressure is given by
$P_{tot}=(P_1+P_2+P_3+ \; \cdots +P_n)\bigg(\dfrac{RT}{V}\bigg)\label{6.6.2a}$
$P_{tot}=\sum_{i=1}^n{n_i}\bigg(\dfrac{RT}{V}\bigg)\label{6.6.2b}$
Equation 6.6.4 restates Equation 6.6.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation 6.6.4 to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example $1$.
Example $1$
Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of $O_2$ and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
1. Calculate the number of moles of $He$ and $O_2$ present.
2. Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.
Solution:
A The number of moles of $He$ is
$n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol$
The number of moles of $O_2$ is
$n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol$
B We can now use the ideal gas law to calculate the partial pressure of each:
$P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm$
$P_{\rm O_2}=\dfrac{n_{\rm O_2} RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm$
The total pressure is the sum of the two partial pressures:
$P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm$
Exercise $1$
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer: $P_{CH_4}=137 \; atm$; $P_{C_2H_6}=13.4\; atm$; $P_{tot}=151\; atm$
Mole Fractions of Gas Mixtures
The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ($X$) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ($n_{tot}$):
$x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\label{6.6.5}$
The mole fraction is a dimensionless quantity between 0 and 1. If $x_A = 1.0$, then the sample is pure $A$, not a mixture. If $x_A = 0$, then no $A$ is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas $A$ to the total pressure of a gas mixture that contains $A$. We can use the ideal gas law to describe the pressures of both gas $A$ and the mixture: $P_A = n_ART/V$ and $P_{tot} = n_tRT/V$. The ratio of the two is thus
$\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \label{6.6.6}$
Rearranging this equation gives
$P_A = x_AP_{tot} \label{6.6.7}$
That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% $N_2$, 21% $O_2$, and 0.9% $Ar$, with trace amounts of gases such as $CO_2$, $H_2O$, and others. This means that 78% of the particles present in the atmosphere are $N_2$; hence the mole fraction of $N_2$ is 78%/100% = 0.78. Similarly, the mole fractions of $O_2$ and $Ar$ are 0.21 and 0.009, respectively. Using Equation 6.6.7, we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of $O_2$ and $Ar$ are 0.21 and 0.009 atm, respectively.
Example $2$
We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.
Inhaled Air / mmHg Exhaled Air / mmHg
$P_{\rm N_2}$ 597 568
$P_{\rm O_2}$ 158 116
$P_{\rm H_2O}$ 0.3 28
$P_{\rm CO_2}$ 5 48
$P_{\rm Ar}$ 8 8
$P_{tot}$ 767 767
Given: pressures of gases in inhaled and exhaled air
Asked for: mole fractions of gases in exhaled air
Strategy:
Calculate the mole fraction of each gas using Equation 6.6.7.
Solution:
The mole fraction of any gas $A$ is given by
$x_A=\dfrac{P_A}{P_{tot}}$
where $P_A$ is the partial pressure of $A$ and $P_{tot}$ is the total pressure. For example, the mole fraction of $CO_2$ is given as:
$x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063$
The following table gives the values of $x_A$ for the gases in the exhaled air.
Gas Mole Fraction
${\rm N_2}$ 0.741
${\rm O_2}$ 0.151
${\rm H_2O}$ 0.037
${\rm CO_2}$ 0.063
${\rm Ar}$ 0.010
Exercise $2$
Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.
Answer
$P_{\rm CO_2}=\rm86\; atm$, $P_{\rm N_2}=\rm2.7\;atm$
Dalton’s Law of Partial Pressures: https://youtu.be/y5-SbspyvBA
Summary
• The partial pressure of each gas in a mixture is proportional to its mole fraction.
The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.05_Dalton%27s_Law_of_Partial_Pressures.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases.
Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a momentum transfer (impulse) from molecules to the walls (Figure $2$).
The momentum transfer to the wall perpendicular to $x$ axis as a molecule with an initial velocity $u_x$ in $x$ direction hits is expressed as:
$\Delta p_x=2mu_x \label{6.7.1}$
The collision frequency, a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume.
$f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{6.7.2}$
The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency.
$P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{6.7.3}$
At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace $u_x^2$ in the expression above with the average value of $u_x^2$, which is denoted by $\overline{u_x^2}$. The overbar designates the average value over all molecules.
The exact expression for pressure is given as :
$P=\dfrac{N}{V}m\overline{u_x^2} \label{6.7.4}$
Finally, we must consider that there is nothing special about $x$ direction. We should expect that $\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}$. Here the quantity $\overline{u^2}$ is called the mean-square speed defined as the average value of square-speed ($u^2$) over all molecules. Since $u^2=u_x^2+u_y^2+u_z^2$ for each molecule, $\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}$. By substituting $\dfrac{1}{3}\overline{u^2}$ for $\overline{u_x^2}$ in the expression above, we can get the final expression for the pressure:
$P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{6.7.5}$
Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.
Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the average translational kinetic energy of the molecules of a gas $(\overline{e_K})$, which can be represented as and states that at a given Kelvin temperature $(T)$, all gases have the same value of
$\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{6.7.6}$
where $N_A$ is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by $N_A$:
$N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{6.7.7}$
where $M$ is the molar mass of the gas molecules and is related to the molecular mass by $M=N_Am$.
By rearranging the equation, we can get the relationship between the root-mean square speed ($u_{\rm rms}$) and the temperature.
The rms speed ($u_{\rm rms}$) is the square root of the sum of the squared speeds divided by the number of particles:
$u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{6.7.8}$
where $N$ is the number of particles and $u_i$ is the speed of particle $i$.
The relationship between $u_{\rm rms}$ and the temperature is given by:
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{6.7.9}$
In this equation, $u_{\rm rms}$ has units of meters per second; consequently, the units of molar mass $M$ are kilograms per mole, temperature $T$ is expressed in kelvins, and the ideal gas constant $R$ has the value 8.3145 J/(K•mol).
The equation shows that $u_{\rm rms}$ of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones.
The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation 6.7.8 tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.
Note
At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.
Example $1$
The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed ($v_{\rm av}$) root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$).
Given: particle speeds
Asked for: average speed ($v_{\rm av}$), root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$)
Strategy:
Use Equation 6.7.6 to calculate the average speed and Equation 6.7.8 to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.
Solution:
The average speed is the sum of the speeds divided by the number of particles:
$v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s$
The rms speed is the square root of the sum of the squared speeds divided by the number of particles:
$v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s$
The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence $v_{\rm m}=6.0$ m/s. The $v_{\rm rms}$ of the particles, which is related to the average kinetic energy, is greater than their average speed.
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in Chapter 14. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure $1$. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure $1$ were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• Pressure versus Volume: At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.
• Volume versus Temperature: Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
• Pressure of Gas Mixtures: Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Example $2$
The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the
1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?
Given: temperatures and volume
Asked for: effect of increase in temperature
Strategy:
Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.
Solution:
1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.
Exercise $2$
A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the
1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?
Answer: a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases
Kinetic-Molecular Theory of Gases: https://youtu.be/9f83XAYfXAg
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T,$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}},$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}.$
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.06_The_Kinetic_Molecular_Theory_of_Gases.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
Diffusion and Effusion
As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment.
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{6.8.1}$
Note
At a given temperature, heavier molecules move more slowly than lighter molecules.
Example $1$
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C).
1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?
Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
Strategy:
1. Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.
2. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.
Solution:
1. The molar mass of 238UF6 is 238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation 6.8.1:
$\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043$
2. rate U235F6rate U238F6=352.04349.03=1.0043
Thus passing UF6 containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% 235UF6.
3. B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: final purity = (initial purity)(separation)n
In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give
$1.0043^n=\dfrac{0.990}{0.00720}=137.50$
4. Taking the logarithm of both sides gives
$n\ln(1.0043)=\ln(137.50)$
$n=\dfrac{\ln(137.50)}{\ln(1.0043)}=1148$
Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Figure $2$ shows a small part of a system that is used to prepare enriched uranium on a large scale.
Exercise $1$
Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.
1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?
Answer: a. ratio of effusion rates = 1.15200; one step gives 0.000154% 3He; b. 96 steps
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 6.8.4 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure $3$ for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Note
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure $4$.
The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).
Note
The denser the gas, the shorter the mean free path.
Example $2$
Calculate the rms speed of a sample of cis-2-butene (C4H8) at 20°C.
Given: compound and temperature
Asked for: rms speed
Strategy:
Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 6.8.5 to calculate the rms speed of the gas.
Solution:
To use Equation 6.8.4, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s$
or approximately 810 mi/h.
Exercise $2$
Calculate the rms speed of a sample of radon gas at 23°C.
Answer: 1.82 × 102 m/s (about 410 mi/h)
The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In Section 10.8, we explore how the behavior of real gases differs from that of ideal gases.
Graham’s law of Diffusion and Effusion: https://youtu.be/9HO-qgh-iGI
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T,$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}},$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}.$
• Graham’s law for effusion: $\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}$
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.07_Effusion_and_Diffusion.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas.
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in Figure $1$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in Figure $1$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected. Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\left(P + \dfrac{an^2}{V^2}\right) (V − nb)=nRT \label{6.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ( (L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{6.9.1}$ —$P + (an^2/V^2$)—corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term—$V − nb$—corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the an2/V2 term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
$n=\dfrac{m}{M}=\rm\dfrac{500\;g}{70.906\;g/mol}=7.052\;mol$
Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure:
$P=\dfrac{nRT}{V}=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L}=43.1\;atm$
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for P gives
$\begin{split}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{split}$
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer: a. 77 atm; b. 67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.10_Real_Gases.txt |
Gases that deviate from ideality are known as Real Gases, which originate from two factors: (1) First, the theory assumes that as pressure increases, the volume of a gas becomes very small and approaches zero. While it does approach a small number, it will not be zero because molecules do occupy space (i.e. have volume) and cannot be compressed. (2) Intermolecular forces do exist in gases. These become increasingly important in low temperatures, when translational (definition of translational, please) molecular motion slows down, almost to a halt. However, at high temperatures, or even normal, every day temperatures, the intermolecular forces are very small and tend to be considered negligible.
5.12 Chemistry in the Atmosphere
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.11%3A_Characteristics_of_Several_Real_Gases.txt |
What is Energy?
Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in a way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it.
The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning.
Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". In the 17th Century, the great mathematician Gottfried Leibnitz (1646-1716) suggested the distinction between vis viva ("live force") and vis mortua ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853).
Kinetic energy and potential energy
Whatever energy may be, there are basically two kinds.
Kinetic energy is associated with the motion of an object, and its direct consequences are part of everyone's daily experience; the faster the ball you catch in your hand, and the heavier it is, the more you feel it. Quantitatively, a body with a mass m and moving at a velocity v possesses the kinetic energy mv2/2.
Example 1
A rifle shoots a 4.25 g bullet at a velocity of 965 m s–1. What is its kinetic energy?
Solution
The only additional information you need here is that
1 J = 1 kg m2 s–2:
KE = ½ × (.00425 kg) (965 m s–1)2 = 1980 J
Potential energy is energy a body has by virtue of its location. But there is more: the body must be subject to a "restoring force" of some kind that tends to move it to a location of lower potential energy. Think of an arrow that is subjected to the force from a stretched bowstring; the more tightly the arrow is pulled back against the string, the more potential energy it has.
More generally, the restoring force comes from what we call a force field— a gravitational, electrostatic, or magnetic field. We observe the consequences of gravitational potential energy all the time, such as when we walk, but seldom give it any thought.
If an object of mass m is raised off the floor to a height h, its potential energy increases by mgh, where g is a proportionality constant known as the acceleration of gravity; its value at the earth's surface is 9.8 m s–2.
Example 2
Find the change in potential energy of a 2.6 kg textbook that falls from the 66-cm height of a table top onto the floor.
Solution
PE = m g h = (2.6 kg)(9.8 m s–2)(0.66 m) = 16.8 kg m2 s–2 = 16.8 J
Similarly, the potential energy of a particle having an electric charge qdepends on its location in an electrostatic field.
"Chemical energy"
Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules. "Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an exothermic chemical reaction, the electrons and nuclei within the reactants undergo rearrangement into products possessing lower energies, and the difference is released to the environment in the form of heat.
Interconversion of potential and kinetic energy
Transitions between potential and kinetic energy are such an intimate part of our daily lives that we hardly give them a thought. It happens in walking as the body moves up and down. Our bodies utilize the chemical energy in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of chemical energy to other forms.
Energy is conserved: it can neither be created nor destroyed. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal.
• When drop a book, its potential energy is transformed into kinetic energy. When it strikes the floor, this transformation is complete. What happens to the energy then? The kinetic energy that at the moment of impact was formerly situated exclusively in the moving book, now becomes shared between the book and the floor, and in the form of randomized thermal motions of the molecular units of which they are made; we can observe this effect as a rise in temperature.
• Much of the potential energy of falling water can be captured by a water wheel or other device that transforms the kinetic energy of the exit water into kinetic energy. The output of a hydroelectric power is directly proportional to its height above the level of the generator turbines in the valley below. At this point, the kinetic energy of the exit water is transferred to that of the turbine, most of which (up to 90 percent in the largest installations) is then converted into electrical energy.
• Will the temperature of the water at the bottom of a water fall be greater than that at the top? James Joule himself predicted that it would be. It has been calculated that at Niagara falls, that complete conversion of the potential energy of 1 kg of water at the top into kinetic energy when it hits the plunge pool 58 meters below will result in a temperature increase of about 0.14 C°. (But there are lots of complications. For example, some of the water breaks up into tiny droplets as it falls, and water evaporates from droplets quite rapidly, producing a cooling effect.)
• Chemical energy can also be converted, at least partially, into electrical energy: this is what happens in a battery. If a highly exothermic reaction also produces gaseous products, the latter may expand so rapidly that the result is an explosion — a net conversion of chemical energy into kinetic energy (including sound).
Thermal energy
Kinetic energy is associated with motion, but in two different ways. For a macroscopic object such as a book or a ball, or a parcel of flowing water, it is simply given by ½ mv2. However, as we mentioned above, when an object is dropped onto the floor, or when an exothermic chemical reaction heats surrounding matter, the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as thermal energy. We observe the effects of this as a rise in the temperature of the surroundings. The temperature of a body is direct measure of the quantity of thermal energy is contains.
Thermal energy is never completely recoverable
Once kinetic energy is thermalized, only a portion of it can be converted back into potential energy. The remainder simply gets dispersed and diluted into the environment, and is effectively lost.
To summarize, then:
• Potential energy can be converted entirely into kinetic energy..
• Potential energy can also be converted, with varying degrees of efficiency,into electrical energy.
• The kinetic energy of macroscopic objects can be transferred between objects (barring the effects of friction).
• Once kinetic energy becomes thermalized, only a portion of it can be converted back into either potential energy or be concentrated back into the kinetic energy of a macroscopic. This limitation, which has nothing to do with technology but is a fundamental property of nature, is the subject of the second law of thermodynamics.
• A device that is intended to accomplish the partial transformation of thermal energy into organized kinetic energy is known as a heat engine.
Energy scales are always arbitrary
You might at first think that a book sitting on the table has zero kinetic energy since it is not moving. But if you think about it, the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and the sun itself is moving away from the other stars in the general expansion of the universe. Since these motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity of the book is measured with respect to the table; on this so-called laboratory coordinate system, the kinetic energy of the book can be considered zero.
We do the same thing with potential energy. If the book is on the table, its potential energy with respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then push the book off the table, its potential energy will be negative after it reaches the floor.
Energy units
Energy is measured in terms of its ability to perform work or to transfer heat. Mechanical work is done when a force f displaces an object by a distance d:
\[w = f × d\]
The basic unit of energy is the joule. One joule is the amount of work done when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec2, so the basic dimensions of the joule are kg m2 s–2. The other two units in wide use. the calorie and the BTU (British thermal unit) are defined in terms of the heating effect on water. Because of the many forms that energy can take, there are a correspondingly large number of units in which it can be expressed, a few of which are summarized below.
1 calorie will raise the temperature of 1 g of water by 1 C°. The “dietary” calorie is actually 1 kcal. An average young adult expends about 1800 kcal per day just to stay alive. (you should know this definition)
1 cal = 4.184 J
1 BTU (British Thermal Unit) will raise the temperature of 1 lb of water by 1F°. 1 BTU = 1055 J
The erg is the c.g.s. unit of energy and a very small one; the work done when a 1-dyne force acts over a distance of 1 cm.
1 J = 107 ergs
1 erg = 1 d-cm = 1 g cm2 s–2
The electron-volt is even tinier: 1 eV is the work required to move a unit electric charge (1 C) through a potential difference of 1 volt. 1 J = 6.24 × 1018 eV
The watt is a unit of power, which measures the rate of energy flow in J sec–1. Thus the watt-hour is a unit of energy. An average human consumes energy at a rate of about 100 watts; the brain alone runs at about 5 watts.
1 J = 2.78 × 10–4watt-hr
1 w-h = 3.6 kJ
The liter-atmosphere is a variant of force-displacement work associated with volume changes in gases. 1 L-atm = 101.325 J
The huge quantities of energy consumed by cities and countries are expressed in quads; the therm is a similar but smaller unit. 1 quad = 1015 Btu = 1.05 × 1018 J
If the object is to obliterate cities or countries with nuclear weapons, the energy unit of choice is the ton of TNT equivalent. 1 ton of TNT = 4.184 GJ
(by definition)
In terms of fossil fuels, we have barrel-of-oil equivalent, cubic-meter-of-natural gas equivalent, and ton-of-coal equivalent.
1 bboe = 6.1 GJ
1 cmge = 37-39 mJ
1 toce = 29 GJ
Heat and Work
Heat and work are both measured in energy units, so they must both represent energy. How do they differ from each other, and from just plain “energy” itself? In our daily language, we often say that "this object contains a lot of heat", but this is gibberish in thermodynamics terms, although it is ok to say that the object is "hot", indicating that its temperature is high. The term "heat" has a special meaning in thermodynamics: it is a process in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a different temperature than its surroundings. Hence, thermal energy can only flow from a higher temperature to a lower temperature. It is this flow that constitutes "heat". Use of the term "flow" of heat recalls the incorrect 18th-century notion that heat is an actual substance called “caloric” that could flow like a liquid.
Note: Heat
We often say that "this object contains a lot of heat," however, this makes no sense since heat represents an energy transfer.
Transfer of thermal energy can be accomplished by bringing two bodies into physical contact (the kettle on top of the stove, or through an electric heating element inside the kettle). Another mechanism of thermal energy transfer is by radiation; a hot object will convey energy to any body in sight of it via electromagnetic radiation in the infrared part of the spectrum. In many cases, both modes will be active.
Work refers to the transfer of energy some means that does not depend on temperature difference. Work, like energy, can take various forms, the most familiar being mechanical and electrical.
• Mechanical work arises when an object moves a distance Δx against an opposing force f: \[w = f Δx\]
• Electrical work is done when a body having a charge q moves through a potential difference ΔV.
Note: Work
A transfer of energy to or from a system by any means other than heat is called “work”.
Work can be completely converted into heat (by friction, for example), but heat can only be partially converted to work. Conversion of heat into work is accomplished by means of a heat engine, the most common example of which is an ordinary gasoline engine. The science of thermodynamics developed out of the need to understand the limitations of steam-driven heat engines at the beginning of the Industrial Age. The Second Law of Thermodynamics, states that the complete conversion of heat into work is impossible. Something to think about when you purchase fuel for your car!
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/06%3A_Thermochemistry/9.1%3A_The_Nature_of_Energy.txt |
Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.
Introduction
Hess's Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess's law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in Gibbs' Energy and Entropy.
Definition: Hess's Law
The heat of any reaction $\Delta{H^°_f}$ for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:
(Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.)
Application
Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water:
$C_{(s)} + 2 H_2O_{(g)} \rightarrow CO_{2\, (g)} + 2 H_{2\, (g)} \tag{2}$
Calorimetry reveals that this reaction requires the input of 90.1 kJ of heat for every mole of $C_{(s)}$ consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, $q > 0$ for an endothermic reaction. When heat is evolved, the reaction is exothermic and $q < 0$ by convention.
It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, $C_{(s)}$, into another, $H_{2(g)}$. To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that
$C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)} \tag{3}$
produces $393.5\, kJ$ for one mole of carbon burned; hence $q=-393.5\, kJ$. The reaction
$2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \tag{4}$
produces 483.6 kJ for two moles of hydrogen gas burned, so q=-483.6 kJ. It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in equation [2], 90.1 kJ, is exactly equal to the difference between the heat evolved, -393.5 kJ, in the combustion of carbon and the heat evolved, -483.6 kJ, in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen, we get
$C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}$
$2 H_2O_{(g)} \rightarrow 2 H_{2(g)} + O_{2(g)}$
$C_{(s)} + O_{2(g)} + 2 H_2O_{(g)} \rightarrow CO_{2(g)} + 2 H_{2(g)} + O_{2(g)} \tag{5}$
Canceling the $O_{2(g)}$ from both sides, since it is net neither a reactant nor product, equation [5] is equivalent to equation [2]. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields equation [2]. And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of equation [2]. By studying many chemical reactions in this way, we discover that this result, known as Hess's Law, is general.
Why it works
A pictorial view of Hess's Law as applied to the heat of equation [2] is illustrative. In figure 1, the reactants C(s) + 2 H2O(g) are placed together in a box, representing the state of the materials involved in the reaction prior to the reaction. The products CO2(g) + 2 H2(g) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing C(s), O2(g), and 2 H2(g). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in equation [3] and equation [4].
This picture of Hess's Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing $C_{(s)}$, $O_{2(g)}$, and 2 $H_{2(g)}$. A consequence of our observation of Hess's Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product (this statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions).
A slightly different view of figure 1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants.
By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, H, so that the change in the function, ΔH = Hproducts - Hreactants, is equal to the heat of reaction q under constant pressure conditions. H, which we call the enthalpy, is a state function, since its value depends only on the state of the materials under consideration, that is, the temperature, pressure and composition of these materials.
The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus a state function, the elevation gain is independent of the path. Now, the existence of an energy state function H is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in subfigure 2.1, with reactants R being converted to products P. We wish to calculate the heat absorbed or released in this reaction, which is ΔH. Since H is a state function, we can follow any path from R to P and calculate ΔH along that path. In subfigure 2.2, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in figure 1, the atoms involved in the reaction are C, H, and O, each of which are represented in the intermediate state in elemental form. We can see in subfigure 2.2 that the ΔH for the overall reaction is now the difference between the ΔH in the formation of the products P from the elements and the ΔH in the formation of the reactants R from the elements.
The ΔH values for formation of each material from the elements are thus of general utility in calculating ΔH for any reaction of interest. We therefore define the standard formation reaction for reactant R, as
elements in standard state R
and the heat involved in this reaction is the standard enthalpy of formation, designated by ΔHf°. The subscript f, standing for "formation," indicates that the ΔH is for the reaction creating the material from the elements in standard state. The superscript ° indicates that the reactions occur under constant standard pressure conditions of 1 atm. From subfigure 2.2, we see that the heat of any reaction can be calculated from
$\Delta{H^°_f} = \Delta{H^°_{f,products}} -\Delta{H^°_{f,reactants}} \tag{6}$
Extensive tables of ΔH°f values (Table T1) have been compiled that allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react.
Example 1
The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants. Enthalpy is an extensive property and hence changes when the size of the sample changes. This means that the enthalpy of the reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction.
H2 (g) + 1/2O2 (g) → H2O (g) ΔH° = -572 kJ
2H2 (g) + O2 (g) → 2H2O (g) ΔH° = -1144kJ
The sign of the reaction enthalpy changes when a process is reversed.
H2 (g) + 1/2O2 (g) → H2O (g) ΔH° = -572 kJ
When switched:
H2O (g) → H2 (g) + 1/2O2 (g) ΔH° = +572 kJ
Since enthalpy is a state function, it is path independent. Therefore, it does not matter what reactions one uses to obtain the final reaction.
Contributors and Attributions
• Shelly Cohen (UCD)
9.7: Present Sources of Energy
Wikipedia Link: Energy Development
9.8: New Energy Sources
Wikipedia Link: Energy Development | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/06%3A_Thermochemistry/9.5%3A_Hess%27s_Law.txt |
As you read the print off this computer screen now, you are reading pages of fluctuating energy and magnetic fields. Light, electricity, and magnetism are all different forms of electromagnetic radiation.
Introduction
Electromagnetic radiation is a form of energy that is produced by oscillating electric and magnetic disturbance, or by the movement of electrically charged particles traveling through a vacuum or matter. The electric and magnetic fields come at right angles to each other and combined wave moves perpendicular to both magnetic and electric oscillating fields thus the disturbance. Electron radiation is released as photons, which are bundles of light energy that travel at the speed of light as quantized harmonic waves. This energy is then grouped into categories based on its wavelength into the electromagnetic spectrum. These electric and magnetic waves travel perpendicular to each other and have certain characteristics, including amplitude, wavelength, and frequency.
General Properties of all electromagnetic radiation:
1. Electromagnetic radiation can travel through empty space. Most other types of waves must travel through some sort of substance. For example, sound waves need either a gas, solid, or liquid to pass through in order to be heard.
2. The speed of light is always a constant. (Speed of light : 2.99792458 x 108 m s-1)
3. Wavelengths are measured between the distances of either crests or troughs. It is usually characterized by the Greek symbol $\lambda$.
Waves and their Characteristics
Fig. 1 & 2: Electromagnetic Waves
Fig. 3: An EM Wave
Amplitude
Amplitude is the distance from the maximum vertical displacement of the wave to the middle of the wave. This measures the magnitude of oscillation of a particular wave. In short, the amplitude is basically the height of the wave. Larger amplitude means higher energy and lower amplitude means lower energy. Amplitude is important because it tells you the intensity or brightness of a wave in comparison with other waves.
Wavelength
Wavelength ($\lambda$) is the distance of one full cycle of the oscillation. Longer wavelength waves such as radio waves carry low energy; this is why we can listen to the radio without any harmful consequences. Shorter wavelength waves such as x-rays carry higher energy that can be hazardous to our health. Consequently lead aprons are worn to protect our bodies from harmful radiation when we undergo x-rays. This wavelength frequently relationship is characterized by:
$c = \lambda\nu$
where
• c is the speed of light,
• $\lambda$ is wavelength, and
• $\nu$ is frequency.
Shorter wavelength means greater frequency, and greater frequency means higher energy. Wavelengths are important in that they tell one what type of wave one is dealing with.
Fig. 4: Different Wavelengths and Frequencies
Remember, Wavelength tells you the type of light and Amplitude tells you about the intensity of the light
Frequency
Frequency is defined as the number of cycles per second, and is expressed as sec-1 or Hertz (Hz). Frequency is directly proportional to energy and can be express as:
$E = h\nu$
where
• E is energy,
• h is Planck's constant, (h= 6.62607 x 10-34 J), and
• $\nu$ is frequency.
Period
Period (T) is the amount of time a wave takes to travel one wavelength; it is measured in seconds (s).
Velocity
The velocity of wave in general is expressed as:
$velocity = \lambda\nu$
For Electromagnetic wave, the velocity in vacuum is $2.99 \times 10^8\;m/s$ or $186,282$ miles/second.
Electromagnetic spectrum
Figure 24.5.1: Electromagnetic spectrum with light highlighted. from Wikipedia.
As a wave’s wavelength increases, the frequency decreases, and as wave’s wavelength decreases, the frequency increases. When electromagnetic energy is released as the energy level increases, the wavelength decreases and frequency decreases. Thus, electromagnetic radiation is then grouped into categories based on its wavelength or frequency into the electromagnetic spectrum. The different types of electromagnetic radiation shown in the electromagnetic spectrum consists of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays. The part of the electromagnetic spectrum that we are able to see is the visible light spectrum.
Fig. 6: Electromagnetic Spectrum with Radiation Types
Radiation Types
Radio Waves are approximately 103 m in wavelength. As the name implies, radio waves are transmitted by radio broadcasts, TV broadcasts, and even cell phones. Radio waves have the lowest energy levels. Radio waves are used in remote sensing, where hydrogen gas in space releases radio energy with a low frequency and is collected as radio waves. They are also used in radar systems, where they release radio energy and collect the bounced energy back. Especially useful in weather, radar systems are used to can illustrate maps of the surface of the Earth and predict weather patterns since radio energy easily breaks through the atmosphere. ;
Microwaves can be used to broadcast information through space, as well as warm food. They are also used in remote sensing in which microwaves are released and bounced back to collect information on their reflections.
Microwaves can be measured in centimeters. They are good for transmitting information because the energy can go through substances such as clouds and light rain. Short microwaves are sometimes used in Doppler radars to predict weather forecasts.
Infrared radiation can be released as heat or thermal energy. It can also be bounced back, which is called near infrared because of its similarities with visible light energy. Infrared Radiation is most commonly used in remote sensing as infrared sensors collect thermal energy, providing us with weather conditions.
This picture represents a snap shot in mid-infrared light.
Visible Light is the only part of the electromagnetic spectrum that humans can see with an unaided eye. This part of the spectrum includes a range of different colors that all represent a particular wavelength. Rainbows are formed in this way; light passes through matter in which it is absorbed or reflected based on its wavelength. Thus, some colors are reflected more than other, leading to the creation of a rainbow.
Color Region
Wavelength (nm)
Violet 380-435
Blue 435-500
Cyan 500-520
Green 520-565
Yellow 565-590
Orange 590-625
Red 625-740
Fig. 7: The color regions of the Visible Spectrum
Fig. 8: Dispersion of Light Through A Prism
Ultraviolet, Radiation, X-Rays, and Gamma Rays are all related to events occurring in space. UV radiation is most commonly known because of its severe effects on the skin from the sun, leading to cancer. X-rays are used to produce medical images of the body. Gamma Rays can used in chemotherapy in order to rid of tumors in a body since it has such a high energy level. The shortest waves, Gamma rays, are approximately 10-12 m in wavelength. Out this huge spectrum, the human eyes can only detect waves from 390 nm to 780 nm.
Equations of Waves
The mathematical description of a wave is:
$y = A\sin(kx - \omega{t})$
where A is the amplitude, k is the wave number, x is the displacement on the x-axis.
$k = \dfrac{2\pi}{\lambda}$
where $\lambda$ is the wavelength. Angular frequency described as:
$\omega = 2\pi \nu = \dfrac{2\pi}{T}$
where $\nu$ is frequency and period (T) is the amount of time for the wave to travel one wavelength.
Interference
An important property of waves is the ability to combine with other waves. There are two type of interference: constructive and destructive. Constructive interference occurs when two or more waves are in phase and and their displacements add to produce a higher amplitude. On the contrary, destructive interference occurs when two or more waves are out of phase and their displacements negate each other to produce lower amplitude.
Figure 9 & 10: Constructive and Destructive Interference
Interference can be demonstrated effectively through the double slit experiment. This experiment consists of a light source pointing toward a plate with one slit and a second plate with two slits. As the light travels through the slits, we notice bands of alternating intensity on the wall behind the second plate. The banding in the middle is the most intense because the two waves are perfectly in phase at that point and thus constructively interfere. The dark bands are caused by out of phase waves which result in destructive interference. This is why you observe nodes on figure 4. In a similar way, if electrons are used instead of light, electrons will be represented both as waves and particles.
Fig. 11 & 12: Double-Slit Interference Experiment
Wave-Particle Duality
Electromagnetic radiation can either acts as a wave or a particle, a photon. As a wave, it is represented by velocity, wavelength, and frequency. Light is an EM wave since the speed of EM waves is the same as the speed of light. As a particle, EM is represented as a photon, which transports energy. When a photon is absorbed, the electron can be moved up or down an energy level. When it moves up, it absorbs energy, when it moves down, energy is released. Thus, since each atom has its own distinct set of energy levels, each element emits and absorbs different frequencies. Photons with higher energies produce shorter wavelengths and photons with lower energies produce longer wavelengths.
Fig. 13: Photon Before and After Emission
Ionizing and Non-Ionizing Radiation
Electromagnetic Radiation is also categorized into two groups based, ionizing and non-ionizing, on the severity of the radiation. Ionizing radiation holds a great amount of energy to remove electrons and cause the matter to become ionized. Thus, higher frequency waves such as the X-rays and gamma-rays have ionizing radiation. However, lower frequency waves such as radio waves, do not have ionizing radiation and are grouped as non-ionizing.
Electromagnetic Radiation and Temperature
Electromagnetic radiation released is related to the temperature of the body. Stephan-Boltzmann Law says that if this body is a black body, one which perfectly absorbs and emits radiation, the radiation released is equal to the temperature raised to the fourth power. Therefore, as temperature increases, the amount of radiation released increases greatly. Objects that release radiation very well also absorb radiation at certain wavelengths very well. This is explained by the Kirchhoff’s Law. Wavelengths are also related to temperature. As the temperature increases, the wavelength of maximum emission decreases.
Problems
1. What is the wavelength of a wave with a frequency of 4.28 Hz?
2. What is the frequency of a wave with a wavelength of 200 cm?
3. What is the frequency of a wave with a wavelength of 500 pm?
4. What is the wavelength of a wave with a frequency of 2.998 × 105 Hz?
5. A radio transmits a frequency of 100 Hz. What is the wavelength of this wave?
Answers:
1. 700m
2. 1.5 × 108 Hz
3. 4.0 × 1017 Hz
4. 100m
5. 2.998 × 106 m
12.02 The Nature of Matter
In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure.By using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists discovered was the electron stream acted the same was as light proving de Broglie correct.
Contributors and Attributions
• Duy Nguyen (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.01_Electromagnetic_Radiation.txt |
A hydrogen discharge tube is a slim tube containing hydrogen gas at low pressure with an electrode at each end. If a high voltage (5000 volts) is applied, the tube lights up with a bright pink glow. If the light is passed through a prism or diffraction grating, it is split into its various colors. This is a small part of the hydrogen emission spectrum. Most of the spectrum is invisible to the eye because it is either in the infrared or the ultraviolet region of the electromagnetic spectrum.
The photograph shows part of a hydrogen discharge tube on the left, and the three most apparent lines in the visible part of the spectrum on the right. (Ignore the "smearing," particularly to the left of the red line. This is caused by flaws in the way the photograph was taken. See note below.)
This photograph is by courtesy of Dr Rod Nave of the Department of Physics and Astronomy at Georgia State University, Atlanta.
Extending hydrogen's emission spectrum into the UV and IR
The hydrogen spectrum is complex, comprising more than the three lines visible to the naked eye. It is possible to detect patterns of lines in both the ultraviolet and infrared regions of the spectrum as well. These fall into a number of "series" of lines named after the person who discovered them. The diagram below shows three of these series, but there are others in the infrared to the left of the Paschen series shown in the diagram.
The diagram is quite complicated. Consider first at the Lyman series on the right of the diagram; this is the broadest series, and the easiest to decipher.
The frequency scale is marked in PHz—petaHertz. Peta means "1015 times". The value 3 PHz is equal to 3 × 1015 Hz. The quantity "hertz" indicates "cycles per second".
The Lyman series is a series of lines in the ultraviolet region. The lines grow closer and closer together as the frequency increases. Eventually, they are so close together that it becomes impossible to see them as anything other than a continuous spectrum. This is suggested by the shaded part on the right end of the series. At one particular point, known as the series limit, the series ends.
In Balmer series or the Paschen series, the pattern is the same, but the series are more compact. In the Balmer series, notice the position of the three visible lines from the photograph further up the page.
Frequency and Wavelength
The hydrogen spectrum is often drawn using wavelengths of light rather than frequencies. Unfortunately, because of the mathematical relationship between the frequency of light and its wavelength, two completely different views of the spectrum are obtained when it is plotted against frequency or against wavelength. The mathematical relationship between frequency and wavelength is the following:
Rearranging this gives equations for either wavelength or frequency:
$\lambda =\dfrac{c}{\nu}$
or
$\nu=\dfrac{c}{\lambda }$
There is an inverse relationship between the two variables—a high frequency means a low wavelength and vice versa.
Drawing the hydrogen spectrum in terms of wavelength
This is what the spectrum looks like plotted in terms of wavelength instead of frequency:
Compare this to the same spectrum in terms of frequency:
When juxtaposed, the two plots form a confusing picture. The remainder of the article employs the spectrum plotted against frequency, because in this spectrum it is much easier visualize what is occurring in the atom.
The Balmer and Rydberg Equations
In an amazing demonstration of mathematical insight, in 1885 Balmer came up with a simple formula for predicting the wavelength of any of the lines in what we now know as the Balmer series. Three years later, Rydberg generalized this so that it was possible to determine the wavelengths of any of the lines in the hydrogen emission spectrum. Rydberg's equation is as follows:
$\dfrac{1}{\lambda}=R_H \left( \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\right)$
where
• $R_H$ is the Rydberg constant.
• $n_1$ and $n_2$ are integers (whole numbers). $n_2$ is always greater than $n_1$. In other words, if $n_1$ is, say, 2 then $n_2$ can be any whole number between 3 and infinity.
The various combinations of numbers that can be substituted into this formula allow the calculation the wavelength of any of the lines in the hydrogen emission spectrum; there is close agreement between the wavelengths generated by this formula and those observed in a real spectrum.
A modified version of the Rydberg equation can be used to calculate the frequency of each of the lines:
The origin of the hydrogen emission spectrum
The lines in the hydrogen emission spectrum form regular patterns and can be represented by a (relatively) simple equation. Each line can be calculated from a combination of simple whole numbers. Why does hydrogen emit light when excited by a high voltage and what is the significance of those whole numbers?
When unexcited, hydrogen's electron is in the first energy level—the level closest to the nucleus. But if energy is supplied to the atom, the electron is excited into a higher energy level, or even removed from the atom altogether. The high voltage in a discharge tube provides that energy. Hydrogen molecules are first broken up into hydrogen atoms (hence the atomic hydrogen emission spectrum) and electrons are then promoted into higher energy levels.
Suppose a particular electron is excited into the third energy level. It would tend to lose energy again by falling back down to a lower level. It can do this in two different ways. It could fall all the way back down to the first level again, or it could fall back to the second level and then, in a second jump, down to the first level.
Assigning particular electron jumps to individual lines in the spectrum
If an electron falls from the 3-level to the 2-level, it must lose an amount of energy exactly equal to the energy difference between those two levels. That energy which the electron loses is emitted as light (which "light" includes UV and IR as well as visible radiation). Each frequency of light is associated with a particular energy by the equation:
The higher the frequency, the higher the energy of the light. If an electron falls from the 3-level to the 2-level, red light is seen. This is the origin of the red line in the hydrogen spectrum. From the frequency of the red light, its energy can be calculated. That energy must be exactly the same as the energy gap between the 3-level and the 2-level in the hydrogen atom.
The last equation can therefore be rewritten as a measure of the energy gap between two electron levels:
The greatest possible fall in energy will therefore produce the highest frequency line in the spectrum. The greatest fall will be from the infinity level to the 1-level. (The significance of the infinity level will be made clear later.)
The next few diagrams are in two parts, with the energy levels at the top and the spectrum at the bottom.
If an electron falls from the 6-level, the difference is slightly less than before, and so the frequency is slightly lower (because of the scale of the diagram, it is impossible to depict the levels beyond 7).
All other possible jumps to the first level make up the whole Lyman series. The spacings between the lines in the spectrum reflect the changes in spacings between the energy levels.
If the same is done for the 2-level, the Balmer series is shown. These energy gaps are all much smaller than in the Lyman series, and so the frequencies produced are also much lower.
