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Learning Objectives
• To understand the trends in properties and reactivity of the group 14 elements.
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure $1$). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater.
Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.)
Although the first glasses were prepared from silica (silicon oxide, SiO2) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na2SiF6 with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country.
Preparation and General Properties of the Group 14 Elements
The natural abundance of the group 14 elements varies tremendously. Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite:
$\mathrm{SiO_2(s)}+\mathrm{3C(s)}\xrightarrow{\Delta}\mathrm{SiC(s)}+\mathrm{2CO(g)} \label{$1$}$
$\mathrm{SiC(s)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{C(graphite)} \label{$2$}$
One allotrope of carbon, diamond, is metastable under normal conditions, with a ΔG°f of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm3 versus 2.2 g/cm3). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices.
The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (part (a) in Figure $2$). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds.
Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl2 to give SiCl4, followed by the fractional distillation of the impure SiCl4 and reduction with H2:
$\mathrm{SiCl_4(l)}+\mathrm{2H_2(g)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{4HCl(g)} \label{$3$}$
Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (part (b) in Figure $2$).
In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices.
Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity:
$\mathrm{PbS(s)}+\frac{3}{2}\mathrm{O_2(g)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\mathrm{SO_2(g)} \label{$4$}$
$\mathrm{PbO(s)}+\mathrm{C(s)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO(g)} \label{$5$}$
or
$\mathrm{PbO(s)}+\mathrm{CO(g)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO_2(g)} \label{$6$}$
By far the single largest use of lead is in lead storage batteries. The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Zeff, the relative stability of the +2 oxidation state increases smoothly from carbon to lead.
The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14.
Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO2 is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO2 is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO2 molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms.
The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH2 group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH2 group into the silicon analogue of n-hexane (Si6H14) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead.
In Table $1$ "Selected Properties of the Group 14 Elements" we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge.
Table $1$: Selected Properties of the Group 14 Elements
Property Carbon Silicon Germanium Tin Lead
*The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated.
X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element.
atomic symbol C Si Ge Sn Pb
atomic number 6 14 32 50 82
atomic mass (amu) 12.01 28.09 72.64 118.71 207.2
valence electron configuration* 2s22p2 3s23p2 4s24p2 5s25p2 6s26p2
melting point/boiling point (°C) 4489 (at 10.3 MPa)/3825 1414/3265 939/2833 232/2602 327/1749
density (g/cm3) at 25°C 2.2 (graphite), 3.51 (diamond) 2.33 5.32 7.27(white) 11.30
atomic radius (pm) 77 (diamond) 111 125 145 154
first ionization energy (kJ/mol) 1087 787 762 709 716
most common oxidation state +4 +4 +4 +4 +4
ionic radius (pm) ≈29 ≈40 53 69 77.5
electron affinity (kJ/mol) −122 −134 −119 −107 −35
electronegativity 2.6 1.9 2.0 2.0 1.8
standard reduction potential (E°, V) (for EO2 → E in acidic solution) 0.21 −0.86 −0.18 −0.12 0.79
product of reaction with O2 CO2, CO SiO2 GeO2 SnO2 PbO
type of oxide acidic (CO2) acidic neutral (CO) amphoteric amphoteric amphoteric
product of reaction with N2 none Si3N4 none Sn3N4 none
product of reaction with X2 CX4 SiX4 GeX4 SnX4 PbX2
product of reaction with H2 CH4 none none none none
The group 14 elements follow the same pattern as the group 13 elements in their periodic properties.
Reactions and Compounds of Carbon
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
$CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{$7$}$
The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
$\mathrm{HCO_2H(l)}\xrightarrow{\mathrm{H_2SO_4(l)}}\mathrm{CO(g)}+\mathrm{H_3O^+(aq)}+\mathrm{HSO^-_4}\label{$8$}$
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
$\mathrm{CO(g)}+\mathrm{Cl_2(g)}\xrightarrow{\Delta}\textrm{Cl}_2\textrm{C=O(g)}\label{$9$}$
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
$\mathrm{C(s)}+\mathrm{2S(g)}\xrightarrow{\Delta}\mathrm{CS_2(g)}\label{$1$0}$
The selenium analog CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
$\pi$ bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C4− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation $\ref{Eq1}$) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
Example $1$
For each reaction, explain why the given product forms.
1. CO(g) + Cl2(g) → Cl2C=O(g)
2. CO(g) + BF3(g) → F3B:C≡O(g)
3. Sr(s) + 2C(s) $\xrightarrow{\Delta}$ SrC2(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Because the carbon in CO is in an intermediate oxidation state (+2), CO can be either a reductant or an oxidant; it is also a Lewis base. The other reactant (Cl2) is an oxidant, so we expect a redox reaction to occur in which the carbon of CO is further oxidized. Because Cl2 is a two-electron oxidant and the carbon atom of CO can be oxidized by two electrons to the +4 oxidation state, the product is phosgene (Cl2C=O).
2. Unlike Cl2, BF3 is not a good oxidant, even though it contains boron in its highest oxidation state (+3). Nor can BF3 behave like a reductant. Like any other species with only six valence electrons, however, it is certainly a Lewis acid. Hence an acid–base reaction is the most likely alternative, especially because we know that CO can use the lone pair of electrons on carbon to act as a Lewis base. The most probable reaction is therefore the formation of a Lewis acid–base adduct.
3. Typically, both reactants behave like reductants. Unless one of them can also behave like an oxidant, no reaction will occur. We know that Sr is an active metal because it lies far to the left in the periodic table and that it is more electropositive than carbon. Carbon is a nonmetal with a significantly higher electronegativity; it is therefore more likely to accept electrons in a redox reaction. We conclude, therefore, that Sr will be oxidized, and C will be reduced. Carbon forms ionic carbides with active metals, so the reaction will produce a species formally containing either C4− or C22−. Those that contain C4− usually involve small, highly charged metal ions, so Sr2+ will produce the acetylide (SrC2) instead.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$
2. C(s) + H2O(l) →
3. NaHCO3(s) + H2SO4(aq) →
Answer
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$ CO2(g)
2. C(s) + H2O(l) → no reaction
3. NaHCO3(s) + H2SO4(aq) → CO2(g) + NaHSO4(aq) + H2O(l)
Reactions and Compounds of the Heavier Group 14 Elements
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Video $1$: Time lapse tin pest reaction.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62− ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
$MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{$1$1}$
In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure $3$). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44− unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure $4$). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na+ ions inside the zeolite cavities are displaced by the more highly charged Mg2+ and Ca2+ ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
$3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{$1$2}$
Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4− ion. They react with aqueous acid to form silicon hydrides such as SiH4:
$Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{$1$3}$
Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure $5$). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Example $2$
For each reaction, explain why the given products form.
1. Pb(s) + Cl2(g) → PbCl2(s)
2. Mg2Si(s) + 4H2O(l) → SiH4(g) + 2Mg(OH)2(s)
3. GeO2(s) + 4OH(aq) → GeO44−(aq) + 2H2O(l)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Lead is a metal, and chlorine is a nonmetal that is a strong oxidant. Thus we can expect a redox reaction to occur in which the metal acts as a reductant. Although lead can form compounds in the +2 and +4 oxidation states, Pb4+ is a potent oxidant (the inert-pair effect). Because lead prefers the +2 oxidation state and chlorine is a weaker oxidant than fluorine, we expect PbCl2 to be the product.
2. This is the reaction of water with a metal silicide, which formally contains the Si4− ion. Water can act as either an acid or a base. Because the other compound is a base, we expect an acid–base reaction to occur in which water acts as an acid. Because Mg2Si contains Si in its lowest possible oxidation state, however, an oxidation–reduction reaction is also a possibility. But water is a relatively weak oxidant, so an acid–base reaction is more likely. The acid (H2O) transfers a proton to the base (Si4−), which can accept four protons to form SiH4. Proton transfer from water produces the OH ion, which will combine with Mg2+ to give magnesium hydroxide.
3. We expect germanium dioxide (GeO2) to be amphoteric because of the position of germanium in the periodic table. It should dissolve in strong aqueous base to give an anionic species analogous to silicate.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. PbO2(s) $\xrightarrow{\Delta}$
2. GeCl4(s) + H2O(l) →
3. Sn(s) + HCl(aq) →
Answer
1. $\mathrm{PbO_2(s)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\frac{1}{2}\mathrm{O_2(g)}$
2. GeCl4(s) + 2H2O(l) → GeO2(s) + 4HCl(aq)
3. Sn(s) + 2HCl(aq) → Sn2+(aq) + H2(g) + 2Cl(aq)
Summary
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22− (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
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21.1: Periodic Trends and Charge Density
Problems
1. List three physical properties that are important in describing the behavior of the main group elements.
2. Arrange K, Cs, Sr, Ca, Ba, and Li in order of
1. increasing ionization energy.
2. increasing atomic size.
3. increasing electronegativity.
1. Arrange Rb, H, Be, Na, Cs, and Ca in order of
1. decreasing atomic size.
2. decreasing magnitude of electron affinity.
1. Which periodic trends are affected by Zeff? Based on the positions of the elements in the periodic table, which element would you expect to have the highest Zeff? the lowest Zeff?
1. Compare the properties of the metals and nonmetals with regard to their electronegativities and preferred oxidation states.
1. Of Ca, Br, Li, N, Zr, Ar, Sr, and S, which elements have a greater tendency to form positive ions than negative ions?
1. Arrange As, O, Ca, Sn, Be, and Sb in order of decreasing metallic character.
1. Give three reasons the chemistry of the second-period elements is generally not representative of their groups as a whole.
1. Compare the second-period elements and their heavier congeners with regard to
1. magnitude of electron affinity.
2. coordination number.
3. the solubility of the halides in nonpolar solvents.
1. The heavier main group elements tend to form extended sigma-bonded structures rather than multiple bonds to other atoms. Give a reasonable explanation for this tendency.
1. What is the diagonal effect? How does it explain the similarity in chemistry between, for example, boron and silicon?
1. Although many of the properties of the second- and third-period elements in a group are quite different, one property is similar. Which one?
1. Two elements are effective additives to solid rocket propellant: beryllium and one other element that has similar chemistry. Based on the position of beryllium in the periodic table, identify the second element.
1. Give two reasons for the inert-pair effect. How would this phenomenon explain why Sn2+ is a better reducing agent than Pb2+?
1. Explain the following trend in electron affinities: Al (−41.8 kJ/mol), Si (−134.1 kJ/mol), P (−72.0 kJ/mol), and S (−200.4 kJ/mol).
1. Using orbital energy arguments, explain why electron configurations with more than four electron pairs around the central atom are not observed for second-period elements.
Answers
1. Cs > Rb > Ca > Na > Be > H
2. H > Na > Rb > Cs > Ca > Be
1. Ca > Be > Sn > Sb > As > O
1. aluminum
1. The magnitude of electron affinity increases from left to right in a period due to the increase in Zeff; P has a lower electron affinity than expected due to its half-filled 3p shell, which requires the added electron to enter an already occupied 3p orbital.
Structure and Reactivity
1. The following table lists the valences, coordination numbers, and ionic radii for a series of cations. Which would you substitute for K+ in a crystalline lattice? Explain your answer.
Metal Charge Coordination Number Ionic Radius (pm)
Li +1 4 76
Na +1 6 102
K +1 6 138
Mg +2 6 72
Ca +2 6 100
Sr +2 6 118
Answer
1. Sr2+; it is the ion with the radius closest to that of K+.
21.2: Group 1: The Alkali Metals
Problems
1. Which of the group 1 elements reacts least readily with oxygen? Which is most likely to form a hydrated, crystalline salt? Explain your answers.
2. The alkali metals have a significant electron affinity, corresponding to the addition of an electron to give the Manion. Why, then, do they commonly lose the ns1electron to form the M+ cation rather than gaining an electron to form M?
3. Lithium is a far stronger reductant than sodium; cesium is almost as strong as lithium, which does not agree with the expected periodic trend. What two opposing properties explain this apparent anomaly? Is the same anomaly found among the alkaline earth metals?
4. Explain why the ionic character of LiCl is less than that of NaCl. Based on periodic trends, would you expect the ionic character of BeCl2 to be greater or less than that of LiCl? Why?
5. Alkali metals and carbon form intercalation compounds with extremely high electrical conductivity. Is this conductivity through the layers or along the layers? Explain your answer.
6. Electrolysis is often used to isolate the lighter alkali metals from their molten halides. Why are halides used rather than the oxides or carbonates, which are easier to isolate? With this in mind, what is the purpose of adding calcium chloride to the alkali metal halide?
7. The only alkali metal that reacts with oxygen to give a compound with the expected stoichiometry is lithium, which gives Li2O. In contrast, sodium reacts with oxygen to give Na2O2, and the heavier alkali metals form superoxides. Explain the difference in the stoichiometries of these products.
8. Classify aqueous solutions of Li2O, Na2O, and CsO2 as acidic, basic, or amphoteric.
9. Although methanol is relatively unreactive, it can be converted to a synthetically more useful form by reaction with LiH. Predict the products of reacting methanol with LiH. Describe the visual changes you would expect to see during this reaction.
10. Lithium reacts with atmospheric nitrogen to form lithium nitride (Li3N). Why do the other alkali metals not form analogous nitrides? Explain why all the alkali metals react with arsenic to form the corresponding arsenides (M3As).
Structure and Reactivity
1. Write a balanced chemical equation to describe each reaction.
1. the electrolysis of fused (melted) sodium chloride
2. the thermal decomposition of KClO3
3. the preparation of hydrogen fluoride from calcium fluoride and sulfuric acid
4. the oxidation of sodium metal by oxygen
1. What products are formed at the anode and the cathode during electrolysis of
1. molten lithium hydride?
2. molten lithium chloride?
3. aqueous sodium fluoride?
Write the corresponding half-reactions for each reaction.
1. Sodium metal is prepared by electrolysis of molten NaCl. If 25.0 g of chlorine gas are produced in the electrolysis of the molten salt using 9.6 A (C/s) of current, how many hours were required for the reaction? What mass of sodium was produced?
1. Sodium peroxide can remove CO2 from the air and replace it with oxygen according to the following unbalanced chemical equation:
Na2O2(s) + CO2(g) → Na2CO3(s) + O2(g)
1. Balance the chemical equation.
2. Identify each oxidation and reduction half-reaction.
3. Assuming complete reaction, what will be the pressure inside a sealed 1.50 L container after reacting excess sodium peroxide with carbon dioxide that was initially at 0.133 atm and 37°C?
1. Predict the products of each chemical reaction and then balance each chemical equation.
1. K(s) + CH3CH2OH(l) →
2. Na(s) + CH3CO2H(l) →
3. NH4Cl(s) + Li(s) →
4. (CH3)2NH(l) + K(s) →
1. Predict the products of each reaction.
1. an alkyl chloride with lithium metal
2. rubidium with oxygen
1. A 655 mg sample of graphite was allowed to react with potassium metal, and 744 mg of product was isolated. What is the stoichiometry of the product?
1. Perchloric acid, which is used as a reagent in a number of chemical reactions, is typically neutralized before disposal. When a novice chemist accidentally used K2CO3 to neutralize perchloric acid, a large mass of KClO4 (Ksp = 1.05 × 10−2) precipitated from solution. What mass of potassium ion is present in 1.00 L of a saturated solution of KClO4?
1. A key step in the isolation of the alkali metals from their ores is selective precipitation. For example, lithium is separated from sodium and potassium by precipitation of Li2CO3 (Ksp = 8.15 × 10−4). If 500.0 mL of a 0.275 M solution of Na2CO3 are added to 500.0 mL of a 0.536 M lithium hydroxide solution, what mass of Li2CO3 will precipitate (assuming no further reactions occur)? What mass of lithium will remain in solution?
Answer
1. 5.54 g Li2CO3; 0.82 g Li+
21.3: Group 2: The Alkaline Earth Metals
Problems
1. The electronegativities of Li and Sr are nearly identical (0.98 versus 0.95, respectively). Given their positions in the periodic table, how do you account for this?
2. Arrange Na, Ba, Cs, and Li in order of increasing Zeff.
3. Do you expect the melting point of NaCl to be greater than, equal to, or less than that of MgCl2? Why?
4. Which of the group 2 elements would you expect to form primarily ionic rather than covalent organometallic compounds? Explain your reasoning.
5. Explain why beryllium forms compounds that are best regarded as covalent in nature, whereas the other elements in group 2 generally form ionic compounds.
6. Why is the trend in the reactions of the alkaline earth metals with nitrogen the reverse of the trend seen for the alkali metals?
7. Is the bonding in the alkaline earth hydrides primarily ionic or covalent in nature? Explain your answer. Given the type of bonding, do you expect the lighter or heavier alkaline earth metals to be better reducing agents?
8. Using arguments based on ionic size, charge, and chemical reactivity, explain why beryllium oxide is amphoteric. What element do you expect to be most similar to beryllium in its reactivity? Why?
9. Explain why the solubility of the carbonates and sulfates of the alkaline earth metals decreases with increasing cation size.
10. Beryllium oxide is amphoteric, magnesium oxide is weakly basic, and calcium oxide is very basic. Explain how this trend is related to the ionic character of the oxides.
11. Do you expect the $\Delta H^\circ_\textrm f$ of CaH2 to be greater than, the same as, or less than that of BaH2? Why or why not?
12. Which of the s-block elements would you select to carry out a chemical reduction on a small scale? Consider cost, reactivity, and stability in making your choice. How would your choice differ if the reduction were carried out on an industrial scale?
Structure and Reactivity
1. Beryllium iodide reacts vigorously with water to produce HI. Write a balanced chemical equation for this reaction and explain why it is violent.
2. Predict the products of each reaction and then balance each chemical equation.
1. Mg(OH)2(aq) + (NH4)3PO4(aq) →
2. calcium carbonate and sulfuric acid →
3. CaCl2(aq) + Na3PO4(aq) →
4. the thermal decomposition of SrCO3
1. Predict the products of each reaction and then balance each chemical equation.
1. Sr(s) + O2(g) →
2. the thermal decomposition of CaCO3(s)
3. CaC2(s) + H2O(l) →
4. RbHCO3(s) + H2SO4(aq) →
1. Indicate whether each pair of substances will react and, if so, write a balanced chemical equation for the reaction.
1. an alkyl chloride and magnesium metal
2. strontium metal and nitrogen
3. magnesium metal and cold water
4. beryllium and nitrogen
1. Using a thermodynamic cycle and information presented in Chapter 7 and Chapter 8, calculate the lattice energy of magnesium nitride (Mg3N2). ($\Delta H^\circ_\textrm f$ for Mg3N2 is −463 kJ/mol, and ΔH° for N(g) + 3e → N3− is +1736 kJ.) How does the lattice energy of Mg3N2 compare with that of MgCl2 and MgO? (See Chapter 25 for the enthalpy of formation values.)
1. The solubility products of the carbonate salts of magnesium, calcium, and strontium are 6.82 × 10−6, 3.36 × 10−9, and 5.60 × 10−10, respectively. How many milligrams of each compound would be present in 200.0 mL of a saturated solution of each? How would the solubility depend on the pH of the solution? Why?
1. The solubility products of BaSO4 and CaSO4 are 1.08 × 10−10 and 4.93 × 10−5, respectively. What accounts for this difference? When 500.0 mL of a solution that contains 1.00 M Ba(NO3)2 and 3.00 M Ca(NO3)2 is mixed with a 2.00 M solution of Na2SO4, a precipitate forms. What is the identity of the precipitate? How much of it will form before the second salt precipitates?
1. Electrolytic reduction is used to produce magnesium metal from MgCl2. The goal is to produce 200.0 kg of Mg by this method.
1. How many kilograms of MgCl2 are required?
2. How many liters of chlorine gas will be released at standard temperature and pressure?
3. How many hours will it take to process the magnesium metal if a total current of 1.00 × 104 A is used?
1. A sample consisting of 20.35 g of finely divided calcium metal is allowed to react completely with nitrogen. What is the mass of the product?
1. What mass of magnesium hydride will react with water to produce 1.51 L of hydrogen gas at standard temperature and pressure?
Answers
3
1. 2Sr(s) + O2(g) → 2SrO(s)
2. $\mathrm{CaCO_3(s)}\xrightarrow\Delta\mathrm{CaO(s)}+\mathrm{CO_2(g)}$
3. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq)
4. 2RbHCO3(s) + H2SO4(aq) → Rb2SO4(aq) + 2CO2(g) + 2H2O(l)
1. The Ba2+ ion is larger and has a lower hydration energy than the Ca2+ ion. The precipitate is BaSO4; 117 g of BaSO4.
1. 25.09 g of Ca3N2
21.4: Group 13: The Boron Family
Conceptual Problems
1. None of the group 13 elements was isolated until the early 19th century, even though one of these elements is the most abundant metal on Earth. Explain why the discovery of these elements came so late and describe how they were finally isolated.
2. Boron and aluminum exhibit very different chemistry. Which element forms complexes with the most ionic character? Which element is a metal? a semimetal? What single property best explains the difference in their reactivity?
3. The usual oxidation state of boron and aluminum is +3, whereas the heavier elements in group 13 have an increasing tendency to form compounds in the +1 oxidation state. Given that all group 13 elements have an ns2np1 electron configuration, how do you explain this difference between the lighter and heavier group 13 elements?
4. Do you expect the group 13 elements to be highly reactive in air? Why or why not?
5. Which of the group 13 elements has the least metallic character? Explain why.
6. Boron forms multicenter bonds rather than metallic lattices with delocalized valence electrons. Why does it prefer this type of bonding? Does this explain why boron behaves like a semiconductor rather than a metal? Explain your answer.
7. Because the B–N unit is isoelectronic with the C–C unit, compounds that contain these units tend to have similar chemistry, although they exhibit some important differences. Describe the differences in physical properties, conductivity, and reactivity of these two types of compounds.
8. Boron has a strong tendency to form clusters. Does aluminum have this same tendency? Why or why not?
9. Explain why a B–O bond is much stronger than a B–C bond.
10. The electron affinities of boron and aluminum are −27 and −42 kJ/mol, respectively. Contrary to the usual periodic trends, the electron affinities of the remaining elements in group 13 are between those of B and Al. How do you explain this apparent anomaly?
11. The reduction potentials of B and Al in the +3 oxidation state are −0.87 V and −1.66 V, respectively. Do you expect the reduction potentials of the remaining elements of group 13 to be greater than or less than these values? How do you explain the differences between the expected values and those given in Table $1$ "Selected Properties of the Group 13 Elements"?
12. The compound Al2Br6 is a halide-bridged dimer in the vapor phase, similar to diborane (B2H6). Draw the structure of Al2Br6 and then compare the bonding in this compound with that found in diborane. Explain the differences.
13. The compound AlH3 is an insoluble, polymeric solid that reacts with strong Lewis bases, such as Me3N, to form adducts with 10 valence electrons around aluminum. What hybrid orbital set is formed to allow this to occur?
Answers
1. The high stability of compounds of the group 13 elements with oxygen required powerful reductants such as metallic potassium to be isolated. Al and B were initially prepared by reducing molten AlCl3 and B2O3, respectively, with potassium.
1. Due to its low electronegativity and small size, boron is an unreactive semimetal rather than a metal.
1. The B–N bond is significantly more polar than the C–C bond, which makes B–N compounds more reactive and generally less stable than the corresponding carbon compounds. Increased polarity results in less delocalization and makes the planar form of BN less conductive than graphite.
1. Partial pi bonding between O and B increases the B–O bond strength.
1. Periodic trends predict that the cations of the heavier elements should be easier to reduce, so the elements should have less negative reduction potentials. In fact, the reverse is observed because the heavier elements have anomalously high Zeff values due to poor shielding by filled (n − 1)d and (n − 2)f subshells.
1. dsp3
Structure and Reactivity
1. Is B(OH)3 a strong or a weak acid? Using bonding arguments, explain why.
2. Using bonding arguments, explain why organoaluminum compounds are expected to be potent Lewis acids that react rapidly with weak Lewis bases.
3. Imagine that you are studying chemistry prior to the discovery of gallium, element 31. Considering its position in the periodic table, predict the following properties of gallium:
1. chemical formulas of its most common oxide, most common chloride, and most common hydride
2. solubility of its oxide in water and the acidity or basicity of the resulting solution
3. the principal ion formed in aqueous solution
1. The halides of Al, Ga, In, and Tl dissolve in water to form acidic solutions containing the hydrated metal ions, but only the halides of aluminum and gallium dissolve in aqueous base to form soluble metal-hydroxide complexes. Show the formulas of the soluble metal–hydroxide complexes and of the hydrated metal ions. Explain the difference in their reactivities.
2. Complete and balance each chemical equation.
1. BCl3(g) + H2(g) $\xrightarrow{\Delta}$
2. 6C2H4(g) + B2H6(g) →
3. B2H6(g) + 3Cl2(g) →
4. B2H6(g) + 2(C2H5)2S(g) →
1. Complete and balance each chemical equation.
1. BBr3(g) + H2(g) $\xrightarrow{\Delta}$
2. BF3(g) + F(g) →
3. LiH(s) + B2H6(g) →
4. B(OH)3(s) + NaOH(aq) →
1. Write a balanced chemical equation for each reaction.
1. the dissolution of Al2O3 in dilute acid
2. the dissolution of Ga2O3 in concentrated aqueous base
3. the dissolution of Tl2O in concentrated aqueous acid
4. Ga(l) + S8(s)
5. Tl(s) + H2S(g)
1. Write a balanced chemical equation for the reaction that occurs between Al and each species.
1. Cl2
2. O2
3. S
4. N2
5. H2O
6. H+(aq)
1. Write a balanced chemical equation that shows how you would prepare each compound from its respective elements or other common compounds.
1. In2I6
2. B(OH)3
3. Ga2O3
4. [Tl(H2O)6]3+
5. Al(OH)4
6. In4C3
1. Write a balanced chemical equation that shows how you would prepare each compound from its respective elements or other common compounds.
1. BCl3
2. InCl3
3. Tl2S
4. Al(OH)3
5. In2O3
6. AlN
1. Diborane is a spontaneously flammable, toxic gas that is prepared by reacting NaBH4 with BF3. Write a balanced chemical equation for this reaction.
1. Draw the Lewis electron structure of each reactant and product in each chemical equation. Then describe the type of bonding found in each reactant and product.
1. 2B(s) + 3X2(g) $\xrightarrow{\Delta}$ 2BX3(g)
2. 4B(s) + 3O2(g) $\xrightarrow{\Delta}$ 2B2O3(s)
3. 2B(s) + N2(g) $\xrightarrow{\Delta}$ 2BN(s)
1. Draw the Lewis electron structure of each reactant and product in each chemical equation. Then describe the type of bonding found in each reactant and product.
1. B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l)
2. B2H6(g) + 6CH3CH=CH2(g) → 2[B(CH2CH2CH3)3](l)
Answers
1. 12BCl3(g) + 18H2(g) $\xrightarrow{\Delta}$ B12(s) + 36HCl(g)
2. 6C2H4(g) + B2H6(g) → 2B(C2H5)3(l)
3. 6B2H6(g) + 18Cl2(g) → B12(s) + 36HCl(g)
4. B2H6(g) + 2(C2H5)2S(g) → 2H3B·S(C2H5)2(l)
1. Al2O3(s) + 6H3O+(aq) + 3H2O(l) → 2Al(H2O)63+(aq)
2. Ga2O3(s) + 2OH(aq) + 3H2O(l) → 2Ga(OH)4(aq)
3. Tl2O(s) + 2H+(aq) + 9H2O(l) → 2Tl(H2O)6+(aq)
4. 16Ga(l) + 3S8(s) → 8Ga2S3(s)
5. 2Tl(s) + H2S(g) → Tl2S(s) + H2(g)
1. 2In(s) + 3I2(s) $\xrightarrow{\Delta}$ In2I6(s)
2. B12(s) + 18Cl2(g) → 12BCl3(l)
BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
1. 4Ga(l) + 3O2(g) $\xrightarrow{\Delta}$ 2Ga2O3(s)
2. 2Tl(s) + 3Cl2(g) → 2TlCl3(s); TlCl3(s) + 6H2O(l) → Tl(H2O6)3+(aq) + 3Cl(aq)
3. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g); 2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Al(OH)4 + 3H2(g) + 2Na+(aq)
4. 4In(s) + 3C(s) $\xrightarrow{\Delta}$ In4C3(s)
21.5: Group 14: The Carbon Family
Conceptual Problems
1. Why is the preferred oxidation state of lead +2 rather than +4? What do you expect the preferred oxidation state of silicon to be based on its position in the periodic table?
2. Carbon uses pπ–pπ overlap to form compounds with multiple bonds, but silicon does not. Why? How does this same phenomenon explain why the heavier elements in group 14 do not form catenated compounds?
3. Diamond is both an electrical insulator and an excellent thermal conductor. Explain this property in terms of its bonding.
4. The lighter chalcogens (group 16) form π bonds with carbon. Does the strength of these π bonds increase or decrease with increasing atomic number of the chalcogen? Why?
5. The heavier group 14 elements can form complexes that contain expanded coordination spheres. How does this affect their reactivity compared with the reactivity of carbon? Is this a thermodynamic effect or a kinetic effect? Explain your answer.
6. Refer to Table $1$ for the values of the electron affinities of the group 14 elements. Explain any discrepancies between these actual values and the expected values based on usual periodic trends.
7. Except for carbon, the elements of group 14 can form five or six electron-pair bonds. What hybrid orbitals are used to allow this expanded coordination? Why does carbon not form more than four electron-pair bonds?
8. Which of the group 14 elements is least stable in the +4 oxidation state? Why?
Structure and Reactivity
1. Predict the products of each reaction and balance each chemical equation.
1. CaC2(s) + HCl(g) →
2. Pb(s) + Br2(l) $\xrightarrow{\Delta}$
3. (CH3)3N(l) + H2O2(aq) →
4. Pb(N3)2(s) $\xrightarrow{\Delta}$
1. Write a balanced chemical equation to indicate how you would prepare each compound.
1. SiF62− from its elements and other common compounds
2. SiO2 from SiCl4
3. GeS2 from its elements
4. Si(CH3)4 from Si
1. Write a balanced chemical equation to indicate how you would prepare each compound.
1. CO2 from CuO
2. methane from Be2C
3. Si(OH)4 from Si
4. Si3N4 from Si
Answers
1. CaC2(s) + 2HCl(g) → CaCl2(s) + C2H2(g)
2. Pb(s) + Br2(l) $\xrightarrow{\Delta}$ PbBr2(s)
3. (CH3)3N(l) + H2O2(aq) → (CH3)3N–O(l) + H2O(l)
4. Pb(N3)2(s) $\xrightarrow{\Delta}$ Pb(s) + 3N2(g)
1. CuO(s) + CO(s) $\xrightarrow{\Delta}$ Cu(s) + CO2(g)
2. Be2C(s) + 4HCl(aq) → 2BeCl2(aq) + CH4(g)
3. Si(s) + 2Cl2(g) → SiCl4(l); SiCl4(l) + 4H2O(l) → Si(OH)4(s) + 4HCl(aq)
4. 3Si(s) + 2N2(g) $\xrightarrow{\Delta}$ Si3N4(s) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.E%3A_Exercises.txt |
• 22.1: Periodic Trends in Bonding
• 22.2: Group 18 - The Noble Gases
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves.
• 22.3: Group 17: The Halogens
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X− ion. Their chemistry is exclusively that of nonmetals.
• 22.4: Group 16: The Oxygen Family
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds.
• 22.5: Group 15: The Nitrogen Family
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect.
• 22.6: Hydrogen: A Unique Element
• 22.E: Exercises
22: Chemistry of The Main-Group Elements II
Periodic trends affect bonding, because of how the elements are arranged on the periodic table. For example elements can be arranged by their electronegative, electron affinity, atomic radius, or ionization energy. Electronegative is the atoms ability to attract other bonded atoms. Electron affinity is an atoms ability to attract another atom. The atomic radius is the radius of an elements atom. Ionization energy is the energy it takes to remove an atom from another atom. Other periodic trends are when the attraction of the atoms for the pair of bonding electrons is different, this is polar covalent bonds. Properties in compounds are used to determine the type of bonding and structure, not just the elements being used. These different properties help group elements to make them either more available or less available for bonding.
Fluorides
It may seem counterintuitive to say that HF is the weakest hydrohalic acid because fluorine has the highest electronegativity. However, the H-F bond is very strong; if the H-X bond is strong, the resulting acid is weak. A strong bond is determined by a short bond length and a large bond dissociation energy. Of all the hydrogen halides, HF has the shortest bond length and largest bond dissociation energy.
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 $1$ ) 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.
Oxides
Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. These are termed as oxides because here, oxygen is in combination with only one element. Based on their acid-base characteristics oxides are classified as acidic, basic, amphoteric or neural:
1. An oxide that combines with water to give an acid is termed as an acidic oxide.
2. The oxide that gives a base in water is known as a basic oxide.
3. An amphoteric solution is a substance that can chemically react as either acid or base.
4. However, it is also possible for an oxide to be neither acidic nor basic, but is a neutral oxide.
There are different properties which help distinguish between the three types of oxides. The term anhydride ("without water") refers to compounds that assimilate H2O to form either an acid or a base upon the addition of water.