The Paschen series is made up of the transitions to the 3-level, but they are omitted to avoid cluttering the diagram.
The significance of the numbers in the Rydberg equation
In the Rydberg equation, n1 and n2 represent the energy levels at either end of the jump that produces a particular line in the spectrum.
• In the Lyman series, $n_1 =1$, because electrons transition to the 1-level to produce lines in the Lyman series.
• In the Balmer series, $n_1 =2$, because electrons fall to the 2-level.
n2 is the level being jumped from. In the case before, in which a red line is produced by electrons falling from the 3-level to the 2-level, n2 is equal to 3.
The significance of the infinity level
The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion.
Using the spectrum to find the ionization energy of hydrogen
When there is no additional energy supplied to it, hydrogen atom's electron is found at the 1-level. This is known as its ground state. If enough energy is supplied to move the electron up to the infinity level, the atom is ionized. The ionization energy per electron is therefore a measure of the difference in energy between the 1-level and the infinity level. In above diagrams, that particular energy jump produces the series limit of the Lyman series.
The frequency of the Lyman series limit can be used to calculate the energy required to promote the electron in one atom from the 1-level to the point of ionization. This energy can then be used to calculate the ionization energy per mole of atoms. A problem with this approach is that the frequency of a series limit is quite difficult to find accurately from a spectrum because the lines are so close together in that region that the spectrum looks continuous.
Finding the frequency of the series limit graphically
The following is a list of the frequencies of the seven most widely spaced lines in the Lyman series, together with the increase in frequency between successive lines.
As the lines become closer together, the increase in frequency is lessened. At the series limit, the gap between the lines is zero. Consequently, if the increase in frequency is plotted against the actual frequency, the curve can be extrapolated to the point at which the increase becomes zero, the frequency of the series limit.
In fact, two graphs can be plotted from the data in the table above. The frequency difference is related to two frequencies. For example, the figure of 0.457 is found by subtracting 2.467 from 2.924. Which of the two values should be plotted against 0.457 does not matter, as long as consistency is maintained—the difference must always be plotted against either the higher or the lower figure. At the limit, the two frequency numbers are the same.
As illustrated in the graph below, plotting both of the possible curves on the same graph makes it easier to decide exactly how to extrapolate the curves. Because these are curves, they are much more difficult to extrapolate than straight lines.
Both lines indicate a series limit at about 3.28 x 1015 Hz.
With this information, it is possible calculate the energy needed to remove a single electron from a hydrogen atom. Recall the equation above:
The energy gap between the ground state and the point at which the electron leaves the atom can be determined by substituting the frequency and looking up the value of Planck's constant from a data book.
$\begin{eqnarray} \Delta E &=& h\nu \ &=& (6.626 \times 10^{-34})(3.28 \times 10^{15}) \ &=& 2.173 \times 10^{-18}\ J \end{eqnarray}$
This is the ionization energy for a single atom. To find the normally quoted ionization energy, this value is multiplied by the number of atoms in a mole of hydrogen atoms (the Avogadro constant) and then dividing by 1000 to convert joules to kilojoules.
$\begin{eqnarray} Ionization\ energy &=& (2.173 \times 10^{-18})( 6.022 \times 10^{23})( \frac{1}{1000}) \ &=& 1310\ kJ\ mol^{-1} \end{eqnarray}$
This compares well with the normally quoted value for hydrogen's ionization energy of 1312 kJ mol-1. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.03_The_Atomic_Spectrum_of_Hydrogen.txt |
Learning Objectives
• Introduce the fundamentals behind the Bohr Atom and demonstrate it can predict the Rydberg's equation for the atomic spectrum of hydrogen
Rutherford's Failed Planetary Atom
Ernest Rutherford had proposed a model of atoms based on the $\alpha$-particle scattering experiments of Hans Geiger and Ernest Marsden. In these experiments helium nuclei ($\alpha$-particles) were shot at thin gold metal foils. Most of the particles were not scattered; they passed unchanged through the thin metal foil. Some of the few that were scattered were scattered in the backward direction; i.e. they recoiled. This backward scattering requires that the foil contain heavy particles. When an $\alpha$-particle hits one of these heavy particles it simply recoils backward, just like a ball thrown at a brick wall. Since most of the α-particles don’t get scattered, the heavy particles (the nuclei of the atoms) must occupy only a very small region of the total space of the atom. Most of the space must be empty or occupied by very low-mass particles. These low-mass particles are the electrons that surround the nucleus.
There are some basic problems with the Rutherford model. The Coulomb force that exists between oppositely charge particles means that a positive nucleus and negative electrons should attract each other, and the atom should collapse. To prevent the collapse, the electron was postulated to be orbiting the positive nucleus. The Coulomb force (discussed below) is used to change the direction of the velocity, just as a string pulls a ball in a circular orbit around your head or the gravitational force holds the moon in orbit around the Earth. The origin for this hypothesis that suggests this perspective is plausible is the similarity of gravity and Coulombic interactions. The expression for the force of gravity between two masses (Newton's Law of gravity) is
$F_{gravity} \propto \dfrac{m_1m_2}{r^2}\label{1.8.1}$
with $m_1$ and $m_2$ representing the mass of object 1 and 2, respectively and $r$ representing the distance between the objects centers
The expression for the Coulomb force between two charged species is
$F_{Coulomb} \propto \dfrac{Q_1Q_2}{r^2}\label{1.8.2}$
with $Q_1$ and $Q_2$ representing the charge of object 1 and 2, respectively and $r$ representing the distance between the objects centers.
However, this analogy has a problem too. An electron going around in a circle is constantly being accelerated because its velocity vector is changing. A charged particle that is being accelerated emits radiation. This property is essentially how a radio transmitter works. A power supply drives electrons up and down a wire and thus transmits energy (electromagnetic radiation) that your radio receiver picks up. The radio then plays the music for you that is encoded in the waveform of the radiated energy.
If the orbiting electron is generating radiation, it is losing energy. If an orbiting particle loses energy, the radius of the orbit decreases. To conserve angular momentum, the frequency of the orbiting electron increases. The frequency increases continuously as the electron collapses toward the nucleus. Since the frequency of the rotating electron and the frequency of the radiation that is emitted are the same, both change continuously to produce a continuous spectrum and not the observed discrete lines. Furthermore, if one calculates how long it takes for this collapse to occur, one finds that it takes about $10^{‑11}$ seconds. This means that nothing in the world based on the structure of atoms could exist for longer than about $10^{-11}$ seconds. Clearly something is terribly wrong with this classical picture, which means that something was missing at that time from the known laws of physics.
Conservative Forces can be explained with Potentials
A conservative force is dependent only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point. When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken. The potential can be constructed as simple derivatives for 1-D forces:
$F = -\dfrac{dV}{dx} \nonumber$
or as gradients in 3-D forces
$F = -\nabla V \nonumber$
where $\nabla$ is the vector of partial derivatives
$\nabla = \left ( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right) \nonumber$
The most familiar conservative forces are gravity and Coloumbic forces.
The Coulomb force law (Equation $\ref{1.8.2}$) comes from the corresponding Coulomb potential (sometimes call electrostatic potential)
$V(r)=\dfrac{kQ_1 Q_2}{r} \label{1.8.5}$
and it can be easily verified that the Coulombic force from this interaction ($F(r)$) is
$F(r)=-\dfrac{dV}{dr} \label{1.8.6}$
As $r$ is varied, the energy will change, so that we have an example of a potential energy curve $V(r)$ (Figure $\PageIndex{2; left}$). If $Q_1$ and $Q_2$ are the same sign, then the curve which is a purely repulsive potential, i.e., the energy increases monotonically as the charges are brought together and decreases monotonically as they are separated. From this, it is easy to see that like charges (charges of the same sign) repel each other.
If the charges are of opposite sign, then the curve appears roughly Figure $\PageIndex{2; right}$ and this is a purely attractive potential. Thus, the energy decreases as the charges are brought together, implying that opposite charges attract
The Bohr Model
It is observed that line spectra discussed in the previous sections show that hydrogen atoms absorb and emit light at only discrete wavelengths. This observation is connected to the discrete nature of the allowed energies of a quantum mechanical system. Quantum mechanics postulates that, in contrast to classical mechanics, the energy of a system can only take on certain discrete values. This leaves us with the question: How do we determine what these allowed discrete energy values are? After all, it seems that Planck's formula for the allowed energies came out of nowhere.
The model we will describe here, due to Niels Bohr in 1913, is an early attempt to predict the allowed energies for single-electron atoms such as $\ce{H}$, $\ce{He^{+}}$, $\ce{Li^{2+}}$, $\ce{Be^{3+}}$, etc. Although Bohr's reasoning relies on classical concepts and hence, is not a correct explanation, the reasoning is interesting, and so we examine this model for its historical significance.
Consider a nucleus with charge $+Ze$ and one electron orbiting the nucleus. In this analysis, we will use another representation of the constant $k$ in Coulomb's law (Equation $\ref{1.8.5}$), which is more commonly represented in the form:
$k=\dfrac{1}{4\pi \epsilon_0} \label{1.8.7}$
where $\epsilon_0$ is known as the permittivity of free space with the numerical value $\epsilon_0 = 8.8541878\times 10^{-12} \ C^2 J^{-1} m^{-1}$.
The total energy of the electron (the nucleus is assumed to be fixed in space at the origin) is the sum of kinetic and potential energies:
$E_{total}=\underset{\text{kinetic energy}}{\dfrac{p^2}{2m_e}} - \underset{\text{potential energy}}{\dfrac{Ze^2}{4\pi \epsilon_0 r}} \nonumber$
The force on the electron is
$\vec{F}=-\dfrac{Ze^2}{4\pi \epsilon_0 r^3}r \nonumber$
and its magnitude is
$F=|\vec{F}|=\dfrac{Ze^2}{4\pi \epsilon_0 r^3}|r|=\dfrac{Ze^2}{4\pi \epsilon_0 r^2} \nonumber$
since $\vec{F}=m_e \vec{a}$, the magnitude, it follows that $|\vec{F}|=m_e |\vec{a}|$. If we assume that the orbit is circular, then the acceleration is purely centripetal, so
$|a|=\dfrac{v^2}{r} \nonumber$
where $v$ is the velocity of the electron. Equating force $|F|$ to $m_e |a|$, we obtain
$\dfrac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\dfrac{v^2}{r} \nonumber$
or
$\dfrac{Ze^2}{4\pi \epsilon_0}=m_e v^2 r \nonumber$
or
$\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=(m_e vr)^2 \label{1.8.14}$
The reason for writing the equation this way is that the quantity $m_e vr$ is the classical orbital angular momentum of the electron. Bohr was familiar with Maxwell's theory of classical electromagnetism and knew that in a classical theory, the orbiting electron should radiate energy away and eventually collapse into the nucleus (Figure 1.8.1 ). He circumvented this problem by following Planck's idea underlying blackbody radiation and positing that the orbital angular momentum $m_e vr$ of the electron could only take on specific values
$m_e vr=n\hbar\label{1.8.15}$
with $n=1,2,3,...$.
Note that the electron must be in motion, so $n=0$ is not allowed.
Substituting Equation $\ref{1.8.15}$ into the Equation $\ref{1.8.14}$, we find
$\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=n^2 (\hbar)^2 \label{1.8.16}$
Equation \ref{1.8.16} implies that orbits could only have certain allowed radii
\begin{align}r_n &= \dfrac{4\pi \epsilon_0 \hbar^2}{Ze^2 m_e}n^2 \ &=\dfrac{a_0}{Z}n^2 \label{1.8.16B} \end{align}
with $n=1,2,3,...$. The collection of constants has been defined to be $a_0$
$a_0=\dfrac{4\pi \epsilon_0 \hbar^2}{e^2 m_e} \label{1.8.17}$
a quantity that is known as the Bohr radius.
We can also calculate the allowed momenta since $m_e vr=n\hbar$, and $p=m_e v$. Thus,
\begin{align}p_n r_n &=n\hbar\[4pt] p_n &=\dfrac{n\hbar}{r_n}\[4pt] &=\dfrac{\hbar Z}{a_0 n} \[4pt] &= \dfrac{Ze^2 m_e}{4\pi \epsilon_0 \hbar n}\end{align} \label{1.8.18}
From $p_n$ and $r_n$, we can calculate the allowed energies from
$E_n=\dfrac{p^2_n}{2m_e}-\dfrac{Ze^2}{4\pi \epsilon_0 r_n} \label{1.8.19}$
Substituting in the expressions for $p_n$ and $r_n$ and simplifying gives
$E_n=-\dfrac{Z^2 e^4 m_e}{32\pi^2 \epsilon_{0}^{2}\hbar^2}\dfrac{1}{n^2}=-\dfrac{e^4 m_e}{8 \epsilon_{0}^{2}h^2}\dfrac{Z^2}{n^2} \label{1.8.20}$
We can redefine a new energy scale by defining the Rydberg as
$1 \ Ry = \dfrac{e^4 m_e}{8\epsilon_{0}^{2} h^2} =2.18\times 10^{-18} \ J. \nonumber$
and this simplifies the allowed energies predicted by the Bohr model (Equation \ref{1.8.20}) as
$E_n=-(2.18\times 10^{-18})\dfrac{Z^2}{n^2} \ J=-\dfrac{Z^2}{n^2} \ R_y \label{1.8.21}$
Hence, the energy of the electron in an atom also is quantized. Equation $\ref{1.8.21}$ gives the energies of the electronic states of the hydrogen atom. It is very useful in analyzing spectra to represent these energies graphically in an energy-level diagram. An energy-level diagram has energy plotted on the vertical axis with a horizontal line drawn to locate each energy level (Figure 1.8.4 ).
These turn out to be the correct energy levels, apart from small corrections that cannot be accounted for in this pseudo-classical treatment. Despite the fact that the energies are essentially correct, the Bohr model masks the true quantum nature of the electron, which only emerges from a fully quantum mechanical analysis.
Exercise 1.8.1
Calculate a value for the Bohr radius using Equation $\ref{1.8.16}$ to check that this equation is consistent with the value 52.9 pm. What would the radius be for $n = 1$ in the $\ce{Li^{2+}}$ ion?
Answer
Starting from Equation \ref{1.8.16} and solving for $r$:
\begin{align*} \dfrac{Ze^2m_er}{4πϵ_0} &=n^2ℏ^2 \ r &=\dfrac{4 n^2 \hbar^2 πϵ_0}{Z e^2 m_e} \end{align*} \nonumber
with
• $e$ is the fundamental charge: $e=1.60217662 \times 10^{-19}C^2$
• $m_e$ is the mass of an electron: $m_e= 9.10938356 \times 10^{-31}kg$
• $\epsilon_o$ is the permittivity of free space: $\epsilon_o = 8.854 \times 10^{-12}C^2N^{-1}m^{-2}$
• $\hbar$ is the reduced planks constant: $\hbar=1.0546 \times 10^{-34}m^2kg/s$
For the ground-state of the hydrogen atom: $Z=1$ and $n=1$.
\begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{e^2m_e} \ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854 \times 10^{-12}C^2N^{-1}m^{-2}}{(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \ &=5.29 \times 10^{-11}m = 52.9\, pm\end{align*} \nonumber
For the ground-state of the lithium +2 ion: $Z=3$ and $n=1$
\begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{3 e^2m_e} \ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854\times10^{-12}C^2N^{-1}m^{-2}}{3(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \ &=1.76 \times 10^{-11}m = 17.6 \,pm\end{align*} \nonumber
As expected, the $\ce{Li^{2+}}$ has a smaller radius than the $\ce{H}$ atoms because of the increased nuclear charge.
Exercise 1.8.2 : Rydberg states
How do the radii of the hydrogen orbits vary with $n$? Prepare a graph showing $r$ as a function of $n$. States of hydrogen atoms with $n = 200$ have been prepared (called Rydberg states). What is the diameter of the atoms in these states?
Answer
This is a straightforward application of Equation of \ref{1.8.16B}. The hydrogen atom has only certain allowable radii and these radii can be predicted from the equation that relates them with each $n$. Note that the electron must be in motion so $n = 0$ is not allowed.
$4 \pi \epsilon_{0}=1.113 \times 10^{-10} \mathrm{C}^{2} \mathrm{J}^{-1} \mathrm{m}^{-1}$ and $\hbar=1.054 \times 10^{-34} \mathrm{J} \mathrm{s},$ also knowing
\begin{aligned} e &=1.602 \times 10^{-19} \mathrm{C} \text { with } \ m_{e} &=9.109 \times 10^{-31} \mathrm{kg} \end{aligned} \nonumber
and $Z$ is the nuclear charge, we use this equation directly. A simplification can be made by taking advantage of the fact that
$a_{0}=\frac{4 \pi \epsilon_{0} \hbar^{2}}{e^{2} m_{e}} \nonumber$
resulting in
$r_{n}=\frac{a_{0}}{Z} n^{2} \nonumber$
where $a_{0}=5.292 \times 10^{-11} \mathrm{m}$ which is the Bohr Radius.
Suppose we want to find the radius where $n=200 . n^{2}=40000$ so plugging in directly we have
\begin{align*} r_{n} &=\frac{\left(5.292 \times 10^{-11}\right)}{(1)}(40000) \[4pt] &=2.117 \times 10^{-6} m \end{align*} \nonumber
for the radius of a hydrogen atom with an electron excited to the $\mathrm{n}=200$ state. The diameter is then $4.234 \times 10^{-6} \mathrm{m}$.
The Wave Argument for Quantization
The above discussion is based off of a classical picture of an orbiting electron with the quantization from the angular momentum (Equation $\ref{1.8.15}$) requirement lifted from Planck's quantization arguments. Hence, only allows certain trajectories are stable (with differing radii). However, as discussed previously, the electron will have a wavelike property also with a de Broglie wavelength $\lambda$
$\lambda = \dfrac{h}{p} \nonumber$
Hence, a larger momentum $p$ implies a shorter wavelength. That means as $n$ increases (Equation $\ref{1.8.21}$), the wavelength must also increase; this is a common feature in quantum mechanics and will be often observed. In the Bohr atom, the circular symmetry and the wave property of the electron requires that the electron waves have an integer number of wavelengths (Figure $\PageIndex{1A}$). If not, then the waves will overlap imperfectly and cancel out (i.e., the electron will cease to exist) as demonstrated in Figure $\PageIndex{1B}$.
A more detailed discussion of the effect of electron waves in atoms will be discuss in the following chapters.
Derivation of the Rydberg Equation from Bohr Model
Given a prediction of the allowed energies of a system, how could we go about verifying them? The general experimental technique known as spectroscopy permits us to probe the various differences between the allowed energies. Thus, if the prediction of the actual energies, themselves, is correct, we should also be able to predict these differences. Let us assume that we are able to place the electron in Bohr's hydrogen atom into an energy state $E_n$ for $n>1$, i.e. one of its so-called excited states. The electron will rapidly return to its lowest energy state, known as the ground state and, in doing so, emit light. The energy carried away by the light is determined by the condition that the total energy is conserved (Figure 1.8.6 ).
Thus, if $n_i$ is the integer that characterizes the initial (excited) state of the electron, and $n_f$ is the final state (here we imagine that $n_f =1$, but is applicable in cases that $n_f <n_i$, i.e., emission)
$E_{nf}=E_{ni}-h\nu \label{1.8.22}$
or
$\nu=\dfrac{E_{ni}-E_{nf}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2} h^3}\left ( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right ) \label{1.8.23}$
We can now identify the Rydberg constant $R_H$ with the ratio of constants on the right hand side of Equation $\ref{1.8.23}$
$R_H = \dfrac {m_ee^4}{8 \epsilon ^2_0 h^3 } \label {2-22}$
Evaluating $R_H$ from the fundamental constants in this formula gives a value within 0.5% of that obtained experimentally from the hydrogen atom spectrum.
Thus, by observing the emitted light, we can determine the energy difference between the initial and final energy levels, which results in the emission spectra discussed in Sections 1.4 and 1.5. Different values of $n_f$ determine which emission spectrum is observed, and the examples shown in the figure are named after the individuals who first observed them. The figure below shows some of the transitions possible for different $n_f$ and $n_i$ values discussed previously.
If the atom absorbs light it ends up in an excited state as a result of the absorption. The absorption is only possible for light of certain frequencies, and again, conservation of energy determines what these frequencies are. If light is absorbed, then the final energy $E_{nf}$ will be related to the initial energy $E_{ni}$ with $n_f >n_i$ by
$E_{nf}=E_{ni}+h\nu \label{1.8.24}$
or
$\nu=\dfrac{E_{nf}-E_{ni}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2}h^3}\left ( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}}\right ) \label{1.8.25}$
Exercise 1.8.3
1. Calculate the energy of a photon that is produced when an electron in a hydrogen atom goes from an orbit with $n = 4$ to an orbit with $n = 1$.
2. What happens to the energy of the photon as the initial value of $n$ approaches infinity?
Answer
a:
\begin{align*} E_{\text{nf}} &= E_{ni} - h\nu \ E_{photon} = h\nu &= E_{nf} - E_{ni}\ &= \frac{Z^2e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\ &=\frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{1^2} - \frac{1}{4^2}\right)\ &=2.18 \times 10^{-18}\left(1 - \frac{1}{16} \right)\ &=2.04 \times 10^{-18} J \end{align*} \nonumber
b:
As $n_i \rightarrow \infty$
\begin{align*} E_{photon} &= \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \frac{1}{n_i^2} &\rightarrow 0\ E_{photon} &\rightarrow \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2}\right)\ \end{align*} \nonumber
Bohr’s proposal explained the hydrogen atom spectrum, the origin of the Rydberg formula, and the value of the Rydberg constant. Specifically it demonstrated that the integers in the Rydberg formula are a manifestation of quantization. The energy, the angular momentum, and the radius of the orbiting electron all are quantized. This quantization also parallels the concept of stable orbits in the Bohr model. Only certain values of $E$, $M$, and $r$ are possible, and therefore the electron cannot collapse onto the nucleus by continuously radiating energy because it can only have certain energies, and it cannot be in certain regions of space. The electron can only jump from one orbit (quantum state) to another. The quantization means that the orbits are stable, and the electron cannot spiral into the nucleus in spite of the attractive Coulomb force.
Although Bohr’s ideas successfully explained the hydrogen spectrum, they failed when applied to the spectra of other atoms. In addition a profound question remained. Why is angular momentum quantized in units of $\hbar$? As we shall see, de Broglie had an answer to this question, and this answer led Schrödinger to a general postulate that produces the quantization of angular momentum as a consequence. This quantization is not quite as simple as proposed by Bohr, and we will see that it is not possible to determine the distance of the electron from the nucleus as precisely as Bohr thought. In fact, since the position of the electron in the hydrogen atom is not at all as well defined as a classical orbit (such as the moon orbiting the earth) it is called an orbital. An electron orbital represents or describes the position of the electron around the nucleus in terms of a mathematical function called a wavefunction that yields the probability of positions of the electron. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.04_The_Bohr_Model.txt |
In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously.
The Nature of Measurement
In order to understand the conceptual background of the Heisenberg Uncertainty Principle it is important to understand how physical values are measured. In almost any measurement that is made, light is reflected off the object that is being measured and processed. The shorter the wavelength of light used, or the higher its frequency and energy, the more accurate the results. For example, when attempting to measure the speed of a tennis ball as it is dropped off of a ledge, photons(measurement of light) are shot off the tennis ball, reflected, and then processed by certain equipment. Because the tennis ball is so large compared to the photons, it is unaffected by the efforts of the observer to measure its physical quantities. However, if a photon is shot at an electron, the minuscule size of the electron and its unique wave-particle duality introduces consequences that can be ignored when taking measurements of macroscopic objects.
Heisenberg himself encountered such limitations as he attempted to measure the position of an electron with a microscope. As noted, the accuracy of any measurement is limited by the wavelength of light illuminating the electron. Therefore, in principle, one can determine the position as accurately as one wishes by using light of very high frequency, or short wave-lengths. However, the collision between such high energy photons of light with the extremely small electron causes the momentum of the electron to be disturbed.
Thus, increasing the energy of the light (and increasing the accuracy of the electron's position measurement), increases such a deviation in momentum. Conversely, if a photon has low energy the collision does not disturb the electron, yet the position cannot be accurately determined. Heisenberg concluded in his famous 1927 paper on the topic,
"At the instant of time when the position is determined, that is, at the instant when the photon is scattered by the electron, the electron undergoes a discontinuous change in momentum. This change is the greater the smaller the wavelength of the light employed, i.e., the more exact the determination of the position. At the instant at which the position of the electron is known, its momentum therefore can be known only up to magnitudes which correspond to that discontinuous change; thus, the more precisely the position is determined, the less precisely the momentum is known..." (Heisenberg, 1927, p. 174-5).
Heisenberg realized that since both light and particle energy are quantized, or can only exist in discrete energy units, there are limits as to how small, or insignificant, such an uncertainty can be. As proved later in this text, that bound ends up being expressed by Planck's Constant, h = 6.626*10-34 J*s.
It is important to mention that The Heisenberg Principle should not be confused with the observer effect. The observer effect is generally accepted to mean that the act of observing a system will influence that which is being observed. While this is important in understanding the Heisenberg Uncertainty Principle, the two are not interchangeable. The error in such thinking can be explained using the wave-particle duality of electromagnetic waves, an idea first proposed by Louis de Broglie. Wave-particle duality asserts that any energy exhibits both particle- and wave-like behavior. As a consequence, in quantum mechanics, a particle cannot have both a definite position and momentum. Thus, the limitations described by Heisenberg are a natural occurrence and have nothing to do with any limitations of the observational system.
Heisenberg’s Uncertainty Principle
It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles.First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio h/λ, where h represents Planck’s constant and λ represents the wavelength of the photon. The position of a photon of light is simply its wavelength, \lambda\).. In order to represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore,
$\Delta{p}=\dfrac{h}{\lambda}$
$\Delta{x}= \lambda$
By substituting $\Delta{x}$ for $\lambda$ in the first equation, we derive
$\Delta{p}=\dfrac{h}{\Delta{x}}$
or,
$\Delta{p}\Delta{x}=h$
Note, we can derive the same formula by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let Δp=mu, and Δx=h/mu (from De Broglie’s expression for the wavelength of a particle). Substituting in Δp for mu in the second equation leads to the very same equation derived above-ΔpΔx=h. This equation was refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as
$\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}$
What this equation reveals is that the more accurately a particle’s position is known, or the smaller Δx is, the less accurately the momentum of the particle Δp is known. Mathematically, this occurs because the smaller Δx becomes, the larger Δp must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known.
Understanding the Uncertainty Principle through Wave Packets and the Slit Experiment
It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously.
Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light.
The Importance of the Heisenberg Uncertainty Principle
Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom.
Problems
1. What aspect of the Bohr model of the atom does the Heisenberg Uncertainty Principle discredit?
2. What is the difference between the Heisenberg Uncertainty Principle and the Observer Effect?
3. A Hydrogen atom has a radius of 0.05nm with a position accuracy of 1.0%. What is the uncertainty in determining the velocity?
4. What is the uncertainty in the speed of a beam of electrons whose position is known with an uncertainty of 10 nm?
5. Using the Uncertainty Principle, find the radius of an atom (in nm) that has an electron with a position accuracy of 3.0% and a known velocity of $2\times 10^9\, m/s$.
Answers:
1.) The Heisenberg Uncertainty Principle discredits the aspect of the Bohr atom model that an electron is constrained to a one-dimensional orbit of a fixed radius around the nucleus.
2.) The Observer Effect means the act of observing a system will influence what is being observed, whereas the Heisenberg Uncertainty Principle has nothing to do with the observer or equipment used during observation. It simply states that a particle behaves both as a wave and a particle and therefore cannot have both a definite momentum and position.
3.) Uncertainty principle: ΔxΔp≥h/4Π
Can be written (x)(m)(v)(%)=h/4Π
(position)(mass of electron)(velocity)(percent accuracy)=(Plank's Constant)/4Π
(.05*10-9 m)(9.11*10-31 kg)(v)(.01)=(6.626*10-34 J*s)/4Π
v=1*108 m/s
4.) Uncertainty principle: ΔxΔp≥h/4Π
(10*10-9 m)(Δp)≥(6.626*10-34 J*s)/4Π
Δp≥5*10-26 (kg*m)/s
5.) Uncertainty principle: ΔxΔp≥h/4Π
Can be written (x)(m)(v)(%)=h/4Π
(position)(mass of electron)(velocity)(percent accuracy)=h/4Π
(radius)(9.11*10-31 kg)(2*109)(.03)=(6.626*10-34 J*s)/4Π
r=9*10-12 m
r=9*10-3 nm | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.05_The_Quantum_Mechanical_Model_of_the_Atom.txt |
A total of four quantum numbers are used to describe completely the movement and trajectories of each electron within an atom. The combination of all quantum numbers of all electrons in an atom is described by a wave function that complies with the Schrödinger equation. Each electron in an atom has a unique set of quantum numbers; according to the Pauli Exclusion Principle, no two electrons can share the same combination of four quantum numbers. Quantum numbers are important because they can be used to determine the electron configuration of an atom and the probable location of the atom's electrons. Quantum numbers are also used to understand other characteristics of atoms, such as ionization energy and the atomic radius.
In atoms, there are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number, $n$, describes the energy of an electron and the most probable distance of the electron from the nucleus. In other words, it refers to the size of the orbital and the energy level an electron is placed in. The number of subshells, or $l$, describes the shape of the orbital. It can also be used to determine the number of angular nodes. The magnetic quantum number, ml, describes the energy levels in a subshell, and ms refers to the spin on the electron, which can either be up or down.
The Principal Quantum Number ($n$)
The principal quantum number, $n$, designates the principal electron shell. Because n describes the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as $n=1$ designates the first principal shell (the innermost shell). The first principal shell is also called the ground state, or lowest energy state. This explains why $n$ can not be 0 or any negative integer, because there exists no atoms with zero or a negative amount of energy levels/principal shells. When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where $n=2$. This is called absorption because the electron is "absorbing" photons, or energy. Known as emission, electrons can also "emit" energy as they jump to lower principle shells, where n decreases by whole numbers. As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on.
$n=1,2,3,4…$
Example $1$
If n = 7, what is the principal electron shell?
Example $2$
If an electron jumped from energy level n = 5 to energy level n = 3, did absorption or emission of a photon occur?
Answer
Emission, because energy is lost by release of a photon.
The Orbital Angular Momentum Quantum Number ($l$)
The orbital angular momentum quantum number $l$ determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number $l$. (For more information about angular nodes, see Electronic Orbitals.) Each value of $l$ indicates a specific s, p, d, f subshell (each unique in shape.) The value of $l$ is dependent on the principal quantum number $n$. Unlike $n$, the value of $l$ can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number ($n-1$):
$l=0, 1, 2, 3, 4…, (n-1)$
Example $3$
If $n = 7$, what are the possible values of $l$?
Answer
Since $l$ can be zero or a positive integer less than ($n-1$), it can have a value of 0, 1, 2, 3, 4, 5 or 6.
Example $4$
If $l = 4$, how many angular nodes does the atom have?
Answer
The number of angular nodes is equal to the value of l, so the number of nodes is also 4.
The Magnetic Quantum Number ($m_l$)
The magnetic quantum number $m_l$ determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number $l$. Given a certain $l$, $m_l$ is an interval ranging from $–l$ to $+l$, so it can be zero, a negative integer, or a positive integer.
$m_l= -l, (-l +1),( -l +2),…, -2, -1, 0, 1, 2, … (l – 1), (l – 2), +l$
Example $5$
Example: If $n=3$, and $l=2$, then what are the possible values of $m_l$?
Answer
Since $m_l$ must range from $–l$ to $+l$, then $m_l$ can be: -2, -1, 0, 1, or 2.
The Electron Spin Quantum Number ($m_s$)
Unlike $n$, $l$, and $m_l$, the electron spin quantum number $m_s$ does not depend on another quantum number. It designates the direction of the electron spin and may have a spin of +1/2, represented by↑, or –1/2, represented by ↓. This means that when $m_s$ is positive the electron has an upward spin, which can be referred to as "spin up." When it is negative, the electron has a downward spin, so it is "spin down." The significance of the electron spin quantum number is its determination of an atom's ability to generate a magnetic field or not. (Electron Spin.)
$m_s= \pm \dfrac{1}{2}$
Example $5$
List the possible combinations of all four quantum numbers when $n=2$, $l=1$, and $m_l=0$.
Answer
The fourth quantum number is independent of the first three, allowing the first three quantum numbers of two electrons to be the same. Since the spin can be +1/2 or =1/2, there are two combinations:
• $n=2$, $l=1$, $m_l =0$, $m_s=+1/2$
• $n=2$, $l=1$, $m_l=0$, $m_s=-1/2$
Example $6$
Can an electron with $m_s=1/2$ have a downward spin?
Answer
No, if the value of $m_s$ is positive, the electron is "spin up."
A Closer Look at Shells, Subshells, and Orbitals
Principal Shells
The value of the principal quantum number n is the level of the principal electronic shell (principal level). All orbitals that have the same n value are in the same principal level. For example, all orbitals on the second principal level have a principal quantum number of n=2. When the value of n is higher, the number of principal electronic shells is greater. This causes a greater distance between the farthest electron and the nucleus. As a result, the size of the atom and its atomic radius increases.
Because the atomic radius increases, the electrons are farther from the nucleus. Thus it is easier for the atom to expel an electron because the nucleus does not have as strong a pull on it, and the ionization energy decreases.
Example $7$
Which orbital has a higher ionization energy, one with $n=3$ or $n=2$?
Answer
The orbital with n=2, because the closer the electron is to the nucleus or the smaller the atomic radius, the more energy it takes to expel an electron.
Subshells
The number of values of the orbital angular number l can also be used to identify the number of subshells in a principal electron shell:
• When n = 1, l= 0 (l takes on one value and thus there can only be one subshell)
• When n = 2, l= 0, 1 (l takes on two values and thus there are two possible subshells)
• When n = 3, l= 0, 1, 2 (l takes on three values and thus there are three possible subshells)
After looking at the examples above, we see that the value of n is equal to the number of subshells in a principal electronic shell:
• Principal shell with n = 1 has one subshell
• Principal shell with n = 2 has two subshells
• Principal shell with n = 3 has three subshells
To identify what type of possible subshells n has, these subshells have been assigned letter names. The value of l determines the name of the subshell:
Name of Subshell Value of $l$
s subshell 0
p subshell 1
d subshell 2
f subshell 3
Therefore:
• Principal shell with n = 1 has one s subshell (l = 0)
• Principal shell with n = 2 has one s subshell and one p subshell (l = 0, 1)
• Principal shell with n = 3 has one s subshell, one p subshell, and one d subshell (l = 0, 1, 2)
We can designate a principal quantum number, n, and a certain subshell by combining the value of n and the name of the subshell (which can be found using l). For example, 3p refers to the third principal quantum number (n=3) and the p subshell (l=1).
Example $8$
What is the name of the orbital with quantum numbers n=4 and l=1?
Answer
Knowing that the principal quantum number n is 4 and using the table above, we can conclude that it is 4p.
Example $9$
What is the name of the oribital(s) with quantum number n=3?
Answer
3s, 3p, and 3d. Because n=3, the possible values of l = 0, 1, 2, which indicates the shapes of each subshell.
Orbitals
The number of orbitals in a subshell is equivalent to the number of values the magnetic quantum number ml takes on. A helpful equation to determine the number of orbitals in a subshell is 2l +1. This equation will not give you the value of ml, but the number of possible values that ml can take on in a particular orbital. For example, if l=1 and ml can have values -1, 0, or +1, the value of 2l+1 will be three and there will be three different orbitals. The names of the orbitals are named after the subshells they are found in:
s orbitals p orbitals d orbitals f orbitals
l 0 1 2 3
ml 0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3
Number of orbitals in designated subshell 1 3 5 7
In the figure below, we see examples of two orbitals: the p orbital (blue) and the s orbital (red). The red s orbital is a 1s orbital. To picture a 2s orbital, imagine a layer similar to a cross section of a jawbreaker around the circle. The layers are depicting the atoms angular nodes. To picture a 3s orbital, imagine another layer around the circle, and so on and so on. The p orbital is similar to the shape of a dumbbell, with its orientation within a subshell depending on ml. The shape and orientation of an orbital depends on l and ml.
To visualize and organize the first three quantum numbers, we can think of them as constituents of a house. In the following image, the roof represents the principal quantum number n, each level represents a subshell l, and each room represents the different orbitals ml in each subshell. The s orbital, because the value of ml can only be 0, can only exist in one plane. The p orbital, however, has three possible values of ml and so it has three possible orientations of the orbitals, shown by Px, Py, and Pz. The pattern continues, with the d orbital containing 5 possible orbital orientations, and f has 7:
Another helpful visual in looking at the possible orbitals and subshells with a set of quantum numbers would be the electron orbital diagram. (For more electron orbital diagrams, see Electron Configurations.) The characteristics of each quantum number are depicted in different areas of this diagram.
Restrictions
• Pauli Exclusion Principle: In 1926, Wolfgang Pauli discovered that a set of quantum numbers is specific to a certain electron. That is, no two electrons can have the same values for n, l, ml, and ms. Although the first three quantum numbers identify a specific orbital and may have the same values, the fourth is significant and must have opposite spins.
• Hund's Rule: Orbitals may have identical energy levels when they are of the same principal shell. These orbitals are called degenerate, or "equal energy." According to Hund's Rule, electrons fill orbitals one at a time. This means that when drawing electron configurations using the model with the arrows, you must fill each shell with one electron each before starting to pair them up. Remember that the charge of an electron is negative and electrons repel each other. Electrons will try to create distance between it and other electrons by staying unpaired. This further explains why the spins of electrons in an orbital are opposite (i.e. +1/2 and -1/2).
• Heisenberg Uncertainty Principle: According to the Heisenberg Uncertainty Principle, we cannot precisely measure the momentum and position of an electron at the same time. As the momentum of the electron is more and more certain, the position of the electron becomes less certain, and vice versa. This helps explain integral quantum numbers and why n=2.5 cannot exist as a principal quantum number. There must be an integral number of wavelengths (n) in order for an electron to maintain a standing wave. If there were to be partial waves, the whole and partial waves would cancel each other out and the particle would not move. If the particle was at rest, then its position and momentum would be certain. Because this is not so, n must have an integral value. It is not that the principal quantum number can only be measured in integral numbers, it is because the crest of one wave will overlap with the trough of another, and the wave will cancel out.
Problems
1. Suppose that all you know about a certain electron is that its principal quantum number is 3. What are the possible values for the other four quantum numbers?
2. Is it possible to have an electron with these quantum numbers: $n=2$, $l=1$, $m_l=3$, $m_s=1/2$? Why or why not?
3. Is it possible to have two electrons with the same $n$, $l$, and $m_l$?
4. How many subshells are in principal quantum level $n=3$?
5. What type of orbital is designated by quantum numbers $n=4$, $l=3$, and $m_l =0$?
Solutions
• When $n=3$, $l=0$, $m_l = 0$, and $m_s=+1/2 \text{ or } -1/2$
• $l=1$, $m_l = -1, 0, or +1$, and $m_s=+1/2 \text{ or } -1/2$
• $l=2$, $m_l = -2, -1, 0, 1, \text{ or }+2$, and $m_s=+1/2 \text{ or } -1/2$
1. No, it is not possible. $m_l=3$ is not in the range of $-l$ to $+l$. The value should be be either -1, 0, or +1.
2. Yes, it is possible to have two electrons with the same $n$, $l$, and $m_l$. The spin of one electron must be +1/2 while the spin of the other electron must be -1/2.