Acidic Oxides
Acidic oxides are the oxides of non-metals (Groups 14-17) and these acid anhydrides form acids with water:
• Sulfurous Acid
$SO_2 + H_2O \rightarrow H_2SO_3 \label{1}$
• Sulfuric Acid
$SO_3 + H_2O \rightarrow H_2SO_4 \label{2}$
• Carbonic Acid
$CO_2 + H_2O \rightarrow H_2CO_3 \label{3}$
Acidic oxides are known as acid anhydrides (e.g., sulfur dioxide is sulfurous anhydride and sulfur trioxide is sulfuric anhydride) and when combined with bases, they produce salts, e.g.,
$SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \label{4}$
Basic Oxides
Generally Group 1 and Group 2 elements form bases called base anhydrides or basic oxides, e.g.,
$K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) \label{5}$
Basic oxides are the oxides of metals. If soluble in water, they react with water to produce hydroxides (alkalies) e.g.,
$CaO + H_2O \rightarrow Ca(OH)_2 \label{6}$
$MgO + H_2O \rightarrow Mg(OH)_2 \label{7}$
$Na_2O + H_2O \rightarrow 2NaOH \label{8}$
These metallic oxides are known as basic anhydrides. They react with acids to produce salts, e.g.,
$MgO + 2HCl \rightarrow MgCl_2 + H_2O \label{9}$
$Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O \label{10}$
Amphoteric Oxides
An amphoteric solution is a substance that can chemically react as either acid or base. For example, when HSO4- reacts with water it will make both hydroxide and hydronium ions:
$HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}$
$HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}$
Amphoteric oxides are metallic oxides, which show both basic as well as acidic properties. When they react with an acid, they produce salt and water, showing basic properties. While reacting with alkalies they form salt and water showing acidic properties, e.g.,
$ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}$
$ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}$
$Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}$
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}$
Amphoteric oxides have both acidic and basic properties. A common example of an amphoteric oxide is aluminum oxide. In general, amphoteric oxides form with metalloids. (see chart below for more detail). Example with acidic properties:
$Al_2O_3 + H_2O \rightarrow 2 Al(OH) + 2H^+ \label{17}$
Example with basic properties:
$Al_2O_3 + H_2O \rightarrow 2Al^{3+} + 3OH^- \label{18}$
Neutral Oxides
Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N2O); nitric oxide (NO), etc., are neutral oxides.
Peroxides and Dioxides
Oxides: Group 1 metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of $M_2O$. With the oyxgen exhibiting an oxidation number of -2.
$4 Li + O_2 \rightarrow 2Li_2O \label{19}$
Peroxides: Often Lithium and Sodium reacts with excess oxygen to produce the peroxide, $M_2O_2$. with the oxidation number of the oxygen equal to -1.
$H_2 + O_2 \rightarrow H_2O_2 \label{20}$
Superoxides: Often Potassium, Rubidium, and Cesium react with excess oxygen to produce the superoxide, $MO_2$. with the oxidation number of the oxygen equal to -1/2.
$Cs + O_2 \rightarrow CsO_2 \label{21}$
A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. They contain more oxygen than the corresponding basic oxide, e.g., sodium, calcium and barium peroxides.
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}$
$Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}$
Dioxides like PbO2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar molecular formulae. These oxides, however, do not give hydrogen peroxide by action with dilute acids. Dioxides on reaction with concentrated HCl yield Cl2 and on reacting with concentrated H2SO4 yield O2.
$PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}$
$2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}$
Compound Oxides
Compound oxides are metallic oxides that behave as if they are made up of two oxides, one that has a lower oxidation and one with a higher oxidation of the same metal, e.g.,
$\textrm{Red lead: } Pb_3O_4 = PbO_2 + 2PbO \label{26}$
$\textrm{Ferro-ferric oxide: } Fe_3O_4 = Fe_2O_3 + FeO \label{27}$
On treatment with an acid, compound oxides give a mixture of salts.
$\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}$
Preparation of Oxides
Oxides can be generated via multiple reactions. Below are a few.
By direct heating of an element with oxygen: Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g.,
$2Mg + O_2 \xrightarrow{Heat} 2MgO$
$2Ca + O_2 \xrightarrow{Heat} 2CaO$
$S + O_2 \xrightarrow{Heat} SO_2$
$P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5$
By reaction of oxygen with compounds at higher temperatures: At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g.,
• sulfides are usually oxidized when heated with oxygen.
$2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$
$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$
• When heated with oxygen, compounds containing carbon and hydrogen are oxidized.
$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$
• By thermal decomposition of certain compounds like hydroxides, carbonates, and nitrates
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$
$2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2$
$Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O$
By oxidation of some metals with nitric acid
$2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2$
$Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O$
By oxidation of some non-metals with nitric acid
$C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O$
Trends in Acid-Base Behavior
The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. For example, in third period, the behavior of oxides changes as follows:
$\underset{\large{Basic}}{\underbrace{Na_2O,\: MgO}}\hspace{20px} \underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px} \underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}$
If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. It may also help to examine the physical properties of oxides, but it is not necessary (Table $1$). Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Na2O and MgO). Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. Cl2O, SO2, P4O10). There is a trend within acid-base behavior: basic oxides are present on the left side of the period and acidic oxides are found on the right side.
Table $1$: Oxides of the s- and p-block elements.basic oxides ( purple), amphoteric oxides (blue), and acidic oxides (red)
1 2 3 14 15 16 17
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
K
Ca
Ga
Ge
As
Se
Br
Rb
Sr
In
Sn
Sb
Te
I
Cs
Ba
Tl
Pb
Bi
Po
At
Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. Thus Al2O3 entails the marking point at which a change over from a basic oxide to acidic oxide occurs. It is important to remember that the trend only applies for oxides in their highest oxidation states. The individual element must be in its highest possible oxidation state because the trend does not follow if all oxidation states are included. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides.
Contributors and Attributions
Binod Shrestha (University of Lorraine) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.1%3A_Periodic_Trends_in_Bonding.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_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.2%3A_Group_18_-_The_Noble_Gases.txt |
Learning Objectives
• To understand the periodic trends and reactivity of the group 17 elements: the halogens.
Because the halogens are highly reactive, none is found in nature as the free element. Hydrochloric acid, which is a component of aqua regia (a mixture of HCl and HNO3 that dissolves gold), and the mineral fluorspar (CaF2) were well known to alchemists, who used them in their quest for gold. Despite their presence in familiar substances, none of the halogens was even recognized as an element until the 19th century.
Because the halogens are highly reactive, none is found in nature as the free element.
Chlorine was the first halogen to be obtained in pure form. In 1774, Carl Wilhelm Scheele (the codiscoverer of oxygen) produced chlorine by reacting hydrochloric acid with manganese dioxide. Scheele was convinced, however, that the pale green gas he collected over water was a compound of oxygen and hydrochloric acid. In 1811, Scheele’s “compound” was identified as a new element, named from the Greek chloros, meaning “yellowish green” (the same stem as in chlorophyll, the green pigment in plants). That same year, a French industrial chemist, Bernard Courtois, accidentally added too much sulfuric acid to the residue obtained from burned seaweed. A deep purple vapor was released, which had a biting aroma similar to that of Scheele’s “compound.” The purple substance was identified as a new element, named iodine from the Greek iodes, meaning “violet.” Bromine was discovered soon after by a young French chemist, Antoine Jérôme Balard, who isolated a deep red liquid with a strong chlorine-like odor from brine from the salt marshes near Montpellier in southern France. Because many of its properties were intermediate between those of chlorine and iodine, Balard initially thought he had isolated a compound of the two (perhaps ICl). He soon realized, however, that he had discovered a new element, which he named bromine from the Greek bromos, meaning “stench.” Currently, organic chlorine compounds, such as PVC (polyvinylchloride), consume about 70% of the Cl2 produced annually; organobromine compounds are used in much smaller quantities, primarily as fire retardants.
Because of the unique properties of its compounds, fluorine was believed to exist long before it was actually isolated. The mineral fluorspar (now called fluorite [CaF2]) had been used since the 16th century as a “flux,” a low-melting-point substance that could dissolve other minerals and ores. In 1670, a German glass cutter discovered that heating fluorspar with strong acid produced a solution that could etch glass. The solution was later recognized to contain the acid of a new element, which was named fluorine in 1812. Elemental fluorine proved to be very difficult to isolate, however, because both HF and F2 are extraordinarily reactive and toxic. After being poisoned three times while trying to isolate the element, the French chemist Henri Moissan succeeded in 1886 in electrolyzing a sample of KF in anhydrous HF to produce a pale green gas (Figure $1$). For this achievement, among others, Moissan narrowly defeated Mendeleev for the Nobel Prize in Chemistry in 1906. Large amounts of fluorine are now consumed in the production of cryolite (Na3AlF6), a key intermediate in the production of aluminum metal. Fluorine is also found in teeth as fluoroapatite [Ca5(PO4)3F], which is formed by reacting hydroxyapatite [Ca5(PO4)3OH] in tooth enamel with fluoride ions in toothpastes, rinses, and drinking water.
The heaviest halogen is astatine (At), which is continuously produced by natural radioactive decay. All its isotopes are highly radioactive, and the most stable has a half-life of only about 8 h. Consequently, astatine is the least abundant naturally occurring element on Earth, with less than 30 g estimated to be present in Earth’s crust at any one time.
Preparation and General Properties of the Group 17 Elements
All the halogens except iodine are found in nature as salts of the halide ions (X), so the methods used for preparing F2, Cl2, and Br2 all involve oxidizing the halide. Reacting CaF2 with concentrated sulfuric acid produces gaseous hydrogen fluoride:
$CaF_{2(s)} + H_2SO_{4(l)} \rightarrow CaSO_{4(s)} + 2HF_{(g)} \label{1}$
Fluorine is produced by the electrolysis of a 1:1 mixture of HF and K+HF2 at 60–300°C in an apparatus made of Monel, a highly corrosion-resistant nickel–copper alloy:
$KHF_2\cdot HF(l) \xrightarrow{electrolysis}F_2(g) + H_2(g) \label{2}$
Fluorine is one of the most powerful oxidants known, and both F2 and HF are highly corrosive. Consequently, the production, storage, shipping, and handling of these gases pose major technical challenges.
Although chlorine is significantly less abundant than fluorine, elemental chlorine is produced on an enormous scale. Fortunately, large subterranean deposits of rock salt (NaCl) are found around the world (Figure $2$), and seawater consists of about 2% NaCl by mass, providing an almost inexhaustible reserve. Inland salt lakes such as the Dead Sea and the Great Salt Lake are even richer sources, containing about 23% and 8% NaCl by mass, respectively. Chlorine is prepared industrially by the chloralkali process, which uses the following reaction:
$2NaCl_{(aq)} +2H_2O_{(l)} \xrightarrow{electrolysis} 2NaOH(aq) + Cl_{2(g)} + H_{2(g)} \label{3}$
Bromine is much less abundant than fluorine or chlorine, but it is easily recovered from seawater, which contains about 65 mg of Br per liter. Salt lakes and underground brines are even richer sources; for example, the Dead Sea contains 4 g of Br per liter. Iodine is the least abundant of the nonradioactive halogens, and it is a relatively rare element. Because of its low electronegativity, iodine tends to occur in nature in an oxidized form. Hence most commercially important deposits of iodine, such as those in the Chilean desert, are iodate salts such as Ca(IO3)2. The production of iodine from such deposits therefore requires reduction rather than oxidation. The process is typically carried out in two steps: reduction of iodate to iodide with sodium hydrogen sulfite, followed by reaction of iodide with additional iodate:
$2IO^−_{3(aq)} + 6HSO^−_{3(aq)} \rightarrow 2I^−_{(aq)} + 6SO^2−_{4(aq)} + 6H^+_{(aq)} \label{4}$
$5I^−_{(aq)} + IO^−_{3(aq)} + 6H^+_{(aq)} \rightarrow 3I_{2(s)} + 3H_2O_{(l)} \label{5}$
Because the halogens all have ns2np5 electron configurations, their chemistry is dominated by a tendency to accept an additional electron to form the closed-shell ion (X). Only the electron affinity and the bond dissociation energy of fluorine differ significantly from the expected periodic trends shown in Table $1$. Electron–electron repulsion is important in fluorine because of its small atomic volume, making the electron affinity of fluorine less than that of chlorine. Similarly, repulsions between electron pairs on adjacent atoms are responsible for the unexpectedly low F–F bond dissociation energy. (As discussed earlier, this effect is also responsible for the weakness of O–O, N–N, and N–O bonds.)
Electrostatic repulsions between lone pairs of electrons on adjacent atoms cause single bonds between N, O, and F to be weaker than expected.
Table $1$: Selected Properties of the Group 17 Elements
Property Fluorine Chlorine Bromine Iodine Astatine
*The configuration shown does not include filled d and f subshells. The values cited are for the six-coordinate anion (X−).
atomic symbol F Cl Br I At
atomic number 9 17 35 53 85
atomic mass (amu) 19.00 35.45 79.90 126.90 210
valence electron configuration* 2s22p5 3s23p5 4s24p5 5s25p5 6s26p5
melting point/boiling point (°C) −220/−188 −102/−34.0 −7.2/58.8 114/184 302/—
density (g/cm3) at 25°C 1.55 (g/L) 2.90 (g/L) 3.10 4.93
atomic radius (pm) 42 79 94 115 127
first ionization energy (kJ/mol) 1681 1251 1140 1008 926
normal oxidation state(s) −1 −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1 (+1, +3, +5, +7) −1, +1
ionic radius (pm) 133 181 196 220
electron affinity (kJ/mol) −328 −349 −325 −295 −270
electronegativity 4.0 3.2 3.0 2.7 2.2
standard reduction potential (E°, V) (X2 → X in basic solution) +2.87 +1.36 +1.07 +0.54 +0.30
dissociation energy of X2(g) (kJ/mol) 158.8 243.6 192.8 151.1 ~80
product of reaction with O2 O2F2 none none none none
type of oxide acidic acidic acidic acidic acidic
product of reaction with N2 none none none none none
product of reaction with H2 HF HCl HBr HI HAt
Because it is the most electronegative element in the periodic table, fluorine forms compounds in only the −1 oxidation state. Notice, however, that all the halogens except astatine have electronegativities greater than 2.5, making their chemistry exclusively that of nonmetals. The halogens all have relatively high ionization energies, but the energy required to remove electrons decreases substantially as we go down the column. Hence the heavier halogens also form compounds in positive oxidation states (+1, +3, +5, and +7), derived by the formal loss of ns and np electrons.
Because ionization energies decrease down the group, the heavier halogens form compounds in positive oxidation states (+1, +3, +5, and +7).
Reactions and Compounds of the Halogens
Fluorine is the most reactive element in the periodic table, forming compounds with every other element except helium, neon, and argon. The reactions of fluorine with most other elements range from vigorous to explosive; only O2, N2, and Kr react slowly. There are three reasons for the high reactivity of fluorine:
1. Because fluorine is so electronegative, it is able to remove or at least share the valence electrons of virtually any other element.
2. Because of its small size, fluorine tends to form very strong bonds with other elements, making its compounds thermodynamically stable.
3. The F–F bond is weak due to repulsion between lone pairs of electrons on adjacent atoms, reducing both the thermodynamic and kinetic barriers to reaction.
With highly electropositive elements, fluorine forms ionic compounds that contain the closed-shell F ion. In contrast, with less electropositive elements (or with metals in very high oxidation states), fluorine forms covalent compounds that contain terminal F atoms, such as SF6. Because of its high electronegativity and 2s22p5 valence electron configuration, fluorine normally participates in only one electron-pair bond. Only a very strong Lewis acid, such as AlF3, can share a lone pair of electrons with a fluoride ion, forming AlF63−.
Oxidative strength decreases down group 17.
The halogens (X2) react with metals (M) according to the general equation
$M_{(s,l)} + nX_{2(s,l,g)} \rightarrow MX_{n(s,l)} \label{6}$
For elements that exhibit multiple oxidation states fluorine tends to produce the highest possible oxidation state and iodine the lowest. For example, vanadium reacts with the halogens to give VF5, VCl4, VBr4, and VI3.
Metal halides in the +1 or +2 oxidation state, such as CaF2, are typically ionic halides, which have high melting points and are often soluble in water. As the oxidation state of the metal increases, so does the covalent character of the halide due to polarization of the M–X bond. With its high electronegativity, fluoride is the least polarizable, and iodide, with the lowest electronegativity, is the most polarizable of the halogens. Halides of small trivalent metal ions such as Al3+ tend to be relatively covalent. For example, AlBr3 is a volatile solid that contains bromide-bridged Al2Br6 molecules. In contrast, the halides of larger trivalent metals, such as the lanthanides, are essentially ionic. For example, indium tribromide (InBr3) and lanthanide tribromide (LnBr3) are all high-melting-point solids that are quite soluble in water.
As the oxidation state of the metal increases, the covalent character of the corresponding metal halides also increases due to polarization of the M–X bond.
All halogens react vigorously with hydrogen to give the hydrogen halides (HX). Because the H–F bond in HF is highly polarized (Hδ+–Fδ−), liquid HF has extensive hydrogen bonds, giving it an unusually high boiling point and a high dielectric constant. As a result, liquid HF is a polar solvent that is similar in some ways to water and liquid ammonia; after a reaction, the products can be recovered simply by evaporating the HF solvent. (Hydrogen fluoride must be handled with extreme caution, however, because contact of HF with skin causes extraordinarily painful burns that are slow to heal.) Because fluoride has a high affinity for silicon, aqueous hydrofluoric acid is used to etch glass, dissolving SiO2 to give solutions of the stable SiF62− ion.
Glass etched with hydrogen flouride.© Thinkstock
Except for fluorine, all the halogens react with water in a disproportionation reaction, where X is Cl, Br, or I:
$X_{2(g,l,s)} + H_2O_{(l)} \rightarrow H^+_{(aq)} + X^−_{(aq)} + HOX_{(aq)} \label{7}$
The most stable oxoacids are the perhalic acids, which contain the halogens in their highest oxidation state (+7). The acid strengths of the oxoacids of the halogens increase with increasing oxidation state, whereas their stability and acid strength decrease down the group. Thus perchloric acid (HOClO3, usually written as HClO4) is a more potent acid and stronger oxidant than perbromic acid. Although all the oxoacids are strong oxidants, some, such as HClO4, react rather slowly at low temperatures. Consequently, mixtures of the halogen oxoacids or oxoanions with organic compounds are potentially explosive if they are heated or even agitated mechanically to initiate the reaction. Because of the danger of explosions, oxoacids and oxoanions of the halogens should never be allowed to come into contact with organic compounds.
Both the acid strength and the oxidizing power of the halogen oxoacids decrease down the group.
The halogens react with one another to produce interhalogen compounds, such as ICl3, BrF5, and IF7. In all cases, the heavier halogen, which has the lower electronegativity, is the central atom. The maximum oxidation state and the number of terminal halogens increase smoothly as the ionization energy of the central halogen decreases and the electronegativity of the terminal halogen increases. Thus depending on conditions, iodine reacts with the other halogens to form IFn (n = 1–7), ICl or ICl3, or IBr, whereas bromine reacts with fluorine to form only BrF, BrF3, and BrF5 but not BrF7. The interhalogen compounds are among the most powerful Lewis acids known, with a strong tendency to react with halide ions to give complexes with higher coordination numbers, such as the IF8 ion:
$IF_{7(l)} + KF_{(s)} \rightarrow KIF_{8(s)} \label{8}$
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7). The interhalogen compounds are also potent oxidants and strong fluorinating agents; contact with organic materials or water can result in an explosion.
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7), but the importance of the higher oxidation states generally decreases down the group.
Example $1$
For each reaction, explain why the given products form.
1. ClF3(g) + Cl2(g) → 3ClF(g)
2. 2KI(s) + 3H2SO4(aq) → I2(aq) + SO2(g) + 2KHSO4(aq) + 2H2O(l)
3. Pb(s) + 2BrF3(l) → PbF4(s) + 2BrF(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. When the reactants have the same element in two different oxidation states, we expect the product to have that element in an intermediate oxidation state. We have Cl3+ and Cl0 as reactants, so a possible product would have Cl in either the +1 or +2 oxidation state. From our discussion, we know that +1 is much more likely. In this case, Cl2 is behaving like a reductant rather than an oxidant.
2. At first glance, this appears to be a simple acid–base reaction, in which sulfuric acid transfers a proton to I to form HI. Recall, however, that I can be oxidized to I2. Sulfuric acid contains sulfur in its highest oxidation state (+6), so it is a good oxidant. In this case, the redox reaction predominates.
3. This is the reaction of a metallic element with a very strong oxidant. Consequently, a redox reaction will occur. The only question is whether lead will be oxidized to Pb(II) or Pb(IV). Because BrF3 is a powerful oxidant and fluorine is able to stabilize high oxidation states of other elements, it is likely that PbF4 will be the product. The two possible reduction products for BrF3 are BrF and Br2. The actual product will likely depend on the ratio of the reactants used. With excess BrF3, we expect the more oxidized product (BrF). With lower ratios of oxidant to lead, we would probably obtain Br2 as the product.
Exercise $1$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. CaCl2(s) + H3PO4(l) →
2. GeO2(s) + HF(aq) →
3. Fe2O3(s) + HCl(g) $\xrightarrow{\Delta}$
4. NaClO2(aq) + Cl2(g) →
Answer
1. CaCl2(s) + H3PO4(l) → 2HCl(g) + Ca(HPO4)(soln)
2. GeO2(s) + 6HF(aq) → GeF62−(aq) + 2H2O(l) + 2H+(aq)
3. Fe2O3(s) + 6HCl(g) $\xrightarrow{\Delta}$ 2FeCl3(s) + 3H2O(g)
4. 2NaClO2(aq) + Cl2(g) → 2ClO2(g) + 2NaCl(aq)
Summary
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group. Fluorine, the most reactive element in the periodic table, has a low F–F bond dissociation energy due to repulsions between lone pairs of electrons on adjacent atoms. Fluorine forms ionic compounds with electropositive elements and covalent compounds with less electropositive elements and metals in high oxidation states. All the halogens react with hydrogen to produce hydrogen halides. Except for F2, all react with water to form oxoacids, including the perhalic acids, which contain the halogens in their highest oxidation state. Halogens also form interhalogen compounds; the heavier halogen, with the lower electronegativity, is the central atom. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.3%3A_Group_17%3A_The_Halogens.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 16 elements: the chalcogens.
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element.
Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility.
Group 16 is the first group in the p block with no stable metallic elements.
Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics.
Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices.
Jöns Jakob Berzelius (1779–1848)
Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium).
The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium.
Preparation and General Properties of the Group 16 Elements
Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2:
$\mathrm{2KClO_3(s)\overset{MnO_2(s)}{\underset{\Delta}\rightleftharpoons}2KCl(s)+3O_2(g)} \label{22.4.1}$
Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure $1$). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2).
Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.”
With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure $1$. As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions.
Table $1$: Selected Properties of the Group 16 Elements
Property Oxygen Sulfur Selenium Tellurium Polonium
*The configuration shown does not include filled d and f subshells.
The values cited for the hexacations are for six-coordinate ions and are only estimated values.
atomic mass (amu) 16.00 32.07 78.96 127.60 209
atomic number 8 16 34 52 84
atomic radius (pm) 48 88 103 123 135
atomic symbol O S Se Te Po
density (g/cm3) at 25°C 1.31 (g/L) 2.07 4.81 6.24 9.20
electron affinity (kJ/mol) −141 −200 −195 −190 −180
electronegativity 3.4 2.6 2.6 2.1 2.0
first ionization energy (kJ/mol) 1314 1000 941 869 812
ionic radius (pm) 140 (−2) 184 (−2), 29 (+6) 198 (−2), 42 (+6) 221 (−2), 56 (+6) 230 (−2), 97 (+4)
melting point/boiling point (°C) −219/−183 115/445 221/685 450/988 254/962
normal oxidation state(s) −2 +6, +4, −2 +6, +4, −2 +6, +4, −2 +2 (+4)
product of reaction with H2 H2O H2S H2Se none none
product of reaction with N2 NO, NO2 none none none none
product of reaction with O2 SO2 SeO2 TeO2 PoO2
product of reaction with X2 O2F2 SF6, S2Cl2, S2Br2 SeF6, SeX4 TeF6, TeX4 PoF4, PoCl2, PoBr2
standard reduction potential (E°, V) (E0 → H2E in acidic solution) +1.23 +0.14 −0.40 −0.79 −1.00
type of oxide acidic acidic amphoteric basic
valence electron configuration* 2s22p4 3s23p4 4s24p4 5s25p4 6s26p4
Reactions and Compounds of Oxygen
As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure:
$2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)} \;\;\; ΔG^o = −119\; kJ/mol \label{1}$
As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds.
Despite the strength of the O=O bond ($D_\mathrm{O_2}$ = 494 kJ/mol), $O_2$ is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in Table $2$. With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens.
Table $2$: Some Properties of O2 and Related Diatomic Species
Species Bond Order Number of Unpaired e O–O Distance (pm)*
*Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175.
O2+ 2.5 1 112
O2 2 2 121
O2 1.5 1 133
O22− 1 0 149
Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid:
$H_2O_{(l)} + SO_{3(g)} \rightarrow H_2SO_{4(aq)} \label{2}$
The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect:
$Al_2O_{3(s)} + 6H^+_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3H_2O_{(l)} \label{3}$
$Al_2O_{3(s)} + 2OH^−_{(aq)} + 3H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} \label{4}$
Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric.
Example $1$
For each reaction, explain why the given products form.
1. Ga2O3(s) + 2OH(aq) + 3H2O(l) → 2Ga(OH)4(aq)
2. 3H2O2(aq) + 2MnO4(aq) + 2H+(aq) → 3O2(g) + 2MnO2(s) + 4H2O(l)
3. KNO3(s) $\xrightarrow{\Delta}$ KNO(s) + O2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Gallium is a metal. We expect the oxides of metallic elements to be basic and therefore not to react with aqueous base. A close look at the periodic table, however, shows that gallium is close to the diagonal line of semimetals. Moreover, aluminum, the element immediately above gallium in group 13, is amphoteric. Consequently, we predict that gallium will behave like aluminum (Equation $\ref{4}$).
2. Hydrogen peroxide is an oxidant that can accept two electrons per molecule to give two molecules of water. With a strong oxidant, however, H2O2 can also act as a reductant, losing two electrons (and two protons) to produce O2. Because the other reactant is permanganate, which is a potent oxidant, the only possible reaction is a redox reaction in which permanganate is the oxidant and hydrogen peroxide is the reductant. Recall that reducing permanganate often gives MnO2, an insoluble brown solid. Reducing MnO4 to MnO2 is a three-electron reduction, whereas the oxidation of H2O2 to O2 is a two-electron oxidation.
3. This is a thermal decomposition reaction. Because KNO3 contains nitrogen in its highest oxidation state (+5) and oxygen in its lowest oxidation state (−2), a redox reaction is likely. Oxidation of the oxygen in nitrate to atomic oxygen is a two-electron process per oxygen atom. Nitrogen is likely to accept two electrons because oxoanions of nitrogen are known only in the +5 (NO3) and +3 (NO2) oxidation states.
Exercise $2$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. SiO2(s) + H+(aq) →
2. NO(g) + O2(g) →
3. SO3(g) + H2O(l) →
4. H2O2(aq) + I(aq) →
Answer
1. SiO2(s) + H+(aq) → no reaction
2. 2NO(g) + O2(g) → 2NO2(g)
3. SO3(g) + H2O(l) → H2SO4(aq)
4. H2O2(aq) + 2I(aq) → I2(aq) + 2OH(aq)
Reactions and Compounds of the Heavier Chalcogens
Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column.
Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (Figure $2$), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group.
As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion.
Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group.
Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure $2$). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known.
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate.
The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid.
The stability of the highest oxidation state of the chalcogens decreases down the column.
Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5.
The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group.
Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous.
Example $2$
For each reaction, explain why the given product forms or no reaction occurs.
1. SO2(g) + Cl2(g) → SO2Cl2(l)
2. SF6(g) + H2O(l) → no reaction
3. 2Se(s) + Cl2(g) → Se2Cl2(l)
Given: balanced chemical equations
Asked for: why the given products (or no products) form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs.
Solution
1. One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds.
1. Sulfur hexafluoride is a nonmetallic halide. Such compounds normally react vigorously with water to produce an oxoacid of the nonmetal and the corresponding hydrohalic acid. In this case, however, we have a highly stable species, presumably because all of sulfur’s available orbitals are bonding orbitals. Thus SF6 is not likely to react with water.
2. Here we have the reaction of a chalcogen with a halogen. The halogen is a good oxidant, so we can anticipate that a redox reaction will occur. Only fluorine is capable of oxidizing the chalcogens to a +6 oxidation state, so we must decide between SeCl4 and Se2Cl2 as the product. The stoichiometry of the reaction determines which of the two is obtained: SeCl4 or Se2Cl2.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. Te(s) + Na(s) $\xrightarrow{\Delta}$
2. SF4(g) + H2O(l) →
3. CH3SeSeCH3(soln) + K(s) →
4. Li2Se(s) + H+(aq) →
Answer
1. Te(s) + 2Na(s) → Na2Te(s)
2. SF4(g) + 3H2O(l) → H2SO3(aq) + 4HF(aq)
3. CH3SeSeCH3(soln) + 2K(s) → 2KCH3Se(soln)
4. Li2Se(s) + 2H+(aq) → H2Se(g) + 2Li+(aq)
Summary
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.4%3A_Group_16%3A_The_Oxygen_Family.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 15 elements: the pnicogens.
Like the group 14 elements, the lightest member of group 15, nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores. Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb2S3), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (Figure $1$). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells.
In the form of its yellow sulfide ore, orpiment (As2S3), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting.
Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!).
Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry.
Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO2), noticed that a gas remained when CO2 was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO3 (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals.
Preparation and General Properties of the Group 15 Elements
Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide:
$\mathrm{2NaN_3(s)\xrightarrow{\Delta}2Na(l)+3N_2(g)} \label{Eq1}$
In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO3 and NaNO3 in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it.
The other three pnicogens are much less abundant: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M2S3), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals.
In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (Table $1$). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns2np3 valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH3 and PH3), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases.
Table $1$: Selected Properties of the Group 15 Elements
Property Nitrogen Phosphorus Arsenic Antimony Bismuth
*The configuration shown does not include filled d and f subshells. For white phosphorus. For gray arsenic. §The values cited are for six-coordinate ions in the indicated oxidation states. The N5+, P5+, and As5+ ions are not known species. ||The chemical form of the elements in these oxidation states varies considerably. For N, the reaction is NO3 + 3H+ + 2e HNO2 + H2O; for P and As, it is $\ce{H3EO4 + 2H^{+} + 2e^{−} → H3EO3 + H2O}$; and for Sb it is $\ce{Sb2O5 + 4e^{-} + 10H^{+} → 2Sb^{3+} + 5H2O}$.
atomic symbol N P As Sb Bi
atomic number 7 15 33 51 83
atomic mass (amu) 14.01 30.97 74.92 121.76 209.98
valence electron configuration* 2s22p3 3s23p3 4s24p3 5s25p3 6s26p3
melting point/boiling point (°C) −210/−196 44.15/281c 817 (at 3.70 MPa)/603 (sublimes) 631/1587 271/1564
density (g/cm3) at 25°C 1.15 (g/L) 1.82 5.75 6.68 9.79
atomic radius (pm) 56 98 114 133 143
first ionization energy (kJ/mol) 1402 1012 945 831 703
common oxidation state(s) −3 to +5 +5, +3, −3 +5, +3 +5, +3 +3
ionic radius (pm)§ 146 (−3), 16 (+3) 212 (−3), 44 (+3) 58 (+3) 76 (+3), 60 (+5) 103 (+3)
electron affinity (kJ/mol) 0 −72 −78 −101 −91
electronegativity 3.0 2.2 2.2 2.1 1.9
standard reduction potential (E°, V) (EV → EIII in acidic solution)|| +0.93 −0.28 +0.56 +0.65
product of reaction with O2 NO2, NO P4O6, P4O10 As4O6 Sb2O5 Bi2O3
type of oxide acidic (NO2), neutral (NO, N2O) acidic acidic amphoteric basic
product of reaction with N2 none none none none
product of reaction with X2 none PX3, PX5 AsF5, AsX3 SbF5, SbCl5, SbBr3, SbI3 BiF5, BiX3
product of reaction with H2 none none none none none
In group 15, the stability of the +5 oxidation state decreases from P to Bi.
Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases.
Reactions and Compounds of Nitrogen
Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp3 hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N3).
Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N2, whose N≡N bond is so strong (DN≡N = 942 kJ/mol) compared with the N–N and N=N bonds (DN–N = 167 kJ/mol; DN=N = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N2. In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive.
Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. At higher temperatures, however, N2 reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances.
Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N2 reacts with H2 to form ammonia, with O2 to form a mixture of NO and NO2, and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX3) are known. Except for NF3, all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH3 rather than N2. Both nitrogen monoxide (NO) and nitrogen dioxide (NO2) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO2 reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO2) and nitric acid (HNO3):
$\ce{2NO2(g) + H2O(l) \rightarrow HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms N2O (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with CO2 and can be represented as N=N+=O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows:
Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements.
At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as Li3N and Ca3N2. These compounds consist of ionic lattices formed by Mn+ and N3− ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity.
Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as Si3N4 and BN, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials.
Ammonia (NH3) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an O2 atmosphere:
$\ce{4NH3(g) + 3O2(g) \rightarrow 2N2(g) + 6H2O(g)} \label{Eq3}$
About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer.
Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid (HN3), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine (N2H4) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers.
B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials.
Example $1$
For each reaction, explain why the given products form when the reactants are heated.
1. Sr(s) + N2O(g) $\xrightarrow{\Delta}$ SrO(s) + N2(g)
2. NH4NO2(s) $\xrightarrow{\Delta}$ N2(g) + 2H2O(g)
3. Pb(NO3)2(s) $\xrightarrow{\Delta}$ PbO2(s) + 2NO2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. As an alkali metal, strontium is a strong reductant. If the other reactant can act as an oxidant, then a redox reaction will occur. Nitrous oxide contains nitrogen in a low oxidation state (+1), so we would not normally consider it an oxidant. Nitrous oxide is, however, thermodynamically unstable (ΔH°f > 0 and ΔG°f > 0), and it can be reduced to N2, which is a stable species. Consequently, we predict that a redox reaction will occur.
2. When a substance is heated, a decomposition reaction probably will occur, which often involves the release of stable gases. In this case, ammonium nitrite contains nitrogen in two different oxidation states (−3 and +3), so an internal redox reaction is a possibility. Due to its thermodynamic stability, N2 is the probable nitrogen-containing product, whereas we predict that H and O will combine to form H2O.