3. There are three subshells in principal quantum level $n=3$.
4. Since $l=3$ refers to the f subshell, the type of orbital represented is 4f (combination of the principal quantum number n and the name of the subshell).
Contributors and Attributions
• Anastasiya Kamenko, Tamara Enriquez (UCD), Mandy Lam (UCD)
• Dr. Craig Fisher (Japan Fine Ceramics Center) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.06%3A_Particle_in_a_Box.txt |
An atom is composed of a nucleus containing neutrons and protons with electrons dispersed throughout the remaining space. Electrons, however, are not simply floating within the atom; instead, they are fixed within electronic orbitals. Electronic orbitals are regions within the atom in which electrons have the highest probability of being found.
Quantum Numbers describing Electronic Orbitals
There are multiple orbitals within an atom. Each has its own specific energy level and properties. Because each orbital is different, they are assigned specific quantum numbers: 1s, 2s, 2p 3s, 3p,4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. The numbers, (n=1,2,3, etc.) are called principal quantum numbers and can only be positive numbers. The letters (s,p,d,f) represent the orbital angular momentum quantum number () and the orbital angular momentum quantum number may be 0 or a positive number, but can never be greater than n-1. Each letter is paired with a specific value:
s: subshell = 0
p: subshell = 1
d: subshell = 2
f: subshell = 3
An orbital is also described by its magnetic quantum number (m). The magnetic quantum number can range from to +. This number indicates how many orbitals there are and thus how many electrons can reside in each atom.
Orbitals that have the same or identical energy levels are referred to as degenerate. An example is the 2p orbital: 2px has the same energy level as 2py. This concept becomes more important when dealing with molecular orbitals. The Pauli exclusion principle states that no two electrons can have the same exact orbital configuration; in other words, the same quantum numbers. However, the electron can exist in spin up (ms = +1/2) or with spin down (ms = -1/2) configurations. This means that the s orbital can contain up to two electrons, the p orbital can contain up to six electrons, the d orbital can contain up to 10 electrons, and the f orbital can contain up to 14 electrons.
s subshell p subshell d subshell f subshell
Table 1: Breakdown and Properties of Subshells
ℓ = 0 ℓ = 1 ℓ = 2 ℓ = 3
m = 0 m= -1, 0, +1 m= -2, -1, 0, +1, +2 m= -3, -2, -1, 0, +1, +2, +3
One s orbital Three p orbitals Five d orbitals Seven f orbitals
2 s orbital electrons 6 p orbital electrons 10 d orbital electrons 14 f orbital electrons
Visualizing Electron Orbitals
As discussed in the previous section, the magnetic quantum number (ml) can range from –l to +l. The number of possible values is the number of lobes (orbitals) there are in the s, p, d, and f subshells. As shown in Table 1, the s subshell has one lobe, the p subshell has three lobes, the d subshell has five lobes, and the f subshell has seven lobes. Each of these lobes is labeled differently and is named depending on which plane the lobe is resting in. If the lobe lies along the x plane, then it is labeled with an x, as in 2px. If the lobe lies along the xy plane, then it is labeled with a xy such as dxy. Electrons are found within the lobes. The plane (or planes) that the orbitals do not fill are called nodes. These are regions in which there is a 0 probability density of finding electrons. For example, in the dyx orbital, there are nodes on planes xz and yz. This can be seen in Figure \(1\).
Radial and Angular Nodes
There are two types of nodes, angular and radial nodes. Angular nodes are typically flat plane (at fixed angles), like those in the diagram above. The quantum number determines the number of angular nodes in an orbital. Radial nodes are spheres (at fixed radius) that occurs as the principal quantum number increases. The total nodes of an orbital is the sum of angular and radial nodes and is given in terms of the \(n\) and \(l\) quantum number by the following equation:
\[ N = n-l -1\]
For example, determine the nodes in the 3pz orbital, given that n = 3 and = 1 (because it is a p orbital). The total number of nodes present in this orbital is equal to n-1. In this case, 3-1=2, so there are 2 total nodes. The quantum number determines the number of angular nodes; there is 1 angular node, specifically on the xy plane because this is a pz orbital. Because there is one node left, there must be one radial node. To sum up, the 3pz orbital has 2 nodes: 1 angular node and 1 radial node. This is demonstrated in Figure 2.
Another example is the 5dxy orbital. There are four nodes total (5-1=4) and there are two angular nodes (d orbital has a quantum number =2) on the xz and zy planes. This means there there must be two radial nodes. The number of radial and angular nodes can only be calculated if the principal quantum number, type of orbital (s,p,d,f), and the plane that the orbital is resting on (x,y,z, xy, etc.) are known.
Electron Configuration within an Orbital
We can think of an atom like a hotel. The nucleus is the lobby where the protons and neutrons are, and in the floors above, we find the rooms (orbitals) with the electrons. The principal quantum number is the floor number, the subshell type lets us know what type of room it is (s being a closet, p being a single room, d having two adjoining rooms, and f being a suit with three rooms) , the magnetic quantum number lets us know how many beds there are in the room, and two electrons can sleep in one bed (this is because each has a different spin; -1/2 and 1/2). For example, on the first floor we have the s orbital. The s orbital is a closet and has one bed in it so the first floor can hold a total of two electrons. The second floor has the room styles s and p. The s is a closet with one bed as we know and the p room is a single with three beds in it so the second floor can hold a total of 8 electrons.
Each orbital, as previously mentioned, has its own energy level associated to it. The lowest energy level electron orbitals are filled first and if there are more electrons after the lowest energy level is filled, they move to the next orbital. The order of the electron orbital energy levels, starting from least to greatest, is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Since electrons all have the same charge, they stay as far away as possible because of repulsion. So, if there are open orbitals in the same energy level, the electrons will fill each orbital singly before filling the orbital with two electrons. For example, the 2p shell has three p orbitals. If there are more electrons after the 1s, and 2s orbitals have been filled, each p orbital will be filled with one electron first before two electrons try to reside in the same p orbital. This is known as Hund's rule.
The way electrons move from one orbital to the next is very similar to walking up a flight of stairs. When walking up stairs, you place one foot on the first stair and then another foot on the second stair. At any point in time, you can either stand with both feet on the first stair, or on the second stair but it is impossible to stand in between the two stairs. This is the way electrons move from one electron orbital to the next. Electrons can either jump to a higher energy level by absorbing, or gaining energy, or drop to a lower energy level by emitting, or losing energy. However, electrons will never be found in between two orbitals.
Problems
1. Which orbital would the electrons fill first? The 2s or 2p orbital?
2. How many d orbitals are there in the d subshell?
3. How many electrons can the p orbital hold?
4. Determine the number of angular and radial nodes of a 4f orbital.
5. What is the shape of an orbital with 4 radial nodes and 1 angular node in the xy plane?
Solutions
1. The 2s orbital would be filled before the 2p orbital because orbitals that are lower in energy are filled first. The 2s orbital is lower in energy than the 2p orbital.
2. There are 5 d orbitals in the d subshell.
3. A p orbital can hold 6 electrons.
4. Based off of the given information, n=4 and =3. Thus, there are 3 angular nodes present. The total number of nodes in this orbital is: 4-1=3, which means there are no radial nodes present.
5. 1 angular node means =1 which tells us that we have a p subshell, specifically the pz orbital because the angular node is on the xy plane. The total number of nodes in this orbital is: 4 radial nodes +1 angular node=5 nodes. To find n, solve the equation: nodes=n-1; in this case, 5=n-1, so n=6. This gives us a: 6pz orbital | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.09%3A_Orbital_Shapes_and_Energies.txt |
Electron Spin or Spin Quantum Number is the fourth quantum number for electrons in atoms and molecules. Denoted as $m_s$, the electron spin is constituted by either upward ($m_s=+1/2$) or downward ($m_s=-1/2$) arrows.
Introduction
In 1920, Otto Stern and Walter Gerlach designed an experiment, which unintentionally led to the discovery that electrons have their own individual, continuous spin even as they move along their orbital of an atom. Today, this electron spin is indicated by the fourth quantum number, also known as the Electron Spin Quantum Number and denoted by ms. In 1925, Samuel Goudsmit and George Uhlenbeck made the claim that features of the hydrogen spectrum that were unexamined might by explained by assuming electrons act as if it has a spin. This spin can be denoted by an arrow pointing up, which is +1/2, or an arrow pointing down, which is -1/2.
The experiment mentioned above by Otto Stern and Walter Gerlach was done with silver which was put in an oven and vaporized. The result was that silver atoms formed a beam that passed through a magnetic field in which it split in two.
An explanation of this is that an electron has a magnetic field due to its spin. When electrons that have opposite spins are put together, there is no net magnetic field because the positive and negative spins cancel each other out. The silver atom used in the experiment has a total of 47 electrons, 23 of one spin type, and 24 of the opposite. Because electrons of the same spin cancel each other out, the one unpaired electron in the atom will determine the spin. There is a high likelihood for either spin due to the large number of electrons, so when it went through the magnetic field it split into two beams.
Brief Explanation of Quantum Numbers
Note: In this module, capital "L" will be used instead of small case "l" for angular momentum quantum number.
A total of four quantum numbers were developed to better understand the movement and pathway of electrons in its designated orbital within an atom.
1. Principal quantum number (n): energy level n = 1, 2, 3, 4, ...
2. Orbital Angular Momentum Quantum Number (L): shape (of orbital) L = 0, 1, 2, 3, ...n-1
3. Magnetic Quantum Number (mL): orientation mL = interval of (-L, +L)
4. Electron Spin Quantum Number (ms): independent of other three quantum numbers because ms is always = –½ or +½
(For more information about the three quantum numbers above, see Quantum Number.)
The lines represent how many orientations each orbital has, (e.g. the s-orbital has one orientation, a p-orbital has three orientations, etc.) and each line can hold up to two electrons, represented by up and down arrows. An electron with an up arrow means it has an electron spin of +$\frac{1}{2}$, and an electron with a down arrow means it has an electron spin of -$\frac{1}{2}$.
Electron Spin
Significance: determines if an atom will or will not generate a magnetic field (For more information, scroll down to Magnetic Spin, Magnetism, and Magnetic Field Lines). Although the electron spin is limited to or –½, certain rules apply when assigning electrons of different spins to fill a subshell (orientations of an orbital) . For more information, scroll down to Assigning Spin Direction.
Magnetic Spin, Magnetism, and Magnetic Field Lines
An atom with unpaired electrons are termed as paramagnetic
• results in a net magnetic field because electrons within the orbital are not stabilized or balanced enough
• atoms are attracted to magnets
An atom with paired electrons are termed as diamagnetic
• results in no magnetic field because electrons are uniform and stabilized within the orbital
• atoms are not attracted to magnets
Applying concepts of magnetism with liquid nitrogen and liquid oxygen:
The magnetic spin of an electron follows in the direction of the magnetic field lines as shown below.
{{media("www.youtube.com/watch?v=uj0DFDfQajw\)
Assigning Spin Direction
An effective visual on how to assign spin directions can be represented by the orbital diagram (shown previously and below.) Restrictions apply when assigning spin directions to electrons, so the following Pauli Exclusion Principle and Hund's Rule help explain this.
When one is filling an orbital, such as the p orbital, you must fill all orbitals possible with one electron spin before assigning the opposite spin. For example, when filling the fluorine, which will have a total of two electrons in the s orbital and a total of five electrons in the p orbital, one will start with the s orbital which will contain two electrons. So, the first electron one assigns will be spin up and the next spin down. Moving on to the three p orbitals that one will start by assigning a spin up electron in each of the three orbitals. That takes up three of the five electrons, so with the remaining two electrons, one returns to the first and second p orbital and assigns the spin down electron. This means there will be one unpaired electron in fluorine so it will be paramagnetic.
Pauli Exclusion Principle
The Pauli exclusion principle declares that there can only be a maximum of two electrons for every one orientation, and the two electrons must be opposite in spin direction; meaning one electron has $m_s = +\frac{1}{2}$ and the other electron has $m_s = -\frac{1}{2}$.
Hund's Rule
Hund's Rule declares that the electrons in the orbital are filled up first by the +$\frac{1}{2}$ spin. Once all the orbitals are filled with unpaired +$\frac{1}{2}$ spins, the orbitals are then filled with the -$\frac{1}{2}$ spin. (See examples below, labeled electronic configuration.)
(For more information on Pauli Exclusion Principle and Hund's Rules, see Electronic Configuration.)
Identifying Spin Direction
1. Determine the number of electrons the atom has.
2. Draw the electron configuration for the atom. See Electronic Configurations for more information.
3. Distribute the electrons, using up and down arrows to represent the electron spin direction.
Example 1: Sulfur
Sulfur - S (16 electrons)
Electronic Configuration:
1s2 2s2 2p6 3s2 3p4 OR [Ne] 3s2 3p4
As shown in the following image, this is a demonstration of a -$\frac{1}{2}$ and a +$\frac{1}{2}$ Electron Spin.
Principal Quantum Number & (s, p, d, f) Orbitals
When given a principal quantum number, n, with either the s, p, d or f-orbital, identify all the possibilities of L, mL and ms.
Example 2
Given 5f, identify all the possibilities of the four quantum numbers.
Solution
In this problem, the principal quantum number is n = 5 (the subshell number placed in front of the orbital, the f-orbital in this case). Since we are looking at the f-obital, therefore L = 3. (Look under "Subshells" in the module Quantum Numbers for more information) Knowing L = 3, we can interpret that mL = 0, $\pm$ 1, $\pm$ 2, $\pm$ 3 since mL = -L,...,-1, 0, 1,...+L. As for ms, since it isn't specified in the problem as to whether it is -$\frac{1}{2}$ or +$\frac{1}{2}$, therefore for this problem, it could be both; meaning that the electron spin quantum number is $\pm$$\frac{1}{2}$.
Example 3
Given 6s and mL= +1, identify all the possibilities of the four quantum numbers.
Solution
The principal quantum number is n = 6. Looking at the s-orbital, we know that L = 0. Knowing that mL = -L,...,-1, 0, 1,...+L, therefore mL = +1 is not possible since in this problem, the interval of mL can only equal to 0 according to the angular momentum quantum number, L.
Example 4
Given 4d and ms = +$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
Solution
The principal quantum number is n = 4. Given that it is a d-orbital, we know that L = 2. Therefore, mL = 0, $\pm$ 1, $\pm$ 2 since mL = -L,...,-1, 0, 1,...+L. For ms, this problem specifically said ms = +$\frac{1}{2}$; meaning that the electron spin quantum number is +$\frac{1}{2}$.
Electron Spin: Where to begin?
First, draw a table labeled n, L, mLand ms, as shown below:
n L mL ms
Then, depending on what the question is asking for, fill in the boxes accordingly. Finally, determine the number of electrons for the given quantum number, n, with regards to L, mL and ms.
Example 5
How many electrons can have n = 5 and L = 1? 6
n L mL ms
5 1 -1 -$\frac{1}{2}$, +$\frac{1}{2}$
0
+1
This problem includes both -$\frac{1}{2}$, +$\frac{1}{2}$, therefore the answer is 6 electrons based on the mL.
Example 6
How many electrons can have n = 5 and ms = -$\frac{1}{2}$? 5
n L mL ms
5 4 $\pm$ 4 -$\frac{1}{2}$
3 $\pm$ 3
2 $\pm$ 2
1 $\pm$ 1
0 0
This problem only wants the Spin Quantum Number to be -$\frac{1}{2}$, the answer is 5 electrons based on the mL.
Example 7
How many electrons can have n = 3, L = 2 and mL = 3? zero
Solution
n L mL ms
3 2 NOT POSSIBLE
Since mL = -L...-1, 0, +1...+L (See Electronic Orbitals for more information), mL is not possible because L = 2, so it is impossible for mL to be equal to 3. So, there is zero electrons.
Example 8
How many electrons can have n = 3, mL = +2 and ms = +$\frac{1}{2}$? 1
Solution
n L mL ms
3 2 -2, +2 +$\frac{1}{2}$
1 $\pm$ 1
0 0
This problem only wants the Spin Quantum Number to be +$\frac{1}{2}$, the answer is 1 electrons based on the mL.
Practice Problems
1. Identify the spin direction (e.g. ms = -$\frac{1}{2}$ or +$\frac{1}{2}$ or $\pm$$\frac{1}{2}$ ) of the outermost electron in a Sodium (Na) atom.
2. Identify the spin direction of the outermost electron in a Chlorine (Cl) atom.
3. Identify the spin direction of the outermost electron in a Calcium (Ca) atom.
4. Given 5p and ms = +$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
5. Given 6f, identify all the possibilities of the four quantum numbers.
6. How many electrons can have n = 4 and L = 1?
7. How many electrons can have n = 4, L = 1, mL = -2 and ms = +$\frac{1}{2}$?
8. How many electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$?
9. How many electrons can have n = 5, L = 4, mL = +3 and ms = -$\frac{1}{2}$?
10. How many electrons can have n = 4, L = 2, mL = $\pm$1 and ms = -$\frac{1}{2}$?
11. How many electrons can have n = 3, L = 3, mL = +2?
Solutions: Check your work!
Problem (1): Sodium (Na) --> Electronic Configuration [Ne] 3s1
Spin direction for the valence electron or ms = +$\frac{1}{2}$
Sodium (Na) with a neutral charge of zero is paramagnetic, meaning that the electronic configuration for Na consists of one or more unpaired electrons.
Problem (2): Chlorine (Cl) --> Electronic Configuration [Ne] 3s2 3p5
Spin direction for the valence electron or ms = +$\frac{1}{2}$
Chlorine (Cl) with a neutral charge of zero is paramagnetic.
Problem (3): Calcium (Ca) --> Electronic Configuration [He] 4s2
Spin direction for the valence electron or ms = $\pm$$\frac{1}{2}$
Whereas for Calcium (Ca) with a neutral charge of zero, it is diamagnetic; meaning that ALL the electrons are paired as shown in the image above.
Problem (4): Given 5p and ms = -$\frac{1}{2}$, identify all the possibilities of the four quantum numbers.
The principal quantum number is n = 5. Given that it is a p-orbital, we know that L = 1. And based on L, mL = 0, $\pm$ 1 since mL = -L,...,-1, 0, 1,...+L. As for ms, this problem specifically says ms = -$\frac{1}{2}$, meaning that the spin direction is -$\frac{1}{2}$, pointing downwards ("down" spin).
Problem (5): Given 6f, identify all the possibilities of the four quantum numbers.
The principal quantum number is n = 6. Given that it is a f-orbital, we know that L = 3. Based on L, mL = 0, $\pm$ 1, $\pm$ 2, $\pm$ 3 since mL = -L,...,-1, 0, 1,...+L. As for ms, since it isn't specified in the problem as to whether it is -$\frac{1}{2}$ or +$\frac{1}{2}$, therefore for this problem, it could be both; meaning that the electron spin quantum number is $\pm$$\frac{1}{2}$.
Problem (6): How many electrons can have n = 4 and L = 1? 6
n L mL ms
4 1 $\pm$ 1 -$\frac{1}{2}$, + $\frac{1}{2}$
0
This problem includes both -$\frac{1}{2}$, +$\frac{1}{2}$, therefore the answer is 6 electrons based on the mL.
Problem (7): How many electrons can have n = 4, L = 1, mL = -2 and ms = +$\frac{1}{2}$? zero
n L mL ms
4 1 NOT POSSIBLE + $\frac{1}{2}$
Since mL = -L...-1, 0, +1...+L, mL is not possible because L = 1, so it is impossible for mL to be equal to 2 when mL MUST be with the interval of -L and +L. So, there is zero electron.
Problem (8): How many electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$? 2
n L mL ms
5 3 $\pm$ 2 + $\frac{1}{2}$
$\pm$ 1
0
This problem only wants the Spin Quantum Number to be +$\frac{1}{2}$ and mL = $\pm$ 2, therefore 2 electrons can have n = 5, L = 3, mL = $\pm$ 2 and ms = +$\frac{1}{2}$.
Problem (9): How many electrons can have n = 5, L = 4 and mL = +3? 2
n L mL ms
5 4 -3, +3 -$\frac{1}{2}$, + $\frac{1}{2}$
$\pm$ 2
$\pm$ 1
0
This problem includes both -$\frac{1}{2}$ and +$\frac{1}{2}$ and given that mL = +3, therefore the answer is 2 electrons.
Problem (10): How many electrons can have n = 4, L = 2 and mL = $\pm$ 1? 4
n L mL ms
4 2 $\pm$ 1 -$\frac{1}{2}$, + $\frac{1}{2}$
0
This problem includes both -$\frac{1}{2}$ and +$\frac{1}{2}$ and given that mL = $\pm$ 1, therefore the answer is 4 electrons.
Problem (11): How many electrons can have n = 3, L = 3, mL = +2 and ms = -$\frac{1}{2}$? zero
n L mL ms
3 3 (NOT POSSIBLE) $\pm$ 2 -$\frac{1}{2}$
$\pm$ 1
0
Since L = n - 1, there is zero electron, not possible because in this problem, n = L = 3.
Contributors and Attributions
• Liza Chu (UCD), Sharon Wei (UCD), Mandy Lam (UCD), Lara Cemo (UCD)
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one electron is assigned as a spin up (+1/2) electron, the other electron must be spin-down (-1/2) electron.
Electrons in the same orbital have the same first three quantum numbers, e.g., $n=1$, $l=0$, $m_l=0$ for the 1s subshell. Only two electrons can have these numbers, so that their spin moments must be either $m_s = -1/2$ or $m_s = +1/2$. If the 1s orbital contains only one electron, we have one $m_s$ value and the electron configuration is written as 1s1 (corresponding to hydrogen). If it is fully occupied, we have two $m_s$ values, and the electron configuration is 1s2 (corresponding to helium). Visually these two cases can be represented as
As you can see, the 1s and 2s subshells for beryllium atoms can hold only two electrons and when filled, the electrons must have opposite spins. Otherwise they will have the same four quantum numbers, in violation of the Pauli Exclusion Principle.
The Aufbau section discussed how electrons fill the lowest energy orbitals first, and then move up to higher energy orbitals only after the lower energy orbitals are full. However, there is a problem with this rule. Certainly, 1s orbitals should be filled before 2s orbitals, because the 1s orbitals have a lower value of $n$, and thus a lower energy. What about filling the three different 2p orbitals? In what order should they be filled? The answer to this question involves Hund's rule.
Hund's rule states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied.
2. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. In visualizing this process, consider how electrons exhibit the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals, they first try to get as far as possible from each other before having to pair up.
Example $1$: Nitrogen Atoms
Consider the correct electron configuration of the nitrogen (Z = 7) atom: 1s2 2s2 2p3
The p orbitals are half-filled; there are three electrons and three p orbitals. This is because the three electrons in the 2p subshell will fill all the empty orbitals first before pairing with electrons in them.
Keep in mind that elemental nitrogen is found in nature typically as molecular nitrogen, $\ce{N2}$, which requires molecular orbitals instead of atomic orbitals as demonstrated above.
Example $2$: Oxygen Atoms
Next, consider oxygen (Z = 8) atom, the element after nitrogen in the same period; its electron configuration is: 1s2 2s2 2p4
Oxygen has one more electron than nitrogen; as the orbitals are all half-filled, the new electron must pair up. Keep in mind that elemental oxygen is found in nature typically as molecular oxygen, $\ce{O_2}$, which has molecular orbitals instead of atomic orbitals as demonstrated above.
Hund's Rule Explained
According to the first rule, electrons always enter an empty orbital before they pair up. Electrons are negatively charged and, as a result, they repel each other. Electrons tend to minimize repulsion by occupying their own orbitals, rather than sharing an orbital with another electron. Furthermore, quantum-mechanical calculations have shown that the electrons in singly occupied orbitals are less effectively screened or shielded from the nucleus. Electron shielding is further discussed in the next section.
For the second rule, unpaired electrons in singly occupied orbitals have the same spins. Technically speaking, the first electron in a sublevel could be either "spin-up" or "spin-down." Once the spin of the first electron in a sublevel is chosen, however, the spins of all of the other electrons in that sublevel depend on that first spin. To avoid confusion, scientists typically draw the first electron, and any other unpaired electron, in an orbital as "spin-up."
Example $3$: Carbon and Oxygen
Consider the electron configuration for carbon atoms: 1s22s22p2: The two 2s electrons will occupy the same orbital, whereas the two 2p electrons will be in different orbital (and aligned the same direction) in accordance with Hund's rule.
Consider also the electron configuration of oxygen. Oxygen has 8 electrons. The electron configuration can be written as 1s22s22p4. To draw the orbital diagram, begin with the following observations: the first two electrons will pair up in the 1s orbital; the next two electrons will pair up in the 2s orbital. That leaves 4 electrons, which must be placed in the 2p orbitals. According to Hund’s rule, all orbitals will be singly occupied before any is doubly occupied. Therefore, two p orbital get one electron and one will have two electrons. Hund's rule also stipulates that all of the unpaired electrons must have the same spin. In keeping with convention, the unpaired electrons are drawn as "spin-up", which gives (Figure 1).
Purpose of Electron Configurations
When atoms come into contact with one another, it is the outermost electrons of these atoms, or valence shell, that will interact first. An atom is least stable (and therefore most reactive) when its valence shell is not full. The valence electrons are largely responsible for an element's chemical behavior. Elements that have the same number of valence electrons often have similar chemical properties.
Electron configurations can also predict stability. An atom is most stable (and therefore unreactive) when all its orbitals are full. The most stable configurations are the ones that have full energy levels. These configurations occur in the noble gases. The noble gases are very stable elements that do not react easily with any other elements. Electron configurations can assist in making predictions about the ways in which certain elements will react, and the chemical compounds or molecules that different elements will form. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.10%3A_Electron_Spin_and_the_Pauli_Principle.txt |
Electrons with more than one atom, such as Helium (He), and Nitrogen (N), are referred to as multi-electron atoms. Hydrogen is the only atom in the periodic table that has one electron in the orbitals under ground state. We will learn how additional electrons behave and affect a certain atom.
Introduction
Let's review some basic concepts. First, electrons repel against each other. Particles with the same charge repel each other, while oppositely charged particles attract each other. For example, a proton, which is positively charged, is attracted to electrons, which are negatively charged. However, if we put two electrons together or two protons together, they will repel one another. Since neutrons lack a charge, they will neither repel nor attract protons or electrons.
Figure 1: a) The two electrons are placed together and repel each other because of the same charge. b) The two protons are repelling each other for the same reason. c) When oppositely charged particles, an electron and a proton, are placed together, they attract each other.
Protons and neutrons are located in an atom's nucleus. Electrons float around the atom in energy levels. Energy levels consist of orbitals and sub-orbitals. The lower the energy level the electron is located at, the closer it is to nucleus. As we go down and to the right of the periodic table, the number of electrons that an element has increases. Since there are more electrons, the atom experiences greater repulsion and electrons will tend to stay as far away from each other as possible. Our main focus is what effects take place when more electrons surround the nucleus. To better understand the following concepts it is a good idea to first review quantum mechanics.
Shielding (Screening)
With more protons in the nucleus, the attractive force for electrons to the nucleus is stronger. Thus, the orbital energy becomes more negative (less energy). Orbital energy also depends on the type of l orbital an electron is located in. The lower the number of l, the closer it is to the nucleus. For example, l=0 is the s orbital. S orbitals are closer to the nucleus than the p orbitals (l=1) that are closer to the nucleus than the d orbitals (l=2) that are closer to the f orbitals (l=3).
More electrons create the shielding or screening effect. Shielding or screening means the electrons closer to the nucleus block the outer valence electrons from getting close to the nucleus. See figure 2. Imagine being in a crowded auditorium in a concert. The enthusiastic fans are going to surround the auditorium, trying to get as close to the celebrity on the stage as possible. They are going to prevent people in the back from seeing the celebrity or even the stage. This is the shielding or screening effect. The stage is the nucleus and the celebrity is the protons. The fans are the electrons. Electrons closest to the nucleus will try to be as close to the nucleus as possible. The outer/valence electrons that are farther away from the nucleus will be shielded by the inner electrons. Therefore, the inner electrons prevent the outer electrons from forming a strong attraction to the nucleus. The degree to which the electrons are being screened by inner electrons can be shown by ns<np<nd<nf where n is the energy level. The inner electrons will be attracted to the nucleus much more than the outer electrons. Thus, the attractive forces of the valence electrons to the nucleus are reduced due to the shielding effects. That is why it is easier to remove valence electrons than the inner electrons. It also reduces the nuclear charge of an atom.
Penetration
Penetration is the ability of an electron to get close to the nucleus. The penetration of ns > np > nd > nf. Thus, the closer the electron is to the nucleus, the higher the penetration. Electrons with higher penetration will shield outer electrons from the nucleus more effectively. The s orbital is closer to the nucleus than the p orbital. Thus, electrons in the s orbital have a higher penetration than electrons in the p orbital. That is why the s orbital electrons shield the electrons from the p orbitals. Electrons with higher penetration are closer to the nucleus than electrons with lower penetration. Electrons with lower penetration are being shielded from the nucleus more.
Radial Probability Distribution
Radial probability distribution is a type of probability to find where an electron is mostly likely going to be in an atom. The higher the penetration, the higher probability of finding an electron near the nucleus. As shown by the graphs, electrons of the s orbital are found closer to the nucleus than the p orbital electrons. Likewise, the lower the energy level an electron is located at, the higher chance it has of being found near the nucleus. The smaller the energy level (n) and the orbital angular momentum quantum number (l) of an electron is, the more likely it will be near the nucleus. As electrons get to higher and higher energy levels, the harder it is to locate it because the radius of the sphere is greater. Thus, the probability of locating an electron will be more difficult. Radial probability distribution can be found by multiplying 4πr², the area of a sphere with a radius of r and R²(r).
Radial Probability Distribution = 4πr² X R²(r)
By using the radial probability distribution equation, we can get a better understanding about an electron's behavior, as shown on Figures 3.1-3.3.
Figures 3.1, 3.2, and 3.3 show that the lower the energy level, the higher the probability of finding the electron close to the nucleus. Also, the lower momentum quantum number gets, the closer it is to the nucleus.
Contributors and Attributions
• Mixue (Michelle) Xie (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.11_Polyelectronic_Atoms.txt |
The periodic law was developed independently by Dmitri Mendeleev and Lothar Meyer in 1869. Mendeleev created the first periodic table and was shortly followed by Meyer. They both arranged the elements by their mass and proposed that certain properties periodically reoccur. Meyer formed his periodic law based on the atomic volume or molar volume, which is the atomic mass divided by the density in solid form. Mendeleev's table is noteworthy because it exhibits mostly accurate values for atomic mass and it also contains blank spaces for unknown elements.
Introduction
In 1804 physicist John Dalton advanced the atomic theory of matter, helping scientists determine the mass of the known elements. Around the same time, two chemists Sir Humphry Davy and Michael Faraday developed electrochemistry which aided in the discovery of new elements. By 1829, chemist Johann Wolfgang Doberiner observed that certain elements with similar properties occur in group of three such as; chlorine, bromine, iodine; calcium, strontium, and barium; sulfur, selenium, tellurium; iron, cobalt, manganese. However, at the time of this discovery too few elements had been discovered and there was confusion between molecular weight and atomic weights; therefore, chemists never really understood the significance of Doberiner's triad.
In 1859 two physicists Robert Willhem Bunsen and Gustav Robert Kirchoff discovered spectroscopy which allowed for discovery of many new elements. This gave scientists the tools to reveal the relationships between elements. Thus in 1864, chemist John A. R Newland arranged the elements in increasing of atomic weights. Explaining that a given set of properties reoccurs every eight place, he named it the law of Octaves.
The Periodic Law
In 1869, Dmitri Mendeleev and Lothar Meyer individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically." Meyer based his laws on the atomic volume (the atomic mass of an element divided by the density of its solid form), this property is called Molar volume.
$\text{Atomic (molar) volume (cm}^3\text{/mol)} = \dfrac{\text{ molar mass (g/ mol)}}{\rho \text{ (cm}^3\text{/g)}} \nonumber$
Mendeleev's Periodic Table
Mendeleev's periodic table is an arrangement of the elements that group similar elements together. He left blank spaces for the undiscovered elements (atomic masses, element: 44, scandium; 68, gallium; 72, germanium; & 100, technetium) so that certain elements can be grouped together. However, Mendeleev had not predicted the noble gases, so no spots were left for them.
In Mendeleev's table, elements with similar characteristics fall in vertical columns, called groups. Molar volume increases from top to bottom of a group3
Example
The alkali metals (Mendeleev's group I) have high molar volumes and they also have low melting points which decrease in the order:
Li (174 oC) > Na (97.8 oC) > K (63.7 oC) > Rb (38.9 oC) > Cs (28.5 oC)
Atomic Number as the Basis for the Periodic Law
Assuming there were errors in atomic masses, Mendeleev placed certain elements not in order of increasing atomic mass so that they could fit into the proper groups (similar elements have similar properties) of his periodic table. An example of this was with argon (atomic mass 39.9), which was put in front of potassium (atomic mass 39.1). Elements were placed into groups that expressed similar chemical behavior.
In 1913 Henry G.J. Moseley did researched the X-Ray spectra of the elements and suggested that the energies of electron orbitals depend on the nuclear charge and the nuclear charges of atoms in the target, which is also known as anode, dictate the frequencies of emitted X-Rays. Moseley was able to tie the X-Ray frequencies to numbers equal to the nuclear charges, therefore showing the placement of the elements in Mendeleev's periodic table. The equation he used:
$\nu = A(Z-b)^2 \nonumber$
with
• $\nu$: X-Ray frequency
• $Z$: Atomic Number
• $A$ and $b$: constants
With Moseley's contribution the Periodic Law can be restated:
Similar properties recur periodically when elements are arranged according to increasing atomic number."
Atomic numbers, not weights, determine the factor of chemical properties. As mentioned before, argon weights more than potassium (39.9 vs. 39.1, respectively), yet argon is in front of potassium. Thus, we can see that elements are arranged based on their atomic number. The periodic law is found to help determine many patterns of many different properties of elements; melting and boiling points, densities, electrical conductivity, reactivity, acidic, basic, valance, polarity, and solubility.
The table below shows that elements increase from left to right accordingly to their atomic number. The vertical columns have similar properties within their group for example Lithium is similar to sodium, beryllium is similar to magnesium, and so on.
Group 1 2 13 14 15 16 17 18
Element Li Be B C N O F Ne
Atomic Number 3 4 5 6 7 8 9 10
Atomic Mass 6.94 9.01 10.81 12.01 14.01 15.99 18.99 20.18
Element Na Mg Al Si P S Cl Ar
Atomic Number 11 12 13 14 15 16 17 18
Atomic Mass 22.99 24.31 26.98 20.09 30.97 32.07 35.45 39.95
Elements in Group 1 (periodic table) have similar chemical properties and are called alkali metals. Elements in Group 2 have similar chemical properties, they are called the alkaline earth metals.
Short form periodic table
The short form periodic table is a table where elements are arranged in 7 rows, periods, with increasing atomic numbers from left to right. There are 18 vertical columns known as groups. This table is based on Mendeleev's periodic table and the periodic law.
Long form Periodic Table
In the long form, each period correlates to the building up of electronic shell; the first two groups (1-2) (s-block) and the last 6 groups (13-18) (p-block) make up the main-group elements and the groups (3-12) in between the s and p blocks are called the transition metals. Group 18 elements are called noble gases, and group 17 are called halogens. The f-block elements, called inner transition metals, which are at the bottom of the periodic table (periods 8 and 9); the 15 elements after barium (atomic number 56) are called lanthanides and the 14 elements after radium (atomic number 88) are called actinides.
Problems
1) The periodic law states that
1. similar properties recur periodically when elements are arranged according to increasing atomic number
2. similar properties recur periodically when elements are arranged according to increasing atomic weight
3. similar properties are everywhere on the periodic table
4. elements in the same period have same characteristics
2) Which element is most similar to Sodium
1. Potassium
2. Aluminum
3. Oxygen
4. Calcium
3) According to the periodic law, would argon be in front of potassium or after? Explain why.
4) Which element is most similar to Calcium?
1. Carbon
2. Oxygen
3. Strontium
4. Iodine
5) Who were the two chemists that came up with the periodic law?
1. John Dalton and Michael Faraday
2. Dmitri Mendeleev and Lothar Meyer
3. Michael Faraday and Lothar Meyer
4. John Dalton and Dmitri Mendeleev
Answers
1. A
2. A
3. Argon would in front of potassium because the periodic law states that the periodic table increases from left to right based on atomic number not atomic weights
4. C
5. B | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.12%3A_The_History_of_the_Periodic_Table.txt |
Aufbau comes from the German word "Aufbauen" which means "to build". In essence when writing electron configurations we are building up electron orbitals as we proceed from atom to atom. As we write the electron configuration for an atom, we will fill the orbitals in order of increasing atomic number. The Aufbau principle originates from the Pauli’s exclusion principle which says that no two fermions (e.g., electrons) in an atom can have the same set of quantum numbers, hence they have to "pile up" or "build up" into higher energy levels. How the electrons build up is a topic of electron configurations.
Example \(1\)
If we follow the pattern across a period from B (Z=5) to Ne (Z=10) the number of electrons increase and the subshells are filled. Here we are focusing on the p subshell in which as we move towards Ne, the p subshell becomes filled.
• B (Z=5) configuration: 1s2 2s2 2p1
• C (Z=6) configuration: 1s2 2s2 2p2
• N (Z=7) configuration: 1s2 2s2 2p3
• O (Z=8) configuration: 1s2 2s2 2p4
• F (Z=9) configuration: 1s2 2s2 2p5
• Ne (Z=10) configuration: 1s2 2s2 2p6
Electron configuration can be described as how electrons are assembled within the orbitals shells and subshells of an atom. It is important to understand what an electron is in order to fully understand the electron configuration. An electron is a sub atomic particle that is associated with a negative charge. Electrons are found outside of the nucleus, as opposed to neutrons (particles with neutral charge,) and protons (particles with positive charge.) Furthermore, electrons are associated with energy, more specifically quantum energy, and exemplify wave-like and particle-like characteristics. The word configuration simply means the arrangement of something. Therefore electron configuration in straightforward language means the arrangement of electrons.
Introduction
In general when filling up the electron diagram, it is customary to fill the lowest energies first and work your way up to the higher energies. Principles and rules such as the Pauli exclusion principle, Hund’s rule, and the Aufbau process are used to determine how to properly configure electrons. The Pauli exclusion rule basically says that at most, 2 electrons are allowed to be in the same orbital. Hund’s rule explains that each orbital in the subshell must be occupied with one single electron first before two electrons can be in the same orbital. Lastly, the Aufbau process describes the process of adding electron configuration to each individualized element in the periodic table. Fully understanding the principles relating to electron configuration will promote a better understanding of how to design them and give us a better understanding of each element in the periodic table. How the periodic table was formed has an intimate correlation with electron configuration. After studying the relationship between electron configuration and the period table, it was pointed out by Niels Bohr that electron configurations are similar for elements within the same group in the periodic table. Groups occupy the vertical rows as opposed to a period which is the horizontal rows in the table of elements.
S, P, D, and F Blocks
• It is easy to see how similar electron configurations are in a group when written out. (Allow “n” to be the principal quantum number.) Lets first take a look at group 1 atoms. Group 1 atoms are the alkali metals. Let n=1. Notice the similar configuration within all the group 1 elements.
Group Element Configuration
1 H 1s1
1 Li [He]2s1
1 Na [Ne]3s1
1 K [Ar]4s1
1 Rb [Kr]5s1
1 Cs [Xe]6s1
1 Fr [Rn]7s1
Now consider group 16 elements. These elements also will also have similar electron configurations to each another because they are in the same group; these elements have 6 valence electrons.
Group Element Configuration
16 O [He]2s22p4
16 S [Ne]3s23p4
16 Se [Ar]3d104s2 4p4
16 Te [Kr]4d105s2 5p4
16 Po [Xe]4f14 5d106s2 6p4
• 1. Question, True or False: Elements in the same period have similar electron configurations.
Answer: False. Elements in the same GROUP have similar electron configurations.
2. Question: What element has the electron configuration [Ar] 4s2 3d10 4p5?
Answer: Bromine
3. Question: What element has the electron configuration [Xe] 4f14 5d10 6s2 6p3?
Answer: Bismuth
4. Question: Demonstrate how elements in a group share similar characteristics by filling in the electron configurations for the Group 18 elements:
Group Element Configuration
18 He
18 Ne
18 Ar
18 Kr
18 Xe
18 Rn
Answer: 1s2, [He]2s22p6, [Ne]3s23p6, [Ar]3d104s24p6, [Kr]4d105s25p6, [Xe]4f145d106s26p6
5. Question: How many valence electrons are there in Iodine?
Answer: Iodine, z=53, group 17. This means there are seven valence electrons.
6. Question: What is the highest number of electrons a 4p subshell can hold?
Answer: 6! Each 3 p orbital can hold 2 electrons so if they are all filled, the answer is 6. You get this by multiplying the three orbitals by 2 electrons per orbital, so 3 multiplied by 2 equals 6.
Make up some practice problems for the future readers.
Contributors and Attributions
• Mariana Gerontides (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.13%3A_The_Aufbau_Principles_and_the_Periodic_Table.txt |
The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration.
Atomic Radius
We can never determine the atomic radius of an atom because there is never a zero probability of finding an electron, and thus never a distinct boundary to the atom. All that we can measure is the distance between two nuclei (internuclear distance). A covalent radius is one-half the distance between the nuclei of two identical atoms. An ionic radius is one-half the distance between the nuclei of two ions in an ionic bond. The distance must be apportioned for the smaller cation and larger anion. A metallic radius is one-half the distance between the nuclei of two adjacent atoms in a crystalline structure. The noble gases are left out of the trends in atomic radii because there is great debate over the experimental values of their atomic radii. The SI units for measuring atomic radii are the nanometer (nm) and the picometer (pm). $1 \, nm = 1 \times 10^{-9}\, m$ and $1\, pm = 1 \times 10^{-12}\, m$.
To explain this trend, the concept of screening and penetration must be understood. Penetration is commonly known as the distance that an electron is from the nucleus. Screening is defined as the concept of the inner electrons blocking the outer electrons from the nuclear charge. Within this concept we assume that there is no screening between the outer electrons and that the inner electrons shield the outer electrons from the total positive charge of the nucleus. In order to comprehend the extent of screening and penetration within an atom, scientists came up with the effective nuclear charge, $Z_{eff}$. The equation for calculating the effective nuclear charge is shown below.
$Z_{eff}= Z - S \nonumber$
In the equation S represents the number of inner electrons that screen the outer electrons. Students can easily find S by using the atomic number of the noble gas that is one period above the element. For example, the S we would use for Chlorine would be 10 (the atomic number of Neon). Z is the total number of electrons in the atom. Since we know that a neutral atom has an identical number of protons and electrons, we can use the atomic number to define Z. For example, Chlorine would have a Z value of 17 (the atomic number of Chlorine). Continuing to use Chlorine as an example, the 10 inner electrons (S) would screen out the positive charge of ten protons. Therefore there would be and effective nuclear charge of 17-10 or +7. The effective nuclear charge shows that the nucleus is pulling the outer electrons with a +7 charge and therefore the outer electrons are pulled closer to the nucleus and the atomic radii is smaller. In summary, the greater the nuclear charge, the greater pull the nucleus has on the outer electrons and the smaller the atomic radii. In contrast, the smaller nuclear charge, the lesser pull the nucleus has on the outer electrons, and the larger atomic radii. Additionally, as the atomic number increases, the effective nuclear charge also increases. Figure 3 depicts the effect that the effective nuclear charge has on atomic radii.
Now we are ready to describe the atomic radius trend in the periodic table. The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases. Moving down a group in the periodic table, the number of filled electron shells increases. In a group, the valence electrons keep the same effective nuclear charge, but now the orbitals are farther from the nucleus. Therefore, the nucleus has less of a pull on the outer electrons and the atomic radii are larger.
We can now use these concept to explain the atomic radius differences of cations and anions. A cation is an atom that has lost one of its outer electrons. Cations have a smaller radius than the atom that they were formed from. With the loss of an electron, the positive nuclear charge out powers the negative charge that the electrons exert. Therefore, the positive nucleus pulls the electrons tighter and the radius is smaller. An anion is an atom that has gained an outer electron. Anions have a greater radius than the atom that they were formed from. The gain of an electron does not alter the nuclear charge, but the addition of an electron causes a decrease in the effective nuclear charge. Therefore, the electrons are held more loosely and the atomic radius is increased.
Ionization Energy (ionization potential)
Expelling an electron from an atom requires enough energy to overcome the magnetic pull of the positive charge of the nucleus. Therefore, ionization energy (I.E. or I) is the energy required to completely remove an electron from a gaseous atom or ion. The Ionization Energy is always positive.
The energy required to remove one valence electron is the first ionization energy, the second ionization energy is the energy required to remove a second valence electron, and so on.
• 1st ionization energy
$\ce{Na(g) -> Na^{+}(g) + e^{-}(g)} \nonumber$
• 2nd ionization energy
$\ce{ Na^{+}(g) -> Na^{2+}(g) + e^{-}} \nonumber$
Ionization energies increase relative to high effective charge. The highest ionization energies are the noble gases because they all have high effective charge due to their octet formation and require a high amount of energy to destroy that stable configuration. The highest amount of energy required occurs with the elements in the upper right hand corner. Additionally, elements in the left corner have a low ionization energy because losing an electron allows them to have the noble gas configuration. Therefore, it requires less energy to remove one of their valence electrons
Table 1: Ionization Energies of certain elements (1st IE, 2nd IE, etc)
Element 1st 2nd 3rd 4th 5th 6th 7th
Na 496 4562
Mg 738 1451 7733
Al 577 1817 2745 11580
Si 786 1577 3232 4356 16090
P 1060 1903 2912 4957 6274 21270
S 999.6 2251 3361 4564 7013 8496 27110
Cl 1256 2297 3822 5158 6542 9362 11020
Ar 1520 2666 3931 5771 7238 8781 12000
These are the ionization energies for the period three elements. Notice how Na after in the second I.E, Mg in the third I.E., Al in the fourth I.E., and so on, all have a huge increase in energy compared to the proceeding one. This occurs because the proceeding configuration was in a stable octet formation; therefore it requires a much larger amount of energy to ionize.
Ionization Energies increase going left to right across a period and increase going up a group. As you go up a group, the ionization energy increases, because there are less electron shielding the outer electrons from the pull of the nucleus. Therefore, it requires more energy to out power the nucleus and remove an electron. As we move across the periodic table from left to right, the ionization energy increases , due to the effective nuclear charge increasing. This is because the larger the effective nuclear charge, the stronger the nucleus is holding onto the electron and the more energy it takes to release an electron.
The ionization energy is only a general rule. There are some instances when this trend does not prove to be correct. These can typically be explained by their electron configuration. For example, Magnesium has a higher ionization energy than Aluminum. Magnesium has an electron configuration of [Ne]3s2. Magnesium has a high ionization energy because it has a filled 3s orbital and it requires a higher amount of energy to take an electron from the filled orbital.
Electron Affinity
Electron affinity (E.A.) is the energy change that occurs when an electron is added to a gaseous atom. Electron affinity can further be defined as the enthalpy change that results from the addition of an electron to a gaseous atom. It can be either positive or negative value. The greater the negative value, the more stable the anion is.
• (Exothermic) The electron affinity is positive
$\ce{X(g) + e^{-} -> X^{-} + Energy} \nonumber$
• (Endothermic) The electron affinity is negative
$\ce{X(g) + e^{-} + Energy -> X^{-}} \nonumber$
It is more difficult to come up with trends that describe the electron affinity. Generally, the elements on the right side of the periodic table will have large negative electron affinity. The electron affinities will become less negative as you go from the top to the bottom of the periodic table. However, Nitrogen, Oxygen, and Fluorine do not follow this trend. The noble gas electron configuration will be close to zero because they will not easily gain electrons.
Electronegativity
Electronegativity is the measurement of an atom to compete for electrons in a bond. The higher the electronegativity, the greater its ability to gain electrons in a bond. Electronegativity will be important when we later determine polar and nonpolar molecules. Electronegativity is related with ionization energy and electron affinity. Electrons with low ionization energies have low electronegativities because their nuclei do not exert a strong attractive force on electrons. Elements with high ionization energies have high electronegativities due to the strong pull exerted by the positive nucleus on the negative electrons. Therefore the electronegativity increases from bottom to top and from left to right.
Metallic Character
The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character.
As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations.
Melting Points: Trends in melting points and molecular mass of binary carbon-halogen compounds and hydrogen halides are due to intermolecular forces. Melting destroys the arrangement of atoms in a solid, therefore the amount of heat necessary for melting to occur depends on the strength of attraction between the atoms. This strength of attraction increases as the number of electrons increase. Increase in electrons increases bonding.
Example: Melting point of HF should be approximately -145 °C based off melting points of HCl, HBr, and HI, but the observed value is -83.6°C.
Heat and electricity conductibility vary regularly across a period. Melting points may increase gradually or reach a peak within a group then reverse direction.
Example: Third period elements Na, Mg, and Al are good conductors of heat and electricity while Si is only a fair conductor and the nonmetals P, S, Cl and Ar are poor conductors.
Redox Potentials
Oxidation Potential
Oxidation is a reaction that results in the loss of an electron. Oxidation potential follows the same trends as the ionization energy. That is because the smaller the ionization energy, the easier it is to remove an electron. (e.g)
$K_{(s)} \rightarrow K^+ + e^- \nonumber$
Reduction Potential
Reduction is a reaction that results in the gaining of an electron. Reduction potentials follow the same trend as the electron affinity. That is because the larger, negative electron affinity, the easier it is to give an electron. Example of Reduction:
$F_{(s)} + e^- \rightarrow F^- \nonumber$
Uses in knowing the Periodic Properties of Elements
1. Predicting greater or smaller atomic size and radial distribution in neutral atoms and ions
2. Measuring and comparing ionization energies
3. Comparing electron affinities and electronegativities
• Predicting redox potential
• Comparing metallic character with other elements; its ability to form cations
• Predicting what reaction may or may not occur due to the trends
• Determining greater cell potential (sum of oxidation and reduction potential) between reactions
• Completing chemical reactions according to trends
The Periodic Table of Elements categorizes like elements together. Dmitri Mendeleev, a Russian scientist, was the first to create a widely accepted arrangement of the elements in 1869. Mendeleev believed that when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically. Although most modern periodic tables are arranged in eighteen groups (columns) of elements, Mendeleev's original periodic table had the elements organized into eight groups and twelve periods (rows).
On the periodic table, elements that have similar properties are in the same groups (vertical). From left to right, the atomic number (z) of the elements increases from one period to the next (horizontal). The groups are numbered at the top of each column and the periods on the left next to each row. The main group elements are groups 1,2 and 13 through 18. These groups contain the most naturally abundant elements, and are the most important for life. The elements shaded in light pink in the table above are known as transition metals. The two rows of elements starting at z=58, are sometimes called inner transition metals and have that have been extracted and placed at the bottom of the table, because they would make the table too wide if kept continuous. The 14 elements following lanthanum (z=57) are called lanthanides, and the 14 following actinium (z=89) are called actinides.
Elements in the periodic table can be placed into two broad categories, metals and nonmetals. Most metals are good conductors of heat and electricity, are malleable and ductile, and are moderate to high melting points. In general, nonmetals are nonconductors of heat and electricity, are nonmalleable solids, and many are gases at room temperature. Just as shown in the table above, metals and nonmetals on the periodic table are often separated by a stairstep diagonal line, and several elements near this line are often called metalloids (Si, Ge, As, Sb, Te, and At). Metalloids are elements that look like metals and in some ways behave like metals but also have some nonmetallic properties. The group to the farthest right of the table, shaded orange, is known as the noble gases. Noble gases are treated as a special group of nonmetals.
Alkali Metals/Alkali Earth Metals
The Alkali metals are comprised of group 1 of the periodic table and consist of Lithium, Sodium, Rubidium, Cesium, and Francium. These metals are highly reactive and form ionic compounds (when a nonmetal and a metal come together) as well as many other compounds. Alkali metals all have a charge of +1 and have the largest atom sizes than any of the other elements on each of their respective periods.
Alkali Earth Metals are located in group 2 and consist of Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium. Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive. These metals may also form ionic and other compounds and have a charge of +2.
Transition Metals
The transition metals range from groups IIIB to XIIB on the periodic table. These metals form positively charged ions, are very hard, and have very high melting and boiling points. Transition metals are also good conductors of electricity and are malleable.
Lanthanides and Actinides
Lanthanides (shown in row ** in chart above) and Actinides (shown in row * in chart above), form the block of two rows that are placed at the bottom of the periodic table for space issues. These are also considered to be transition metals. Lanthanides are form the top row of this block and are very soft metals with high boiling and melting points. Actinides form the bottom row and are radioactive. They also form compounds with most nonmetals. To find out why these elements have their own section, check out the electron configurations page.
Metalloids
As mentioned in the introduction, metalloids are located along the staircase separating the metals from the nonmetals on the periodic table. Boron, silicon, germanium, arsenic, antimony, and tellurium all have metal and nonmetal properties. For example, Silicon has a metallic luster but is brittle and is an inefficient conductor of electricity like a nonmetal. As the metalloids have a combination of both metallic and nonmetal characteristics, they are intermediate conductors of electricity or "semiconductors".
Halogens
Halogens are comprised of the five nonmetal elements Flourine, Chlorine, Bromine, Iodine, and Astatine. They are located on group 17 of the periodic table and have a charge of -1. The term "halogen" means "salt-former" and compounds that contain one of the halogens are salts. The physical properties of halogens vary significantly as they can exist as solids, liquids, and gases at room temperature. However in general, halogens are very reactive, especially with the alkali metals and earth metals of groups 1 and 2 with which they form ionic compounds.
Noble Gases
The noble gases consist of group 18 (sometimes reffered to as group O) of the periodic table of elements. The noble gases have very low boiling and melting points and are all gases at room temperature. They are also very nonreactive as they already have a full valence shell with 8 electrons. Therefore, the noble gases have little tendency to lose or gain electrons.
Useful Relationships from the Periodic Table
The periodic table of elements is useful in determining the charges on simple monoatomic ions. For main-group elements, those categorized in groups 1, 2, and 13-18, form ions they lose the same number of electrons as the corresponding group number to which they fall under. For example, K atoms (group 1) lose one electron to become K+ and Mg atoms (group 2) lose two electrons to form Mg2+. The other main-group elements found in group 13 and higher form more than one possible ion.
The elements in groups 3-12 are called transition elements, or transition metals. Similar to the main-group elements described above, the transition metals form positive ions but due to their capability of forming more than two or more ions of differing charge, a relation between the group number and the charge is non-existent.
Exercise $1$
Arrange these elements according to decreasing atomic size: Na, C, Sr, Cu, Fr
Answer
Fr, Sr, Cu, Na, C
Exercise $1$
Arrange these elements according to increasing negative E. A.: Ba, F, Si, Ca, O
Answer
Ba, Ca, Si, O, F
Exercise $1$
Arrange these elements according to increasing metallic character: Li, S, Ag, Cs, Ge
Answer
Li, S, Ge, Ag, Cs
Exercise $1$
Which reaction do you expect to have the greater cell potential?
1. $\ce{2Na(s) + Cl2(g)→ 2NaCl(s) or 2Cs(s) +Cl2(g) → 2RbCl(s)$
2. $\ce{2Na(s) + Cl2(g)→ 2NaCl(s) or Be(s) + Cl2(g) → BeCl2(s)}$
Answer
Second equation
First equation
Exercise $1$
Which equation do you expect to occur?
1. $\ce{I2(s) + 2Br (aq) → Br2(l) + 2I(aq)}$
2. $\ce{Cl2(g) + 2I (aq) → I2 (s) + 2Cl(aq)}$
Answer
5. A) Yes
B) No
Problems
*Highlight Answer:_____ to view answers.
1. An element that is an example of a metalloid is (a) S; (b) Zn; (c) Ge; (d) Re; (e) none of these
Answer: (c) Ge
2. In the periodic table, the vertical (up and down) columns are called (a) periods; (b) transitions; (c) families/groups; (d) metalloids; (e) none of these.
Answer: (c) families/groups
3. Why are noble gases inert (nonreactive)?
Answer: Noble gases are inert because they already have a full valence electron shell and have little tendency to gain or lose electrons.
4. What are compounds that contain a halogen called?
Answer: Salts
5. Lanthanides and Actinides are: (a) alkali earth metals; (b) transition metals; (c) metalloids; (d) alkali metals; (e) none of these
Answer: (b) transition metals | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.15%3A_Periodic_Trends_in_Atomic_Properties.txt |
Alkali metals are the chemical elements found in Group 1 of the periodic table. The alkali metals include: lithium, sodium, potassium, rubidium, cesium, and francium. Although often listed in Group 1 due to its electronic configuration, hydrogen is not technically an alkali metal since it rarely exhibits similar behavior. The word "alkali" received its name from the Arabic word "al qali," meaning "from ashes", which since these elements react with water to form hydroxide ions, creating alkaline solutions (pH>7).
• Group 1: Properties of Alkali Metals
This page discusses the trends in some atomic and physical properties of the Group 1 elements - lithium, sodium, potassium, rubidium and cesium. Sections below cover the trends in atomic radius, first ionization energy, electronegativity, melting and boiling points, and density.
• Group 1: Reactivity of Alkali Metals
Alkali metals are among the most reactive metals. This is due in part to their larger atomic radii and low ionization energies. They tend to donate their electrons in reactions and have an oxidation state of +1. These metals are characterized by their soft texture and silvery color. They also have low boiling and melting points and are less dense than most elements. All these characteristics can be attributed to these elements' large atomic radii and weak metallic bonding.
• Chemistry of Hydrogen (Z=1)
Hydrogen is one of the most important elements in the world. It is all around us. It is a component of water (H2O), fats, petroleum, table sugar (C6H12O6), ammonia (NH3), and hydrogen peroxide (H2O2)—things essential to life, as we know it. This module will explore several aspects of the element and how they apply to the world.
• Chemistry of Lithium (Z=3)
Chlorine is a halogen in Lithium is a rare element found primarily in molten rock and saltwater in very small amounts. It is understood to be non-vital in human biological processes, although it is used in many drug treatments due to its positive effects on the human brain. Because of its reactive properties, humans have utilized lithium in batteries, nuclear fusion reactions, and thermonuclear weapons.
• Chemistry of Sodium (Z=11)
Sodium is metallic element found in the first group of the periodic table. As the sixth most abundant element in the Earth's crust, sodium compounds are commonly found dissolved in the oceans, in minerals, and even in our bodies.
• Chemistry of Potassium (Z=19)
In its pure form, potassium has a white-sliver color but it quickly oxidizes upon exposure to air, tarnishing in minutes if it is not stored under oil or grease. Potassium is essential to several aspects of plant, animal, and human life and is thus mined, manufactured, and consumed in huge quantities around the world.
• Chemistry of Rubidium (Z=37)
Rubidium (Latin: rubidius = red) is similar in physical and chemical characteristics to potassium, but much more reactive. It is the seventeenth most abundant element and was discovered by its red spectral emission in 1861 by Bunsen and Kirchhoff. Its melting point is so low you could melt it in your hand if you had a fever (39°C). But that would not be a good idea because it would react violently with the moisture in your skin.
• Chemistry of Cesium (Z=55)
Cesium is so reactive that it will even explode on contact with ice! It has been used as a "getter" in the manufacture of vacuum tubes (i.e., it helps remove trace quantities of remaining gases). An isotope of cesium is used in the atomic clocks.
• Chemistry of Francium (Z=87)
Francium is the last of the known alkali metals and does not occur to any significant extent in nature. All known isotopes are radioactive and have short half-lives (22 minutes is the longest). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/07%3A_Atomic_Structure_and_Periodicity/12.16%3A_The_Properties_of_a_Group%3A_The_Alkali_Metals.txt |
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium when one or more species in the reaction is shared with another reaction. This results in a shifitng of the equilibrium properties.
Introduction
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\nonumber$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
\begin{align*} \ce{NaCl &\rightleftharpoons Na^{+}} + \color{Green} \ce{Cl^{-}}\[4pt] \ce{KCl &\rightleftharpoons K^{+}} + \color{Green} \ce{Cl^{-}} \[4pt] \ce{CaCl_2 &\rightleftharpoons Ca^{2+}} + \color{Green} \ce{2 Cl^{-}}\[4pt] \ce{AlCl_3 &\rightleftharpoons Al^{3+}} + \color{Green} \ce{3 Cl^{-}}\[4pt] \ce{AgCl & \rightleftharpoons Ag^{+}} + \color{Green} \ce{Cl^{-}} \end{align*}
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\ce{[Na^{+}] = [Ca^{2+}] = [H^{+}] = 0.10\, \ce M}. \nonumber$
but
\begin{align*} \ce{[Cl^{-}]} &= 0.10 \, \ce{(due\: to\: NaCl)}\[4pt] &+ 0.20\, \ce{(due\: to\: CaCl_2)} \[4pt] &+ 0.10\, \ce{(due\: to\: HCl)} \[4pt] &= 0.40\, \ce{M} \end{align*}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}$
Le Chatelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$
Consider the lead(II) ion concentration in this saturated solution of $\ce{PbCl2}$. The balanced reaction is
$\ce{ PbCl2 (s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \label{Ex1.1}$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s\nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^{-}]^2 \[4pt] &= s \times (2s)^2 \[4pt] 1.7 \times 10^{-5} &= 4s^3 \[4pt] s^3 &= \dfrac{1.7 \times 10^{-5}}{4} \[4pt] &= 4.25 \times 10^{-6} \[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \[4pt] &= 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{align*}
The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression in Equation \ref{Ex1.1}. What happens to that equilibrium if extra chloride ions are added? According to Le Chatelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
A Simple Example
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$\ce{[Pb^{2+}]} = s \label{2}\nonumber$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$\ce{[Cl^{-} ]} = 0.100\; M \label{3}\nonumber$
The rest of the mathematics looks like this:
\begin{align*} K_{sp}& = [Pb^{2+}][Cl^-]^2 \[4pt] & = s \times (0.100)^2 \[4pt] 1.7 \times 10^{-5} & = s \times 0.00100 \end{align*}
therefore:
\begin{align*} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \[4pt] & = 1.7 \times 10^{-3} \, \text{M} \end{align*}
Finally, compare that value with the simple saturated solution:
Original solution:
$\ce{[Pb^{2+}]} = 0.0162 \, M \label{5}\nonumber$
Solution in 0.100 M $\ce{NaCl}$ solution:
$\ce{[Pb^{2+}]} = 0.0017 \, M \label{6}\nonumber$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3A}$
The common ion effect of $\ce{H3O^{+}}$ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3B}$
Consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \dfrac{[\ce{NH_4^{+}}][\ce{OH^{-}}]}{[\ce{NH_3}]} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing $Q$ to decrease towards $K$.
Common Ion Effect on Solubility
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $4$
Consider the reaction:
$\ce{ PbCl_2(s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \nonumber$
What happens to the solubility of $\ce{PbCl2(s)}$ when 0.1 M $\ce{NaCl}$ is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5} \nonumber$
Identify the common ion: $\ce{Cl^{-}}$
Notice: $Q_{sp} > K_{sp}$ The addition of $\ce{NaCl}$ has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of $\ce{PbCl2(s)}$ is equivalent to the concentration of $\ce{Pb^{2+}}$ produced because they are in a 1:1 ratio.
Because $K_{sp}$ for the reaction is $1.7 \times 10^{-5}$, the overall reaction would be
$(s)(2s)^2= 1.7 \times 10^{-5}. \nonumber$
Solving the equation for $s$ gives $s= 1.62 \times 10^{-2}\, \text{M}$. The coefficient on $\ce{Cl^{-}}$ is 2, so it is assumed that twice as much $\ce{Cl^{-}}$ is produced as $\ce{Pb^{2+}}$, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because $\ce{Na^{+}}$ and $\ce{Cl^{-}}$ are in a 1:1 ratio in the ionic salt, $\ce{NaCl}$. Therefore, the overall molarity of $\ce{Cl^{-}}$ would be $2s + 0.1$, with $2s$ referring to the contribution of the chloride ion from the dissociation of lead chloride.
\begin{align*} Q_{sp} &= [\ce{Pb^{2+}}][\ce{Cl^{-}}]^2 \[4pt] &= 1.8 \times 10^{-5} \[4pt] &= (s)(2s + 0.1)^2 \[4pt] s &= [Pb^{2+}] \[4pt] &= 1.8 \times 10^{-3} M \[4pt] 2s &= [\ce{Cl^{-}}] \[4pt] &\approx 0.1 M \end{align*}
Notice that the molarity of $\ce{Pb^{2+}}$ is lower when $\ce{NaCl}$ is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that $[\ce{Cl^{-}}]$ is approximately 0.1 M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for $\ce{PbCl2(s)}$ is greater than the equilibrium constant because of the added $\ce{Cl^{-}}$. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.01%3A_Solutions_of_Acids_or_Bases_Containing_a_Common_Ion.txt |
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Origin of the Henderson-Hasselbalch Equation
Where the Henderson-Hasselbalch approximation comes from
$HA + H_2O \rightleftharpoons H_3O^+ + A^- \label{1}$
where,
• $A^-$ is the conjugate base
• $HA$ is the weak acid
We know that $K_a$ is equal to the products over the reactants and, by definition, H2O is essentially a pure liquid that we consider to be equal to one.
$K_a = [H_3O^+][A^-] \label{2}$
Take the $-\log$ of both sides:
$-\log \; K_a = -\log([H_3O^+][A^-]) \label{3}$
$-\log \; K_a = -\log[H_3O^+] \; -\log[A^-] \label{4}$
Using the following two relationships:
$-\log[K_a] = pK_a \label{5}$
$-\log[H_3O^+] = pH \label{6}$
We can simplify the above equation:
$pK_a = pH - \log[A^-] \label{7}$
If we add $\log[A^-]$ to both sides, we get the Henderson-Hasselbalch approximation:
$pH = pK_a + \log[A^-] \label{8}$
This approximation is only valid when:
1. The conjugate base / acid falls between the values of 0.1 and 10
2. The molarity of the buffers exceeds the value of the Ka by a factor of at least 100
Example $1$
Suppose we needed to make a buffer solution with a pH of 2.11. In the first case, we would try and find a weak acid with a pKa value of 2.11. However, at the same time the molarities of the acid and the its salt must be equal to one another. This will cause the two molarities to cancel; leaving the $\log [A^-]$ equal to $\log(1)$ which is zero.
$pH = pK_a + \log[A^-] = 2.11 + \log(1) = 2.11 \nonumber$
This is a very unlikely scenario, however, and you won't often find yourself with Case #1
Example $2$
What mass of $NaC_7H_5O_2$ must be dissolved in 0.200 L of 0.30 M HC7H5O2 to produce a solution with pH = 4.78? (Assume solution volume is constant at 0.200L)
Solution
$HC_7H_5O_2 + H_20 \rightleftharpoons H_3O^+ + C_7H_5O_2$
$K_a =6.3 \times 10^{-5}$
$K_a = \dfrac{[H_3O^+][C_7H_5O_2]}{[HC_7H_5O_2]} = 6.3 \times 10^{-5}$
$[H_3O^+] = 10^{-pH} = 10^{-4.78} = 16.6 \times 10^{-6}\;M\;[HC_7H_5O_2] = 0.30\;M\;[C_7H_5O_2] =$
$[C_7H_5O_2^-] = K_a \times \dfrac{[HC_7H_5O_2]}{[H_3O^+]} \nonumber$
$1.14 \; M = 6.3 \times 10^{-5} \times \dfrac{0.30}{16.6 \times 10^{-6}} \nonumber$
Mass = 0.200 L x 1.14 mol C7H5O2- / 1L x 1mol NaC7H5O2 / 1 mol C7H5O2- x 144 g NaC7H5O2 / 1 mol NaC7H5O2 = 32.832 g NaC7H5O2
8.04: Buffer Capacity
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Blood as a Buffer
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
• Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.
• How Does A Buffer Maintain pH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
• Introduction to Buffers
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Thumbnail: Simulated titration of an acidified solution of a weak acid (pKa = 4.7) with alkali. (Public Domain; Lasse Havelund). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.02%3A_Buffered_Solutions.txt |
$K_a$ is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. A big $K_a$ value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. A small $K_a$ will indicate that you are working with a weak acid and that it will only partially dissociate into ions.
General Guide to Solving Problems involving $K_a$
Generally, the problem usually gives an initial acid concentration and a $K_a$ value. From there you are expected to know:
1. How to write the $K_a$ formula
2. Set up in an ICE table based on the given information
3. Solve for the concentration value, x.
4. Use x to find the equilibrium concentration.
5. Use the concentration to find pH.
How to write the $K_a$ formula
The general formula of an acid dissociating into ions is
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1}$
with
• $HA$ is the acid,
• $A^-$ is the conjugate base and
• $H_3O^+$ is the the hydronium ion
By definition, the $K_a$ formula is written as the products of the reaction divided by the reactants of the reaction
$K_a = \dfrac{[Products]}{[Reactants]} \label{2}$
Based off of this general template, we plug in our concentrations from the chemical equation. The concentrations on the right side of the arrow are the products and the concentrations on the left side are the reactants. Using this information, we now can plug the concentrations in to form the $K_a$ equation. We then write:
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{3}$
The concentration of the hydrogen ion ($[H^+]$) is often used synonymously with the hydrated hydronium ion ($[H_3O^+]$).
To find a concentration of hydronium ions in solution from a pH, we use the formula:
$[H_3O^+]= 10^{-pH}$
This can be flipped to calculate pH from hydronium concentration:
$pH = -\log[H_3O^+]$
• An acidic solution is one that has an excess of $H_3O^+$ ions compared to $OH^-$ ions.
• An basic (or alkaline) solution is one that has an excess of $OH^-$ ions compared to $H_3O^+$ ions.
• A neutral solution is one that has equal concentrations of $OH^-$ ions and $H_3O^+$ ions.
At 25 °C, we can correlate whether a solution is acidic, basic, or neutral based off of the measured pH of the solutions:
• pH = 7 is neutral
• pH > 7 is basic
• pH < 7 is acidic
However, these relationships are not valid at temperatures outside 25 °C.
cid
Calculate the pH of a weak acid solution of 0.2 M HOBr, given:
$HOBr + H_2O \rightleftharpoons H_3O^+ + OBr^-$
$K_a = 2 \times 10^{-9}$
Solution
Step 1: The ICE Table
Since we were given the initial concentration of HOBr in the equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for H3O+ and OBr-, we can assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. M stands for molarity.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration
Equilibrium Concentration
Because we started off without an initial concentration of H3O+ and OBr-, it has to come from somewhere. In the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the HOBr box.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.2 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][OBr-]}{[HOBr-]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.2 - x)}$
Step 4: Set the new equation equal to the given Ka
$2 \times 10^{-9} = \dfrac{(x)(x)}{(0.2 - x)}$
Step 5: Solve for x
$x^2 + (2 \times 10^{-9})x - (4 \times 10^{-10}) = 0$
To solve for x, we use the quadratic formula
• $a=1$
• $b=2 \times 10^{-9}$
• $c=-4 \times 10^{-10}$
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-2 \times 10^{-9} \pm \sqrt{(2 \times10^{-9})^2 - 4(1)(-4 \times 10^{-10})}}{2(1)}$
$x=2.0 \times 10^{-5}$
Step 6: Plug x back into the ICE table to find the concentration
$x= [H_3O^+] = 2 \times 10^{-5} \; M$
Step 7: Use the formula using the concentration to find pH
$pH = -\log[H_3O^+] = -\log(2 \times 10^{-5}) = -(-4.69) = 4.69$
$pH= 4.69$
Example $2$: Concentrated Solution of Acetic Acid (Vineger)
For acetic acid, HC2H3O2, the $K_a$ value is $1.8 \times 10^{-5}$. Calculate the concentration of H3O+ in a 0.3 M solution of HC2H3O2.
Solution
Step 1: The ICE Table
Since we were given the initial concentration of HC2H3O2 in the original equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for $H_3O^+$ and $C_2H_3O_2^-$, we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration
Equilibrium Concentration
Because we started off without any initial concentration of H3O+ and C2H3O2-, is has to come from somewhere. For the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the $HC_2H_3O_2$ box.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.3 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][C_2H_3O_2]}{[HC_2H_3O_2]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.3 - x)}$
Step 4: Set the new equation equal to the given Ka
$1.8 x 10^{-5} = \dfrac{(x)(x)}{(0.3 - x)}$
Step 5: Solve for x
$(x^2)+ (1.8 \times 10^{-5}x)-(5.4 \times 10^{-6})$
To solve for x, we use the quadratic formula
• $a=1$
• $b=1.8 \times 10^{-5}$
• $c=-5.4 \times 10^{-6}$
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-1.8 \times 10^{-5} \pm \sqrt{(1.8 \times10^{-5})^2 - 4(1)(-5.4 \times 10^{-6})}}{2(1)}$
$x=0.0023$
Step 6: Plug x back into the ICE table to find the concentration
$x= [H_3O^+] = 0.0023\; M$
Example $3$: Concentrated Solution of Benzoic Acid
Find the equilibrium concentration of HC7H5O2from a 0.43 M solution of Benzoic Acid, HC7H5O2.
Solution
Given: $K_a$ for HC7H5O2= 6.4 x 10-5
Step 1: The ICE Table
HC7H5O2 H3O+ C7H5O2-
Initial Concentration 0.43 M 0 0
Change in Concentration -x +x +x
Equilibrium Concentration (0.43-x)M x M x M
Step 2: Create the $K_a$ equation using this equation :$K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][C_7H_5O_2-]}{[HC_7H_5O_2]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.43 - x)}$
Step 4: Set the new equation equal to the given Ka
$6.4 x 10^{-5} = \dfrac{(x)(x)}{(0.43 - x)}$
Step 5: Solve for x.
x=0.0052
Step 6: Plug x back into the ICE table to find the concentration
[HC7H5O2]= (0.43-x)M
[HC7H5O2]= (0.43-0.0052)M
Answer: [HC7H5O2]= 0.425 M
Example $4$: Concentrated Solution of Hypochlorous acid
For a 0.2 M solution of Hypochlorous acid, calculate all equilibrium concentrations.
Solution
Given: $K_a = 3.5 \times 10^{-8}$
Step 1: The ICE Table
HOCl H3O+ OCl-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.2 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][OCl-]}{[HOCl-]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.2 - x)}$
Step 4: Set the new equation equal to the given Ka
$3.5 x 10^{-8} = \dfrac{(x)(x)}{(0.2 - x)}$
Step 5: Solve for x
x=8.4 x 10-5
Step 6: Plug x back into the ICE table to find the concentration
[HOCl]= [(.2)-(8.4 x 10-5)]=.199
[H3O+]=8.4 x 10-5
[OCl-]=8.4 x 10-5
Example $5$: pH
Calculate the pH from the equilibrium concentrations of [H3O+] in Example $4$.
Solution
Given:
[HOCl]=0.199
[H3O+]=8.4 x 10-5
[OCl-]=8.4 x 10-5
Step 1: Use the formula using the concentration of [H3O+] to find pH
$pH = -\log[H3O+] = -\log(8.4 x 10^{-5}) = 4.08$
Contributors and Attributions
• Kellie Berman (UCD), Alysia Kreitem (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.05%3A_Titrations_and_pH_Curves.txt |
pH indicators are weak acids that exist as natural dyes and indicate the concentration of H+ (\(H_3O^+\)) ions in a solution via color change. A pH value is determined from the negative logarithm of this concentration and is used to indicate the acidic, basic, or neutral character of the substance you are testing.
Introduction
pH indicators exist as liquid dyes and dye-infused paper strips. They are added to various solutions to determine the pH values of those solutions. Whereas the liquid form of pH indicators is usually added directly to solutions, the paper form is dipped into solutions and then removed for comparison against a color/pH key.
pH 3 4 5 6 7 8 9 10
Color
Very Acidic Acidic Neutral Basic Very Basic
See Figure 1 and 2 to see a color range (1) of a universal indicator (2).
The Implications of the Indicated pH via the Equation
Recall that the value of pH is related to the concentration of H+ (\(H_3O^+\)) of a substance. pH itself is approximated as the cologarithm or negative logarithm of the \(H^+\) ion concentration (Figure 3).
\[pH \approx -log[H_3O^+] \tag{3} \]
A pH of 7 indicates a neutral solution like water. A pH less than 7 indicates an acidic solution and a pH greater than 7 indicates a basic solution. Ultimately, the pH value indicates how much H+ has dissociated from molecules within a solution. The lower the pH value, the higher concentration of H+ ions in the solution and the stronger the acid. Likewise, the higher the pH value, the lower the concentration of H+ ions in the solution and the weaker the acid.
How the Color Change of the Indicator Happens
The color change of a pH indicator is caused by the dissociation of the H+ ion from the indicator itself. Recall that pH indicators are not only natural dyes but also weak acids. The dissociation of the weak acid indicator causes the solution to change color. The equation for the dissociation of the H+ ion of the pH indicator is show below (Figure 4).
\[HIn + H_2O \rightleftharpoons H_3O^+ + In^- \tag{4} \]
with
• \(HIn\) is the acidic pH indicator and
• \(In^-\) is the conjugate base of the pH indicator
It is important here to note that the equation expressed in figure 4 is in equilibrium, meaning Le Chatelier's principle applies to it. Thus, as the concentration of \(H_3O^+\) (H+) increases or decreases, the equilibrium shifts to the left or right accordingly. An increase in the \(HIn\) acid concentration causes the equilibrium to shift to the right (towards products), whereas an increase of the \(In^-\) base concentration causes the equilibrium to shift to the left (towards reactants).
pH Ranges of pH Indicators
pH indicators are specific to the range of pH values one wishes to observe. For example, common indicators such as phenolphthalein, methyl red, and bromothymol blue are used to indicate pH ranges of about 8 to 10, 4.5 to 6, and 6 to 7.5 accordingly. On these ranges, phenolphthalein goes from colorless to pink, methyl red goes from red to yellow, and bromothymol blue goes from yellow to blue. For universal indicators, however, the pH range is much broader and the number of color changes is much greater. See figures 1 and 2 in the introduction for visual representations. Usually, universal pH indicators are in the paper strip form.