3. Again, this is probably a thermal decomposition reaction. If one element is in an usually high oxidation state and another in a low oxidation state, a redox reaction will probably occur. Lead nitrate contains the Pb2+ cation and the nitrate anion, which contains nitrogen in its highest possible oxidation state (+5). Hence nitrogen can be reduced, and we know that lead can be oxidized to the +4 oxidation state. Consequently, it is likely that lead(II) nitrate will decompose to lead(IV) oxide and nitrogen dioxide when heated. Even though PbO2 is a powerful oxidant, the release of a gas such as NO2 can often drive an otherwise unfavorable reaction to completion (Le Chatelier’s principle). Note, however, that PbO2 will probably decompose to PbO at high temperatures.
Exercise 23.3.1
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$
2. NH4NO3(s) $\xrightarrow{\Delta}$
3. Sr(s) + N2(g) →
Answer
1. NO(g) + H2O(l) $\xrightarrow{\Delta}$ no reaction
2. NH4NO3(s) $\xrightarrow{\Delta}$ N2O(g) + 2H2O(g)
3. 3Sr(s) + N2(g) → Sr3N2(s)
Reactions and Compounds of the Heavier Pnicogens
Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P4 tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors.
As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group.
The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O2, whereas arsenic burns in pure O2 if ignited, and antimony and bismuth react with O2 only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO3. Only bismuth behaves like a metal, dissolving in HNO3 to give the hydrated Bi3+ cation.
The reactivity of the heavier group 15 elements decreases as we go down the column.
The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp3 or d2sp3 hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl5) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H3AsO4). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base:
$\ce{AsF5(soln) + F^{−}(soln) \rightarrow AsF^{−}6(soln)} \label{Eq4}$
In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth.
Except for BiF3, which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, $\ce{H3PO3}$ and $\ce{H3AsO3}$, where E is P or As:
$\ce{EX3(l) + 3H2O(l) \rightarrow H3EO3(aq) + 3HX(aq)} \label{Eq5}$
Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers.
Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C.
With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P4O6 and As4O6 have cage structures formed by inserting an oxygen atom into each edge of the P4 or As4 tetrahedron (part (a) in Figure $2$), and they behave like typical nonmetal oxides. For example, P4O6 reacts with water to form phosphorous acid (H3PO3). Consistent with its position between the nonmetal and metallic oxides, Sb4O6 is amphoteric, dissolving in either acid or base. In contrast, Bi2O3 behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi3+ ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E4O10 in the +5 oxidation state (part (b) in Figure $2$. In contrast, Bi2O5 is so unstable that there is no absolute proof it exists.
The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P4S3 (part (c) in Figure $2$), to layered or ribbon structures, such as Sb2S3 and Bi2S3, which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M4P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP15) are lower melting and less thermally stable because they contain catenated Pn units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth.
The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{Bi(s) +\frac{3}{2}Br(l)\rightarrow BiBr_3(s)}$
2. 2(CH3)3As(l) + O2(g) → 2(CH3)3As=O(s)
3. PBr3(l) + 3H2O(l) → H3PO3(aq) + 3HBr(aq)
4. As(s) + Ga(s) $\xrightarrow{\Delta}$ GaAs(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Bromine is an oxidant, and bismuth is a metal that can be oxidized. Hence a redox reaction is likely to occur. To identify the product, recall that bismuth can form compounds in either the +3 or +5 oxidation state. The heaviest pnicogen, bismuth is rather difficult to oxidize to the +5 oxidation state because of the inert-pair effect. Hence the product will probably be bismuth(III) bromide.
2. Trimethylarsine, with a lone pair of electrons on the arsenic atom, can act as either a Lewis base or a reductant. If arsenic is oxidized by two electrons, then oxygen must be reduced, most probably by two electrons to the −2 oxidation state. Because As(V) forms strong bonds to oxygen due to π bonding, the expected product is (CH3)3As=O.
3. Phosphorus tribromide is a typical nonmetal halide. We expect it to react with water to produce an oxoacid of P(III) and the corresponding hydrohalic acid.Because of the strength of the P=O bond, phosphorous acid (H3PO3) is actually HP(O)(OH)2, which contains a P=O bond and a P–H bond.
4. Gallium is a metal with a strong tendency to act as a reductant and form compounds in the +3 oxidation state. In contrast, arsenic is a semimetal. It can act as a reductant to form compounds in the +3 or +5 oxidation state, or it can act as an oxidant, accepting electrons to form compounds in the −3 oxidation state. If a reaction occurs, then a binary compound will probably form with a 1:1 ratio of the elements. GaAs is an example of a III-V compound, many of which are used in the electronics industry.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. PCl5(s) + H2O(l) →
2. Bi2O5(s) $\xrightarrow{\Delta}$
3. Ca3P2(s) + H+(aq) →
4. NaNH2(s) + PH3(soln) →
Answer
1. PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
2. Bi2O5(s) $\xrightarrow{\Delta}$ Bi2O3(s) + O2(g)
3. Ca3P2(s) + 6H+(aq) → 2PH3(g) + 3Ca2+(aq)
4. NaNH2(s) + PH3(soln) → NaPH2(s) + NH3(soln)
Summary
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp3 or d2sp3 hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.5%3A_Group_15%3A_The_Nitrogen_Family.txt |
Learning Objectives
• To describe the physical and chemical properties of hydrogen and predict its reactivity.
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s1 electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in Group 17 because the addition of a single electron to a hydrogen atom completes its valence shell.
Although hydrogen has an ns1 electron configuration, its chemistry does not resemble that of the Group 1 metals.
Isotopes of Hydrogen
Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table $1$ "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is protium (1H or H), followed by deuterium (2H or D), which has an additional neutron. The rarest isotope of hydrogen is tritium (3H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to 3He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes.
Table $1$: The Isotopes of Hydrogen
Protium Deuterium Tritium
symbol $\mathrm{_1^1H}$ $\mathrm{_1^2H}$ $\mathrm{_1^3H}$
neutrons 0 1 2
mass (amu) 1.00783 2.0140 3.01605
abundance (%) 99.9885 0.0115 ~10−17
half-life (years) 12.32
boiling point of X2 (K) 20.28 23.67 25
melting point/boiling point of X2O (°C) 0.0/100.0 3.8/101.4 4.5/?
The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H2, D2, and T2 differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction.
Harold Urey (1893–1981)
Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium (2H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth.
Because the normal boiling point of D2O is 101.4°C (compared to 100.0°C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D+) has twice the mass of a proton (H+), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H2 over D2 or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D2O (1.1044 g/cm3 at 25°C) is greater than that of H2O (0.99978 g/cm3). Heavy water was an important constituent of early nuclear reactors.
Because deuterons diffuse so much more slowly, D2O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D+ is so slow compared with that of H+ because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions.
Bonding in Hydrogen and Hydrogen-Containing Compounds
The 1s1 electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure $1$):
1. Losing its electron to form a proton (H+) with an empty 1s orbital. The proton is a Lewis acid that can accept a pair of electrons from another atom to form an electron-pair bond. In the acid–base reactions, the proton always binds to a lone pair of electrons on an atom in another molecule to form a polar covalent bond. If the lone pair of electrons belongs to an oxygen atom of a water molecule, the result is the hydronium ion (H3O+).
2. Accepting an electron to form a hydride ion (H), which has a filled 1s2 orbital. Hydrogen reacts with relatively electropositive metals, such as the alkali metals (group 1) and alkaline earth metals (group 2), to form ionic hydrides, which contain metal cations and H ions.
3. Sharing its electron with an electron on another atom to form an electron-pair bond. With a half-filled 1s1 orbital, the hydrogen atom can interact with singly occupied orbitals on other atoms to form either a covalent or a polar covalent electron-pair bond, depending on the electronegativity of the other atom.
Hydrogen can also act as a bridge between two atoms. One familiar example is the hydrogen bond, an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure $2$). An example of this kind of interaction is the hydrogen bonding network found in water (Figure $2$). Hydrogen can also form a three-center bond (or electron-deficient bond), in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds.
Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp3 orbitals on two boron atoms (Figure $3$). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures.
Hydrogen can lose its electron to form H+, accept an electron to form H, share its electron, hydrogen bond, or form a three-center bond.
Synthesis, Reactions, and Compounds of Hydrogen
The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure $4$). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K).
The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid:
$M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}$
Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base:
$\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}$
Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D2O:
$MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}$
On an industrial scale, H2 is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H2 known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst:
$\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}$
Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H2(g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities.
Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework.
Summary and Key Takeaway
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H+), a hydride ion (H), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.6%3A_Hydrogen%3A_A_Unique_Element.txt |
22.1: Periodic Trends in Bonding
Problems
1. Can an oxide be neither acidic nor basic?
2. $Rb + O_2\: (excess) \rightarrow \:?$
3. $Na + O_2 \rightarrow \:?$
4. BaO2 is which of the following: hydroxide, peroxide, or superoxide?
5. What is an amphoteric solution?
6. Why is it difficult to obtain oxygen directly from water?
Solutions
1. Yes, an example is carbon monoxide (CO). CO doesn’t produce a salt when reacted with an acid or a base.
2. $Rb + O_2 \; (excess) \rightarrow RbO_2$
With the presence of excess oxygen, Rubidium forms a superoxide. Please review section regarding basic oxides above for more detail.
3. $2 Na + O_2 \rightarrow Na_2O$
Note: The problem does not specify that the oxygen was in excess, so it cannot be a peroxide. Please review section regarding basic oxides for more detail.
4. BaO2 is a peroxide. Barium has an oxidation state of +2 so the oxygen atoms have oxidation state of -1. As a result, the compound is a peroxide, but more specifically referred to as barium peroxide.
5. An amphoteric solution is a substance that can chemically react as either acid or base. See section above on Properties of Amphoteric Oxides for more detail.
6. Water as such is a neutral stable molecule. It is difficult to break the covalent O-H bonds easily. Hence, electrical energy through the electrolysis process is applied to separate dioxygen from water. When a small amount of acid is added to water ionization is initiated which helps in electrochemical reactions as follows.
$[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4$
At cathode:
$[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4$
At anode:
$4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^-$
Net reaction:
$2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g)$
Oxygen can thus be obtained from acidified water by its electrolysis.
22.2: Group 18: The Noble Gases
Conceptual Problems
1. The chemistry of the noble gases is largely dictated by a balance between two competing properties. What are these properties? How do they affect the reactivity of these elements?
2. Of the group 18 elements, only krypton, xenon, and radon form stable compounds with other atoms and then only with very electronegative elements. Why?
3. Give the type of hybrid orbitals used by xenon in each species.
1. XeF2
2. XeF4
3. XeO3
4. XeOF4
5. XeO4
6. XeO64−
1. Which element is the least metallic—B, Ga, Tl, Pb, Ne, or Ge?
2. Of Br, N, Ar, Bi, Se, He, and S, which would you expect to form positive ions most easily? negative ions most easily?
3. Of BCl3, BCl4, CH4, H3N·BF3, PCl3, PCl5, XeO3, H2O, and F, which species do you expect to be
1. electron donors?
2. electron acceptors?
3. neither electron donors nor acceptors?
4. both electron donors and acceptors?
1. Of HCl, HClO4, HBr, H2S, HF, KrF2, and PH3, which is the strongest acid?
2. Of CF4, NH3, NF3, H2O, OF2, SiF4, H2S, XeF4, and SiH4, which is the strongest base?
Structure and Reactivity
1. Write a balanced chemical equation showing how you would prepare each compound from its elements and other common compounds.
1. XeF2
2. XeF4
3. XeF6
4. XeOF4
5. XeO3
1. Write a balanced chemical equation showing how you would make each compound.
1. XeF2 from Xe gas
2. NaXeF7 from its elements
3. RnO3 from Rn
1. In an effort to synthesize XeF6, a chemist passed fluorine gas through a glass tube containing xenon gas. However, the product was not the one expected. What was the actual product?
2. Write a balanced chemical equation to describe the reaction of each species with water.
1. B2H6
2. F2
3. C4+
1. Using heavy water (D2O) as the source of deuterium, how could you prepare each compound?
1. LiAlD4
2. D2SO4
3. SiD4
4. DF
2. Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. Al2O3(s) in OH(aq)
2. Ar(g) + F2(g)
3. PI3(s) + H2O(l)
4. H3PO3(l) + OH(aq)
5. Bi(s) + excess Br2(l)
Answers
1. Xe(g) + F2(g) → XeF2(s)
2. Xe(g) + 2F2(g) → XeF4(s)
3. Xe(g) + 3F2(g) → XeF6(s)
4. 2XeF6(s) + SiO2(s) → 2XeOF4(l) + SiF4(l)
5. XeF6(s) + 3H2O(l) → XeO3(s) + 6HF(aq)
1. SiF4; SiO2(s) + 2F2(g) → SiF4(l) + 2O2(g)
1. 2Na(s) + 2D2O(l) → D2(g) + 2NaOD(aq)
2Li(s) + D2(g) → 2LiD(s)
4LiD(s) + AlCl3(soln) → LiAlD4(s) + 3LiCl(soln)
1. D2O(l) + SO3(g) → D2SO4(l)
2. SiCl4(l) + LiAlD4(s) [from part (a)] → SiD4(g) + LiCl(s) + AlCl3(s)
3. CaF2(s) + D2SO4(l) [from part (b)] → 2DF(g) + CaSO4(s)
22.3: Group 17: The Halogens
Conceptual Problems
1. The lightest elements of groups 15, 16, and 17 form unusually weak single bonds. Why are their bonds so weak?
2. Fluorine has an anomalously low F–F bond energy. Why? Why does fluorine form compounds only in the −1 oxidation state, whereas the other halogens exist in multiple oxidation states?
3. Compare AlI3, InCl3, GaF3, and LaBr3 with respect to the type of M–X bond formed, melting point, and solubility in nonpolar solvents.
4. What are the formulas of the interhalogen compounds that will most likely contain the following species in the indicated oxidation states: I (+3), Cl (+3), I (−1), Br (+5)?
5. Consider this series of bromides: AlBr3, SiBr4, and PBr5. Does the ionic character of the bond between the Br atoms and the central atom decrease or increase in this series?
6. Chromium forms compounds in the +6, +3, and +2 oxidation states. Which halogen would you use to produce each oxidation state? Justify your selections.
7. Of ClF7, BrF5, IF7, BrF3, ICl3, IF3, and IF5, which one is least likely to exist? Justify your selection.
Answers
1. Electrostatic repulsions between lone pairs on adjacent atoms decrease bond strength.
1. Ionic character decreases as Δχ decreases from Al to P.
1. ClF7
Structure and Reactivity
1. SiF4 reacts easily with NaF to form SiF62−. In contrast, CF4 is totally inert and shows no tendency to form CF62− under even extreme conditions. Explain this difference.
2. Predict the products of each reaction and then balance each chemical equation.
1. Xe(g) + excess F2(g) →
2. Se(s) + Cl2(g) →
3. SO2(g) + Br2(g) →
4. NaBH4(s) + BF3(soln) →
1. Write a balanced chemical equation for the reaction of aqueous HF with
1. SiO2.
2. Na2CO3.
3. CaO.
1. Oxyhalides of sulfur, such as the thionyl halides (SOX2, where X is F, Cl, or Br), are well known. Because the thionyl halides react vigorously with trace amounts of water, they are used for dehydrating hydrated metal salts. Write a balanced chemical equation to show the products of reaction of SOCl2 with water.
1. Write a balanced chemical equation describing each reaction.
1. the burning of sulfur in a chlorine atmosphere
2. the dissolution of iodine in a potassium iodide solution
3. the hydrolysis of PCl3
4. the preparation of HF from calcium fluoride and sulfuric acid
5. the thermal decomposition of KClO3
6. the oxidation of sulfide ion by elemental iodine
1. Write the complete Lewis electron structure, the type of hybrid used by the central atom, and the number of lone pair electrons present on the central atom for each compound.
1. CF4
2. PCl3
3. XeF4
Answers
1. Carbon has no low energy d orbitals that can be used to form a set of d2sp3 hybrid orbitals. It is also so small that it is impossible for six fluorine atoms to fit around it at a distance that would allow for formation of strong C–F bonds.
1. SiO2(s) + 6HF(aq) → SiF62−(aq) + 2H+(aq) + 2H2O(l)
2. Na2CO3(s) + 2HF(aq) → CO2(g) + 2NaF(aq) + H2O(l)
3. CaO(s) + 2HF(aq) → CaF2(s) + H2O(l)
1. S8(s) + 4Cl2(g) → 4S2Cl2(l)
2. I2(s) + KI(aq) → I3(aq) + K+(aq)
3. PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq)
4. CaF2(s) + H2SO4(aq) → 2HF(aq) + CaSO4(s)
5. 2KClO3(s) $\xrightarrow{\Delta}$ 2KCl(s) + 3O2(g)
6. 8S2−(aq) + 8I2(aq) → S8(s) + 16I(aq)
22.4: Group 16: The Oxygen Family
Conceptual Problems
1. Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Why?
2. Classify each oxide as basic, acidic, amphoteric, or neutral.
1. CaO
2. SO2
3. NO
4. Rb2O
5. PbO2
1. Classify each oxide as basic, acidic, amphoteric, or neutral.
1. BaO
2. Br2O
3. SnO
4. B2O3
5. Sb2O3
1. Polarization of an oxide affects its solubility in acids or bases. Based on this, do you expect RuO2 to be an acidic, a basic, or a neutral oxide? Is the compound covalent? Justify your answers.