Graphing pH vs. the H+ (\(H_3O^+\)) Concentration
It is important to note that the pH scale is a logarithmic scale: hence an increase of 1 pH unit corresponds to a ten times increase of \(H_3O^+\). For example, a solution with a pH of 3 will have an H+ (\(H_3O^+\)) concentration ten times greater than that of a solution with a pH of 4. As pH is the negative logarithm of the H+ (\(H_3O^+\)) concentration of a foreign substance, the lower the pH value, the higher the concentration of H+ (\(H_3O^+\)) ions and the stronger the acid. Additionally, the higher the pH value, the lower the H+ (\(H_3O^+\)) concentration and the stronger the base.
Indicators in Nature
pH indicators can be used in a variety of ways, including measuring the pH of farm soil, shampoos, fruit juices, and bodies of water. Additionally, pH indicators can be found in nature, so therefore their presence in plants and flowers can indicate the pH of the soil from which they grow.
Hydrangeas
Nature contains several natural pH indicators as well: for example, some flower petals (especially Roses and Hydrangeas), certain fruits (cherries, strawberries) and leaves can change color if the pH of the soil changes. See figure 7.
(7)
Lemon juice
In the lemon juice experiment, the pH paper turns from blue to vivid red, indicating the presence of \(H_3O^+\) ions: lemon juice is acidic. Refer to the table of Universal Indicator Color change (figure 1 in the introduction) for clarification.
Cleaning Detergent
The household detergent contained a concentrated solution of sodium bicarbonate, commonly known as baking soda. As shown, the pH paper turns a dark blue: baking soda (in solution) is basic.Refer to the table of Universal Indicator Color change (figure 1 in the introduction) for clarification.
Here is a closer look of the pH papers before and after dipping them in the lemon juice and cleaning detergent (Figure 10):
neutral acidic neutral basic
Figure 10:
Cabbage Juices
Here is a simple demonstration that you could try in the lab or at home to get a better sense of how indicator paper works. Make sure to always wear safety glasses and gloves when performing an experiment!
Materials
• 1 cabbage
• cooking pot
• white paper coffee filters
• strainer
• water
• a bowl
Procedure
1. Peal the cabbage leaves and place them into the pot.
2. Add water into the pot, making sure the water covers the cabbage entirely.
3. Place the pot on the stove and allow to cook at medium heat for about 30 to 35 minutes.
4. Allow it to cool, then pour contents into the bowl using the strainer.
5. Soak your coffee filters in the cabbage juice for about 25 to 30 minutes.
6. Allow the filters to fully dry, then cut them into strips.
7. Now start your pH testing (starts out blue, changes to green [basic], and red [acidic]).
Practice Problems
1. A hair stylist walks into a store and wants to buy a shampoo with slightly acidic/neutral pH for her hair. She finds 5 brands that she really likes, but since she never took any introductory chemistry classes, she is unsure about which one to purchase. The first has a pH of 3.6, the second of 13, the third of 8.2, the fourth of 6.8 and the fifth of 9.7. Which one should she buy?
Answer: The brand that has a pH of 6.8 since it's under 7 (neutral) but very close to it, making it slightly acidic.
2. You decide to test the pH of your brand new swimming pool on your own. The instruction manual advises to keep it between 7.2-7.6. Shockingly, you realize it's set at 8.3! Horrified, you panic and are unsure whether you should add some basic or acidic chemicals in your pool (being mindful of the dose, of course. Those specific chemicals are included in the set, so no need to worry about which one you have to use and (eek!) if they are legal for public use). Which one should you add?
Answer: Since the goal is to lower the pH to its ideal value, we must add acidic solution to the pool.
3. Let's say the concentration of Hydronium ions in an aqueous solution is 0.033 mol/L. What is the corresponding pH of this solution, and based on your answer identify whether the solution is acidic, basic or neutral.
Answer: Using the formula \(pH \approx -log[H_3O+]\)
pH= -log[0.033]= 1.48 : The solution is highly acidic!
4. Now let's do the inverse: Say you have a solution with a pH of 9.4. What is the H30+ ions concentration?
Answer: [\(H_3O^+\)]= 10-9.4= 3.98E-10 mol/L. Seem too low to be true? Think again, if the pH is >7, the solution will be basic, hence the hydronium ions will be low compared to the hydroxide (OH- ions).
5. A more trickier one: 0.00026 moles of acetic acid are added to 2.5 L of water. What is the pH of the solution?
Answer: M=n/L : Macetic acid= 0.00026/2.5 =1.04E-4 mol/L
pH= -log[1.04E-4]= 3.98
• www.krampf.com/experiments/Science_Experiment16.html
• There are many common household products and garden plants that can be used or made into pH indicators. For more information on these common house hold indicators visit http://chemistry.about.com/cs/acidsandbases/a/aa060703a.htm
• commons.wikimedia.org/wiki/Ca...ndicator_paper | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.06%3A_Acid-Base_Indicators.txt |
Considering the relation between solubility and $K_{sq}$ is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants ($K_{sq}$) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the $K_{sq}$ of a slightly soluble solute from its solubility.
Introduction
Solubility is the ability of a substance to dissolve. The two participants in the dissolution process are the solute and the solvent. The solute is the substance that is being dissolved, and the solvent is the substance that is doing the dissolving. For example, sugar is a solute and water is a solvent. Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place.
The solubility product constant ($K_{sq}$) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. For example, the higher the $K_{sq}$, the more soluble the compound is. $K_{sq}$ is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. $K_{sq}$ is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.)
Example 1: Barium Carbonate
Consider the compound barium carbonate BaCO3 (an ionic compound that is not very soluble):
$\ce{BaCO_{3(s)} \rightleftharpoons Ba^{2+} (aq)+CO_3^{2-} (aq) } \nonumber$
Solution
First, write down the equilibrium constant expression:
$K_c = \dfrac{[\ce{Ba^{2+}}][\ce{CO_3^{2-}}]}{[\ce{BaCO_3}]} \nonumber$
The activity of solid $\ce{BaCO3}$ is 1, and considering that the concentrations of these ions are small, the activities of the ions are approximated to their molar concentrations. $K_{sq}$ is therefore equal to the product of the ion concentrations:
\begin{align*} K_{sp} &= [\ce{Ba^{2+}}][\ce{CO_3^{2-}}] \[4pt] &= 5.1 \times 10^{-9} \end{align*}
How are $K_{sp}$ and Solubility Related?
The relation between solubility and the solubility product constants is that one can be used to derive the other. In other words, there is a relationship between the solute's molarity and the solubility of the ions because $K_{sq}$ is the product of the solubility of each ion in moles per liter.
For example, to find the $K_{sq}$ of a slightly soluble compound from its solubility, the solubility of each ion must be converted from mass per volume to moles per liter to find the molarity of each ion. These numbers can then be substituted into the $K_{sq}$ formula, which is the product of the solubility of each ion. An example of this process is given below:
Example 2: Lead Iodide
Suppose the aqueous solubility for compound PbI2 is 0.54 grams/100 ml at 25 °C and calculate the $K_{sq}$ of PbI2 at 25°C.
$PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^-_{(aq)} \nonumber$
Solution
a. Convert 0.54 grams of $PbI_2$ to moles:
$0.54\ grams \times \dfrac{1\ mol\ PbI_2}{461.0\ grams} = 0.001171\ mol\ PbI_2 \nonumber$
b. Convert ml to L:
$\dfrac{100\ mL}{1000\ L} = 0.100\ L \nonumber$
c. Find the molarity:
$\dfrac{0.001171\ mol}{0.100\ L} = 0.01171\ M\ PbI_2 \nonumber$
d. Now find the molarity of each ion by using the stoichiometric ratio (remember there are two I- ions for each Pb2+ ion):
$\begin{eqnarray} [Pb^{2+}] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{1\ mol\ Pb}{1\ mol\ PbI_2} \ &=& 0.011714\ M\ Pb^{2+} \ [I^-] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{2\ mol\ I^-}{1\ mol\ PbI_2} \ &=& 0.23427\ M\ I^- \end{eqnarray}$
e. Finally, plug in the molarity to find $K_{sq}$:
$\begin{eqnarray} K_{sp} &=& [Pb^{2+}][I^-]^2 \ &=& (0.011714\ M)(0.023427\ M)^2 \ &=& 6.4 \times 10^{-6} \end{eqnarray}$
*This relation facilitates solving for the molar solubility of the ionic compounds when the $K_{sq}$ is given to us. The process involves working backwards from $K_{sq}$ to the molarity of the ionic compound.*
Example 3
Suppose the $K_{sq}$ at 25 0C is 8.5 x 10-17 for the compound AgI. What is the molar solubility?
$AgI_{(s)} \rightleftharpoons Ag^+_{(aq)} + I^-_{(aq)}\nonumber$
Solution
a. Let "g" represent the number of moles:
$\begin{eqnarray} K_{sp} &=& [Ag^{2+}][I^-] \ &=& g^2 \ &=& 8.5 \times 10^{-17} \end{eqnarray}$
b. Solve for "g":
$\begin{eqnarray} g^2 &=& 8.5 \times 10^{-17} \ g &=& (8.5 \times 10^{-17})^{\dfrac{1}{2}} \ &=& 9.0 \times 10^{-9} \end{eqnarray}$
The molar solubility of AgI is 9.0 x 10-9 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.08%3A_Solubility_Equilibria_and_the_Solubility_Product.txt |
Learning Objectives
• Calculate ion concentrations to maintain a heterogeneous equilibrium.
• Calculate pH required to precipitate a metal hydroxide.
• Design experiments to separate metal ions in a solution of mixtures of metals.
Chemical Separation by Precipitation
A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$, $\ce{Br-}$, $\ce{SO4^2-}$, $\ce{CO3^2-}$, $\ce{S^2-}$, $\ce{Cr2O4^2-}$, $\ce{PO4^2-}$, $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions.
There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.
What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.
• All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$, $\ce{ClO3-}$, $\ce{NO2-}$, $\ce{HCOO-}$, and $\ce{CH3COO-}$.
• All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$, $\ce{Hg2^2+}$, and $\ce{Pb^2+}$. $\ce{CaF2}$, $\ce{BaF2}$, and $\ce{PbF2}$ are also insoluble.
• All sulfates are soluble, except those of $\ce{Ba^2+}$, $\ce{Sr^2+}$, and $\ce{Pb^2+}$. The doubly charged sulfates are usually less soluble than halides and nitrates.
• Most singly charge cations $\ce{K+}$, $\ce{Na+}$, $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.
These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.
Chemical Separation of Metal Ions
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in the Handbook of this website. On the Handbook Menu, clicking the item Salts Ksp will give the Ksp's of some salts. In the first two examples, we show how barium and strontium can be separated as chromate.
Example 1
The Ksp for strontium chromate is 3.6E-5 and the Ksp for barium chromate is 1.2E-10. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Solution
Since the Ksp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Qsp for $\ce{BaCrO4}$ also increases from zero to Ksp of $\ce{BaCrO4}$, at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Qsp for $\ce{SrCrO4}$ increases to Ksp of $\ce{SrCrO4}$; it then precipitates.
Let us write the equilibrium equations and data down to help us think. Further, let x be the concentration of chromate to precipitate $\ce{Sr^2+}$, and y be that to precipitate $\ce{Ba^2+}$. According to the definition of Ksp we have,
$\begin{array}{cccccl} \ce{SrCrO4 &\rightarrow &Sr^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 3.6\times 10^{-5}\ &&0.30 &&x & \end{array}$
$x = \dfrac{\textrm{3.6e-5}}{0.30} = 1.2 \times 10^{-4} M$
$\begin{array}{cccccl} \ce{BaCrO4 &\rightarrow &Ba^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 1.2 \times 10^{-10}\ &&0.30 &&y & \end{array}$
$y = \dfrac{\textrm{1.2e-10}}{0.30} = 4.0 \times 10^{-10} \;M$
The Ksp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained a little less than 1.2e-4 M, $\ce{Sr^2+}$ ions will remain in the solution.
DISCUSSION
In reality, to control the increase of $\ce{[CrO4^2- ]}$ is very difficult.
Example 2
The Ksp for strontium chromate is $3.6\times 10^{-5}$ and the Ksp for barium chromate is $1.2\times 10^{-10}$. Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$. Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.
Solution
From the solution given in Example 1, $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated as follows.
$\ce{[Ba^2+]\,} \textrm{3.6\times 10^{-4} = 1.2\times 10^{-10}}$
The Ksp of $\ce{BaCrO4}$.
Thus,
$\ce{[Ba^2+]} = \textrm{3.33e-7 M}$
Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$, the mole ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is 0.30 / 3.33e-7 = 9.0e5. In other words, the amount of $\ce{Ba^2+}$ ion in the solid is only 1e-6 (1 ppm) of all metal ions, providing that all the solid was removed when $\ce{[CrO4^2- ]} = \textrm{3.6e-4 M}$.
DISCUSSION
The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.
Example 3
What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts Ksp.
Salt Ksp Salt Ksp
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Ag2SO4}$ $1.4\times 10^{-5}$
$\ce{Hg2Cl2}$ $1.3\times 10^{-18}$ $\ce{BaSO4}$ $1.1\times 10^{-10}$
$\ce{PbCl2}$ $1.7 \times 10^{-5}$ $\ce{CaSO4}$ $2.4\times 10^{-5}$
$\ce{PbSO4}$ $6.3\times 10^{-7}$
$\ce{SrSO4}$ $3.2\times 10^{-7}$
Because the Ksp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$, apppears a good choice of negative ions for their separation.
The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.
DISCUSSION
Find more detailed information about the solubility of lead chloride as a function of temperature.
Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$?
Questions
1. Iron(II) hydroxide is only sparingly soluble in water at 25oC; its Ksp is $7.9 \times 10^{-16}$. Calculate the solubility of iron(II) hydroxide in a buffer solution with pH = 7.00.
2. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. What value of $\ce{[CrO4^2- ]}$ will result in a maximum separation of these two ions?
3. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. Calculate the $\ce{[Ca^2+]/[Ba^2+]}$ ratio in the solution when $\ce{[CrO4^2- ]}$ is maintained at $1.2 \times 10^{-3}\; M$.
Solutions
1. Answer $\ce{[Fe^2+]} = \textrm{0.079 M}$
Consider...
$\ce{[OH- ]} = 10^{(-14+7)} = \textrm{1.00e-7 (buffer)}$.
$\ce{[Fe^2+]} (\textrm{1.00e-7})^2 = K_{\ce{sp}}$; $\ce{[Fe^2+]} =\: ?$
This $\ce{Fe^2+}$ concentration is low; it is not very soluble in a neutral solution (pH = 7).
What is $\ce{[Fe^2+]}$ in a solution whose pH = 6.00?
2. Answer $\ce{[CrO4^2- ]} = \textrm{2.37e-3 M}$
Consider...
Solid $\ce{BaCrO4}$ will form first as $\ce{[CrO4^2- ]}$ increases. The maximum $\ce{[CrO4^2- ]}$ to precipitate $\ce{CaCrO4}$ is estimated as follows.
$\ce{[CrO4^2- ]} = \dfrac{\textrm{7.1e-4}}{0.30} = \textrm{2.37e-3 M}$
Estimate $\ce{[Ba^2+]}$ when $\ce{[CrO4^2- ]} = \textrm{2.3e-3 M}$, slightly below the maximum concentration.
3. Answer $\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \textrm{3e6}$
Consider...
$\ce{[Ba^2+]} = \dfrac{\textrm{1.2e-10}}{\textrm{1.2e-3}} = \textrm{1e-7}$;
$\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \dfrac{0.3}{\textrm{1e-7}} =\: ?$ The ratio of three million is large! | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Aqueous_Solutions_Equilibria/8.09%3A_Precipitation_and_Qualitative_Analysis_Complex_Ion_Equilibria.txt |
In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding theory by modeling metals as a mixture of atomic cores (atomic cores = positive nuclei + inner shell of electrons) and valence electrons. Metallic bonds occur among metal atoms. Whereas ionic bonds join metals to non-metals, metallic bonding joins a bulk of metal atoms. A sheet of aluminum foil and a copper wire are both places where you can see metallic bonding in action.
Metals tend to have high melting points and boiling points suggesting strong bonds between the atoms. Even a soft metal like sodium (melting point 97.8°C) melts at a considerably higher temperature than the element (neon) which precedes it in the Periodic Table. Sodium has the electronic structure 1s22s22p63s1. When sodium atoms come together, the electron in the 3s atomic orbital of one sodium atom shares space with the corresponding electron on a neighboring atom to form a molecular orbital - in much the same sort of way that a covalent bond is formed.
The difference, however, is that each sodium atom is being touched by eight other sodium atoms - and the sharing occurs between the central atom and the 3s orbitals on all of the eight other atoms. Each of these eight is in turn being touched by eight sodium atoms, which in turn are touched by eight atoms - and so on and so on, until you have taken in all the atoms in that lump of sodium. All of the 3s orbitals on all of the atoms overlap to give a vast number of molecular orbitals that extend over the whole piece of metal. There have to be huge numbers of molecular orbitals, of course, because any orbital can only hold two electrons.
The electrons can move freely within these molecular orbitals, and so each electron becomes detached from its parent atom. The electrons are said to be delocalized. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalized electrons (Figure $1$).
This is sometimes described as "an array of positive ions in a sea of electrons". If you are going to use this view, beware! Is a metal made up of atoms or ions? It is made of atoms. Each positive center in the diagram represents all the rest of the atom apart from the outer electron, but that electron has not been lost - it may no longer have an attachment to a particular atom, but it's still there in the structure. Sodium metal is therefore written as $\ce{Na}$, not $\ce{Na^+}$.
Example $1$: Metallic bonding in magnesium
Use the sea of electrons model to explain why Magnesium has a higher melting point (650 °C) than sodium (97.79 °C).
Solution
If you work through the same argument above for sodium with magnesium, you end up with stronger bonds and hence a higher melting point.
Magnesium has the outer electronic structure 3s2. Both of these electrons become delocalized, so the "sea" has twice the electron density as it does in sodium. The remaining "ions" also have twice the charge (if you are going to use this particular view of the metal bond) and so there will be more attraction between "ions" and "sea".
More realistically, each magnesium atom has 12 protons in the nucleus compared with sodium's 11. In both cases, the nucleus is screened from the delocalized electrons by the same number of inner electrons - the 10 electrons in the 1s2 2s2 2p6 orbitals. That means that there will be a net pull from the magnesium nucleus of 2+, but only 1+ from the sodium nucleus.
So not only will there be a greater number of delocalized electrons in magnesium, but there will also be a greater attraction for them from the magnesium nuclei. Magnesium atoms also have a slightly smaller radius than sodium atoms, and so the delocalized electrons are closer to the nuclei. Each magnesium atom also has twelve near neighbors rather than sodium's eight. Both of these factors increase the strength of the bond still further.
Note: Transition metals tend to have particularly high melting points and boiling points. The reason is that they can involve the 3d electrons in the delocalization as well as the 4s. The more electrons you can involve, the stronger the attractions tend to be.
Bulk properties of metals
Metals have several qualities that are unique, such as the ability to conduct electricity and heat, a low ionization energy, and a low electronegativity (so they will give up electrons easily to form cations). Their physical properties include a lustrous (shiny) appearance, and they are malleable and ductile. Metals have a crystal structure but can be easily deformed. In this model, the valence electrons are free, delocalized, mobile, and not associated with any particular atom. This model may account for:
• Conductivity: Since the electrons are free, if electrons from an outside source were pushed into a metal wire at one end (Figure $2$), the electrons would move through the wire and come out at the other end at the same rate (conductivity is the movement of charge).
• Malleability and Ductility: The electron-sea model of metals not only explains their electrical properties but their malleability and ductility as well. The sea of electrons surrounding the protons acts like a cushion, and so when the metal is hammered on, for instance, the overall composition of the structure of the metal is not harmed or changed. The protons may be rearranged but the sea of electrons with adjust to the new formation of protons and keep the metal intact. When one layer of ions in an electron sea moves along one space with respect to the layer below it, the crystal structure does not fracture but is only deformed (Figure $3$).
• Heat capacity: This is explained by the ability of free electrons to move about the solid.
• Luster: The free electrons can absorb photons in the "sea," so metals are opaque-looking. Electrons on the surface can bounce back light at the same frequency that the light hits the surface, therefore the metal appears to be shiny.
However, these observations are only qualitative, and not quantitative, so they cannot be tested. The "Sea of Electrons" theory stands today only as an oversimplified model of how metallic bonding works.
In a molten metal, the metallic bond is still present, although the ordered structure has been broken down. The metallic bond is not fully broken until the metal boils. That means that boiling point is actually a better guide to the strength of the metallic bond than melting point is. On melting, the bond is loosened, not broken. The strength of a metallic bond depends on three things:
1. The number of electrons that become delocalized from the metal
2. The charge of the cation (metal).
3. The size of the cation.
A strong metallic bond will be the result of more delocalized electrons, which causes the effective nuclear charge on electrons on the cation to increase, in effect making the size of the cation smaller. Metallic bonds are strong and require a great deal of energy to break, and therefore metals have high melting and boiling points. A metallic bonding theory must explain how so much bonding can occur with such few electrons (since metals are located on the left side of the periodic table and do not have many electrons in their valence shells). The theory must also account for all of a metal's unique chemical and physical properties.
Expanding the Range of Bonding Possible
Previously, we argued that bonding between atoms can classified as range of possible bonding between ionic bonds (fully charge transfer) and covalent bonds (fully shared electrons). When two atoms of slightly differing electronegativities come together to form a covalent bond, one atom attracts the electrons more than the other; this is called a polar covalent bond. However, simple “ionic” and “covalent” bonding are idealized concepts and most bonds exist on a two-dimensional continuum described by the van Arkel-Ketelaar Triangle (Figure $4$).
Bond triangles or van Arkel–Ketelaar triangles (named after Anton Eduard van Arkel and J. A. A. Ketelaar) are triangles used for showing different compounds in varying degrees of ionic, metallic and covalent bonding. In 1941 van Arkel recognized three extreme materials and associated bonding types. Using 36 main group elements, such as metals, metalloids and non-metals, he placed ionic, metallic and covalent bonds on the corners of an equilateral triangle, as well as suggested intermediate species. The bond triangle shows that chemical bonds are not just particular bonds of a specific type. Rather, bond types are interconnected and different compounds have varying degrees of different bonding character (for example, polar covalent bonds).
Using electronegativity - two compound average electronegativity on x-axis of Figure $4$.
$\sum \chi = \dfrac{\chi_A + \chi_B}{2} \label{sum}$
and electronegativity difference on y-axis,
$\Delta \chi = | \chi_A - \chi_B | \label{diff}$
we can rate the dominant bond between the compounds. On the right side of Figure $4$ (from ionic to covalent) should be compounds with varying difference in electronegativity. The compounds with equal electronegativity, such as $\ce{Cl2}$ (chlorine) are placed in the covalent corner, while the ionic corner has compounds with large electronegativity difference, such as $\ce{NaCl}$ (table salt). The bottom side (from metallic to covalent) contains compounds with varying degree of directionality in the bond. At one extreme is metallic bonds with delocalized bonding and at the other are covalent bonds in which the orbitals overlap in a particular direction. The left side (from ionic to metallic) is meant for delocalized bonds with varying electronegativity difference.
The Three Extremes in bonding
In general:
• Metallic bonds have low $\Delta \chi$ and low average $\sum\chi$.
• Ionic bonds have moderate-to-high $\Delta \chi$ and moderate values of average $\sum \chi$.
• Covalent bonds have moderate to high average $\sum \chi$ and can exist with moderately low $\Delta \chi$.
Example $2$
Use the tables of electronegativities (Table A2) and Figure $4$ to estimate the following values
• difference in electronegativity ($\Delta \chi$)
• average electronegativity in a bond ($\sum \chi$)
• percent ionic character
• likely bond type
for the selected compounds:
1. $\ce{AsH}$ (e.g., in arsine $AsH$)
2. $\ce{SrLi}$
3. $\ce{KF}$.
Solution
a: $\ce{AsH}$
• The electronegativity of $\ce{As}$ is 2.18
• The electronegativity of $\ce{H}$ is 2.22
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{2.18 + 2.22}{2} \[4pt] &= 2.2 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 2.18 - 2.22 \[4pt] &= 0.04 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (10% or less)
• The bonding is in the middle of a covalent bond and a metallic bond
b: $\ce{SrLi}$
• The electronegativity of $\ce{Sr}$ is 0.95
• The electronegativity of $\ce{Li}$ is 0.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.95 + 0.98}{2} \[4pt] &= 0.965 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 0.98 - 0.95 \[4pt] &= 0.025 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (~3% or less)
• The bonding is likely metallic.
c: $\ce{KF}$
• The electronegativity of $\ce{K}$ is 0.82
• The electronegativity of $\ce{F}$ is 3.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.82 + 3.98}{2} \[4pt] &= 2.4 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= | 0.82 - 3.98 | \[4pt] &= 3.16 \end{align*}
• From Figure $4$, the bond is fairly polar and has a high ionic character (~75%)
• The bonding is likely ionic.
Exercise $2$
Contrast the bonding of $\ce{NaCl}$ and silicon tetrafluoride.
Answer
$\ce{NaCl}$ is an ionic crystal structure, and an electrolyte when dissolved in water; $\Delta \chi =1.58$, average $\sum \chi =1.79$, while silicon tetrafluoride is covalent (molecular, non-polar gas; $\Delta \chi =2.08$, average $\sum \chi =2.94$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.01%3A_Types_of_Chemical_Bonds.txt |
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7.
What if two atoms of equal electronegativity bond together?
Consider a bond between two atoms, A and B. If the atoms are equally electronegative, both have the same tendency to attract the bonding pair of electrons, and so it will be found on average half way between the two atoms:
To get a bond like this, A and B would usually have to be the same atom. You will find this sort of bond in, for example, H2 or Cl2 molecules. Note: It's important to realize that this is an average picture. The electrons are actually in a molecular orbital, and are moving around all the time within that orbital. This sort of bond could be thought of as being a "pure" covalent bond - where the electrons are shared evenly between the two atoms.
What if B is slightly more electronegative than A?
B will attract the electron pair rather more than A does.
That means that the B end of the bond has more than its fair share of electron density and so becomes slightly negative. At the same time, the A end (rather short of electrons) becomes slightly positive. In the diagram, "$\delta$" (read as "delta") means "slightly" - so $\delta+$ means "slightly positive".
A polar bond is a covalent bond in which there is a separation of charge between one end and the other - in other words in which one end is slightly positive and the other slightly negative. Examples include most covalent bonds. The hydrogen-chlorine bond in HCl or the hydrogen-oxygen bonds in water are typical.
If B is a lot more electronegative than A, then the electron pair is dragged right over to B's end of the bond. To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed. The bond is then an ionic bond rather than a covalent bond.
A "spectrum" of bonds
The implication of all this is that there is no clear-cut division between covalent and ionic bonds. In a pure covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do.
Summary
• No electronegativity difference between two atoms leads to a pure non-polar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Patterns of electronegativity in the Periodic Table
The distance of the electrons from the nucleus remains relatively constant in a periodic table row, but not in a periodic table column. The force between two charges is given by Coulomb’s law.
$F=k\dfrac{Q_1Q_2}{r^2}$
In this expression, Q represents a charge, k represents a constant and r is the distance between the charges. When r = 2, then r2= 4. When r = 3, then r2 = 9. When r = 4, then r2 = 16. It is readily seen from these numbers that, as the distance between the charges increases, the force decreases very rapidly. This is called a quadratic change.
The result of this change is that electronegativity increases from bottom to top in a column in the periodic table even though there are more protons in the elements at the bottom of the column. Elements at the top of a column have greater electronegativities than elements at the bottom of a given column.
The overall trend for electronegativity in the periodic table is diagonal from the lower left corner to the upper right corner. Since the electronegativity of some of the important elements cannot be determined by these trends (they lie in the wrong diagonal), we have to memorize the following order of electronegativity for some of these common elements.
F > O > Cl > N > Br > I > S > C > H > metals
The most electronegative element is fluorine. If you remember that fact, everything becomes easy, because electronegativity must always increase towards fluorine in the Periodic Table.
Note
This simplification ignores the noble gases. Historically this is because they were believed not to form bonds - and if they do not form bonds, they cannot have an electronegativity value. Even now that we know that some of them do form bonds, data sources still do not quote electronegativity values for them.
The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right in a row in the periodic table. This effect only holds true for a row in the periodic table because the attraction between charges falls off rapidly with distance. The chart shows electronegativities from sodium to chlorine (ignoring argon since it does not does not form bonds).
As you go down a group, electronegativity decreases. (If it increases up to fluorine, it must decrease as you go down.) The chart shows the patterns of electronegativity in Groups 1 and 7.
Explaining the patterns in electronegativity
The attraction that a bonding pair of electrons feels for a particular nucleus depends on:
• the number of protons in the nucleus;
• the distance from the nucleus;
• the amount of screening by inner electrons.
Why does electronegativity increase across a period?
Consider sodium at the beginning of period 3 and chlorine at the end (ignoring the noble gas, argon). Think of sodium chloride as if it were covalently bonded.
Both sodium and chlorine have their bonding electrons in the 3-level. The electron pair is screened from both nuclei by the 1s, 2s and 2p electrons, but the chlorine nucleus has 6 more protons in it. It is no wonder the electron pair gets dragged so far towards the chlorine that ions are formed. Electronegativity increases across a period because the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly.
Why does electronegativity fall as you go down a group?
As you go down a group, electronegativity decreases because the bonding pair of electrons is increasingly distant from the attraction of the nucleus. Consider the hydrogen fluoride and hydrogen chloride molecules:
The bonding pair is shielded from the fluorine's nucleus only by the 1s2 electrons. In the chlorine case it is shielded by all the 1s22s22p6 electrons. In each case there is a net pull from the center of the fluorine or chlorine of +7. But fluorine has the bonding pair in the 2-level rather than the 3-level as it is in chlorine. If it is closer to the nucleus, the attraction is greater.
Diagonal relationships in the Periodic Table
At the beginning of periods 2 and 3 of the Periodic Table, there are several cases where an element at the top of one group has some similarities with an element in the next group. Three examples are shown in the diagram below. Notice that the similarities occur in elements which are diagonal to each other - not side-by-side.
For example, boron is a non-metal with some properties rather like silicon. Unlike the rest of Group 2, beryllium has some properties resembling aluminum. And lithium has some properties which differ from the other elements in Group 1, and in some ways resembles magnesium. There is said to be a diagonal relationship between these elements. There are several reasons for this, but each depends on the way atomic properties like electronegativity vary around the Periodic Table. So we will have a quick look at this with regard to electronegativity - which is probably the simplest to explain.
Explaining the diagonal relationship with regard to electronegativity
Electronegativity increases across the Periodic Table. So, for example, the electronegativities of beryllium and boron are:
Be 1.5
B 2.0
Electronegativity falls as you go down the Periodic Table. So, for example, the electronegativities of boron and aluminum are:
B 2.0
Al 1.5
So, comparing Be and Al, you find the values are (by chance) exactly the same. The increase from Group 2 to Group 3 is offset by the fall as you go down Group 3 from boron to aluminum. Something similar happens from lithium (1.0) to magnesium (1.2), and from boron (2.0) to silicon (1.8). In these cases, the electronegativities are not exactly the same, but are very close.
Similar electronegativities between the members of these diagonal pairs means that they are likely to form similar types of bonds, and that will affect their chemistry. You may well come across examples of this later on in your course. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.02_Electronegativity.txt |
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Introduction
When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge.
Dipole Moment
When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, an electric dipole is established. The size of a dipole is measured by its dipole moment ($\mu$). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals $3.34 \times 10^{-30}\; C\, m$). The dipole moment of a molecule can be calculated by Equation $\ref{1}$:
$\vec{\mu} = \sum_i q_i \, \vec{r}_i \label{1}$
where
• $\vec{\mu}$ is the dipole moment vector
• $q_i$ is the magnitude of the $i^{th}$ charge, and
• $\vec{r}_i$ is the vector representing the position of $i^{th}$ charge.
The dipole moment acts in the direction of the vector quantity. An example of a polar molecule is $\ce{H_2O}$. Because of the lone pair on oxygen, the structure of $\ce{H_2O}$ is bent (via VSEPR theory), which means that the vectors representing the dipole moment of each bond do not cancel each other out. Hence, water is polar.
The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. Table A2 shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together.
It is relatively easy to measure dipole moments: just place a substance between charged plates (Figure $2$); polar molecules increase the charge stored on the plates, and the dipole moment can be obtained (i.e., via the capacitance of the system). Nonpolar $\ce{CCl_4}$ is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly. In general, polar molecules will align themselves: (1) in an electric field, (2) with respect to one another, or (3) with respect to ions (Figure $2$).
Equation $\ref{1}$ can be simplified for a simple separated two-charge system like diatomic molecules or when considering a bond dipole within a molecule
$\mu_{diatomic} = Q \times r \label{1a}$
This bond dipole is interpreted as the dipole from a charge separation over a distance $r$ between the partial charges $Q^+$ and $Q^-$ (or the more commonly used terms $δ^+$ - $δ^-$); the orientation of the dipole is along the axis of the bond. Consider a simple system of a single electron and proton separated by a fixed distance. When the proton and electron are close together, the dipole moment (degree of polarity) decreases. However, as the proton and electron get farther apart, the dipole moment increases. In this case, the dipole moment is calculated as (via Equation $\ref{1a}$):
\begin{align*} \mu &= Qr \nonumber \[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align*}
The Debye characterizes the size of the dipole moment. When a proton and electron are 100 pm apart, the dipole moment is $4.80\; D$:
\begin{align*} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \[4pt] &= 4.80\; D \label{3} \end{align*}
$4.80\; D$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. If the charge separation is increased then the dipole moment increases (linearly):
• If the proton and electron are separated by 120 pm: $\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \label{4a}$
• If the proton and electron are separated by 150 pm: $\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \label{4b}$
• If the proton and electron are separated by 200 pm: $\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \label{4c}$
Example $1$: Water
The water molecule in Figure $1$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference in electronegativity is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of the O-H bond =1.5 D, so the net dipole moment is
$\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber$
Polarity and Structure of Molecules
The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar). If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide (Figure $\PageIndex{3a}$). This is a linear molecule and each C=O bond is, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygen atoms a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. the vector addition of the dipoles equals zero) and the overall molecule has a zero dipole moment ($\mu=0$).
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
For $AB_n$ molecules, where $A$ is the central atom and $B$ are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid.
Example $3$: $\ce{C_2Cl_4}$
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and $\ce{Cl_2C=CCl_2}$ does not have a net dipole moment.
Example $3$: $\ce{CH_3Cl}$
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
\begin{align*} \mu &= \dfrac{178}{100}(4.80\; D) \nonumber \[4pt] &= 8.54\; D \nonumber \end{align*}
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
Table $1$: Relationship between Bond length, Electronegativity and Dipole moments in simple Diatomics
Compound Bond Length (Å) Electronegativity Difference Dipole Moment (D)
HF 0.92 1.9 1.82
HCl 1.27 0.9 1.08
HBr 1.41 0.7 0.82
HI 1.61 0.4 0.44 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.03_Bond_Polarity_and_Dipole_Moments.txt |
The electron configuration of an atomic species (neutral or ionic) allows us to understand the shape and energy of its electrons. Many general rules are taken into consideration when assigning the "location" of the electron to its prospective energy state, however these assignments are arbitrary and it is always uncertain as to which electron is being described. Knowing the electron configuration of a species gives us a better understanding of its bonding ability, magnetism and other chemical properties.
Introduction
The electron configuration is the standard notation used to describe the electronic structure of an atom. Under the orbital approximation, we let each electron occupy an orbital, which can be solved by a single wavefunction. In doing so, we obtain three quantum numbers (n,l,ml), which are the same as the ones obtained from solving the Schrödinger's equation for Bohr's hydrogen atom. Hence, many of the rules that we use to describe the electron's address in the hydrogen atom can also be used in systems involving multiple electrons. When assigning electrons to orbitals, we must follow a set of three rules: the Aufbau Principle, the Pauli-Exclusion Principle, and Hund's Rule.
The wavefunction is the solution to the Schrödinger equation. By solving the Schrödinger equation for the hydrogen atom, we obtain three quantum numbers, namely the principal quantum number (n), the orbital angular momentum quantum number (l), and the magnetic quantum number (ml). There is a fourth quantum number, called the spin magnetic quantum number (ms), which is not obtained from solving the Schrödinger equation. Together, these four quantum numbers can be used to describe the location of an electron in Bohr's hydrogen atom. These numbers can be thought of as an electron's "address" in the atom.
Notation
To help describe the appropriate notation for electron configuration, it is best to do so through example. For this example, we will use the iodine atom. There are two ways in which electron configuration can be written:
I: 1s22s22p63s23p64s23d104p65s24d105p5
or
I: [Kr]5s24d105p5
In both of these types of notations, the order of the energy levels must be written by increased energy, showing the number of electrons in each subshell as an exponent. In the short notation, you place brackets around the preceding noble gas element followed by the valence shell electron configuration. The periodic table shows that kyrpton (Kr) is the previous noble gas listed before iodine. The noble gas configuration encompases the energy states lower than the valence shell electrons. Therefore, in this case [Kr]=1s22s22p63s23p64s23d104p6.
Quantum Numbers
Principal Quantum Number (n)
The principal quantum number n indicates the shell or energy level in which the electron is found. The value of n can be set between 1 to n, where n is the value of the outermost shell containing an electron. This quantum number can only be positive, non-zero, and integer values. That is, n=1,2,3,4,..
For example, an Iodine atom has its outmost electrons in the 5p orbital. Therefore, the principle quantum number for Iodine is 5.
Orbital Angular Momentum Quantum Number (l)
The orbital angular momentum quantum number, l, indicates the subshell of the electron. You can also tell the shape of the atomic orbital with this quantum number. An s subshell corresponds to l=0, a p subshell = 1, a d subshell = 2, a f subshell = 3, and so forth. This quantum number can only be positive and integer values, although it can take on a zero value. In general, for every value of n, there are n values of l. Furthermore, the value of l ranges from 0 to n-1. For example, if n=3, l=0,1,2.
So in regards to the example used above, the l values of Iodine for n = 5 are l = 0, 1, 2, 3, 4.
Magnetic Quantum Number (ml)
The magnetic quantum number, ml, represents the orbitals of a given subshell. For a given l, ml can range from -l to +l. A p subshell (l=1), for instance, can have three orbitals corresponding to ml = -1, 0, +1. In other words, it defines the px, py and pzorbitals of the p subshell. (However, the ml numbers don't necessarily correspond to a given orbital. The fact that there are three orbitals simply is indicative of the three orbitals of a p subshell.) In general, for a given l, there are 2l+1 possible values for ml; and in a n principal shell, there are n2 orbitals found in that energy level.
Continuing on from out example from above, the ml values of Iodine are ml = -4, -3, -2, -1, 0 1, 2, 3, 4. These arbitrarily correspond to the 5s, 5px, 5py, 5pz, 4dx2-y2, 4dz2, 4dxy, 4dxz, and 4dyz orbitals.
Spin Magnetic Quantum Number (ms)
The spin magnetic quantum number can only have a value of either +1/2 or -1/2. The value of 1/2 is the spin quantum number, s, which describes the electron's spin. Due to the spinning of the electron, it generates a magnetic field. In general, an electron with a ms=+1/2 is called an alpha electron, and one with a ms=-1/2 is called a beta electron. No two paired electrons can have the same spin value.
Out of these four quantum numbers, however, Bohr postulated that only the principal quantum number, n, determines the energy of the electron. Therefore, the 3s orbital (l=0) has the same energy as the 3p (l=1) and 3d (l=2) orbitals, regardless of a difference in l values. This postulate, however, holds true only for Bohr's hydrogen atom or other hydrogen-like atoms.
When dealing with multi-electron systems, we must consider the electron-electron interactions. Hence, the previously described postulate breaks down in that the energy of the electron is now determined by both the principal quantum number, n, and the orbital angular momentum quantum number, l. Although the Schrödinger equation for many-electron atoms is extremely difficult to solve mathematically, we can still describe their electronic structures via electron configurations.
General Rules of Electron Configuration
There are a set of general rules that are used to figure out the electron configuration of an atomic species: Aufbau Principle, Hund's Rule and the Pauli-Exclusion Principle. Before continuing, it's important to understand that each orbital can be occupied by two electrons of opposite spin (which will be further discussed later). The following table shows the possible number of electrons that can occupy each orbital in a given subshell.
subshell number of orbitals total number of possible electrons in each orbital
s 1 2
p 3 (px, py, pz) 6
d 5 (dx2-y2, dz2, dxy, dxz, dyz) 10
f 7 (fz3, fxz2, fxyz, fx(x2-3y2), fyz2, fz(x2-y2), fy(3x2-y2)
14
Using our example, iodine, again, we see on the periodic table that its atomic number is 53 (meaning it contains 53 electrons in its neutral state). Its complete electron configuration is 1s22s22p63s23p64s23d104p65s24d105p5. If you count up all of these electrons, you will see that it adds up to 53 electrons. Notice that each subshell can only contain the max amount of electrons as indicated in the table above.
Aufbau Principle
The word 'Aufbau' is German for 'building up'. The Aufbau Principle, also called the building-up principle, states that electron's occupy orbitals in order of increasing energy. The order of occupation is as follows:
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Another way to view this order of increasing energy is by using Madelung's Rule:
Figure 1. Madelung's Rule is a simple generalization which
dictates in what order electrons should be filled in the
orbitals,
however there are exceptions such as
copper and chromium.
This order of occupation roughly represents the increasing energy level of the orbitals. Hence, electrons occupy the orbitals in such a way that the energy is kept at a minimum. That is, the 7s, 5f, 6d, 7p subshells will not be filled with electrons unless the lower energy orbitals, 1s to 6p, are already fully occupied. Also, it is important to note that although the energy of the 3d orbital has been mathematically shown to be lower than that of the 4s orbital, electrons occupy the 4s orbital first before the 3d orbital. This observation can be ascribed to the fact that 3d electrons are more likely to be found closer to the nucleus; hence, they repel each other more strongly. Nonetheless, remembering the order of orbital energies, and hence assigning electrons to orbitals, can become rather easy when related to the periodic table.
To understand this principle, let's consider the bromine atom. Bromine (Z=35), which has 35 electrons, can be found in Period 4, Group VII of the periodic table. Since bromine has 7 valence electrons, the 4s orbital will be completely filled with 2 electrons, and the remaining five electrons will occupy the 4p orbital. Hence the full or expanded electronic configuration for bromine in accord with the Aufbau Principle is 1s22s22p63s23p64s23d104p5. If we add the exponents, we get a total of 35 electrons, confirming that our notation is correct.
Hund's Rule
Hund's Rule states that when electrons occupy degenerate orbitals (i.e. same n and l quantum numbers), they must first occupy the empty orbitals before double occupying them. Furthermore, the most stable configuration results when the spins are parallel (i.e. all alpha electrons or all beta electrons). Nitrogen, for example, has 3 electrons occupying the 2p orbital. According to Hund's Rule, they must first occupy each of the three degenerate p orbitals, namely the 2px orbital, 2py orbital, and the 2pz orbital, and with parallel spins (Figure 2). The configuration below is incorrect because the third electron occupies does not occupy the empty 2pz orbital. Instead, it occupies the half-filled 2px orbital. This, therefore, is a violation of Hund's Rule (Figure 2).
Figure 2. A visual representation of the Aufbau Principle and Hund's Rule. Note that the filling of electrons in each orbital
(px, py and pz) is arbitrary as long as the electrons are singly filled before having two electrons occupy the same orbital.
(a)This diagram represents the correct filling of electrons for the nitrogen atom. (b) This diagramrepresents the incorrect
filling of the electrons for the nitrogen atom.
Pauli-Exclusion Principle
Wolfgang Pauli postulated that each electron can be described with a unique set of four quantum numbers. Therefore, if two electrons occupy the same orbital, such as the 3s orbital, their spins must be paired. Although they have the same principal quantum number (n=3), the same orbital angular momentum quantum number (l=0), and the same magnetic quantum number (ml=0), they have different spin magnetic quantum numbers (ms=+1/2 and ms=-1/2).
Electronic Configurations of Cations and Anions
The way we designate electronic configurations for cations and anions is essentially similar to that for neutral atoms in their ground state. That is, we follow the three important rules: Aufbau Principle, Pauli-exclusion Principle, and Hund's Rule. The electronic configuration of cations is assigned by removing electrons first in the outermost p orbital, followed by the s orbital and finally the d orbitals (if any more electrons need to be removed). For instance, the ground state electronic configuration of calcium (Z=20) is 1s22s22p63s23p64s2. The calcium ion (Ca2+), however, has two electrons less. Hence, the electron configuration for Ca2+ is 1s22s22p63s23p6. Since we need to take away two electrons, we first remove electrons from the outermost shell (n=4). In this case, all the 4p subshells are empty; hence, we start by removing from the s orbital, which is the 4s orbital. The electron configuration for Ca2+ is the same as that for Argon, which has 18 electrons. Hence, we can say that both are isoelectronic.
The electronic configuration of anions is assigned by adding electrons according to Aufbau Principle. We add electrons to fill the outermost orbital that is occupied, and then add more electrons to the next higher orbital. The neutral atom chlorine (Z=17), for instance has 17 electrons. Therefore, its ground state electronic configuration can be written as 1s22s22p63s23p5. The chloride ion (Cl-), on the other hand, has an additional electron for a total of 18 electrons. Following Aufbau Principle, the electron occupies the partially filled 3p subshell first, making the 3p orbital completely filled. The electronic configuration for Cl- can, therefore, be designated as 1s22s22p63s23p6. Again, the electron configuration for the chloride ion is the same as that for Ca2+ and Argon. Hence, they are all isoelectronic to each other.
Problems
1. Which of the princples explained above tells us that electrons that are paired cannot have the same spin value?
2. Find the values of n, l, ml, and ms for the following:
a. Mg
b. Ga
c. Co
3. What is a possible combination for the quantum numbers of the 5d orbital? Give an example of an element which has the 5d orbital as it's most outer orbital.
4. Which of the following cannot exist (there may be more than one answer):
a. n = 4; l = 4; ml = -2; ms = +1/2
b. n = 3; l = 2; ml = 1; ms = 1
c. n = 4; l = 3; ml = 0; ms = +1/2
d. n = 1; l = 0; ml = 0; ms = +1/2
e. n = 0; l = 0; ml = 0; ms = +1/2
5. Write electron configurations for the following:
a. P
b. S2-
c. Zn3+
Answers
1. Pauli-exclusion Principle
2. a. n = 3; l = 0, 1, 2; ml = -2, -1, 0, 1, 2; ms can be either +1/2 or -1/2
b. n = 4; l = 0, 1, 2, 3; ml = -3, -2, -1, 0, 1, 2, 3; ms can be either +1/2 or -1/2
c. n = 3; l = 0, 1, 2; ml = -2, -1, 0, 1, 2, 3; ms can be either +1/2 or -1/2
3. n = 5; l = 3; ml = 0; ms = +1/2. Osmium (Os) is an example.
4. a. The value of l cannot be 4, because l ranges from (0 - n-1)
b. ms can only be +1/2 or -1/2
c. Okay
d. Okay
e. The value of n cannot be zero.
5. a. 1s22s22p63s23p3
b. 1s22s22p63s23p6
c. 1s22s22p63s23p64s23d7
Contributors and Attributions
• Lannah Lua, Andrew Iskandar (University of California Davis, Undergraduate) Mary Magsombol (University of California Davis) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.04_Ions%3A_Electron_Configurations_and_Sizes.txt |
Ionic bonding is the complete transfer of valence electron(s) between atoms and is a type of chemical bond that generates two oppositely charged ions. It is observed because metals with few electrons in its outer-most orbital. By losing those electrons, these metals can achieve noble-gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in its valence shell tend to readily accept electrons to achieve its noble gas configuration.
Introduction
In ionic bonding, electrons are transferred from one atom to another resulting in the formation of positive and negative ions. The electrostatic attractions between the positive and negative ions hold the compound together. The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of their electrostatic attraction. At the most ideal inter-atomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ions.
At a simple level, a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have. One may well have been left with the strong impression that when other atoms react, they try to organize things such that their outer levels are either completely full or completely empty.
Example: Bonding in NaCl
Sodium Chloride:
• Sodium (2,8,1) has 1 electron more than a stable noble gas structure (2,8). If it gave away that electron it would become more stable.
• Chlorine (2,8,7) has 1 electron short of a stable noble gas structure (2,8,8). If it could gain an electron from somewhere it too would become more stable.
The answer is obvious. If a sodium atom gives an electron to a chlorine atom, both become more stable.
The sodium has lost an electron, so it no longer has equal numbers of electrons and protons. Because it has one more proton than electron, it has a charge of 1+. If electrons are lost from an atom, positive ions are formed. Positive ions are sometimes called cations.
The chlorine has gained an electron, so it now has one more electron than proton. It therefore has a charge of 1-. If electrons are gained by an atom, negative ions are formed. A negative ion is sometimes called an anion.
The nature of the bond
The sodium ions and chloride ions are held together by the strong electrostatic attractions between the positive and negative charges. You need one sodium atom to provide the extra electron for one chlorine atom, so they combine together 1:1. The formula is therefore NaCl.
Example 1: Bonding in MgO
Magnesium Oxide:
Again, noble gas structures are formed, and the magnesium oxide is held together by very strong attractions between the ions. The ionic bonding is stronger than in sodium chloride because this time you have 2+ ions attracting 2- ions. The greater the charge, the greater the attraction. The formula of magnesium oxide is MgO.
2
Calcium Chloride:
This time you need two chlorines to use up the two outer electrons in the calcium. The formula of calcium chloride is therefore CaCl2.
O
Potassium Oxide:
Again, noble gas structures are formed. It takes two potassiums to supply the electrons the oxygen needs. The formula of potassium oxide is K2O.
Some Stable Ions do not have Noble Gas Configurations
You may have come across some of the following ions, which are all perfectly stable, but not one of them has a noble gas structure.
Fe3+ [Ar]3d5
Cu2+ [Ar]3d9
Zn2+ [Ar]3d10
Ag+ [Kr]4d10
Pb2+ [Xe]4f145d106s2
What needs modifying is the view that there is something magic about noble gas structures. There are far more ions which do not have noble gas structures than there are which do.
• Noble gases (apart from helium) have an outer electronic structure ns2np6. Apart from some elements at the beginning of a transition series (scandium forming Sc3+ with an argon structure, for example), all transition metal elements and any metals following a transition series (like tin and lead in Group 4, for example) will have structures like those above.
• That means that the only elements to form positive ions with noble gas structures (apart from odd ones like scandium) are those in groups 1 and 2 of the Periodic Table and aluminum in group 3 (boron in group 3 does not form ions).
• Negative ions are tidier! Those elements in Groups 5, 6 and 7 which form simple negative ions all have noble gas structures.
If elements are not aiming for noble gas structures when they form ions, what decides how many electrons are transferred? The answer lies in the energetics of the process by which the compound is made. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.05_Formation_of_Binary_Ionic_Compounds.txt |
Electrostatic potential maps, also known as electrostatic potential energy maps, or molecular electrical potential surfaces, illustrate the charge distributions of molecules three dimensionally. These maps allow us to visualize variably charged regions of a molecule. Knowledge of the charge distributions can be used to determine how molecules interact with one another.
Introduction
Electrostatic potential maps are very useful three dimensional diagrams of molecules. They enable us to visualize the charge distributions of molecules and charge related properties of molecules. They also allow us to visualize the size and shape of molecules. In organic chemistry, electrostatic potential maps are invaluable in predicting the behavior of complex molecules.
The first step involved in creating an electrostatic potential map is collecting a very specific type of data: electrostatic potential energy. An advanced computer program calculates the electrostatic potential energy at a set distance from the nuclei of the molecule. Electrostatic potential energy is fundamentally a measure of the strength of the nearby charges, nuclei and electrons, at a particular position.
To accurately analyze the charge distribution of a molecule, a very large quantity of electrostatic potential energy values must be calculated. The best way to convey this data is to visually represent it, as in an electrostatic potential map. A computer program then imposes the calculated data onto an electron density model of the molecule derived from the Schrödinger equation. To make the electrostatic potential energy data easy to interpret, a color spectrum, with red as the lowest electrostatic potential energy value and blue as the highest, is employed to convey the varying intensities of the electrostatic potential energy values.
Analogous System
Electrostatic potential maps involve a number of basic concepts. The actual process of mapping the electrostatic potentials of a molecule, however, involves factors that complicate these fundamental concepts. An analogous system will be employed to introduce these basic concepts.
Imagine that there is a special type of mine. This mine is simply an explosive with some charged components on top of it. The circles with positive and negative charges in them are the charged components. If the electric field of the electric components are significantly disturbed, the mine triggers and explodes. The disarming device is positively charged. To disarm the mine, the disarming device must take the path of least electric resistance and touch the first charged mine component on this path. Deviating from this minimal energy path will cause a significant disturbance and the mine will explode. The specific charged components within the mine are known.
Q. How do you disarm the following mine?
The mine with positive charge and negative charge.
A. Touch the bottom most portion of negatively charged component, red, with the disarming device.
Introduction to Coulomb's Law and Electrostatic Energy
Coulomb's Law Formula
$F=k \dfrac{q_aq_b}{r^2}$
with
• Molecular electrostatic potential maps also illustrate information about the charge distribution of a molecule. Electrostatic potential maps convey information about the charge distribution of a molecule because of the properties of the nucleus and nature of electrostatic potential energy. For simplicity, consider moving a positively charged test charge along the spherical isosurface of an atom. The positively charged nucleus emits a radially constant electric field. A region of higher than average electrostatic potential energy indicates the presence of a stronger positive charge or a weaker negative charger. Given the consistency of the nucleuses positive charge, the higher potential energy value indicates the absence of negative charges, which would mean that there are fewer electrons in this region. The converse is also true. Thus a high electrostatic potential indicates the relative absence of electrons and a low electrostatic potential indicates an abundance of electrons. This property of electrostatic potentials can be extrapolated to molecules as well.
Here is a simplified visual representation of the relationship between charge distribution and electrostatic potential. Keep in mind the equation used to find the electrostatic potential.
$\text{Total Electrostatic Potential Energy= \sum_i \text{Electrostatic Potential Energy}$
where
$\text{Potential Energy}=K \dfrac{q_1q_2}{r}$
and $K= \text{Coulomb's Constant}$.
Contributors and Attributions
• Thomas Bottyan | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.06_Partial_Ionic_Character_of_Covalent_Bonds.txt |
Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability. Nonmetals will readily form covalent bonds with other nonmetals in order to obtain stability, and can form anywhere between one to three covalent bonds with other nonmetals depending on how many valence electrons they posses. Although it is said that atoms share electrons when they form covalent bonds, they do not usually share the electrons equally.
Introduction
Only when two atoms of the same element form a covalent bond are the shared electrons actually shared equally between the atoms. When atoms of different elements share electrons through covalent bonding, the electron will be drawn more toward the atom with the higher electronegativity resulting in a polar covalent bond. When compared to ionic compounds, covalent compounds usually have a lower melting and boiling point, and have less of a tendency to dissolve in water. Covalent compounds can be in a gas, liquid, or solid state and do not conduct electricity or heat well. The types of covalent bonds can be distinguished by looking at the Lewis dot structure of the molecule. For each molecule, there are different names for pairs of electrons, depending if it is shared or not. A pair of electrons that is shared between two atoms is called a bond pair. A pair of electrons that is not shared between two atoms is called a lone pair.
Octet Rule
The Octet Rule requires all atoms in a molecule to have 8 valence electrons--either by sharing, losing or gaining electrons--to become stable. For Covalent bonds, atoms tend to share their electrons with each other to satisfy the Octet Rule. It requires 8 electrons because that is the amount of electrons needed to fill a s- and p- orbital (electron configuration); also known as a noble gas configuration. Each atom wants to become as stable as the noble gases that have their outer valence shell filled because noble gases have a charge of 0. Although it is important to remember the "magic number", 8, note that there are many Octet rule exceptions.
Example: As you can see from the picture below, Phosphorus has only 5 electrons in its outer shell (bolded in red). Argon has a total of 8 electrons (bolded in red), which satisfies the Octet Rule. Phosphorus needs to gain 3 electrons to fulfill the Octet Rule. It wants to be like Argon who has a full outer valence shell.
More examples can be found here.
Single Bonds
A single bond is when two electrons--one pair of electrons--are shared between two atoms. It is depicted by a single line between the two atoms. Although this form of bond is weaker and has a smaller density than a double bond and a triple bond, it is the most stable because it has a lower level of reactivity meaning less vulnerability in losing electrons to atoms that want to steal electrons.
Example 1: HCl
Below is a Lewis dot structure of Hydrogen Chloride demonstrating a single bond. As we can see from the picture below, Hydrogen Chloride has 1 Hydrogen atom and 1 Chlorine atom. Hydrogen has only 1 valence electron whereas Chlorine has 7 valence electrons. To satisfy the Octet Rule, each atom gives out 1 electron to share with each other; thus making a single bond.
Double Bonds
A Double bond is when two atoms share two pairs of electrons with each other. It is depicted by two horizontal lines between two atoms in a molecule. This type of bond is much stronger than a single bond, but less stable; this is due to its greater amount of reactivity compared to a single bond.
2
Below is a Lewis dot structure of Carbon dioxide demonstrating a double bond. As you can see from the picture below, Carbon dioxide has a total of 1 Carbon atom and 2 Oxygen atoms. Each Oxygen atom has 6 valence electrons whereas the Carbon atom only has 4 valence electrons. To satisfy the Octet Rule, Carbon needs 4 more valence electrons. Since each Oxygen atom has 3 lone pairs of electrons, they can each share 1 pair of electrons with Carbon; as a result, filling Carbon's outer valence shell (Satisfying the Octet Rule).
Triple Bond
A Triple bond is when three pairs of electrons are shared between two atoms in a molecule. It is the least stable out of the three general types of covalent bonds. It is very vulnerable to electron thieves!
Example 3: Acetylene
Below is a Lewis dot structure of Acetylene demonstrating a triple bond. As you can see from the picture below, Acetylene has a total of 2 Carbon atoms and 2 Hydrogen atoms. Each Hydrogen atom has 1 valence electron whereas each Carbon atom has 4 valence electrons. Each Carbon needs 4 more electrons and each Hydrogen needs 1 more electron. Hydrogen shares its only electron with Carbon to get a full valence shell. Now Carbon has 5 electrons. Because each Carbon atom has 5 electrons--1 single bond and 3 unpaired electrons--the two Carbons can share their unpaired electrons, forming a triple bond. Now all the atoms are happy with their full outer valence shell.
Polar Covalent Bond
A Polar Covalent Bond is created when the shared electrons between atoms are not equally shared. This occurs when one atom has a higher electronegativity than the atom it is sharing with. The atom with the higher electronegativity will have a stronger pull for electrons (Similiar to a Tug-O-War game, whoever is stronger usually wins). As a result, the shared electrons will be closer to the atom with the higher electronegativity, making it unequally shared. A polar covalent bond will result in the molecule having a slightly positive side (the side containing the atom with a lower electronegativity) and a slightly negative side (containing the atom with the higher electronegativity) because the shared electrons will be displaced toward the atom with the higher electronegativity. As a result of polar covalent bonds, the covalent compound that forms will have an electrostatic potential. This potential will make the resulting molecule slightly polar, allowing it to form weak bonds with other polar molecules. One example of molecules forming weak bonds with each other as a result of an unbalanced electrostatic potential is hydrogen bonding, where a hydrogen atom will interact with an electronegative hydrogen, fluorine, or oxygen atom from another molecule or chemical group.
Example: Water, Sulfide, Ozone, etc.
As you can see from the picture above, Oxygen is the big buff creature with the tattoo of "O" on its arm. The little bunny represents a Hydrogen atom. The blue and red bow tied in the middle of the rope, pulled by the two creatures represents--the shared pair of electrons--a single bond. Because the Hydrogen atom is weaker, the shared pair of electrons will be pulled closer to the Oxygen atom.
Nonpolar Covalent Bond
A Nonpolar Covalent Bond is created when atoms share their electrons equally. This usually occurs when two atoms have similar or the same electron affinity. The closer the values of their electron affinity, the stronger the attraction. This occurs in gas molecules; also known as diatomic elements. Nonpolar covalent bonds have a similar concept as polar covalent bonds; the atom with the higher electronegativity will draw away the electron from the weaker one. Since this statement is true--if we apply this to our diatomic molecules--all the atoms will have the same electronegativity since they are the same kind of element; thus, the electronegativities will cancel each other out and will have a charge of 0 (i.e., a nonpolar covalent bond).
Examples of gas molecules that have a nonpolar covalent bond: Hydrogen gas atom, Nitrogen gas atoms, etc.
As you can see from the picture above, Hydrogen gas has a total of 2 Hydrogen atoms. Each Hydrogen atom has 1 valence electron. Since Hydrogen can only fit a max of 2 valence electrons in its orbital, each Hydrogen atom only needs 1 electron. Each atom has 1 valence electron, so they can just share, giving each atom two electrons each.
Problems
1. Determine the type(s) of bond(s) in
• Benzene (C6H6)
• NO3 (Nitrate)
• F2(Fluorine gas)
Solution:
2. Write the electron configuration and determine how many electrons are needed to achieve the nearest noble-gas configuration for the following:
• Arsenic (As)
• Silicon (Si)
• Tellurium (Te)
Solution:
3. Determine which molecules are polar and which molecules are nonpolar for the following:
• Oxygen gas (O2)
• Hydrochloric acid (HCl)
• Carbon dioxide (CO2)
Solution:
4. Which of the following statements are true? (There can be more than one true statement.)
1. A covalent bond is the same as a ionic bond.
2. The Octet rule only applys to molecules with covalent bonds.
3. A molecule is polar if the shared electrons are equally shared.
4. A molecule is nonpolar if the shared electrons are are equally shared.
5. Methane gas (CH4) has a nonpolar covalent bond because it is a gas.
Solution: Only d) is true.
5. Match each atom or molecule with its corresponding letter(s):
• Nitrogen gas
• Argon
• Carbon monoxide
• Hydrogen gas
a) Nonpolar covalent bond
b) Polar covalent bond
c) Follows the Octet Rule
d) Noble gas
e) Two lone pairs
f) Single bond
Solution:
• Nitrogen gas: a), c), e)
• Argon: c), d)
• Carbon monoxide: b), c), e)
• Hydrogen gas: c), f)
Contributors and Attributions
• Camy Fung (UCD), Nima Mirzaee (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.07_The_Covalent_Chemical_Bond%3A_A_Model.txt |
Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When atoms combine to make a compound, energy is always given off, and the compound has a lower overall energy.
When a chemical reaction occurs, molecular bonds are broken and other bonds are formed to make different molecules. For example, the bonds of two water molecules are broken to form hydrogen and oxygen.
$2H_2O \rightarrow 2H_2 + O_2$
Energy is always required to break a bond, which is known as bond energy. While the concept may seem simple, bond energy serves a very important purpose in describing the structure and characteristics of a molecule. It can be used to determine which Lewis Dot Structure is most suitable when there are multiple Lewis Dot Structures.
Energy is always required to break a bond. Energy is released when a bond is made.
Although each molecule has its own characteristic bond energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. We can calculate a more general bond energy by finding the average of the bond energies of a specific bond in different molecules to get the average bond energy.
Table 1: Average Bond Energies (kJ/mol)
Single Bonds Multiple Bonds
H—H
432
N—H
391
I—I
149
C = C
614
H—F
565
N—N
160
I—Cl
208
C ≡ C
839
H—Cl
427
N—F
272
I—Br
175
O = O
495
H—Br
363
N—Cl
200
C = O*
745
H—I
295
N—Br
243
S—H
347
C ≡ O
1072
N—O
201
S—F
327
N = O
607
C—H
413
O—H
467
S—Cl
253
N = N
418
C—C
347
O—O
146
S—Br
218
N ≡ N
941
C—N
305
O—F
190
S—S
266
C ≡ N
891
C—O
358
O—Cl
203
C = N
615
C—F
485
O—I
234
Si—Si
340
C—Cl
339
Si—H
393
C—Br
276
F—F
154
Si—C
360
C—I
240
F—Cl
253
Si—O
452
C—S
259
F—Br
237
Cl—Cl
239
Cl—Br
218
Br—Br
193
*C == O(CO2) = 799
When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond. This correlates with bond order and bond length. When the Bond order is higher, bond length is shorter, and the shorter the bond length means a greater the Bond Energy because of increased electric attraction. In general, the shorter the bond length, the greater the bond energy.
The average bond energies in Table T3 are the averages of bond dissociation energies. For example the average bond energy of O-H in H2O is 464 kJ/mol. This is due to the fact that the H-OH bond requires 498.7 kJ/mol to dissociate, while the O-H bond needs 428 kJ/mol.
$\dfrac{498.7\; kJ/mol + 428\; kJ/mol}{2}=464\; kJ/mol$
When more bond energies of the bond in different molecules that are taken into consideration, the average will be more accurate. However,
• Average bonds values are not as accurate as a molecule specific bond-dissociation energies.
• Double bonds are higher energy bonds in comparison to a single bond (but not necessarily 2-fold higher).
• Triple bonds are even higher energy bonds than double and single bonds (but not necessarily 3-fold higher).
Bond Breakage and Formation
When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change.
In some reactions, the energy of the products is lower than the energy of the reactants. Thus, in the course of the reaction, the substances lose energy to the surrounding environment. Such reactions are exothermic and can be represented by an energy-level diagram in Figure 1 (left). In most cases, the energy is given off as heat (although a few reactions give off energy as light). In chemical reactions where the products have a higher energy than the reactants, the reactants must absorb energy from their environment to react. These reactions are endothermic and can be represented by an energy-level diagrams like Figure 1 (right).
Technically Temperature is Neither a Reactant nor Product
It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" ($\Delta{H} > 0$), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" ($\Delta{H} < 0$), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below.
Exothermic and endothermic reactions can be thought of as having energy as either a "product" of the reaction or a "reactant." Exothermic reactions releases energy, so energy is a product. Endothermic reactions require energy, so energy is a reactant.
Example $1$: Exothermic vs. Endothermic
Is each chemical reaction exothermic or endothermic?
1. $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(ℓ)} + \text{135 kcal}$
2. $N_{2(g)} + O_{2(g)} + \text{45 kcal} \rightarrow 2NO_{(g)}$
Solution
No calculates are required to address this question. Just look at where the "heat" is in the chemical reaction.
1. Because energy is released; this reaction is exothermic.
2. Because energy is absorbed; this reaction is endothermic.
Exercise $1$
If the bond energy for H-Cl is 431 kJ/mol. What is the overall bond energy of 2 moles of HCl?
Answer
Simply multiply the average bond energy of H-Cl by 2. This leaves you with 862 kJ (Table T3).
Example $2$: Generation of Hydrogen Iodide
What is the enthalpy change for this reaction and is it endothermic or exothermic?
$H_2(g)+I_2(g) \rightarrow 2HI(g)$
Solution
First look at the equation and identify which bonds exist on in the reactants.
• one H-H bond and
• one I-I bond
Now do the same for the products
• two H-I bonds
Then identify the bond energies of these bonds from the table above:
• H-H bonds: 436 kJ/mol
• I-I bonds: 151 kJ/mol
The sum of enthalpies on the reaction side is:
436 kJ/mole + 151 kJ/mole = 587 kJ/mol.
This is how much energy is needed to break the bonds on the reactant side. Then we look at the bond formation which is on the product side:
• 2 mol H-I bonds: 297 kJ/mol
The sum of enthalpies on the product side is:
2 x 297 kJ/mol= 594 kJ/mol
This is how much energy is released when the bonds on the product side are formed. The net change of the reaction is therefore
587-594= -7 kJ/mol.
Since this is negative, the reaction is exothermic.
Example $2$: Decomposition of Water
Using the bond energies given in the chart above, find the enthalpy change for the thermal decomposition of water:
$2H_2O (g) \rightarrow 2H_2 + O_2 (g)$
Is the reaction written above exothermic or endothermic? Explain.
Solution
The enthalpy change deals with breaking two mole of O-H bonds and the formation of 1 mole of O-O bonds and two moles of H-H bonds (Table T3).
• The sum of the energies required to break the bonds on the reactants side is 4 x 460 kJ/mol = 1840 kJ/mol.
• The sum of the energies released to form the bonds on the products side is
• 2 moles of H-H bonds = 2 x 436.4 kJ/mol = 872.8 kJ/mol
• 1 moles of O=O bond = 1 x 498.7 kJ/mil = 498.7 kJ/mol
which is an output (released) energy = 872.8 kJ/mol + 498.7 kJ/mol = 1371.5 kJ/mol.
Total energy difference is 1840 kJ/mol – 1371.5 kJ/mol = 469 kJ/mol, which indicates that the reaction is endothermic and that 469 kJ of heat is needed to be supplied to carry out this reaction.
This reaction is endothermic since it requires energy in order to create bonds.
Summary
Energy is released to generate bonds, which is why the enthalpy change for breaking bonds is positive. Energy is required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate octet electronic configurations). The enthalpy change is negative because the system is releasing energy when forming bond.
Contributors and Attributions
• Kim Song (UCD), Donald Le (UCD)
13.09 The Localized Electron Bonding Model
Valence bond (VB) theory assumes that all bonds are localized bonds formed between two atoms by the donation of an electron from each atom. This is actually an invalid assumption because many atoms bond using delocalized electrons. In molecular oxygen VB theory predict that there are no unpaired electrons. VB theory does a good job of qualitatively describing the shapes of covalent compounds. While Molecular Orbital (MO) theory is good for understanding bonding in general. It is more difficult to learn, but predicts the actual properties of molecules better than VB theory. MO theory actually predicts electron transitions because of the differences in the energy levels of orbitals in the molecule. MO theory has been more correct in numerous instances and for this reason it is preferred.
Valence Bond theory describes covalent bond formation as well as the electronic structure of molecules. The theory assumes that electrons occupy atomic orbitals of individual atoms within a molecule, and that the electrons of one atom are attracted to the nucleus of another atom. This attraction increases as the atoms approach one another until the atoms reach a minimum distance where the electron density begins to cause repulsion between the two atoms. This electron density at the minimum distance between the two atoms is where the lowest potential energy is acquired, and it can be considered to be what holds the two atoms together in a chemical bond. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.08_Covalent_Bond_Energies_and_Chemical_Reactions.txt |
Lewis Structures are visual representations of the bonds between atoms and illustrate the lone pairs of electrons in molecules. They can also be called Lewis dot diagrams and are used as a simple way to show the configuration of atoms within a molecule. Between 1916 and 1919, Gilbert Newton Lewis, Walther Kossel, and Irving Langmuir came up with a theory to explain chemical bonding. This theory would be later called Lewis Theory and it is based on the following principles:
1. Valence electrons, or the electrons in the outermost electron shell, have an essential role in chemical bonding.
2. Ionic bonds are formed between atoms when electrons are transferred from one atom to another. Ionic bond is a bond between nonmetals and metals .
3. Covalent bonds are formed between atoms when pairs of electrons are shared between atoms. A covalent bond is between two nonmetals.
4. Electrons are transferred/shared so that each atom may reach a more stable electron configuration i.e. the noble gas configuration which contains 8 valence electrons. This is called octet rule.
Lewis Symbols and Lewis Structures
A Lewis Symbol for an element is composed of a chemical symbol surrounded by dots that are used to represent valence electrons. An example of a Lewis symbol is shown below with the element Carbon, which has the electron configuration of 1s22s22p2:
This Lewis symbol shows that carbon has four valence electrons in its outer orbital and these four electrons play a major role in bonding of carbon molecules.
Lewis symbols differ slightly for ions. When forming a Lewis symbol for an ion, the chemical symbol is surrounded by dots that are used to represent valence electrons, and the whole structure is placed in square brackets with superscript representing the charge of the ion. An example of a Lewis symbol for the cation and anion of Carbon is shown below:
Cation of Carbon
Anion of Carbon
Constructing Lewis Structures
To construct Lewis Structures one can generally abide by the following steps:
1. Find how many valence electrons (N) are in the molecule that needs to be shown on the Lewis Structure by using the periodic table. Find the charge, add an electron for every negative charge and remove an electron for every positive charge.
2. Draw out the single bonds and initial framework, called the skeleton, of the molecule.
3. Complete the octets around the non-central atoms i.e. the terminal atoms by using the lone-pairs of electrons.
4. Compare the number of electrons currently depicted to the number needed (N) in the central atom and add electrons to it if less the number is less than N.
5. If there are extra lone-pair electrons and the octet rule is not filled for the central atom, use the extra electrons to form double or triple bonds around the central atom.
6. Check the formal charge of each atom (Formal Charge explained below).
When constructing the structures keep in mind the following:
• The dots surrounding the chemical symbol are the valence electrons, and each dash represents one covalent bond (consisting of two valence electrons)
• Hydrogen is always terminal in the structure
• The atom with the lowest ionization energy is typically the central atom in the structure
• The octet rule means there are 8 valence electrons around the atoms, but for hydrogen the maximum is 2 electrons
Lewis Structures can differ based on whether the electrons are shared through ionic or covalent bonds.
Example 1: Ionic Bonding in NaCl
An example of ionic bonding can be seen below in the instance of the reaction of Sodium and Chlorine:
$2Na_{(s)} + 2Cl_{2 (g)} \rightarrow 2NaCl_{s}$
Sodium has one valence electron and Chlorine has seven valence electrons; the two elements together form the noble gas configuration. The Chlorine atom takes the valence electron from the Sodium atom leaving the Chlorine atom with one extra electron and thus negatively charged and the Sodium atom without an electron and thus positively charged. The two atoms then become ions and because of their opposite charges the ions are held together in an ionic bond.
An example of covalent bonding can be seen below with the reaction of Hydrogen and Fluorine:
+=
Hydrogen has one valence electron and Fluorine has seven valence electrons; together the elements form the noble gas configuration. The Hydrogen atom shares its electron with Fluorine atom so that the Hydrogen atom has 2 electrons and the Fluorine atom has 8 electrons. Therefore both atoms have their outermost shells completely filled.
Example 2: The Chlorate Ion
Try the Chlorate ion: (ClO3-)
SOLUTION
First, lets find the how many valence electrons chlorate has:
ClO3- : 7 e-(from Cl) + 3(6) e-(from 3 O atoms) + 1 (from the total charge of -1) = 26
There are 26 valence electrons.
Next lets draw the basic framework of the molecule:
The molecule uses covalent bonds to hold together the atoms to the central Chlorine. The remaining electrons become non-bonding electrons. Since 6 electrons were used for the bonds, the 20 others become those un-bonding electrons to complete the octet:
The oxygen atom's shells fill up with 18 electrons, and the other 2 complete Chlorine's octet.
Example 3: Formaldehyde
Constructing the Lewis Structure of the formaldehyde (H2CO) molecule.
SOLUTION
First find the valence electrons:
H2CO: 2(1) e- (from the H atoms) + 4 e- (from the C atom) + 6 e- (from the O atom)
There are 12 valence electrons. Next draw out the framework of the molecule:
To satisfy the octet of Carbon, one of the pairs of electrons on Oxygen must be moved to create a double bond with Carbon. Therefore our Lewis Structure would look as it does below:
The Hydrogen atoms are each filled up with their two electrons and both the Carbon and the Oxygen atoms' octets are filled.
Formal Charge
The charge on each atom in a molecule is called the formal charge. The formal charge can be calculated if the electrons in the bonds of the molecule are equally shared between atoms. This is not the same thing as the net charge of the ion.
In calculating formal charge, the following steps can be extremely helpful:
1. Determine the number of valence electrons that should be present for each atom in the structure.
2. Count the electrons around each atom in the structure (each lone pair = 2 electrons, each single bond =1 electron, each double bond = 2 electrons, each triple bond = 3 electrons).
3. Subtract the number of valence electrons that should be present (from step 1) from the electrons counted in step 2 for each atom. This is the formal charge for each atom.
4. Check that the formal charges add up to equal the overall charge of the molecule.
Formal charge = (number of valence electrons) - (number of non-bonding electrons + 1/2 number of bonding electrons)
In Lewis structures, the most favorable structure has the smallest formal charge for the atoms, and negative formal charges tend to come from more electronegative atoms.
An example of determining formal charge can be seen below with the nitrate ion, NO3-:
• The double bonded O atom has 6 electrons: 4 non-bonding and 2 bonding (one electron for each bond). Since O should have 6 electrons, the formal charge is 0.
• The two singly bonded O atoms each have 7 electrons: 6 non-bonding and 1 bonding electron. Since O should have 6 electrons, and there is one extra electron, those O atoms each have formal charges of -1.
• The N atom has 4 electrons: 4 bonding and 0 non-bonding electrons. Since N should have 5 electrons and there are only 4 electrons for this N, the N atom has a formal charge of +1.
• The charges add up to the overall charge of the ion. 0 + (-1) + (-1) + 1 = -1. Thus, these charges are correct, as the overall charge of nitrate is -1.
Resonance
There are times when more than one acceptable Lewis structure can be drawn for a molecule and no single structure can represent the molecule entirely. When this occurs the molecule/ion is said to have resonance. The combination of the various plausible Lewis structures is called a resonance hybrid.
Some rules for drawing resonance structures are as follows:
1. The same number of electrons must be present for all resonance structures.
2. The octet rule must be obeyed.
3. Nuclei (or the chemical symbol in the structure's representation) cannot be rearranged; only the valence electrons differ for resonance structures.
Example 4: Nitrate Ion
Consider the nitrate ion, NO3-
• Each structure differs based on the movement of two electrons.
• From structure 1 to structure 2 one of the lone pairs on the blue O moves to form a double bond with N. One of the electrons shared in the double bond between the red O and N then moves to be a lone pair on the red O because the N cannot have 10 electrons surrounding it.
• This is the same process that occurs from structure 2 to structure 3 except the changes occur in the green O and blue O.
Problems
1. Draw the Lewis structure for water.
2. Find the formal charge for each of the atoms in water.
3. Draw the Lewis structure for ammonia.
4. Find the formal charge for each of the atoms in ammonia.
Answers:
1.
2. All atoms in water have a formal charge of 0.
3.
4. The formal charge for each H is 0 and for N is 0. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.10%3A_Lewis_Structures.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.11%3A_Resonance.txt |
Three cases can be constructed that do not follow the Octet Rule, and as such, they are known as the exceptions to the Octet Rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions. The Octet Rule is violated in these three scenarios:
1. When there are an odd number of valence electrons
2. When there are too few valence electrons
3. When there are too many valence electrons
Reminder: Always use the Octet Rule when drawing Lewis Dot Structures, these exceptions will only occur when necessary.