2. Arrange CrO3, Al2O3, Sc2O3, and BaO in order of increasing basicity.
3. As the atomic number of the group 16 elements increases, the complexity of their allotropes decreases. What factors account for this trend? Which chalcogen do you expect to polymerize the most readily? Why?
4. Arrange H3BO3, HIO4, and HNO2 in order of increasing acid strength.
5. Of OF2, SO2, P4O6, SiO2, and Al2O3, which is most ionic?
6. Of CO2, NO2, O2, SO2, Cl2O, H2O, NH3, and CH4, which do you expect to have the
1. most polar covalent bond(s)?
2. least polar covalent bond(s)?
1. Of Na2O2, MgO, Al2O3, and SiO2, which is most acidic?
1. Give an example of
1. a covalent hydride that engages in strong hydrogen bonding.
2. an amphoteric oxide.
1. The Si–O bond is shorter and stronger than expected. What orbitals are used in this bond? Do you expect Si to interact with Br in the same way? Why or why not?
Answers
1. Oxygen has the second highest electronegativity of any element; consequently, it prefers to share or accept electrons from other elements. Only with fluorine does oxygen form compounds in positive oxidation states.
1. basic
2. acidic
3. amphoteric
4. acidic
5. amphoteric
1. CrO3 < Al2O3 < Sc2O3 < BaO
1. H3BO3 < HNO2 < HIO4
1. Most polar: H2O; least polar: O2
1. H2O, HF, or NH3
2. SnO or Al2O3
Structure and Reactivity
1. Considering its position in the periodic table, predict the following properties of selenium:
1. chemical formulas of its most common oxide, most common chloride, and most common hydride
2. solubility of its hydride in water, and the acidity or basicity of the resulting solution
3. the principal ion formed in aqueous solution
1. Using arguments based on electronegativity, explain why ZnO is amphoteric. What product would you expect when ZnO reacts with an aqueous
1. acid?
2. base?
1. Write a balanced chemical equation for the reaction of sulfur with
1. O2(g).
2. S2−(aq).
3. F2(g).
4. HNO3(aq).
Answer
1. S8 + 8O2 $\xrightarrow{\Delta}$ 8SO2(g)
2. S8(s) + 8S2−(aq) → 8S22−(aq)
3. S8(s) + 24F2(g) → 8SF6(g)
4. S8(s) + 48HNO3(aq) → 8H2SO4(aq) + 48NO2(g) + 16H2O(l)
22.5: Group 15: The Nitrogen Family
Conceptual Problems
1. Nitrogen is the first diatomic molecule in the second period of elements. Why is N2 the most stable form of nitrogen? Draw its Lewis electron structure. What hybrid orbitals are used to describe the bonding in this molecule? Is the molecule polar?
2. The polymer (SN)n has metallic luster and conductivity. Are the constituent elements in this polymer metals or nonmetals? Why does the polymer have metallic properties?
3. Except for NF3, all the halides of nitrogen are unstable. Explain why NF3 is stable.
4. Which of the group 15 elements forms the most stable compounds in the +3 oxidation state? Explain why.
5. Phosphorus and arsenic react with the alkali metals to produce salts with the composition M3Z11. Compare these products with those produced by reaction of P and As with the alkaline earth metals. What conclusions can you draw about the types of structures favored by the heavier elements in this part of the periodic table?
Structure and Reactivity
1. PF3 reacts with F2 to produce PF5, which in turn reacts with F to give salts that contain the PF6 ion. In contrast, NF3 does not react with F2, even under extreme conditions; NF5 and the NF6 ion do not exist. Why?
2. Red phosphorus is safer to handle than white phosphorus, reflecting their dissimilar properties. Given their structures, how do you expect them to compare with regard to reactivity, solubility, density, and melting point?
3. Bismuth oxalate [Bi2(C2O4)3] is a poison. Draw its structure and then predict its solubility in H2O, dilute HCl, and dilute HNO3. Predict its combustion products. Suggest a method to prepare bismuth oxalate from bismuth.
4. Small quantities of NO can be obtained in the laboratory by reaction of the iodide ion with acidic solutions of nitrite. Write a balanced chemical equation that represents this reaction.
5. Although pure nitrous acid is unstable, dilute solutions in water are prepared by adding nitrite salts to aqueous acid. Write a balanced chemical equation that represents this type of reaction.
6. Metallic versus nonmetallic behavior becomes apparent in reactions of the elements with an oxidizing acid, such as HNO3. Write balanced chemical equations for the reaction of each element of group 15 with nitric acid. Based on the products, predict which of these elements, if any, are metals and which, if any, are nonmetals.
7. Predict the product(s) of each reaction and then balance each chemical equation.
1. P4O10(s) + H2O(l) →
2. AsCl3(l) + H2O(l) →
3. Bi2O3(s) + H2O(l) →
4. Sb4O6(s) + OH(aq) →
5. (C2H5)3Sb(l) + O2(g) $\xrightarrow{\Delta}$
6. SbCl3(s) + LiAlH4(soln) →
7. Ca(s) + N2O(g) $\xrightarrow{\Delta}$
1. Write a balanced chemical equation to show how you would prepare each compound.
1. H3PO4 from P
2. Sb2O5 from Sb
3. SbH3 from Sb
Answers
1. NaNO2(s) + HCl(aq) → HNO2(aq) + NaCl(aq)
1. P4O10(s) + 6H2O(l) → 4H3PO4(aq)
2. AsCl3(l) + 3H2O(l) → H3AsO3(aq) + 3HCl(aq)
3. Bi2O3(s) + 3H2O(l) → 2Bi(OH)3(s)
4. Sb4O6(s) + 4OH(aq) + 2H2O(l) → 4H2SbO3(aq)
5. 2(C2H5)3Sb(l) + 21O2(g) $\xrightarrow{\Delta}$ 12CO2(g) + 15H2O(g) + Sb2O3(s)
6. 4SbCl3(s) + 3LiAlH4(soln) → 4SbH3(g) + 3LiCl(soln) + 3AlCl3(soln)
7. Ca(s) + N2O(g) $\xrightarrow{\Delta}$ CaO(s) + N2(g)
22.6: Hydrogen: A Unique Element
Problems
1. Some periodic tables include hydrogen as a group 1 element, whereas other periodic tables include it as a group 17 element. Refer to the properties of hydrogen to propose an explanation for its placement in each group. In each case, give one property of hydrogen that would exclude it from groups 1 and 17.
2. If there were a planet where the abundances of D2O and H2O were reversed and life had evolved to adjust to this difference, what would be the effects of consuming large amounts of H2O?
3. Describe the bonding in a hydrogen bond and the central B–H bond in B2H7. Why are compounds containing isolated protons unknown?
4. With which elements does hydrogen form ionic hydrides? covalent hydrides? metallic hydrides? Which of these types of hydrides can behave like acids?
5. Indicate which elements are likely to form ionic, covalent, or metallic hydrides and explain your reasoning:
1. Sr
2. Si
3. O
4. Li
5. B
6. Be
7. Pd
8. Al
1. Which has the higher ionization energy—H or H? Why?
1. The electronegativities of hydrogen, fluorine, and iodine are 2.20, 3.98, and 2.66, respectively. Why, then, is HI a stronger acid than HF?
1. If H2O were a linear molecule, would the density of ice be less than or greater than that of liquid water? Explain your answer.
1. In addition to ion–dipole attractions, hydrogen bonding is important in solid crystalline hydrates, such as Na4XeO6·8H2O. Based on this statement, explain why anhydrous Na4XeO6 does not exist.
Answers
1. H has one electron in an s orbital, like the group 1 metals, but it is also one electron short of a filled principal shell, like the group 17 elements. Unlike the alkali metals, hydrogen is not a metal. Unlike the halogens, elemental hydrogen is not a potent oxidant.
1. ionic; it is an alkaline earth metal.
2. covalent; it is a semimetal.
3. covalent; it is a nonmetal.
4. ionic; it is an alkali metal.
5. covalent; it is a semimetal.
6. covalent; it is a period 2 alkaline earth metal.
7. metallic; it is a transition metal.
8. covalent; it is a group 13 metal.
1. Hydrogen bonding with waters of hydration will partially neutralize the negative charge on the terminal oxygen atoms on the XeO64− ion, which stabilizes the solid.
Structure and Reactivity
1. One of the largest uses of methane is to produce syngas, which is a source of hydrogen for converting nitrogen to ammonia. Write a complete equation for formation of syngas from methane and carbon dioxide. Calculate ΔG° for this reaction at 298 K and determine the temperature at which the reaction becomes spontaneous.
1. An alternative method of producing hydrogen is the water–gas shift reaction:
CO(g) + H2O(g) → CO2(g) + H2(g)
Calculate ΔG° for this reaction at 298 K and determine the temperature at which the reaction changes from spontaneous to nonspontaneous (or vice versa).
1. Predict the products of each reaction at 25°C and then balance each chemical equation.
1. CsH(s) + D2O(l) →
2. CH3CO2H(l) + D2O(l) →
3. H3PO4(aq) + D2O(l) →
4. NH2CH2CO2H(s) + D2O(l) →
5. NH4Cl(s) + D2O(l) →
1. Using heavy water (D2O) as the source of deuterium, how could you conveniently prepare
1. D2SO4?
2. LiD?
1. What are the products of reacting NaH with D2O? Do you expect the same products from reacting NaD and H2O? Explain your answer.
1. A 2.50 g sample of zinc metal reacts with 100.0 mL of 0.150 M HCl. What volume of H2 (in liters) is produced at 23°C and 729 mmHg?
1. A chemical reaction requires 16.8 L of H2gas at standard temperature and pressure. How many grams of magnesium metal are needed to produce this amount of hydrogen gas?
1. Seawater contains 3.5% dissolved salts by mass and has an average density of 1.026 g/mL. The volume of the ocean is estimated to be 1.35 × 1021 L. Using the data in Table $1$, calculate the total mass of deuterium in the ocean.
1. From the data in Table $1$, determine the molarity of DOH in water. Do you expect the molarity of D2O in water to be similar? Why or why not?
1. From the data in Table $1$, calculate how many liters of water you would have to evaporate to obtain 1.0 mL of TOD (tritium-oxygen-deuterium). The density of TOD is 1.159 g/mL. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.E%3A_Exercises.txt |
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
Properties and Trends in Transition Metals
The elements of the second and third rows of the Periodic Table show gradual changes in properties across the table from left to right as expected. Electrons in the outer shells of the atoms of these elements have little shielding effects resulting in an increase in effective nuclear charge due to the addition of protons in the nucleus. Consequently, the effects on atomic properties are: smaller atomic radius, increased first ionization energy, enhanced electronegativity and more nonmetallic character. This trend continues until one reaches calcium (Z=20). There is an abrupt break at this point. The next ten elements called the first transition series are remarkably similar in their physical and chemical properties. This general similarity in properties has been explained in terms of their relatively small difference in effective nuclear charge over the series. This occurs because each additional electron enters the penultimate 3d shell providing an effective shield between the nucleus and the outer 4s shell.
What is a Transition Metal?
Thus, the transition elements can be defined as those in which the d electron shells are being filled and so we generally ignore Sc and Zn where Sc(III) is d0 and Zn(II) is d10.
It is useful, at the beginning, to identify the physical and chemical properties of transition elements which differ from main group elements (s-block). Properties of transition elements include:
• have large charge/radius ratio;
• are hard and have high densities;
• have high melting and boiling points;
• form compounds which are often paramagnetic;
• show variable oxidation states;
• form coloured ions and compounds;
• form compounds with profound catalytic activity;
• form stable complexes.
Table $1$ : Summary of select physical properties of transition elements:
Element Group density
/g cm-3
m. p.
/ °C
b.p.
/ °C
radius
/ pm
free atom
configuration
ionization energy
/ kJ mol-1
Uses
Sc 3 2.99 1541 2831 164 [Ar] 3d14s2 631
Ti 4 4.50 1660 3287 147 [Ar]3d24s2 658 -engines/aircraft industry-density is 60% of iron
V 5 5.96 1890 3380 135 [Ar]3d34s2 650 -stainless steel, 19% Cr, 9% Ni the rest Fe
Cr 6 7.20 1857 2670 129 [Ar]3d54s1 653 -alloys eg with C steel, the most significant use
Mn 7 7.20 1244 1962 137 [Ar]3d54s2 717 -alloys eg with Cu
Fe 8 7.86 1535 2750 126 [Ar]3d64s2 759 -alloys eg with C steel, the most significant use
Co 9 8.90 1495 2870 125 [Ar]3d74s2 758 -alloys eg with Cr and W for hardened drill bits
Ni 10 8.90 1455 2730 125 [Ar]3d84s2 737 -alloys Fe/Ni armor plating, resists corrosion
Cu 11 8.92 1083 2567 128 [Ar]3d104s1 746 -high electrical conductivity (2nd to Ag), wiring
Zn 12 7.14 420 907 137 [Ar]3d104s2 906
Densities and Metallic Radii
The transition elements are much denser than the s-block elements and show a gradual increase in density from scandium to copper. This trend in density can be explained by the small and irregular decrease in metallic radii coupled with the relative increase in atomic mass.
Melting and Boiling points
The melting points and the molar enthalpies of fusion of the transition metals are both high in comparison to main group elements. This arises from strong metallic bonding in transition metals which occurs due to delocalization of electrons facilitated by the availability of both d and s electrons.
Ionization Energies
In moving across the series of metals from scandium to zinc a small change in the values of the first and second ionization energies is observed. This is due to the build-up of electrons in the immediately underlying d-sub-shells that efficiently shields the 4s electrons from the nucleus and minimizing the increase in effective nuclear charge $Z_{eff}$ from element to element. The increases in third and fourth ionization energy values are more rapid. However, the trends in these values show the usual discontinuity half way along the series. The reason is that the five d electrons are all unpaired, in singly occupied orbitals. When the sixth and subsequent electrons enter, the electrons have to share the already occupied orbitals resulting in inter-electron repulsions, which would require less energy to remove an electron. Hence, the third ionization energy curve for the last five elements is identical in shape to the curve for the first five elements, but displaced upwards by 580 kJ mol-1.
Oxidation States
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 $5$. 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+.
Electronic Configurations
The electronic configuration of the atoms of the first row transition elements are basically the same. It can be seen in the Table above that there is a gradual filling of the 3d orbitals across the series starting from scandium. This filling is, however, not regular, since at chromium and copper the population of 3d orbitals increase by the acquisition of an electron from the 4s shell. This illustrates an important generalization about orbital energies of the first row transition series. At chromium, both the 3d and 4s orbitals are occupied, but neither is completely filled in preference to the other. This suggests that the energies of the 3d and 4s orbitals are relatively close for atoms in this row.
In the case of copper, the 3d level is full, but only one electron occupies the 4s orbital. This suggests that in copper the 3d orbital energy is lower than the 4s orbital. Thus the 3d orbital energy has passed from higher to lower as we move across the period from potassium to zinc. However, the whole question of preference of an atom to adopt a particular electronic configuration is not determined by orbital energy alone. In chromium it can be shown that the 4s orbital energy is still below the 3d which suggests a configuration [Ar] 3d44s2. However due to the effect of electronic repulsion between the outer electrons the actual configuration becomes [Ar]3d54s1 where all the electrons in the outer orbitals are unpaired. It should be remembered that the factors that determine electronic configuration in this period are indeed delicately balanced.
Redox Couple E°/V
Mn2+(aq.)/Mn(s) -1.18
H+(aq.)/H2(g) 0.00
This shows that elemental Mn is a stronger reductant than molecular hydrogen and hence should be able to displace hydrogen gas from 1 mol dm-3 hydrochloric acid.
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+.
Chemical Reactivity
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.
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:
$\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V}$
$\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V}$
$\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V}$
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+ | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.1%3A_General_Properties_of_Transition_Metals.txt |
Learning Objectives
• Discuss how coordination chemistry can be applied in metallurgic processes.
• Understanding the thermodynamics of Pyrometallurgy
The Earth formed from the same cloud of matter that formed the Sun, but the planets acquired different compositions during the formation and evolution of the solar system. In turn, the natural history of the Earth caused parts of this planet to have differing concentrations of the elements. The mass of the Earth is approximately 5.98×1024 kg. In bulk, by mass, it is composed mostly of iron (32.1%), oxygen (30.1%), silicon (15.1%), magnesium (13.9%), sulfur (2.9%), nickel (1.8%), calcium (1.5%), and aluminum (1.4%); with the remaining 1.2% consisting of trace amounts of other elements. Figure $1$ illustrates the relative atomic-abundance of the chemical elements in Earth's upper continental crust, which is relatively accessible for measurements and estimation.
Many of the elements shown in the graph are classified into (partially overlapping) categories:
1. rock-forming elements (major elements in green field, and minor elements in light green field);
2. rare earth elements (lanthanides, La-Lu, and Y; labeled in blue);
3. major industrial metals (global production >~3×107 kg/year; labeled in red);
4. precious metals (labeled in purple);
5. the nine rarest "metals" (the six platinum group elements and Au, Re, and Te) are in the yellow field.