Exception 1: Species with Odd Numbers of Electrons
The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be the nitrogen (II) oxide molecule ($NO$). Nitrogen atom has 5 valence electrons while the oxygen atom has 6 electrons. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in. The formula to find a formal charge is:
Formal Charge= [# of valence e- the atom would have on its own] - [# of lone pair electrons on that atom] - [# of bonds that atom participates in]
No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons (Figure 1). If you need more information about formal charges, see Lewis Structures. If we were to consider the nitrogen monoxide cation ($NO^+$ with ten valence electrons, then the following Lewis structure would be constructed:
Nitrogen normally has five valence electrons. In Figure 1, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure 1, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide:
Free Radicals
There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. To emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with ${\cdot}OH$, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted ${\cdot}Cl$. Interestingly, molecules with an odd number of Valence electrons will always be paramagnetic.
Exception 2: Incomplete Octets
The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, $BH_3$ (Borane).
If one was to make a Lewis structure for $BH_3$ following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure 3):
The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.
Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure 4).
If you look Figure 4, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure 5):
Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.
This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure 5, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table ($\chi=4.0$). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.
However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure 6:
None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure 4), the one with the double bond (Figure 5), and the one with the ionic bond (Figure 6). The most contributing structure is probably the incomplete octet structure (due to Figure 5 being basically impossible and Figure 6 not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.
As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .
Example: $BF_3$
Draw the Lewis structure for boron trifluoride (BF3).
Solution
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:
3. Add octets to outer atoms:
4. Add extra electrons (24-24=0) to central atom:
5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:
6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
• BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:
Exception 3: Expanded Valence Shells
More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, $PCl_5$ is a legitimate compound (whereas $NCl_5$) is not:
Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:
Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.
One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.
The sulfate ion, SO4-2. is an ion that prefers an expanded octet structure. A strict adherence to the octet rule forms the following Lewis structure:
Example 3: The $ICl_4^-$ Ion
Draw the Lewis structure for $ICl_4^-$ ion.
Solution
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:
3. Add octet of electrons to outer atoms:
4. Add extra electrons (36-32=4) to central atom:
5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)
Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies.
Practice Problems
1. Draw the Lewis structure for the molecule I3-.
2. Draw the molecule ClF3.
3. The central atom for an expanded octet must have an atomic number larger than what?
4. Draw the Lewis structure for the molecule NO2.
5. Which Lewis structure is more likely?
or
Answers
1.
2.
3. 10 (Sodium and higher)
4.
5.
13.13: Molecular Structure: The VSEPR Model
Valence Shell Electron Pair Repulsion (VSPER) theory is used to predict the geometric shape of the molecules based on the electron repulsive force. There are some limitation to VSEPR.
Introduction
The shapes of the molecules is determined mainly by the electrons surrounding the central atom. Therefore, VSEPR theory gives simple directions on how to predict the shape of the molecules. The VSEPR model combines the original ideas of Sidwick and Powell and further development of Nyholm and Gillespie.
How VSEPR works
In a molecule EXn, the valence shell electron pair around the central atom E and the E-X single bonds are very important due to the repulsion in which determine the shape of the molecule. The repulsions decrease in order of: lone pair-lone pair, lone pair-bonding pair, bonding pair-bonding pair. At the same time, the repulsion would decrease in order of: triple bond-single bond, double bond-single bond, and single bond-single bond if the central atom E has multiple bonds. The difference between the electronegativities of E and X also determine the repulsive force between the bonding pairs. If electron-electron repulsive force is less, then more electron density is drawn away from the central atom E.
Shape determination:
VSEPR model works better for simple halides of the p-block elements but can also be used with other substituents. It does not take steric factors, size of the substituents into account. Therefore, the shape of the molecules are arranged so that the energy is minimized. For example:
• BeCl2 has minimum energy when it is a linear molecule.
• BCl3 takes the shape of trigonal planar.
Lone pair electrons are also taken into account. When lone pair electrons are present, the "parent structure" are used as a guideline for determining the shape..
Problems
1. What is VSEPR used in chemistry?
It is used to predict the molecular shape of molecules
2. How to predict a molecule structure using VSEPR theory?
First step is to count the total number of valence electrons. After the total number of electrons is determined, this number is divided by two to give the total number of electron pairs. With the electron pairs of the molecule, the shape of the molecule is determined based on the table shown above.
3. What is the shape of PF5 ?
It is trigonal bipyramidal because it has total of 20 electron pairs. Each Fluorine atom give 1 electron to the Phosphorus central atom which creates total of 5 pairs. Also, each Fluorine atom has 3 electron pairs. With the presence of 5 Fluorine atom, there are 15 more electron pairs so there are 20 electron pairs total.
14.02: The Molecular Orbital Model
Hybridization is the idea that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory. In order to explore this idea further, we will utilize three types of hydrocarbon compounds to illustrate sp3, sp2, and sp hybridization.
Contributors and Attributions
• Jennifer Lau (all images are made by me) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.12%3A_Exceptions_to_the_Octet_Rule.txt |
Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Entropy is also the subject of the Second and Third laws of thermodynamics, which describe the changes in entropy of the universe with respect to the system and surroundings, and the entropy of substances, respectively.
• ‘Disorder’ in Thermodynamic Entropy
Boltzmann’s sense of “increased randomness” as a criterion of the final equilibrium state for a system compared to initial conditions was not wrong.; it was his surprisingly simplistic conclusion: if the final state is random, the initial system must have been the opposite, i.e., ordered. “Disorder” was the consequence, to Boltzmann, of an initial “order” not — as is obvious today — of what can only be called a “prior, lesser but still humanly-unimaginable, large number of accessible microstate
• Microstates
Dictionaries define “macro” as large and “micro” as very small but a macrostate and a microstate in thermodynamics aren't just definitions of big and little sizes of chemical systems. Instead, they are two very different ways of looking at a system. A microstate is one of the huge number of different accessible arrangements of the molecules' motional energy* for a particular macrostate.
• Simple Entropy Changes - Examples
Several Examples are given to demonstrate how the statistical definition of entropy and the 2nd law can be applied. Phase Change, gas expansions, dilution, colligative properties and osmosis.
• Statistical Entropy
Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities.
• Statistical Entropy - Mass, Energy, and Freedom
The energy or the mass of a part of the universe may increase or decrease, but only if there is a corresponding decrease or increase somewhere else in the universe. The freedom in that part of the universe may increase with no change in the freedom of the rest of the universe. There might be decreases in freedom in the rest of the universe, but the sum of the increase and decrease must result in a net increase.
• The Molecular Basis for Understanding Simple Entropy Change
10.05: Entropy and the Second Law of Thermodynamics
The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative.
Introduction
Why is it that when you leave an ice cube at room temperature, it begins to melt? Why do we get older and never younger? And, why is it whenever rooms are cleaned, they become messy again in the future? Certain things happen in one direction and not the other, this is called the "arrow of time" and it encompasses every area of science. The thermodynamic arrow of time (entropy) is the measurement of disorder within a system. Denoted as $\Delta S$, the change of entropy suggests that time itself is asymmetric with respect to order of an isolated system, meaning: a system will become more disordered, as time increases.
Major players in developing the Second Law
• Nicolas Léonard Sadi Carnot was a French physicist, who is considered to be the "father of thermodynamics," for he is responsible for the origins of the Second Law of Thermodynamics, as well as various other concepts. The current form of the second law uses entropy rather than caloric, which is what Sadi Carnot used to describe the law. Caloric relates to heat and Sadi Carnot came to realize that some caloric is always lost in the motion cycle. Thus, the thermodynamic reversibility concept was proven wrong, proving that irreversibility is the result of every system involving work.
• Rudolf Clausius was a German physicist, and he developed the Clausius statement, which says "Heat generally cannot flow spontaneously from a material at a lower temperature to a material at a higher temperature."
• William Thompson, also known as Lord Kelvin, formulated the Kelvin statement, which states "It is impossible to convert heat completely in a cyclic process." This means that there is no way for one to convert all the energy of a system into work, without losing energy.
• Constantin Carathéodory, a Greek mathematician, created his own statement of the second low arguing that "In the neighborhood of any initial state, there are states which cannot be approached arbitrarily close through adiabatic changes of state."
Probabilities
If a given state can be accomplished in more ways, then it is more probable than the state that can only be accomplished in a fewer/one way.
Assume a box filled with jigsaw pieces were jumbled in its box, the probability that a jigsaw piece will land randomly, away from where it fits perfectly, is very high. Almost every jigsaw piece will land somewhere away from its ideal position. The probability of a jigsaw piece landing correctly in its position, is very low, as it can only happened one way. Thus, the misplaced jigsaw pieces have a much higher multiplicity than the correctly placed jigsaw piece, and we can correctly assume the misplaced jigsaw pieces represent a higher entropy.
Derivation and Explanation
To understand why entropy increases and decreases, it is important to recognize that two changes in entropy have to considered at all times. The entropy change of the surroundings and the entropy change of the system itself. Given the entropy change of the universe is equivalent to the sums of the changes in entropy of the system and surroundings:
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}=\dfrac{q_{sys}}{T}+\dfrac{q_{surr}}{T} \label{1}$
In an isothermal reversible expansion, the heat q absorbed by the system from the surroundings is
$q_{rev}=nRT\ln\frac{V_{2}}{V_{1}}\label{2}$
Since the heat absorbed by the system is the amount lost by the surroundings, $q_{sys}=-q_{surr}$.Therefore, for a truly reversible process, the entropy change is
$\Delta S_{univ}=\dfrac{nRT\ln\frac{V_{2}}{V_{1}}}{T}+\dfrac{-nRT\ln\frac{V_{2}}{V_{1}}}{T}=0 \label{3}$
If the process is irreversible however, the entropy change is
$\Delta S_{univ}=\frac{nRT\ln \frac{V_{2}}{V_{1}}}{T}>0 \label{4}$
If we put the two equations for $\Delta S_{univ}$together for both types of processes, we are left with the second law of thermodynamics,
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\geq0 \label{5}$
where $\Delta S_{univ}$ equals zero for a truly reversible process and is greater than zero for an irreversible process. In reality, however, truly reversible processes never happen (or will take an infinitely long time to happen), so it is safe to say all thermodynamic processes we encounter everyday are irreversible in the direction they occur.
The second law of thermodynamics can also be stated that "all spontaneous processes produce an increase in the entropy of the universe".
Gibbs Free Energy
Given another equation:
$\Delta S_{total}=\Delta S_{univ}=\Delta S_{surr}+\Delta S{sys} \label{6}$
The formula for the entropy change in the surroundings is $\Delta S_{surr}=\Delta H_{sys}/T$. If this equation is replaced in the previous formula, and the equation is then multiplied by T and by -1 it results in the following formula.
$-T \, \Delta S_{univ}=\Delta H_{sys}-T\, \Delta S_{sys} \label{7}$
If the left side of the equation is replaced by $G$, which is know as Gibbs energy or free energy, the equation becomes
$\Delta G_{}=\Delta H-T\Delta S \label{8}$
Now it is much simpler to conclude whether a system is spontaneous, non-spontaneous, or at equilibrium.
• $\Delta H$ refers to the heat change for a reaction. A positive $\Delta H$ means that heat is taken from the environment (endothermic). A negative $\Delta H$ means that heat is emitted or given the environment (exothermic).
• $\Delta G$ is a measure for the change of a system's free energy in which a reaction takes place at constant pressure ($P$) and temperature ($T$).
According to the equation, when the entropy decreases and enthalpy increases the free energy change, $\Delta G_{}$, is positive and not spontaneous, and it does not matter what the temperature of the system is. Temperature comes into play when the entropy and enthalpy both increase or both decrease. The reaction is not spontaneous when both entropy and enthalpy are positive and at low temperatures, and the reaction is spontaneous when both entropy and enthalpy are positive and at high temperatures. The reactions are spontaneous when the entropy and enthalpy are negative at low temperatures, and the reaction is not spontaneous when the entropy and enthalpy are negative at high temperatures. Because all spontaneous reactions increase entropy, one can determine if the entropy changes according to the spontaneous nature of the reaction (Equation $\ref{8}). Table \(1$: Matrix of Conditions Dictating Spontaneity
Case $\Delta H$ $\Delta S$ $\Delta G$ Answer
high temperature - + - Spontaneous
low temperature - + - Spontaneous
high temperature - - + Nonspontaneous
low temperature - - - Spontaneous
high temperature + + - Spontaneous
low temperature + + + Nonspontaneous
high temperature + - + Nonspontaneous
low temperature + - + Nonspontaneous
Example $1$
Lets start with an easy reaction:
$2 H_{2(g)}+O_{2(g)} \rightarrow 2 H_2O_{(g)}$
The enthalpy, $\Delta H_{}$, for this reaction is -241.82 kJ, and the entropy, $\Delta S_{}$, of this reaction is -233.7 J/K. If the temperature is at 25º C, then there is enough information to calculate the standard free energy change, $\Delta G_{}$.
The first step is to convert the temperature to Kelvin, so add 273.15 to 25 and the temperature is at 298.15 K. Next plug $\Delta H_{}$, $\Delta S_{}$, and the temperature into the $\Delta G=\Delta H-T \Delta S_{}$.
$\Delta G$= -241.8 kJ + (298.15 K)(-233.7 J/K)
= -241.8 kJ + -69.68 kJ (Don't forget to convert Joules to Kilojoules)
= -311.5 kJ
Example $2$
Here is a little more complex reaction:
$2 ZnO_{(s)}+2 C_{(g)} \rightarrow 2 Zn_{(s)}+2 CO_{(g)}$
If this reaction occurs at room temperature (25º C) and the enthalpy, $\Delta H_{}$, and standard free energy, $\Delta G_{}$, is given at -957.8 kJ and -935.3 kJ, respectively. One must work backwards somewhat using the same equation from Example 1 for the free energy is given.
-935.3 kJ = -957.8 kJ + (298.15 K) ($\Delta S_{}$)
22.47 kJ = (298.15 K) ($\Delta S_{}$) (Add -957.8 kJ to both sides)
0.07538 kJ/K = $\Delta S_{}$ (Divide by 298.15 K to both sides)
Multiply the entropy by 1000 to convert the answer to Joules, and the new answer is 75.38 J/K.
Example $3$
For the following dissociation reaction
$O_{2(g)} \rightarrow 2 O_{(g)}$
under what temperature conditions will it occurs spontaneously?
Solution
By simply viewing the reaction one can determine that the reaction increases in the number of moles, so the entropy increases. Now all one has to do is to figure out the enthalpy of the reaction. The enthalpy is positive, because covalent bonds are broken. When covalent bonds are broken energy is absorbed, which means that the enthalpy of the reaction is positive. Another way to determine if enthalpy is positive is to to use the formation data and subtract the enthalpy of the reactants from the enthalpy of the products to calculate the total enthalpy. So, if the temperature is low it is probable that $\Delta H_{}$ is more than $T*\Delta S_{}$, which means the reaction is not spontaneous. If the temperature is large then $T*\Delta S_{}$ will be larger than the enthalpy, which means the reaction is spontaneous.
Example $4$
The following reaction
$CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$
occurs spontaneously under what temperature conditions? The enthalpy of the reaction is -40 kJ.
Solution
One may have to calculate the enthalpy of the reaction, but in this case it is given. If the enthalpy is negative then the reaction is exothermic. Now one must find if the entropy is greater than zero to answer the question. Using the entropy of formation data and the enthalpy of formation data, one can determine that the entropy of the reaction is -42.1 J/K and the enthalpy is -41.2 kJ. Because both enthalpy and entropy are negative, the spontaneous nature varies with the temperature of the reaction. The temperature would also determine the spontaneous nature of a reaction if both enthalpy and entropy were positive. When the reaction occurs at a low temperature the free energy change is also negative, which means the reaction is spontaneous. However, if the reaction occurs at high temperature the reaction becomes nonspontaneous, for the free energy change becomes positive when the high temperature is multiplied with a negative entropy as the enthalpy is not as large as the product.
Example $5$
Under what temperature conditions does the following reaction occurs spontaneously ?
$H_{2(g)} + I_{(g)} \rightleftharpoons 2 HI_{(g)}$
Solution
Only after calculating the enthalpy and entropy of the reaction is it possible for one can answer the question. The enthalpy of the reaction is calculated to be -53.84 kJ, and the entropy of the reaction is 101.7 J/K. Unlike the previous two examples, the temperature has no affect on the spontaneous nature of the reaction. If the reaction occurs at a high temperature, the free energy change is still negative, and $\Delta G_{}$ is still negative if the temperature is low. Looking at the formula for spontaneous change one can easily come to the same conclusion, for there is no possible way for the free energy change to be positive. Hence, the reaction is spontaneous at all temperatures.
Application of the Second Law
The second law occurs all around us all of the time, existing as the biggest, most powerful, general idea in all of science.
Explanation of Earth's Age
When scientists were trying to determine the age of the Earth during 1800s they failed to even come close to the value accepted today. They also were incapable of understanding how the earth transformed. Lord Kelvin, who was mentioned earlier, first hypothesized that the earth's surface was extremely hot, similar to the surface of the sun. He believed that the earth was cooling at a slow pace. Using this information, Kelvin used thermodynamics to come to the conclusion that the earth was at least twenty million years, for it would take about that long for the earth to cool to its current state. Twenty million years was not even close to the actual age of the Earth, but this is because scientists during Kelvin's time were not aware of radioactivity. Even though Kelvin was incorrect about the age of the planet, his use of the second law allowed him to predict a more accurate value than the other scientists at the time.
Evolution and the Second Law
Some critics claim that evolution violates the Second Law of Thermodynamics, because organization and complexity increases in evolution. However, this law is referring to isolated systems only, and the earth is not an isolated system or closed system. This is evident for constant energy increases on earth due to the heat coming from the sun. So, order may be becoming more organized, the universe as a whole becomes more disorganized for the sun releases energy and becomes disordered. This connects to how the second law and cosmology are related, which is explained well in the video below.
Problems
1. Predict the entropy changes of the converse of SO2 to SO3: 2 SO2 (g) + O2 (g) --> 2 SO3 (g)
2. True/False: $\Delta G$ > 0, the process is spontaneous
3. State the conditions when $\Delta G$ is nonspontaneous.
4. True/False: A nonspontaneous process cannot occur with external intervention.
Answers
1. Entropy decreases
2. False
3. Case 3, Case 6, Case 7, Case 8 (Table above)
4. True
References
1. Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, California: University Science Books, 2005.
2. How the Earth Was Made. Dir. Peter Chin. A&E Home Video, 2008. DVD.
3. Petrucci, Ralph H., William S. Harwood, F. G. Herring, and Jeffry D. Madura. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007. 791-796.
Conditional content (Pro member)
Contributors and Attributions
• Konstantin Malley, Ravneet Singh (UCD), Tianyu Duan (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/10%3A_Entropy_Gibbs_Energy_and_Spontaneity/10.01%3A_Spontaneous_Processes.txt |
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
10.08: Entropy Changes in Chemical Reactions
Entropy is another important aspect of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored. We sometimes speak of the energy in a system as being "partitioned" or divided into various "states". How this energy is divided up is the concern of entropy.
By way of analogy, picture a set of mailboxes. You may have a wall of them in your dormitory or your apartment building. The mailboxes are of several different sizes: maybe there are a few rows of small ones, a couple of rows of medium sized ones, and a row of big mailboxes on the bottom.
Instead of putting mail in these boxes, we're going to use them to hold little packages of energy. Later on, you might take the energy packages out of your own mailbox and use them to take a trip to the mall or the gym. But how does the mail get to your mailbox in the first place?
The energy packages don't arrive in your molecular dormitory with addresses on them. The packages come in different sizes, because they contain different amounts of energy, but other than that there is no identifying information on them.
Some of the packages don't fit into some of the mailboxes, because some of the packages are too big and some of the mailboxes are smaller than the others. The energy packages need to go into mailboxes that they will fit into.
Still, there are an awful lot of mailboxes that most of the energy packages could still fit into. There needs to be some system of deciding where to put all of these packages. It turns out that, in the molecular world, there is such a system, and it follows a pretty simple rule. When a whole pile of energy packages arrive, the postmaster does her best to put one package into every mailbox. Then, when every mailbox has one, she starts putting a second one into each box, and so on.
It didn't have to be that way. It could have been the case that all the energy was simply put into the first couple of mailboxes and the rest were left empty. In other words, the rule could have been that all the energy must be sorted into the same place, instead of being spread around. But that's not how it is.
• Energy is always partitioned into the maximum number of states possible.
Entropy is the sorting of energy into different modes or states. When energy is partitioned or sorted into additional states, entropy is said to increase. When energy is bundled into a smaller number of states, entropy is said to decrease. Nature's bias is towards an increase in entropy.
This is a fundamental law of the universe; there is no reason that can be used to explain why nature prefers high entropy to low entropy. Instead, increasing entropy is itself the basic reason for a wide range of things that happen in the universe.
Entropy is popularly described in terms of "disorder". That can be a useful idea, although it doesn't really describe what is happening energetically.
A better picture of entropy can be built by looking at how a goup of molecules might sort some energy that is added to them. In other words, what are some examples of "states" in which energy can be sorted?
If you get more energy -- maybe by eating breakfast -- one of the immediate benefits is being able to increase your physical activity. You have more energy to move around, to run, to jump. A similar situation is true with molecules.
Molecules have a variety of ways in which they can move, if they are given some energy. They can zip around; this kind of motion is usually called translation. They can tumble and roll; this kind of motion is referred to as rotation. Also, they can wiggle, letting their bonds get longer and shorter by moving individual atoms around a little bit. This type of motion is called vibration.
When molecules absorb extra energy, they may be able to sort the energy into rotational, vibrational and translational states. This only works with energy packages of a certain size; other packages would be sorted into other kinds of states. However, these are just a few examples of what we mean by states.
Okay, so energy is stored in states, and it is sorted into the maximum possible number of states. But how does entropy change in a reaction? We know that enthalpy may change by breaking or forming certain bonds, but how does the energy get sorted again?
The changes in internal entropy during a reaction are often very small. In other words, the energy remaining at the end of the reaction gets sorted more or less the way it was before the reaction. However, there are some very common exceptions.
The most common case in which internal entropy changes a lot is when the number of molecules involved changes between the start of the reaction and the end of the reaction. Maybe two molecules react together to form one, new molecule. Maybe one molecule splits apart to make two, new molecules.
If one molecule splits apart in the reaction, entropy generally increases. Two molecules can rotate, vibrate and translate (or tumble, wiggle and zip around) independently of each other. That means the number of states available for partitioning energy increases when one molecule splits into two.
• Entropy generally increases when a reaction produces more molecules than it started with.
• Entropy generally decreases when a reaction produces fewer molecules than it started with.
Apart from a factor like a change in the number of molecules involved, internal entropy changes are often fairly subtle. They are not as easy to predict as enthalpy changes.
Nevertheless, there may sometimes be a trade-off between enthalpy and entropy. If a reaction splits a molecule into two, it seems likely that an increase in enthalpy will be involved, so that the bond that held the two pieces together can be broken. That's not favourable. However, when that happens, we've just seen that there will be an increase in entropy, because energy can then be sorted into additional modes in the two, independent molecules.
So we have two different factors to balance. There is a tool we often use to decide which factor wins out. It's called free energy, and we will look at it next. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/10%3A_Entropy_Gibbs_Energy_and_Spontaneity/10.06%3A_The_Effect_of_Temperature_on_Spontaneity.txt |
Entropy and enthalpy are two of the basic factors of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored.
• A reaction is favored if enthalpy decreases: There is a bias in nature toward decreasing enthalpy in a system. Reactions can happen when enthalpy is transferred to the surroundings.
• A reaction is favored if entropy increases: There is also a bias in nature toward increasing entropy in a system. Reactions can happen when entropy increases.
Consider the cartoon reaction below. Red squares are being converted to green circles, provided the reaction proceeds from left to right as shown.
Whether or not the reaction proceeds to the right depends on the balance between enthalpy and entropy. There are several combinations possible.
In one case, maybe entropy increases when the red squares turn into green circles, and the enthalpy decreases. If we think of the balance between these two factors, we come to a simple conclusion. Both factors tilt the balance of the reaction to the right. In this case, the red squares will be converted into green circles.
Alternatively, maybe entropy decreases when the red squares turn into green circles, and enthalpy increases. If we think of the balance between these two factors, we come to another simple conclusion. Both factors tilt the balance of the reaction to the left. In this case, the red squares will remain just as they are.
Having two factors may lead to complications. For example, what if enthalpy decreases, but so does entropy? Does the reaction happen, or doesn't it? In that case, we may need quantitation to make a decision. How much does the enthalpy decrease? How much does the entropy decrease? If the effect of the enthalpy decrease is greater than that of the entropy decrease, the reaction may still go forward.
The combined effects of enthalpy and entropy are often combined in what is called "free energy." Free energy is just a way to keep track of the sum of the two effects. Mathematically, the symbol for the internal enthalpy change is "ΔH" and the symbol for the internal entropy change is "ΔS." Free energy is symbolized by "ΔG," and the relationship is given by the following expression:
\[ \Delta G = \Delta H - T \Delta S \]
\(T\) in this expression stands for the temperature (in Kelvin, rather than Celsius or Fahrenheit). The temperature acts as a scaling factor in the expression, putting the entropy and enthalpy on equivalent footing so that their effects can be compared directly.
How do we use free energy? It works the same way we were using enthalpy earlier (that's why the free energy has the same sign as the enthalpy in the mathematical expression, whereas the entropy has an opposite sign). If free energy decreases, the reaction can proceed. If the free energy increases, the reaction can't proceed.
• A reaction is favored if the free energy of the system decreases.
• A reaction is not favored if the free energy of the system increases.
Because free energy takes into consideration both the enthalpy and entropy changes, we don't have to consider anything else to decide if the reaction occurs. Both factors have already been taken into account.
Remember the terms "endothermic" and "exothermic" from our discussion of enthalpy. Exothermic reactions were favored (in which enthalpy decreases). Endothermic ones were not. In free energy terms, we say that exergonic reactions are favored(in which free energy decreases). Endergonic ones (in which free energy increases) are not.
Problem TD4.1.
Imagine a reaction in which the effects of enthalpy and entropy are opposite and almost equally balanced, so that there is no preference for whether the reaction proceeds or not. Looking at the expression for free energy, how do you think the situation will change under the following conditions:
1. the temperature is very cold (0.09 K)
2. the temperature is very warm (500 K)
Problem TD4.2.
Which of the following reaction profiles describe reactions that will proceed? Which ones describe reactions that will not proceed?
How Entropy Rules Thermodynamics
Sometimes it is said that entropy governs the universe. As it happens, enthalpy and entropy changes in a reaction are partly related to each other. The reason for this relationship is that if energy is added to or released from the system, it has to be partitioned into new states. Thus, an enthalpy change can also have an effect on entropy.
Specifically, the internal enthalpy change that we discussed earlier has an effect on the entropy of the surroundings. So far, we have just considered internal entropy changes.
• In an exothermic reaction, the external entropy (entropy of the surroundings) increases.
• In an endothermic reaction, the external entropy (entropy of the surroundings) decreases.
Free energy takes into account both the entropy of the system and the entropy changes that arise because of heat exchange with the surroundings. Together, the system and the surroundings are called "the universe". That's because the system is just everything involved in the reaction, and the surroundings are everything that isn't involved in the reaction.
Enthalpy changes in the system lead to additional partitioning of energy. We might visualize that with the mailbox analogy we used for entropy earlier. In this case, each molecule has its own set of mailboxes, into which it sorts incoming energy.
Looked at in this way, thermodynamics boils down to one major consideration, and that is the combined entropy of both the system and its surroundings (together known as the universe).
• For a reaction to proceed, the entropy of the universe must increase. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/10%3A_Entropy_Gibbs_Energy_and_Spontaneity/10.09%3A_Gibbs_Energy_and_Reactions.txt |
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
10.11: Gibbs Energy and Equilibrium
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
10.12: Free Energy and Work
The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/10%3A_Entropy_Gibbs_Energy_and_Spontaneity/10.10%3A_The_Dependence_of_Free_Energy_on_Pressure.txt |
Skills to Develop
• To understand the basics of voltaic cells
• To connect voltage from a voltaic cell to underlying REDOX chemistry
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant 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, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$Zn_{(s)} + Br_{2(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Br^−_{(aq)} \label{20.3.1}$
The half-reactions are as follows:
reduction half-reaction:
$Br_{2(aq)} + 2e^− \rightarrow 2Br^−_{(aq)} \label{20.3.2}$
oxidation half-reaction:
$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{20.3.3}$
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 $1$). 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.
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.
There are two types of electrochemical cells: galvanic cells and electrolytic cells. 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. A galvanic (voltaic) cell 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 cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $1$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). 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.
Voltaic (Galvanic) 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:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{20.3.4}$
We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. 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 $2$: The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Single Compartment. When a zinc rod 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.
This same reaction can be carried out using the galvanic cell illustrated in Figure $\PageIndex{3a}$. 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 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 (Figure $\PageIndex{3b}$). 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 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) 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 (Figure $\PageIndex{3a}$). 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 $1$.
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 $1$
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:
$3Sn_{(s)} + 2NO^-_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$
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:
1. 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.
2. 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+.
1. 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.
2. 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 $1$
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:
$2MnO^−_{4(aq)} + 5Sn^{2+}_{(aq)} + 16H^+_{(aq)} \rightarrow 2Mn^{2+}_{(aq)} + 5Sn^{4+}_{(aq)} + 8H_2O_{(l)}$
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.
Cell Potential
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. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. 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 $1$ 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 potential (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.
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 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:
$Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{20.4.1}$
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 (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:
$E°_{cell} = E°_{cathode} − E°_{anode} \label{20.4.2}$
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 $\ref{20.4.2}$, 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.
The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials:
$E°_{cell} = E°_{cathode} − E°_{anode}$
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) 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 $2$). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
$2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{20.4.3}$
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure $3$ 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:
$Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{20.4.4}$
The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)}\;\;\; E°_{cathode}=0 V \label{20.4.5}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)}+2e^−\;\;\; E°_{anode}=−0.76\; V \label{20.4.6}$
• overall: $Zn_{(s)}+2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)}+H_{2(g)} \label{20.4.7}$
$E°_{cell}=E°_{cathode}−E°_{anode}=0.76\; V$
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 potential 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
$E°_{cell}: 0 \,V − (−0.76\, V) = 0.76\, V$
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.
E° values do NOT depend on the stoichiometric coefficients for a half-reaction, because it is an intensive property.
Summary
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. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.1%3A_Galvanic_Cells.txt |
Learning Objectives
• Determine standard cell potentials for oxidation-reduction reactions
• Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices
The cell potential results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.4.1 and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is
$\ce{2H+}(aq,\: 1\:M)+\ce{2e-}⇌\ce{H2}(g,\:1\: \ce{atm}) \hspace{20px} E°=\mathrm{0\: V} \nonumber$
E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures.
A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure $2$). In cell notation, the reaction is
$\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\:1\:M)║\ce{Cu^2+}(aq,\:1\:M)│\ce{Cu}(s) \nonumber$
Electrons flow from the anode to the cathode. The reactions, which are reversible, are
\begin{align*} &\textrm{Anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq) + \ce{2e-}\ &\textrm{Cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\ &\overline{\textrm{Overall: }\ce{Cu^2+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{Cu}(s)} \end{align*} \nonumber
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber$
$\mathrm{+0.34\: V}=E^\circ_{\ce{Cu^2+/Cu}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Cu^2+/Cu}}−0=E^\circ_{\ce{Cu^2+/Cu}} \nonumber$
Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure $2$, where
$\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\: 1\:M)║\ce{Ag+}(aq,\: 1\:M)│\ce{Ag}(s) \nonumber$
Electrons flow from left to right, and the reactions are
\begin{align*} &\textrm{anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\ &\overline{\textrm{overall: }\ce{2Ag+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2Ag}(s)} \end{align*} \nonumber
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber$
$\mathrm{+0.80\: V}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Ag+/Ag}}−0=E^\circ_{\ce{Ag+/Ag}} \nonumber$
It is important to note that the potential is not doubled for the cathode reaction.
The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, $E^\circ_\ce{cell}$, for any cell. For example, for the following cell:
$\ce{Cu}(s)│\ce{Cu^2+}(aq,\:1\:M)║\ce{Ag+}(aq,\:1\:M)│\ce{Ag}(s) \nonumber$
\begin{align*} &\textrm{anode (oxidation): }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\ &\overline{\textrm{overall: }\ce{Cu}(s)+\ce{2Ag+}(aq)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)} \end{align*} \nonumber
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Cu^2+/Cu}}=\mathrm{0.80\: V−0.34\: V=0.46\: V} \nonumber$
Again, note that when calculating $E^\circ_\ce{cell}$, standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table $1$. A more complete list is provided in Tables P1 or P2.
Table $1$: Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
$\ce{F2}(g)+\ce{2e-}⟶\ce{2F-}(aq)$ +2.866
$\ce{PbO2}(s)+\ce{SO4^2-}(aq)+\ce{4H+}(aq)+\ce{2e-}⟶\ce{PbSO4}(s)+\ce{2H2O}(l)$ +1.69
$\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)$ +1.507
$\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s)$ +1.498
$\ce{Cl2}(g)+\ce{2e-}⟶\ce{2Cl-}(aq)$ +1.35827
$\ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-}⟶\ce{2H2O}(l)$ +1.229
$\ce{Pt^2+}(aq)+\ce{2e-}⟶\ce{Pt}(s)$ +1.20
$\ce{Br2}(aq)+\ce{2e-}⟶\ce{2Br-}(aq)$ +1.0873
$\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s)$ +0.7996
$\ce{Hg2^2+}(aq)+\ce{2e-}⟶\ce{2Hg}(l)$ +0.7973
$\ce{Fe^3+}(aq)+\ce{e-}⟶\ce{Fe^2+}(aq)$ +0.771
$\ce{MnO4-}(aq)+\ce{2H2O}(l)+\ce{3e-}⟶\ce{MnO2}(s)+\ce{4OH-}(aq)$ +0.558
$\ce{I2}(s)+\ce{2e-}⟶\ce{2I-}(aq)$ +0.5355
$\ce{NiO2}(s)+\ce{2H2O}(l)+\ce{2e-}⟶\ce{Ni(OH)2}(s)+\ce{2OH-}(aq)$ +0.49
$\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)$ +0.34
$\ce{Hg2Cl2}(s)+\ce{2e-}⟶\ce{2Hg}(l)+\ce{2Cl-}(aq)$ +0.26808
$\ce{AgCl}(s)+\ce{e-}⟶\ce{Ag}(s)+\ce{Cl-}(aq)$ +0.22233
$\ce{Sn^4+}(aq)+\ce{2e-}⟶\ce{Sn^2+}(aq)$ +0.151
$\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g)$ 0.00
$\ce{Pb^2+}(aq)+\ce{2e-}⟶\ce{Pb}(s)$ −0.1262
$\ce{Sn^2+}(aq)+\ce{2e-}⟶\ce{Sn}(s)$ −0.1375
$\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s)$ −0.257
$\ce{Co^2+}(aq)+\ce{2e-}⟶\ce{Co}(s)$ −0.28
$\ce{PbSO4}(s)+\ce{2e-}⟶\ce{Pb}(s)+\ce{SO4^2-}(aq)$ −0.3505
$\ce{Cd^2+}(aq)+\ce{2e-}⟶\ce{Cd}(s)$ −0.4030
$\ce{Fe^2+}(aq)+\ce{2e-}⟶\ce{Fe}(s)$ −0.447
$\ce{Cr^3+}(aq)+\ce{3e-}⟶\ce{Cr}(s)$ −0.744
$\ce{Mn^2+}(aq)+\ce{2e-}⟶\ce{Mn}(s)$ −1.185
$\ce{Zn(OH)2}(s)+\ce{2e-}⟶\ce{Zn}(s)+\ce{2OH-}(aq)$ −1.245
$\ce{Zn^2+}(aq)+\ce{2e-}⟶\ce{Zn}(s)$ −0.7618
$\ce{Al^3+}(aq)+\ce{3e-}⟶\ce{Al}(s)$ −1.662
$\ce{Mg^2+}(aq)+\ce{2e-}⟶\ce{Mg}(s)$ −2.372
$\ce{Na+}(aq)+\ce{e-}⟶\ce{Na}(s)$ −2.71
$\ce{Ca^2+}(aq)+\ce{2e-}⟶\ce{Ca}(s)$ −2.868
$\ce{Ba^2+}(aq)+\ce{2e-}⟶\ce{Ba}(s)$ −2.912
$\ce{K+}(aq)+\ce{e-}⟶\ce{K}(s)$ −2.931
$\ce{Li+}(aq)+\ce{e-}⟶\ce{Li}(s)$ −3.04
Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.
Example $1$: Cell Potentials from Standard Reduction Potentials
What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.
Solution
Using Table $1$, the reactions involved in the galvanic cell, both written as reductions, are
$\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \nonumber$
$\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s) \hspace{20px} E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V} \nonumber$
Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:
\begin{align*} &\textrm{Anode (oxidation): }\ce{Ni}(s)⟶\ce{Ni^2+}(aq)+\ce{2e-} \hspace{20px} E^\circ_\ce{anode}=E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V}\ &\textrm{Cathode (reduction): }\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_\ce{cathode}=E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \end{align*} \nonumber
The least common factor is six, so the overall reaction is
$\ce{3Ni}(s)+\ce{2Au^3+}(aq)⟶\ce{3Ni^2+}(aq)+\ce{2Au}(s)$
The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.
$E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{1.498\: V−(−0.257\: V)=1.755\: V} \nonumber$
From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.
Exercise $1$
A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.
Answer
$\ce{Mg}(s)+\ce{2Ag+}(aq)⟶\ce{Mg^2+}(aq)+\ce{2Ag}(s) \hspace{20px} E^\circ_\ce{cell}=\mathrm{0.7996\: V−(−2.372\: V)=3.172\: V} \nonumber$
Summary
Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.
Key Equations
• $E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}$
Glossary
standard cell potential $(E^\circ_\ce{cell})$
the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode
standard hydrogen electrode (SHE)
the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials
standard reduction potential (E°)
the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.2%3A_Standard_Reduction_Potential.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 $\ce{Co(s)}$ with $\ce{Ni^{2+}(aq)}$ to form $\ce{Ni(s)}$ and $\ce{Co^{2+}(aq)}$ occurs spontaneously, but if we reduce the concentration of $\ce{Ni^{2+}}$ by a factor of 100, so that $\ce{[Ni^{2+}]}$ 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 & Gibbs 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), 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); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1}$
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), named after the English physicist and chemist Michael Faraday (1791–1867):
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2}
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 ($w_{max}$) is equal to the product of the cell potential ($E^°_{cell}$) and the total charge transferred during the reaction ($nF$):
$w_{max} = nFE_{cell} \label{20.5.3}$
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.
The change in free energy ($\Delta{G}$) is also a measure of the maximum amount of work that can be performed during a chemical process ($ΔG = w_{max}$). Consequently, there must be a relationship between the potential of an electrochemical cell and $\Delta{G}$; this relationship is as follows:
$\Delta{G} = −nFE_{cell} \label{20.5.4}$
A spontaneous redox reaction is therefore characterized by a negative value of $\Delta{G}$ and a positive value of $E^°_{cell}$, 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:
$\Delta{G^°} = −nFE^°_{cell} \label{20.5.5}$
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example $1$
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: $ΔG^o$ for the reaction and spontaneity
Strategy:
1. From the relevant half-reactions and the corresponding values of $E^o$, write the overall reaction and calculate $E^°_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate $ΔG^o$. If $ΔG^o$ is negative, then the reaction is spontaneous.