Most metals are found as types of rock in the Earth's crust. These ores contain sufficient minerals with important elements including metals that can be economically extracted from the rock. Metal ores are generally oxides, sulfides, silicates (Table $1$) or "native" metals (such as native copper) that are not commonly concentrated in the Earth's crust, or "noble" metals (not usually forming compounds) such as gold (Figure $2$). The ores must be processed to extract the metals of interest from the waste rock and from the ore minerals.
Figure $1$: Composition of Earth's mantle in weight percent
Element Amount Compound Amount
O 44.8
Mg 22.8 SiO2 46
Si 21.5 MgO 37.8
Fe 5.8 FeO 7.5
Ca 2.3 Al2O3 4.2
Al 2.2 CaO 3.2
Na 0.3 Na2O 0.4
K 0.03 K2O 0.04
Sum 99.7 Sum 99.1
Extractive metallurgy is a branch of metallurgical engineering wherein process and methods of extraction of metals from their natural mineral deposits are studied. The field is a materials science, covering all aspects of the types of ore, washing, concentration, separation, chemical processes and extraction of pure metal and their alloying to suit various applications, sometimes for direct use as a finished product, but more often in a form that requires further working to achieve the given properties to suit the applications. The field of ferrous and non-ferrous extractive metallurgy have specialties that are generically grouped into the categories of mineral processing, hydrometallurgy, pyrometallurgy, and electrometallurgy based on the process adopted to extract the metal. Several processes are used for extraction of same metal depending on occurrence and chemical requirements.
It takes multiple steps to extract the "important" element from the ore:
1. First, the ore must be separated from unwanted rocks.
2. Then, the minerals need to be separated out of the ore
3. Since most minerals are not pure metals, further separation methods are required.
Most minerals are chemical compounds that contain metals and other elements.
Concentration
After mining, large pieces of the ore feed are broken through crushing and/or grinding. This step creates particles that are either mostly valuable or mostly waste. TConcentrating the particles of value in a form supporting separation enables the desired metal to be removed from waste products. Froth flotation is one process for separating minerals from the surrounding worthless material by taking advantage of differences in their hydrophobicity. whereby hydrophobicity differences between valuable minerals other material are increased through the use of surfactants and wetting agents (Figure $3$). Flotation is used for the separation of a large range of sulfides, carbonates and oxides prior to further refinement. Phosphates and coal are also upgraded (purified) by flotation technology.
Other ways to concentration minerals include approached based on density (Froth Flotation) used if ore density is less than density of impurities, or via melting points, or magnetic properties (Magnetic separation). This can be used with magnetic ores (i.e Fe3O4) that are passed through a magnetic field, which attracts the magnetic ore and ignore the nonmagnetic impurities.
Roasting
Ore is crushed and heated to a high temp using a strong blast of hot air. The process converts the ores to their oxides which can then be reduced. Consider the natural occurring ores of zinc ($ZnS$ (sphalerite) and $ZnCO_3$ (smithsonite). When roasted, smithsonite decomposes to $ZnO_{(s)}$ and $CO_{2(g)}$ $\ref{Roast1}$ and the hot air involved in roasting sphalerite oxidizes it $\ref{Roast2}$ to produces $ZnO_{(s)}$ and $SO_{2(g)}$.
$ZnCO_{3(s)} \overset{\Delta}{\longrightarrow} ZnO_{(s)} + CO_{2(g)} \label{Roast1}$
$2 ZnS_{(s)} + 3 O_{2(g)} \overset{\Delta}{\longrightarrow} 2 ZnO_{(s)} + 2 SO_{2(g)} \label{Roast2}$
Reduction
$C_{(s)}$ and $CO_{(g)}$ are often used as reducing agents in simultaneous reactions. Oxides of Cr, V, and Mn are reduced using Al. During the reduction process the metal oxide is heated to a temperature above its boiling point in order to vaporize it and condense as a liquid. For example, $ZnO_{(s)}$ that is produced in the roasting process is combined with $C_{(s)}/CO_{(g)}$ mixture and heated at a 1100 oC. Two reduction reactions are viable at this temperature with either carbon (Equation $\ref{Reduction1}$) or carbon monoxide ((Equation $\ref{Reduction2}$) acting the reducing agent.
$ZnO_{(s)} +C_{(s)} \overset{\Delta}{\longrightarrow} Zn_{(g)} + CO_{(g)} \label{Reduction1}$
$ZnO_{(s)} + CO_{(g)} \overset{\Delta}{\longrightarrow} Zn_{(g)} + CO_{2(g)} \label{Reduction2}$
Refining
The impurities contained in the metal product of the roasting/reduction process are removed. For example, the $Zn(l)$ produced in the reduction process often contains impurities of Cd and Pb. Though fractional distillation of Zn(l) will work in the refining process, a more commonly used and efficient method is electrolysis. The $ZnO_{(s)}$ produced from roasting is dissolved in $H_2SO_{4(aq)}$ and $Zn_{(s)}$ powder is added allowing the impurities to be displaced. The solution is then electrolyzed and to get a pure metallic $Zn_{(s)}$ as a result.
Zone Refining
A rod containing the desired pure metal and impurities is passed through a series of heating coils and cooled again. The process isolates the impurities that concentrate in the molten zones, leaving the portions behind them somewhat more pure. This process is repeated until impurities are moved to the end of the rod and cut off, resulting in an almost completely pure metal rod.
Thermodynamics of Pyrometallurgy
Pyrometallurgy is the processes of roasting an ore a high temperatures and then reducing its oxide product. Its characteristics include:
1. large amount of waste as a product of concentration.
2. high energy consumption to maintain high temps.
3. gaseous emissions that must be controlled (i.e CO2 and SO2)
The oldest, and still the most common smelting process for oxide ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal. The basic reactions are:
$MO + C \rightleftharpoons M + CO \label{2.1}$
$MO + ½ O_2 \rightleftharpoons M + ½ CO_2 \label{2.2}$
$MO + CO \rightleftharpoons M + CO_2 \label{2.3}$
Each of these can be regarded as a pair of coupled reactions in which the metal M and the carbon are effectively competing for the oxygen atom. Using reaction $\ref{2.1}$ as an example, it can be broken down into the following two parts:
$MO \rightleftharpoons M + ½ O_2 \;\;\; ΔG^o > 0 \label{2.4}$
$C + ½ O_2 \rightleftharpoons CO \;\;\; ΔG^o < 0 \label{2.5}$
At ordinary environmental temperatures, reaction $\ref{2.4}$ is always spontaneous in the reverse direction (that is why ores form in the first place!), so ΔG° of Reaction $\ref{2.4}$ will be positive. ΔG° for reaction $\ref{2.5}$ is always negative, but at low temperatures it will not be sufficiently negative to drive $\ref{2.4}$.
The smelting process depends on the different ways in which the free energies of reactions like $\ref{2.4}$ and $\ref{2.4}$ vary with the temperature. This temperature dependence is almost entirely dominated by the TΔS° term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δng, the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always accompanied by a large increase in the system volume so ΔS for this step is always positive and the reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that take up oxygen vary, however (Table $1$).
Table $2$: The temperature dependences of the reactions
Reaction
Δng
dG°)/dT
C + ½ O2 → CO
0.5
<0
C + O2 → CO2
0
0
CO + ½ O2 →CO2
–0.5
>0
A plot of the temperature dependences of the free energies of these reactions, superimposed on similar plots for the oxygen removal reactions $\ref{2.4}$ is called an Ellingham diagram. For a given oxide MO to be smeltable, the temperature must be high enough that reaction $\ref{2.4}$ falls below that of at least one of the oxygen-consuming reactions. The slopes of the lines on this diagram are determined by the sign of the entropy change.
Figure $4$: Ellingham diagram. An ore can be reduced by carbon only if its Gibbs free energy of formation falls below that of one of the carbon reduction reactions (blue lines.) Practical refining temperatures are generally limited to about 1500°K.
Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were mainly those such as copper and lead, which can be obtained by smelting at the relatively low temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the development of technology capable of producing higher temperatures, such as the blast furnace. Smelting of aluminum oxide by carbon requires temperatures too high to be practical; commercial production of aluminum is accomplished by electrolysis of the molten ore.
In pyrometallurgy, an ore is heated with a reductant to obtain the metal. Theoretically, it should be possible to obtain virtually any metal from its ore by using coke as the reductant. For example, the reduction of calcum metral from $CaO_{(s)}$:
$\mathrm{CaO(s) + C(s)\xrightarrow{\Delta}Ca(l) + CO(g)} \label{23.2.1}$
Unfortunately, many transition metals (e.g., Titanium) react with carbon to form stable binary carbides:
$Ti + C \rightarrow TiC.$
Consequently, more expensive reductants (such as hydrogen, aluminum, magnesium, or calcium) must be used to obtain these metals. Many metals that occur naturally as sulfides can be obtained by heating the sulfide in air, as shown for lead in the following equation:
$\mathrm{PbS(s) + O_2(g) \xrightarrow{\Delta}Pb(l) + SO_2(g)} \label{23.2.2}$
The reaction is driven to completion by the formation of $SO_2$, a stable gas.
Iron Alloys of Cr, V, and Mn
1. Ferrochrome ( alloy of Fe and Cr) can be produced in the reduction of Cr in an Fe compound containing its chromite ore, Fe(CrO2)2.
2. Ferrovanadium ( alloy of Fe and V) can be produced in the reduction of V that occurs when adding an Fe compound to a V oxide (V2O5).
3. Ferromanganese (alloy of Fe and Mn) can be produced in the reduction of Mn that occurs when adding an Fe compound to a Mn oxide (MnO2).
Titanium Production
Titanium is used in the military and aircraft industry b/c of its low density and ability to maintain its strength are high temps.
Titanium is produced using the following steps:
1. Conversion or rutile or $TiO_2$ to $TiCl_4$ with $C_{(s)}$ and $Cl_2$ under high heat (> 800 oC ) $TiO_{2(s)} + 2C_{(s)} + 2Cl_{(2(g)} \xrightarrow{\Delta}Pb(l) TiCl_{4(g)} + 2CO_{(g)}$
2. Reduction of $TiCl_4$ to $Ti$ with a good reducing agent (i.e., $Mg$ in the Kroll process) carried out at high heat (> 1000 oC ). $TiCl_{4(g)} + 2Mg_{(l)} \xrightarrow{\Delta} Ti_{(s)} + 2MgCl_{2(s)}$
3. Both $Ti_{(s)}$ and $MgCl_{2(s)}$ are electrolyzed to produces $Cl_2$, $Mg$ and titanium sponge (Ti enriched mass). The $Mg$ and $MgCL_2$ are removed via vacuum distillation (reducing of pressure with heat to preferential vaporize volatile metal that are then condenses in a separate container).
4. The spong is further treated and allowed with other metal to be sold commercially.
The Kroll process is slow, expensive, and proposes a health/safety risk due to the high temperature vacuum distillation.
Note: alternative means of producing titanium
There are alternative means of producing titanium including the electrolysis of TiO2 pellets via the following procedure:
1. The pellets are placed at the cathode end(usually Ti(s) or graphite) of an electrolytic cell.
2. They are dissolved in a molten CaCl2(l) electrolyte.
3. The O2- anions are discharged as O2(g) at the anode(graphite).
4. The Ti4+ cation is reduced at the cathode resulting in a tatanium sponge
Copper Production
Cu ores often contain iron sulfides and to achieve a pure(uncontaminated) Cu product, the ores undergo the following processes :
Step 1: Concentration of ore by flotation (Figure $6$.
Step 2: Conversion of Fe sulfide ores to Fe oxides through roasting (e.g., roasting of Chalcopyrite ore )
$2 CuFeS)2 + 3 O)2 \overset{\Delta}{\longrightarrow} 2 FeO + 2 CuS + 2 SO_2$
Step 3: Smelting: mixing with coke/charcoal and heated at high temperatures (i.e 800 oC) for copper to keep Cu from remaining in sulfide form). Impurities are further removed by adding flux if not already present in ore (if gangue is acidic basic flux is added, if gangue is basic acidic flux (i.e., SiO2) is used. Smelting separates the material to separate into 2 layers (copper matte: bottom layer containing molten sulfides of Fe and Cu and silicate slag: top layer formed by reaction of oxides of Fe, Ca, and Al w/ SiO2).
$FeO_{(s)} + SiO_{2(s)} \overset{\Delta}{\longrightarrow} FeSiO_{3(l)}$
Step 4: Conversion: Air is blown through Cu matte in a separate furnace. This converts the remaining iron sulfide to iron oxide followed by the formation of FeSiO3(l) slag. The slag is poured off and the process is repeated resulting in blister copper with bubbles of SO2(g) still present.
$2CuS_{(l)} + 3O_{2(g)} \overset{\Delta}{\longrightarrow} 2Cu_2O_{(l)} + 2SO_{2(g)}$
$2Cu_2O_{(l)} + Cu_2S_{(l)} \overset{\Delta}{\longrightarrow} 6Cu(l) +SO_{2(g)}$
Step 5: Blister Cu is further refined through electrolysis to achieve a high-purity Cu product.
Hydrometallurgical Process
The method of refining zinc described above is an example of a hydrometallurgical process and the hydrometallurgical process eliminates the need to control gaseous emissions that are often produced in roasting. Hydrometallurgy is the process of extraction and refining that involves the use of water and aqueous solutions. It is carried out at moderate temps and is generally carried out in three steps:
1. Leaching: metal ions are extracted from their ore by water/acids/bases/salt solutions. Redox reactions that occur are often essential.
2. Purification and concentration: Impurities are separated either by absorption on the surface of activated charcoal, ion exchange, or water evaporation; leaving behind a more concentrated solution.
3. Precipitation: the process of electrolysis is often used to precipitate the desired metal ions in an ionic solid or reduce them to their free metal.
Summary
Extractive metallurgy is the practice of removing valuable metals from an ore and refining the extracted raw metals into purer form. The field of extractive metallurgy encompasses many specialty sub-disciplines, including mineral processing, hydrometallurgy, pyrometallurgy, and electrometallurgy. Especially in hydrometallurgy, the coordination chemistry of the metals involved plays a large role in their solubility and reactivity as the ore is refined into precious metal. Many methods of extractive metallurgy may be applied to get a pure metal from its naturally occurring ore. First ores must be concentrated and separated from their earthly impurities. Concentration of ores may be achieved by froth flotation in which an ore is contained in a froth that floats at the top of a water vat. After the ore has been concentrated it must be roasted to a high temperature allowing for the desired metal to be oxidized. Once the metal is oxidized it is reduced by a reducing agent. Reduction may require the metal oxide to be heated at a high temperature. The metal must be further refined to remove all of the impurities that it may contain. Refining a metal can be accomplished by electrolysis or zone refining. Certain ores such as those of copper and titanium require special treatment before they can be refined. The extraction of iron ore is particularly important for commercial uses because of the mass production of its alloy, steel.
Problems
1. Describe 2 methods of ore concentration and list the conditions under which they are used.
2. Explain the conditions under which an ore might be roasted.
3. Explain the conditions under which an are might be electrolyzed.
4. What types of reactions occur at the cathode and anode in an electrolyzed solution?
5. How might the transition metal, zinc, be refined?
6. Explain the process of zone refining and how it may make an extracted metal more pure.
7. Explain how a change in free energy is related to extractive metallurgy. What specific process or reaction does it pertain to?
8. Give a brief explanation of why an ore might be heated to high temperatures using the principles of Thermodynamics.
9. What are the characteristics of a pyrometallurgical process and how do they compare with those of a hydrometallurgical process?
10. What are the applications of pig iron and how might it be refined to make pure iron.
11. What are the reactions that take place in the process of making steel?
12. How might an alloy of differing metals effect the properties of steel?
13. Explain 2 alternative methods of extractive metallurgy and how they differ from traditional methods. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.2%3A_Principles_of_Extractive_Metallurgy.txt |
Learning Objectives
• Outline the general approach for the metallurgy of iron into steel
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 overall reaction for the production of iron in a blast furnace is as follows:
$\mathrm{Fe_2O_3(s) +3C(s)\xrightarrow{\Delta}2Fe(l) +3CO(g)} \label{23.2.3}$
The actual reductant is CO, which reduces Fe2O3 to give Fe(l) and CO2(g) (Equation $\ref{23.2.3}$); the CO2 is then reduced back to CO by reaction with excess carbon. As the ore, lime, and coke drop into the furnace (Figure $1$), any silicate minerals in the ore react with the lime to produce a low-melting mixture of calcium silicates called slag, which floats on top of the molten iron. Molten iron is then allowed to run out the bottom of the furnace, leaving the slag behind. Originally, the iron was collected in pools called pigs, which is the origin of the name pig iron.
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 $2$) 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:
$\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g)$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $2$. 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:
$\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)$
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 $3$).
Steel
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. However, there is not just one substance called steel - they are a family of alloys of iron with carbon or various metals.
Impurities in the iron from the Blast Furnace include carbon, sulfur, phosphorus and silicon, which have to be removed.