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 $E^o$. From Table P2, we can find the reduction and oxidation half-reactions and corresponding $E^o$ values:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber
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 $E^o$ is not affected:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber
B
We can now calculate ΔG° using Equation $\ref{20.5.5}$. Because six electrons are transferred in the overall reaction, the value of $n$ is 6:
\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \& =-15.6 \times 10^4\textrm{ J} \ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber
Thus $ΔG^o$ is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise $1$
Use the data in Table P2 to calculate $ΔG^o$ for the reduction of ferric ion by iodide:
$\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 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 $\ce{Fe^{3+}(aq)}$ to $\ce{Fe(s)}$ is not listed in the table, but two related reductions are given:
$\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6}$
$\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7}$
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 $E^o$ is not a state function. However, because $ΔG^o$ is a state function, the sum of the $ΔG^o$ values for the individual reactions gives us $ΔG^o$ for the overall reaction, which is proportional to both the potential and the number of electrons ($n$) transferred. To obtain the value of $E^o$ for the overall half-reaction, we first must add the values of $ΔG^o (= −nFE^o)$ for each individual half-reaction to obtain $ΔG^o$ for the overall half-reaction:
\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9}$
Three electrons ($n = 3$) are transferred in the overall reaction, so substituting into Equation $\ref{20.5.5}$ and solving for $E^o$ gives the following:
\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber
This value of $E^o$ 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.
Values of $E^o$ for half-reactions cannot be added to give $E^o$ for the sum of the half-reactions; only values of $ΔG^o = −nFE^°_{cell}$ for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between $\Delta{G^°}$ and the equilibrium constant $K$, to obtain a relationship between $E^°_{cell}$ and $K$. Recall that for a general reaction of the type $aA + bB \rightarrow cC + dD$, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta{G°} = −RT \ln K \label{20.5.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation $\ref{20.5.5}$), we can write
$−nFE^°_{cell} = −RT \ln K \label{20.5.12}$
Rearranging this equation,
$E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}$
For $T = 298\, K$, Equation $\ref{20.5.12B}$ can be simplified as follows:
\begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align}
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 $2$
Use the data in Table P2 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:
1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and $E^o_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Use Equation $\ref{20.5.13}$ to solve for $\log K$ and then $K$.
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber
B Two electrons are transferred in the overall reaction, so $n = 2$. Solving Equation $\ref{20.5.13}$ for log K and inserting the values of $n$ and $E^o$,
\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \[4pt] K & =2.3\times10^{69}\end{align*} \nonumber
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 $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of $\ce{Sn^{2+}(aq)}$ with oxygen to produce $\ce{Sn^{4+}(aq)}$ and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber$
Answer
$5.7 \times 10^{72}$
Figure $1$ 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 $\Delta{G}$ and the reaction quotient $Q$.
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
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°. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.3%3A_Cell_Potential_Electrical_Work_and_Gibbs_Energy.txt |
Learning Objectives
• Relate cell potentials to Gibbs energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, $\Delta{G}$, is given as follows:
$\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1}$
We also know that $ΔG = −nFE_{cell}$ (under non-standard conditions) and $ΔG^o = −nFE^o_{cell}$ (under standard conditions). Substituting these expressions into Equation $\ref{Eq1}$, we obtain
$−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2}$
Dividing both sides of this equation by $−nF$,
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}$
Equation $\ref{Eq3}$ is called the Nernst equation, 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 ($ΔG = 0$), then Equation $\ref{Eq3}$ reduces to Equation $\ref{Eq31}$ and $\ref{Eq32}$ because $Q = K$, and there is no net transfer of electrons (i.e., Ecell = 0).
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31}$
since
$E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32}$
Substituting the values of the constants into Equation $\ref{Eq3}$ 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 $Q$):
$E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4}$
The Power of the Nernst Equation
The Nernst Equation ($\ref{Eq3}$) can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation $\ref{Eq4}$ 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 $\ref{Eq4}$ 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 $1$
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
$\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber$
Calculate $E_{cell}$ 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, $P_\mathrm{Cl_2}$ = 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 $\ref{Eq4}$ instead of Equation $\ref{Eq3}$. The overall reaction involves the net transfer of two electrons:
$2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber$
$2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber$
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation $\ref{Eq4}$,
\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ & =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \ & =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber
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 $1$
Molecular oxygen will not oxidize $MnO_2$ to permanganate via the reaction
$\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber$
Calculate $E_{cell}$ for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, $P_\mathrm{O_2}$= 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 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:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5}$
The reaction quotient is therefore $Q = [Zn^{2+}]/[Cu^{2+}]$. 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 $\ref{Eq4}$:
\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\ & =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6}
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 $1$. 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:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \ E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ \log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \ Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7}
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.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
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:
$\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8}$
cathode:
$\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9}$
anode:
$\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10}$
Overall
$\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11}$
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:
\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align*} \nonumber
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 cell. 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 $2$
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:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. 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:
$\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
\begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align*} \nonumber
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 $2$
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 $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber$
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 ($K_{sp}$) of sparingly soluble substances. As you learned previously, 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 $1$, which is designed to measure the solubility product of silver chloride:
$K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber$
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:
\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber
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:
\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align}
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 $\ref{Eq122}$ for $K_{sp}$,
\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt] K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber
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 $3$: Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure $1$, 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: Ksp
Strategy:
1. From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
2. Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation $\ref{Eq12}$ and solve for Ksp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of $\ce{Pb^{2+}}$ 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:
\begin{align*}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \ [\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber
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
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \ \textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \ -7.77 & =\log K_\textrm{sp} \ 1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber
Exercise $3$
A concentration cell similar to the one described in Example $3$ 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 $3$, 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 below.
Example $4$: Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
$\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber$
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:
1. Write the overall cell reaction.
2. 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 P2):
Zn(s) + 2H2+(aq) → Zn2+(aq) + H2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation $\ref{Eq4}$), we can calculate [H+] under nonstandard conditions:
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \ \textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \ 16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \ 8.46 &=-\log[\mathrm{H^+}] \ 8.5 &=\mathrm{pH}\end{align*} \nonumber
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise $4$
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:
$Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber$
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb2+ in the groundwater.
Answer
$1.2 \times 10^{−9}\; M$
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.4%3A_Dependence_of_Cell_Potential_on_Concentration.txt |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cell 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 $1$). 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.
The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry.
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 cell is actually a “wet cell”: the electrolyte is an acidic water-based paste containing $MnO_2$, $NH_4Cl$, $ZnCl_2$, graphite, and starch (part (a) in Figure $1$). The half-reactions at the anode and the cathode can be summarized as follows:
• cathode (reduction):
$\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber$
The $\ce{Zn^{2+}}$ ions formed by the oxidation of $\ce{Zn(s)}$ at the anode react with $\ce{NH_3}$ formed at the cathode and $\ce{Cl^{−}}$ ions present in solution, so the overall cell reaction is as follows:
• overall reaction:
$\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3}$
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 $\ce{MnO2}$ 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 $\ce{Zn}$ anode reacts spontaneously with $\ce{NH4Cl}$ in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline battery 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 (reduction)
$\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber$
• overall reaction:
$\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber$
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 "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either $\ce{HgO}$ or $\ce{Ag2O}$ as the oxidant rather than $\ce{MnO2}$ in Figure $\PageIndex{1b}$).
The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown):
• cathode (mercury battery): $\ce{HgO(s) + H2O(l) + 2e^{−} -> Hg(l) + 2OH^{−}(aq)} \nonumber$
• Anode (mercury battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• overall reaction (mercury battery): $\ce{Zn(s) + 2HgO(s) -> 2Hg(l) + ZnO(s)} \nonumber$ with $E_{cell} = 1.35 \,V$.
• cathode reaction (silver battery): $\ce{Ag2O(s) + H2O(l) + 2e^{−} -> 2Ag(s) + 2OH^{−}(aq)} \nonumber$
• anode (silver battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• Overall reaction (silver battery): $\ce{Zn(s) + 2Ag2O(s) -> 2Ag(s) + ZnO(s)} \nonumber$ with $E_{cell} = 1.6 \,V$.
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 $\ce{Hg}$ and $\ce{Ag}$.
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 battery. The anode is lithium metal, and the cathode is a solid complex of $I_2$. 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 (reduction):
$I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11}$
• anode (oxidation):
$2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12}$
• overall:
$2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a}$
with $E_{cell} = 3.5 \, V$
As shown in part (c) in Figure $1$, 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 $TiS_2$, 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–cadmium, 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 $2$, 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 (reduction):
$2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13}$
• anode (oxidation):
$Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14}$
• overall:
$Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15}$
$E_{cell} = 1.4 V$
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:
$NiO(OH)_{(s)} + MH \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16}$
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.
Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators"
Lead–Acid (Lead Storage) Battery
The lead–acid battery 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 $3$, 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 ($PbO_2$). 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 $H_2SO_4$). 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 (reduction):
$PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17}$
with $E^°_{cathode} = 1.685 \; V$
• anode (oxidation):
$Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18}$
with $E^°_{anode} = −0.356 \; V$
• overall:
$Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19}$
and $E^°_{cell} = 2.041 \; V$
As the cell is discharged, a powder of $PbSO_4$ 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. This is often done with the use of a hydrometer.
A hydrometer can be used to test the specific gravity of each cell as a measure of its state of charge (www.youtube.com/watch?v=SRcOqfL6GqQ).
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and $PbSO_4$ is converted back to metallic lead and $PbO_2$. If the battery is recharged too vigorously, however, electrolysis of water can occur:
$2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX}$
This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the $PbSO_4$ or $PbO_2$ 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 $4$. The electrode reactions are as follows:
• cathode (reduction):
$O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20}$
• anode (oxidation):
$2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21}$
• overall:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22}$
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 $O_2 (g)$ 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 $O_2 (g)$.
Summary
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.5%3A_Batteries.txt |
Learning Objectives
• To understand the process of corrosion.
Corrosion 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.
Corrosion is a REDOX 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 ($\ce{Fe2O3 \cdot xH2O}$), commonly known as rust, that does not provide a tight protective film (Figure $1$). 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 will not rust because of the absence of water even if the solvent is saturated with oxygen.
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:
• at cathode: $\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber$ with $E^o_{SRP}=1.23\; V$.
• at anode: $\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber$ with $E^o_{SRP} = −0.45\; V$.
• overall: $\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3}$ with $E^o_{cell} = 1.68\; V$.
The $\ce{Fe^{2+}}$ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing $\ce{Fe^{3+}}$, as represented in the following equation:
$\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4}$
The sign and magnitude of $E^o_{cell}$ for the corrosion process (Equation $\ref{Eq3}$) 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 $2$).
Prophylactic Protection
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. 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 consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy).
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 rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^{2+}}$ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in Table P2 show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. 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 $3$). 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.
Cathodic Protection
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 $\ce{Zn}$ (E° = −0.76 V for $\ce{Zn^{2+} + 2e^{−} -> Zn}$) 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:
$\underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5}$
$\underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6}$
$\underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7}$
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 electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure $4$). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example $1$
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 $E^o°_{cell}$?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, $E^o°_{cell}$, and preventive measures
Strategy:
1. Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate $E^o°_{cell}$.
2. Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution
1. A According to Table P2, both copper and tin are less active metals than iron (i.e., they have higher positive values of $E^o°_{cell}$ 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 $\ce{O2}$ is reduced, and the iron screws will act as anodes at which iron dissolves:
\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber
Over time, the iron screws will dissolve, and the boat will fall apart.
1. 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 $\ref{Eq7}$).
Exercise $1$
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 a
Not unless you plan to sell the house very soon because the $\ce{Cu/Fe}$ pipe joints will lead to rapid corrosion.
Answer b
Any existing $\ce{Pb/Fe}$ joints should be examined carefully for corrosion of the iron pipes due to the $\ce{Pb–Fe}$ junction; the less active $\ce{Pb}$ will have served as the cathode for the reduction of $\ce{O2}$, promoting oxidation of the more active $\ce{Fe}$ nearby.
Summary
Corrosion is a galvanic process that can be prevented using cathodic protection. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.6%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 electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
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 $\,1\; M\, Cd^{2+}$ 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 (Figure $\PageIndex{1a}$).
The overall reaction is as follows:
$\ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber$
with $E°_{cell} = 0.74\; V$
This reaction is thermodynamically spontaneous as written ($ΔG^o < 0$):
\begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber
In this direction, the system is acting as a galvanic cell.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
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) (Figure $\PageIndex{1b}$). 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:
• half-reaction at the cathode:
$\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}$
with $E^°_{cathode} = −0.40 \, V$
• half-reaction at the anode:
$\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}$
with $E^°_{anode} = 0.34 \, V$
• Overall Reaction:
$\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}$
with $E^°_{cell} = −0.74 \: V$
Because $E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ 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 $1$.
Table $1$: Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
Ecell > 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 $\ce{NaCl}$, for example, and an electrical potential is applied, $\ce{Cl^{-}}$ is oxidized at the anode, and $\ce{Na^{+}}$ is reduced at the cathode. The overall reaction is as follows:
$\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}$
This is the reverse of the formation of $\ce{NaCl}$ 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 $\ce{NaCl}$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $\ce{NaCl}$ in a Downs cell (Figure $2$). In this specialized cell, $\ce{CaCl2}$ (melting point = 772°C) is first added to the $\ce{NaCl}$ 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:
$\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7}$
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 P2, 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 $1$
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:
1. List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 7.5, determine which species will be reduced and which species will be oxidized.
2. 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 $1$
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 $3$).
The reactions that occur are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}$
• anode: $2H_2O_{(l)} → O_{2(g)} + 4H^+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}$
• overall: $2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}$
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 $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}
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 overvoltage, 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.
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 electroplating, 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 $4$.
The half-reactions in electroplating a fork, for example, with silver are as follows:
• cathode (fork): $\ce{Ag^{+}(aq) + e^{−} -> Ag(s)} \quad E°_{cathode} = 0.80 V\ \nonumber$
• anode (silver bar): $\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \quad E°_{anode} = 0.80 V \nonumber$
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 $E^o_{cell} = 0\, V$, 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
$\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber$
1 mol of electrons reduces 1 mol of $\ce{Ag^{+}}$ to $\ce{Ag}$ metal. In contrast, in the reaction
$\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber$
1 mol of electrons reduces only 0.5 mol of $\ce{Cu^{2+}}$ to $\ce{Cu}$ metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 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 ($q$ in coulombs) transferred is the product of the current ($I$ in amperes) and the time ($t$, in seconds):
$q = I \times t \label{20.9.14}$
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 $\ce{CuSO4}$ for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \[4pt] &=\mathrm{220\;A\cdot s} \[4pt] &=\textrm{220 C} \end{align*} \nonumber
The number of moles of electrons transferred to $\ce{Cu^{2+}}$ is therefore
\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber
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 $2$
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:
1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. 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:
$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$
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 = (1.85 × 102mol e)(96,485 C/mol e) = 1.78 × 103 C / mole
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber
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 $2$
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
Electroplating: Electroplating(opens in new window) [youtu.be]
Summary
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. Relationship of charge, current and time:
$q = I \times t \nonumber$
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/11%3A_Electrochemistry/11.7%3A_Electrolysis.txt |
Chemical kinetics is the study of rates of chemical processes and includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.
15: Chemical Kinetics
Learning Objectives
• To determine the reaction rate of a reaction.
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 $1$; 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 $1$. 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.
$\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}$
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 calculated for the reaction A → B using Equation $\ref{Eq1}$ is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
Reaction rates generally decrease with time as reactant concentrations decrease.
A Video Discussing Average Reaction Rates. Video Link: Introduction to Chemical Reaction Kinetics(opens in new window) [youtu.be] (opens in new window)
Determining the Reaction Rate of Hydrolysis of Aspirin
We can use Equation $\ref{Eq1}$ 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, as shown in Figure $2$.
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 $1$ and are shown in the graph in Figure $3$.
Table $1$: Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
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 data in Table $1$ 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).
The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align*} \nonumber
The reaction rate can also be calculated 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:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \&=2\times10^{-5}\textrm{ M/h}\end{align*} \nonumber
If the reaction rate is calculated during the last interval given in Table $1$(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
\begin{align*}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \&=8.2\times10^{-6}\textrm{ M/h}\end{align*} \nonumber
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. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide:
$\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2}$
The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from 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 explained in Chapter 10, the volume of CO2 gas formed depends 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:
$\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3}$
The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation $\ref{Eq3}$ so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with 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 with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation $\ref{Eq2}$) corresponds to sucrose, so the reaction rate is generally defined as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4}$
Example $1$: Decomposition Reaction I
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)} \nonumber$
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:
1. 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.
2. 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, define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B The balanced chemical equation shows 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 the rate of change of [N2O5] and [NO2] must be divided 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, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Exercise $1$: Contact Process I
The contact process is used in the manufacture of sulfuric acid. A key step in this process is the reaction of $SO_2$ with $O_2$ to produce $SO_3$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}$
Instantaneous Rates of Reaction
The instantaneous rate of a reaction 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. Comparing this to calculus, 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.
The distinction between the instantaneous and average rates of a reaction is 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 the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its 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 typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Chemical kinetics generally focuses 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.
Example $2$: Decomposition Reaction II
Using the reaction shown in Example $1$, calculate the reaction rate from the following data taken at 56°C:
$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber$
calculate the reaction rate from the following data taken at 56°C:
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:
1. 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.
2. Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example $1$, the reaction rate can be evaluated using any of three expressions:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t} \nonumber$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})} \nonumber$
B Substituting actual values into the expression,
$\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}$
Similarly, NO2 can be used to calculate the reaction rate:
$\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s} \nonumber$
Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used:
$\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s} \nonumber$
Again, this is the same value 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 $2$: Contact Process II
Using the data in the following table, calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
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
Summary
In this Module, the quantitative determination of a reaction rate is demonstrated. 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. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• General definition of rate for A → B: $\textrm{rate}=\frac{\Delta [\textrm B]}{\Delta t}=-\frac{\Delta [\textrm A]}{\Delta t} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/15%3A_Chemical_Kinetics/15.1%3A_Reaction_Rates.txt |
Learning Objectives
• To understand the meaning of the rate law.
The factors that 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.
Rate Laws
Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called rate laws, which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data.
Rate laws may be written from either of two different but related perspectives. A differential rate law 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 law 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); integrated rate laws are discussed in more detail later. The integrated rate law is derived by using calculus to integrate the differential rate law. Whether using 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:
$aA + bB \rightarrow cC + dD \label{14.3.1}$
the experimentally determined rate law usually has the following form:
$\text{rate} = k[A]^m[B]^n \label{14.3.2}$
The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value 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.
Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.
The reaction rate thus 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 order, the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation $\ref{14.3.2}$ tells us that Equation $\ref{14.3.1}$ 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.
The orders of the reactions (e.g. n and m) are not related to the stoichiometric coefficients in the balanced chemical (e.g., a and b).
To illustrate how chemists interpret a differential rate law, consider 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:
$(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} \rightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} \label{14.3.3}$
Combining the rate expression in Equation $\ref{14.3.2}$ with the definition of average reaction rate
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t} \nonumber$
gives a general expression for the differential rate law:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^m[\textrm B]^n \label{14.3.4}$
Inserting the identities of the reactants into Equation $\ref{14.3.4}$ gives the following expression for the differential rate law for the reaction:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{(CH_3)_3CBr}]}{\Delta t}=k[\mathrm{(CH_3)_3CBr}]^m[\mathrm{H_2O}]^n \label{14.3.5}$
Experiments 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. Therefore, m and n in Equation $\ref{14.3.4}$ are 1 and 0, respectively, and,
$\text{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] \label{14.3.6}$
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. The reaction orders state in practical terms 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 working 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:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{CH_3Br}]}{\Delta t}=k'[\mathrm{CH_3Br}] \label{14.3.7}$
This reaction also has an overall reaction order of 1, but the rate constant in Equation $\ref{14.3.7}$ 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 alter reaction conditions to study the mechanics of 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. In contrast, for methyl bromide, the differential rate law becomes
$\text{rate} =k″[CH_3Br][OH^−] \nonumber$
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, providing clues as to how the reactions differ on a molecular level.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber$
is second order in NO2 and zero order in CO at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber$
The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
$\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber$
has been experimentally determined to be $rate = k[NO]^2[H_2]$. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
• order in NO = 2
• order in H2 = 1
• overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
$\ce{rate}=k[\ce{CH3OH}] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
• order in CH3OH = 1
• order in CH3CH2OCOCH3 = 0
• overall order = 1
Example $2$: Differential Rate Laws
Below are 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. $\mathrm{2HI(g)}\xrightarrow{\textrm{Pt}}\mathrm{H_2(g)}+\mathrm{I_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{HI}]}{\Delta t} \right )=k[\textrm{HI}]^2 \nonumber$
2. $\mathrm{2N_2O(g)}\xrightarrow{\Delta}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=k \nonumber$
3. $\mathrm{cyclopropane(g)}\rightarrow\mathrm{propane(g)} \ \textrm{rate}=-\frac{\Delta[\mathrm{cyclopropane}]}{\Delta t}=k[\mathrm{cyclopropane}] \nonumber$
Given: balanced chemical equations and differential rate laws
Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration
Strategy:
1. 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.
2. Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Add all exponents to obtain the overall reaction order.
3. 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)]:
$k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}} \nonumber$
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.
1. 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.
1. 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 $2$
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. \begin{align}\textrm{CH}_3\textrm N\textrm{=NCH}_3\textrm{(g)}\rightarrow\mathrm{C_2H_6(g)}+\mathrm{N_2(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\textrm{CH}_3\textrm N\textrm{=NCH}_3]}{\Delta t} \&=k[\textrm{CH}_3\textrm N\textrm{=NCH}_3]\end{align}
2. \begin{align}\mathrm{2NO_2(g)}+\mathrm{F_2(g)}\rightarrow\mathrm{2NO_2F(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\mathrm{F_2}]}{\Delta t}=-\frac{1}{2}\left ( \frac{\Delta[\mathrm{NO_2}]}{\Delta t} \right ) \&=k[\mathrm{NO_2}][\mathrm{F_2}]\end{align}
Answer a
s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
Answer b
M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.
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 mechanisms 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 $A + B \rightarrow products$, for example, we need to determine k and the exponents m and n in the following equation:
$\text{rate} = k[A]^m[B]^n \label{14.4.11}$
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 $A + B \rightarrow products$ are given in Table $1$.
Table $1$: Rate Data for a Hypothetical Reaction of the Form $A + B \rightarrow products$
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
The general rate law for the reaction is given in Equation $\ref{14.4.11}$. 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 $3$.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber$
Inserting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n} \nonumber$
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.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}$
Substituting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n} \nonumber$
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[A]2[B]0 = k[A]2
We can now calculate the rate constant by inserting the data from any row of Table $3$ into the experimentally determined rate law and solving for $k$. Using Experiment 2, we obtain
19 × 10−3 M/min = k(0.75 M)2
3.4 × 10−2 M−1·min−1 = k
You should verify that using data from any other row of Table $1$ gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
Example $3$
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 $O_2$ to give $NO_2$, 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:
$2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber$
Determine the rate law for the reaction and calculate the rate constant.
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:
1. 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.
2. 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[NO]2[O2]
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
Alternatively, using Experiment 2 gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
The difference is minor and associated with significant digits and likely experimental error in making the table.
The overall reaction order $(m + n) = 3$, so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise $3$
The peroxydisulfate ion (S2O82) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$S_2O^{2−}_{8(aq)} + 3I^−_{(aq)} \rightarrow 2SO^{2−}_{4(aq)} + I^−_{3(aq)} \nonumber$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
kinetics data for this reaction at 25°C.
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
A Video Discussing Initial Rates and Rate Law Expressions. Video Link: Initial Rates and Rate Law Expressions(opens in new window) [youtu.be]
Summary
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 exponent 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/15%3A_Chemical_Kinetics/15.2_Rate_Laws%3A_An_Introduction.txt |
In studying a chemical reaction, it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism by which it takes place, the rate at which it occurs, and the equilibrium toward which it proceeds. According to the law of mass action, the rate of a chemical reaction at a constant temperature depends only on the concentrations of the substances that influence the rate. The substances that influence the rate of reaction are usually one or more of the reactants, but can occasionally include products. Catalysts, which do not appear in the balanced overall chemical equation, can also influence reaction rate. The rate law is experimentally determined and can be used to predict the relationship between the rate of a reaction and the concentrations of reactants.
• 3.3.1: Order of Reaction Experiments
This is an introduction to some of the experimental methods used in school laboratories to find orders of reaction. There are two fundamentally different approaches to this: investigating what happens to the initial rate of the reaction as concentrations change, and following a particular reaction to completion and processing the results from that single reaction.
• 3.3.2: Rate Laws
Reactions are often monitored by some form of spectroscopy. In spectroscopy, "light" or some other frequency of electromagnetic radiation passes through a reaction sample. The light interacts with the molecules in the sample, which absorb particular frequencies of light. Less light exits the sample than the amount that entered it; the amount that exits is measured by a detector on the other side.
• 3.3.3: Reaction Order
The reaction order is the relationship between the concentrations of species and the rate of a reaction.
15.6: Reaction Mechanisms
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:
$\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1}$
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 reaction, 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.
The overall sequence of elementary reactions is the mechanism of the 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.
$\ce{NO2(g) + CO(g) -> NO(g) + CO2 (g)} \label{14.6.2}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of $\ce{NO2}$ with a molecule of $\ce{CO}$ 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:
$rate = k[\ce{NO2}]^2 \label{14.6.3}$
The fact that the reaction is second order in $[\ce{NO2}]$ and independent of $[\ce{CO}]$ tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be
$rate = k[\ce{NO2}][\ce{CO}]. \nonumber$
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
two-step mechanism
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\textrm{slow}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{elementary reaction}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\rightarrow\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{elementary reaction}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$ $\textrm{overall reaction}$
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 $\ce{NO3}$ molecule is an intermediate 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.
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 molecularity 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 $1$). 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
$rate = k[A]. \nonumber$
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 $1$. For a bimolecular elementary reaction of the form A + B → products, the general rate law is
$rate = k[A][B]. \nonumber$
Table $1$: 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
For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).
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 step, 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.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
rate laws for each elementary reaction in our example as well as for the overall reaction.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\mathrm{k_1}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{rate}=k_1[\mathrm{NO_2}]^2\textrm{ (predicted)}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{rate}=k_2[\mathrm{NO_3}][\mathrm{CO}]\textrm{ (predicted)}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\xrightarrow{k}\mathrm{NO}+\mathrm{CO_2}$ $\textrm{rate}=k[\mathrm{NO_2}]^2\textrm{ (observed)}$
The experimentally determined rate law for the reaction of $NO_2$ 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 $k_1$. 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 $1$: A Reaction with an Intermediate
In an alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{k_1}\mathrm{N_2O_4}$
$\textrm{step 2}$ $\underline{\mathrm{N_2O_4}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO}+\mathrm{NO_2}+\mathrm{CO_2}}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$
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:
1. Determine the rate law for each elementary reaction in the reaction.
2. 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 $1$
Iodine monochloride ($\ce{ICl}$) reacts with $\ce{H2}$ as follows:
$\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber$
The experimentally determined rate law is $rate = k[\ce{ICl}][\ce{H2}]$. 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: $\ce{HI}$ is an intermediate.)
Answer
Solutions to Exercise 14.6.1
$\textrm{step 1}$ $\mathrm{ICl}+\mathrm{H_2}\xrightarrow{k_1}\mathrm{HCl}+\mathrm{HI}$ $\mathrm{rate}=k_1[\mathrm{ICl}][\mathrm{H_2}]\,(\textrm{slow})$
$\textrm{step 2}$ $\underline{\mathrm{HI}+\mathrm{ICl}\xrightarrow{k_2}\mathrm{HCl}+\mathrm{I_2}}$ $\mathrm{rate}=k_2[\mathrm{HI}][\mathrm{ICl}]\,(\textrm{fast})$
$\textrm{sum}$ $\mathrm{2ICl}+\mathrm{H_2}\rightarrow\mathrm{2HCl}+\mathrm{I_2}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Example $2$ : Nitrogen Oxide Reacting with Molecular Hydrogen
Assume the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process:
the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process
$\textrm{step 1}$ $\mathrm{NO}+\mathrm{NO}\xrightarrow{k_1}\mathrm{N_2O_2}$ $\textrm{(fast)}$
$\textrm{step 2}$ $\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}$ $\textrm{(slow)}$
$\textrm{step 3}$ $\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}$ $\textrm{(fast)}$
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:
$\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed}$
Answer
• Step 1: $rate = k_1[\ce{NO}]^2$
• Step 2: $rate = k_2[\ce{N_2O_2}][\ce{H_2}]$
• Step 3: $rate = k_3[\ce{N_2O}][\ce{H_2}]$
The overall reaction is then
$\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber$
• Rate Determining Step : #2
• Yes, because the rate of formation of $[\ce{N_2O_2}] = k_1[\ce{NO}]^2$. Substituting $k_1[\ce{NO}]^2$ for $[\ce{N_2O_2}]$ in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where $k = k_1k_2$.
Reaction Mechanism (Slow step followed by fast step): Reaction Mechanism (Slow step Followed by Fast Step)(opens in new window) [youtu.be] (opens in new window)
Summary
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/15%3A_Chemical_Kinetics/15.3%3A_Determining_the_Form_of_the_Rate_Law.txt |
Learning Objectives
• Explain steady state and steady-state approximation.
• Derive a rate law when a mechanism is given but the rate determining step is not identified.
• Derive a general expression of the rate law using the steady-state approximation.
• Make appropriate assumptions so that the derived rate law agrees with the observed rate law.
• Give expressions for the producing rate of an intermediate.
• Give expressions for the consuming rate of an intermediate.
• Express concentration of intermediate in terms of concentration of reactants.
• Eliminate concentrations of intermediates using concentrations of reactants.
• Derive a rate law from the many elementary steps.
• Discuss the derived rate law.
The Steady-State Approximation
When a reaction mechanism has several steps of comparable rates, the rate-determining step is often not obvious. However, there is an intermediate in some of the steps. An intermediate is a species that is neither one of the reactants, nor one of the products. The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains the same in a duration of the reaction.
Definition: Intermediates
An intermediate is a species that is neither one of the reactants, nor one of the products. It transiently exists during the course of the reaction.
When a reaction involves one or more intermediates, the concentration of one of the intermediates remains constant at some stage of the reaction. Thus, the system has reached a steady-state. The concentration of one of the intermediates, $[Int]$, varies with time as shown in Figure $1$. At the start and end of the reaction, [Int] does vary with time.
$\dfrac{d[Int]}{dt}= 0 \nonumber$
When a reaction mechanism has several steps with comparable rates, the rate-determining step is not obvious. However, there is an intermediate in some of the steps. The steady-state approximation implies that you select an intermediate in the reaction mechanism, and calculate its concentration by assuming that it is consumed as quickly as it is generated. In the following, an example is given to show how the steady-state approximation method works.
Example $1$
Use the steady-state approximation to derive the rate law for this reaction
$\ce{2 N2O5 \rightarrow 4 NO2 + O2}\nonumber$
assuming it follows the following three-step mechanism:
\begin{align} \ce{N_2O_5} &\underset{\Large{k_{\textrm b}}}{\overset{\Large{k_{\textrm f}}}\rightleftharpoons} \ce{NO_2 + NO_3} \tag{step 1} \[4pt] \ce{NO3 + NO2} &\ce{->[\large{k_2}] NO + NO2 + O2} \tag{step 2} \[4pt] \ce{NO3 + NO} & \ce{->[\Large{k_3}] 2 NO2} \tag{step 3} \end{align}
Solution
In these steps, $\ce{NO}$ and $\ce{NO3}$ are intermediates. You have
$\ce{production\: rate\: of\: NO} = k_2 \ce{[NO3] [NO2]}$
$\ce{consumption\: rate\: of\: NO} = k_3 \ce{[NO3] [NO]}$
A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption. Thus, we have
$k_2 \ce{[NO3] [NO2]} = k_3 \ce{[NO3] [NO]}$
and solving for $\ce{[NO]}$ gives the result,
$\ce{[NO]} = \dfrac{k_2 \ce{[NO3] [NO2]}}{k_3 \ce{[NO3]}} \tag{1}$
For the other intermediate $\ce{NO3}$,
$\ce{production\: rate\: of\: NO3} = k_{\ce f} \ce{[N2O5]}$
$\ce{consumption\: rate\: of\: NO3} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}$
Applying the steady-state assumption gives:
$k_{\ce f} \ce{[N2O5]} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}$
Thus,
$\ce{[NO3]} = \dfrac{k_{\ce f} \ce{[N2O5]}}{k_2\ce{[NO2]} + k_3\ce{[NO]} + k_{\ce b}\ce{[NO2]}}\tag{2}$
Let's review the three equations (steps) in the mechanism:
Step i. is at equilibrium and thus can not give a rate expression.
Step ii. leads to the production of some products, and the active species $\ce{NO}$ causes further reaction in step iii. This consideration led to a rate expression from step ii. as:
$\ce{\dfrac{d[O2]}{dt}} = k_2 \ce{[NO3] [NO2]} \tag{3}$
Substituting (1) in (2) and then in (3) gives
$\ce{\dfrac{d[O2]}{dt}}= \dfrac{k_{\ce f} k_2 \ce{[N2O5]}}{k_{\ce b} + 2 k_2} = \ce{k [N2O5]}$
where $\ce{k} = \dfrac{k_{\ce f} k_2}{k_{\ce b} + 2 k_2}$.
This is the differential rate law, and it agrees with the experimental results. Carry out the above manipulation yourself on a piece of paper. Simply reading the above will not lead to solid learning yet.
This page gives another example to illustrate the technique of deriving rate laws using the steady-state approximation. The reaction considered here is between $\ce{H2}$ and $\ce{I2}$ gases.
Example $1$
For the reaction:
$\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI_{\large{(g)}}}$
what mechanisms might be appropriate? Derive a rate law from the proposed mechanism.
Solution
Well, this question does not have a simple answer, and there is no way to prove one over another for its validity. Beginning chemistry students will not be asked to propose a mechanism, but you will be asked to derive the rate law from the proposed mechanism.
First of all, you should be able to express the rate of reaction in terms of the concentration changes,
$rate = - \ce{\dfrac{d[H2]}{dt}} = - \ce{\dfrac{ d[I2]}{dt}} = \ce{\dfrac{1}{2}\dfrac{d[HI]}{dt}}$
Look at the overall reaction equation again to see its relationship and the rate expressions.
Proposing a mechanism
In order to propose a mechanism, we apply the following reasoning. Since the bonding between $\ce{I-I}$ is weak, we expect $\ce{I2}$ to dissociate into atoms or radicals. These radicals are active, and they react with $\ce{H2}$ to produce the products. Thus we propose the three-step mechanism:
1. $\ce{I_{2\large{(g)}}} \xrightarrow{\Large{k_1}} \ce{2 I_{\large{(g)}}}$
2. $\ce{2 I_{\large{(g)}}} \xrightarrow{\Large{k_2}} \ce{I_{2\large{(g)}}}$
3. $\ce{H_{2\large{(g)}} + 2 I_{\large{(g)}}} \xrightarrow{\Large{k_3}} \ce{2 HI_{\large{(g)}}}$
Which step would you use to write the differential rate law?
Since only step iii. gives the real products, we expect you to recognize that step iii. hints the rate law to be:
$rate = k_3 \ce{[H2] [I]^2}$
However, this is not a proper rate law, because $\ce{I}$ is an intermediate, not a reactant. So, you have to express $\ce{[I]}$ or $\ce{[I]^2}$ in terms of the concentration of reactants. To do this, we use the steady-state approximation and write out the following relationships:
$\textrm{rate of producing I} = 2 k_1 \ce{[I2]}$
$\textrm{rate of consuming I} = 2 k_2 \ce{[I]^2} + 2 k_3 \ce{[H2] [I]^2}$
$\textrm{producing rate of I} = \textrm{consuming rate of I}$
Thus,
$\ce{[I]^2} = \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}$
Substituting this for $\ce{[I]^2}$ into the rate expression, you have
\begin{align} rate &= k_3 \ce{[H2]} \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}\ &= \dfrac{k_1 k_3 \ce{[H2] [I2]}}{k_2 + k_3 \ce{[H2]}} \end{align}
Discussion
If step iii. is slow, then $k_3$ and $k_2 >> k_3 \ce{[H2]}$. The rate law is reduced to
$\ce{rate} = \ce{k [H2] [I2]}$,
where $\ce{k} = \dfrac{k_1 k_3}{k_2}$. (Work this out on paper yourself; reading the above derivation does not lead to learning.)
Since the rate law is first order with respect to both reactants, one may argue that the rate law also supports a one-step mechanism,
$\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI}$
This elementary step is the same as the overall reaction.
Suppose we use a large quantity of $\ce{H2}$ compared to $\ce{I2}$, then the change in $\ce{[H2]}$ is insignificant. For example, if $\mathrm{[H_2] = 10}$, and $\mathrm{[I_2] = 0.1}$ initially, $\ce{[H2]}$ remains essentially 10 (9.9 with only one significant figure). In other words, $\ce{[H2]}$ hardly changed when the reaction ended. Thus,
$k_3 \ce{[H2]} >> k_3$
and the rate law becomes:
$rate = k_1 \ce{[I2]}$.
Thus, the reaction is a pseudo first order reaction, due to the large quantity of one reactant. The results suggest iii. a fast step (due to large quantity of $\ce{H2}$), and i. the rate determining step. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/15%3A_Chemical_Kinetics/15.7%3A_The_Steady-State_Approximation.txt |
Learning Objectives
• To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.
Catalysts are 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 $1$). 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.
A catalyst affects Ea, not ΔE.
Heterogeneous Catalysis
In heterogeneous catalysis, 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 $2$, 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 $2$ 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 $1$. 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 $1$: 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 $\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}$ $\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}$
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
Homogeneous Catalysis
In homogeneous catalysis, 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 $2$), 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 $2$: 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/PR3complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene
Enzymes
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 substrate.
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 $3$).
Enzyme inhibitors 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. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/15%3A_Chemical_Kinetics/15.9%3A_Catalysis.txt |
Learning Objectives
• To describe the intermolecular forces in liquids.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids.
Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures.
In this section, we explicitly consider three kinds of intermolecular interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding, and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(2\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r3, so doubling the distance between the dipoles decreases the strength of the interaction by 23, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships Between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Video Discussing Dipole Intermolecular Forces. Source: Dipole Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(1\)
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise \(1\)
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(2\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table \(2\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(3\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure \(3\), tends to become more pronounced as atomic and molecular masses increase (Table \(2\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(4\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Video Discussing London/Dispersion Intermolecular Forces. Source: Dispersion Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(2\)
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise \(2\)
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cage like structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Video Discussing Hydrogen Bonding Intermolecular Forces. Source: Hydrogen Bonding Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(3\)
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A. Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B. The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise \(3\)
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and NH3;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example \(4\): Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise \(4\)
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example \(5\)
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
1. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
2. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
3. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise \(6\)
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
Summary
Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/r3, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules; their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cage like structure that is less dense than liquid water. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/16%3A_Liquids_and_Solids/16.01%3A_Intermolecular_Forces.txt |
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