• Removal of sulfur: Sulfur has to be removed first in a separate process. Magnesium powder is blown through the molten iron and the sulfur reacts with it to form magnesium sulfide. This forms a slag on top of the iron and can be removed. $Mg + S \rightarrow MgS \label{127}$
• Removal of carbon: The still impure molten iron is mixed with scrap iron (from recycling) and oxygen is blown on to the mixture. The oxygen reacts with the remaining impurities to form various oxides. The carbon forms carbon monoxide. Since this is a gas it removes itself from the iron! This carbon monoxide can be cleaned and used as a fuel gas.
• Removal of other elements: Elements like phosphorus and silicon react with the oxygen to form acidic oxides. These are removed using quicklime (calcium oxide) which is added to the furnace during the oxygen blow. They react to form compounds such as calcium silicate or calcium phosphate which form a slag on top of the iron.
Table $1$: Special Steels
iron mixed with special properties uses include
stainless steel chromium and nickel resists corrosion cutlery, cooking utensils, kitchen sinks, industrial equipment for food and drink processing
titanium steel titanium withstands high temperatures gas turbines, spacecraft
manganese steel manganese very hard rock-breaking machinery, some railway track (e.g. points), military helmets
Cast iron has already been mentioned above. This section deals with the types of iron and steel which are produced as a result of the steel-making process.
• Wrought iron: If all the carbon is removed from the iron to give high purity iron, it is known as wrought iron. Wrought iron is quite soft and easily worked and has little structural strength. It was once used to make decorative gates and railings, but these days mild steel is normally used instead.
• Mild steel: Mild steel is iron containing up to about 0.25% of carbon. The presence of the carbon makes the steel stronger and harder than pure iron. The higher the percentage of carbon, the harder the steel becomes. Mild steel is used for lots of things - nails, wire, car bodies, ship building, girders and bridges amongst others.
• High carbon steel: High carbon steel contains up to about 1.5% of carbon. The presence of the extra carbon makes it very hard, but it also makes it more brittle. High carbon steel is used for cutting tools and masonry nails (nails designed to be driven into concrete blocks or brickwork without bending). High carbon steel tends to fracture rather than bend if mistreated.
• Special steels: These are iron alloyed with other metals (Table $1$).
Video: You can watch an animation of steelmaking that walks you through the process (https://www.youtube.com/watch?v=nqzJk2836V8). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.3%3A_Metallurgy_of_Iron_and_Steel.txt |
Learning Objectives
• To use periodic trends to understand the chemistry of the transition metals.
As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member.
Group 3 (Sc, Y, La, and Ac)
As shown in Table $1$, the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns2(n − 1)d1 valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas:
$2M_{(s)} + 6H_2O_{(l)} \rightarrow 2M(OH)_{3(s)} + 3H_{2(g)}\label{Eq1}$
The chemistry of the group 3 metals is almost exclusively that of the M3+ ion; the elements are powerful reductants.
Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M3+(aq).
Table $1$: Some Properties of the Elements of Groups 3, 4, and 5
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
3 Sc 21 4s23d1 1.36 162 1541 2.99
Y 39 5s24d1 1.22 180 1522 4.47
La 57 6s25d1 1.10 183 918 6.15
Ac 89 7s26d1 1.10 188 1051 10.07
4 Ti 22 4s23d2 1.54 147 1668 4.51
Zr 40 5s24d2 1.33 160 1855 6.52
Hf 72 6s25d24f14 1.30 159 2233 13.31
5 V 23 4s23d3 1.63 134 1910 6.00
Nb 41 5s24d3 1.60 146 2477 8.57
Ta 73 6s25d34f14 1.50 146 3017 16.65
The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX3. The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M2O3), which react with H2O or CO2 to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters.
Group 4 (Ti, Zr, and Hf)
Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns2(n − 1)d2 valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti:
$2FeTiO_{3(s)} + 6C_{(s)} + 7Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + 2FeCl_{3(g)} + 6CO_{(g)}\label{Eq2}$
followed by reduction of the tetrachlorides with an active metal such as Mg.
The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states.
In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO2-containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines.
Consistent with the periodic trends shown in Figure 23.2, the group 4 metals become denser, higher melting, and more electropositive down the column (Table $1$). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns2(n − 1)d2 valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti2+ are usually powerful reductants. In fact, the Ti2+(aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas.
Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX4), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl2 is an ionic salt, whereas TiCl4 is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO2) and dichalcogenides (MY2). Industrially, TiO2, which is used as a white pigment in paints, is prepared by reacting TiCl4 with oxygen at high temperatures:
$TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} + 2Cl_{2(g)}\label{Eq3}$
The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure $1$). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH2), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB2), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity.
Group 5 (V, Nb, and Ta)
Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores.
Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V4C3, which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V2O5, an important catalyst for the industrial conversion of SO2 to SO3 in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb3Zr and Nb3Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones.
The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4.
As indicated in Table $1$, the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M2O5), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H2O)6]2+; the blue-green [V(H2O)6]3+ ion is acidic, dissociating to form small amounts of the [V(H2O)5(OH)]2+ ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H2O)4VO]2+, which contains a formal V=O bond (Figure $2$). Consistent with its covalent character, V2O5 is acidic, dissolving in base to give the vanadate ion ([VO4]3−), whereas both Nb2O5 and Ta2O5 are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric.
Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO2]+ ion; V(IV) is the blue vanadyl ion [VO]2+; and V(III) and V(II) exist as the hydrated V3+ (blue-green) and V2+ (violet) ions, respectively.
Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY2), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools.
Group 6 (Cr, Mo, and W)
As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS2) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr2O4), which is oxidized to the soluble [CrO4]2− ion under basic conditions and reduced successively to Cr2O3 and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr2O3 in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO3, which is then reduced with hydrogen to give the metal.
The metals become increasing polarizable across the d block.
Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (Table $2$). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr2+, which is a powerful reductant, to CrO3, a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known.
Table $2$: Some Properties of the Elements of Groups 6 and 7
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
6 Cr 24 4s13d5 1.66 128 1907 7.15
Mo 42 5s14d5 2.16 139 2623 10.20
W 74 6s25d44f14 1.70 139 3422 19.30
7 Mn 25 4s23d5 1.55 127 1246 7.30
Tc 43 5s24d5 2.10 136 2157 11.50
Re 75 6s25d54f14 1.90 137 3186 20.80
The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states.
As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF6 is unstable at temperatures above −100°C, whereas MoF6 and WF6 are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX3 (X is Cl, Br, or I), while molybdenum forms MoCl5, MoBr4, and MoI3, and tungsten gives WCl6, WBr5, and WI4.
Both Mo and W react with oxygen to form the covalent trioxides (MoO3 and WO3), but Cr reacts to form only the so-called sesquioxide (Cr2O3). Chromium will form CrO3, which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO4]2−). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr2O3) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo8O26]4− ion, whose structure is as follows:
An isopolymolybdate cluster. The [Mo8O26]4− ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution.
Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS2; it has a layered structure similar to that of TiS2 (Figure $1$), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS2 and WS2 are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS2 particles.
As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits.
Group 7 (Mn, Tc, and Re)
Continuing across the periodic table, we encounter the group 7 elements (Table $2$). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, 99mTc (m for metastable), has become an important biomedical tool for imaging internal organs. Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline.
All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (Figure $3$). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d5 electron configuration, the aqueous Mn3+ ion, with a 3d4 valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound.
Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn2+(aq) ion is pale pink; Mn(OH)3, which contains Mn(III), is a dark brown solid; MnO2, which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO4]2− and the purple permanganate ion [MnO4], respectively.
Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F2 gives only MnF3, a high-melting, red-purple solid, whereas Re reacts with F2 to give ReF7, a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl2, a weaker oxidant than F2, gives MnCl2 and ReCl6. Reaction of Mn with oxygen forms only Mn3O4, a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe3O4). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M2O7), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn2O7, an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO4]2− is a potent oxidant, whereas [TcO4] and [ReO4] are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS2, as well as more complex heptasulfides (M2S7). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties.
The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced.
Groups 8, 9, and 10
In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert.
The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3.
Group 8 (Fe, Ru, and Os)
The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age.
Group 9 (Co, Rh, and Ir)
Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores.
Group 10 (Ni, Pd, and Pt)
Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold.
Over 2000 years ago, the Bactrian civilization in Western Asia used a 75:25 alloy of copper and nickel for its coins. A modern US nickel has the same composition, but a modern Canadian nickel is nickel-plated steel and contains only 2.5% nickel by mass.
Some properties of the elements in groups 8–10 are summarized in Table $3$. As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state.
Table $3$: Some Properties of the Elements of Groups 8, 9, and 10
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
8 Fe 26 4s23d6 1.83 126 1538 7.87
Ru 44 5s14d7 2.20 134 2334 12.10
Os 76 6s25d64f14 2.20 135 3033 22.59
9 Co 27 4s23d7 1.88 125 1495 8.86
Rh 45 5s14d8 2.28 134 1964 12.40
Ir 77 6s25d74f14 2.20 136 2446 22.50
10 Ni 28 4s23d8 1.91 124 1455 8.90
Pd 46 4d10 2.20 137 1555 12.00
Pt 78 6s25d84f14 2.20 139 1768 21.50
We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF2. In contrast to Fe, Ru and Os form a series of fluorides up to RuF6 and OsF7. The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe2O3, and the mixed-valent Fe3O4 (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO2) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block.
Higher oxidation states become less stable across the d-block, but more stable down a group.
Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S22−), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S2−). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character.
The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe3C), which is used to strengthen steel. At high temperatures, Fe3C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH0.5. Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H2 but essentially impermeable to all other gases, including He. Consequently, diffusion of H2 through Pd is an effective method for separating hydrogen from other gases.
Group 11 (Cu, Ag, and Au)
The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times.
This 1 kg gold nugget was found in Australia; in 2005, it was for sale in Hong Kong at an asking price of more than US\$64,000.
Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe2). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer.
The Chuquicamata copper mine in northern Chile, the world’s largest open-pit copper mine, is 4.3 km long, 3 km wide, and 825 m deep. Each gigantic truck in the foreground (and barely visible in the lower right center) can hold 330 metric tn (330,000 kg) of copper ore.
Some properties of the coinage metals are listed in Table $4$. The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns1(n − 1)d10 valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s25d10 valence electron configuration.
Table $4$: Some Properties of the Elements of Groups 11 and 12
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
11 Cu 29 4s13d10 1.90 128 1085 8.96
Ag 47 5s14d10 1.93 144 962 10.50
Au 79 6s15d104f14 2.40 144 1064 19.30
12 Zn 30 4s23d10 1.65 134 420 7.13
Cd 48 5s24d10 1.69 149 321 8.69
Hg 80 6s25d104f14 1.90 151 −38.8 13.53
All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O2 at high temperatures to produce Cu2O and with sulfur to form Cu2S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag2S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO3 and in hot concentrated H2SO4, while Au dissolves in the 3:1 HCl:HNO3 mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN)2] ions, a reaction that is used to separate gold from its ores.
Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au.
All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au3+ and Cu2+). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF2. In contrast, all the gold trihalides (AuX3) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements.
Group 12 (Zn, Cd, and Hg)
We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries.
As shown in Table $4$, the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns2(n − 1)d10 valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg22+.
The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble.
All the possible group 12 dihalides (MX2) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl2). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d10 subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO3 and H2SO4. All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur.
Example $1$
For each reaction, explain why the indicated products form.
1. TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(aq)
2. WO3(s) + 3C(s) + 3Cl2(g) $\xrightarrow{\Delta}$ WCl6(s) + 3CO(g)
3. Sc2O3(s) + 2OH(aq) + 3H2O(l) → 2[Sc(OH)4](aq)
4. 2KMnO4(aq) + 2H2SO4(l) → Mn2O7(l) + 2KHSO4(soln) + H2O(soln)
5. 4CrCl2(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) + 8Cl(aq) + 2H2O(l)
Given: balanced chemical equation
Asked for: why the indicated products form
Strategy:
Refer to the periodic trends in this section.
Solution:
1. The most stable oxidation state for Ti is +4, and neither reactant is a particularly strong oxidant or reductant; hence a redox reaction is unlikely. Similarly, neither reactant is a particularly strong acid or base, so an acid–base reaction is unlikely. Because TiCl4 contains Ti in a relatively high oxidation state (+4), however, it is likely to be rather covalent in character, with reactivity similar to that of a semimetal halide such as SiCl4. Covalent halides tend to hydrolyze in water to produce the hydrohalic acid and either the oxide of the other element or a species analogous to an oxoacid.
2. This reaction involves the oxide of a group 6 metal in its highest oxidation state (WO3) and two elements, one of which is a reductant (C) and the other an oxidant (Cl2). Consequently, some sort of redox reaction will occur. Carbon can certainly react with chlorine to form CCl4, and WO3 is a potential source of oxygen atoms that can react with carbon to produce CO, which is very stable. If CO is one of the products, then it seems likely that the other product will contain the metal and chlorine. The most likely combination is WCl6 (leaving the oxidation state of the metal unchanged).
3. One of the reactants is a strong base (OH), so an acid–base reaction is likely if the other reactant is acidic. Because oxides like Sc2O3, in which the metal is in an intermediate oxidation state, are often amphoteric, we expect Sc2O3 to dissolve in base to form a soluble hydroxide complex.
4. Concentrated sulfuric acid is both an oxidant and a strong acid that tends to protonate and dehydrate other substances. The permanganate ion already contains manganese in its highest possible oxidation state (+7), so it cannot be oxidized further. A redox reaction is impossible, which leaves an acid–base reaction as the most likely alternative. Sulfuric acid is likely to protonate the terminal oxygen atoms of permanganate, allowing them to be lost as water.
5. Molecular oxygen is an oxidant, so a redox reaction is likely if the other reactant can be oxidized. Because chromous chloride contains chromium in its lowest accessible oxidation state, a redox reaction will occur in which Cr2+ ions are oxidized and O2 is reduced. In the presence of protons, the reduction product of O2 is water, so we need to determine only the identity of the oxidation product of Cr2+. Chromium forms compounds in two common higher oxidation states: the Cr3+ ion, which is the most stable, and the [Cr2O7]2− ion, which is a more powerful oxidant than O2. We therefore predict that the reaction will form Cr3+(aq) and water.
Exercise $1$
Predict the products of each reactions and then balance each chemical equation.
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$
2. Na2Cr2O7(aq) + H2SO4(l) →
3. FeBr2(aq) + O2(g) $\xrightarrow{\mathrm{H^+}}$
4. VBr4(l) + H2O(l) →
5. ZrO2(s) + C(s) + Cl2(g) $\xrightarrow{\Delta}$
Answer
1. Cr2+(aq) + Fe3+(aq) $\xrightarrow{\mathrm{H^+}}$ Cr3+(aq) + Fe2+(aq)
2. Na2Cr2O7(aq) + 2H2SO4(l) → 2NaHSO4(soln) + H2O(soln) + 2CrO3(s)
3. 4FeBr2(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l) + 8Br(aq)
4. VBr4(l) + H2O(l) → VO2+(aq) + 4Br(aq) + 2H+(aq)
5. ZrO2(s) + 2C(s) + 2Cl2(g) $\xrightarrow{\Delta}$ ZrCl4(s) + 2CO(g)
Summary
The elements tend to become more polarizable going across the d block and higher oxidation states become less stable; higher oxidation states become more stable going down a group. The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M2O3). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr2O3 is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.4%3A_First-Row_Transition_Metal_Elements_-_Scandium_to_Manganese.txt |
The Iron Triad is composed of three elements: iron (Fe), cobalt (Co), and nickel (Ni), which share similar chemical and physical characteristics. They are found adjacent to each other in period 4 of the periodic table.
Introduction
The Iron Triad is known for possessing ferromagnetic elements similar to gadolinium (Gd), and neodymium (Nd). These types of ferromagnetic elements have the ability to create a large magnetic pole due to their unpaired electrons. When one of these elements is inside an environment where the temperature is at its individual Curie Point (Tc), however, the specific paramagnetic lining of atoms is broken down by the energy within the element and ferromagnetism is lost. The (Tc)'s for iron, cobalt, and nickel are 768°C, 1121°C, and 354°C respectively and are taken advantage of to make use of these elements in industry. In addition, elements in the iron triad are commonly combined with carbon and each other to create various types of alloys. It is because of these magnetic properties and use in alloys that the three elements are typically grouped together and labeled as the "Iron Triad". While very similar in magnetic properties and reaction, these elements are also very unique and used differently in both nature and industry.
Iron
Iron (Fe) is a transition metal with an atomic weight of 55.845 and an atomic number of 26. Its most common and comfortable oxidation state is +3 and is usually a shiny silver color. While it can be found in the sun and the stars, it is extremely common on planet earth and its abundance in the earth’s crust is 4.7%. The pure metal form of iron is rarely found due to its ability to easily react with other metals and environments.
Iron's most common use in worldwide industry is to be used in alloys and steels for everyday use, some which even contain other elements in the Triad. By combining metals like iron with carbon to make alloys, very strong materials can be made that are used for steel housings, supports in television tubes, and other important products. Pure iron is available commercially in the United States and production of about 500 million tons of Iron is created every year throughout the world.
From a biological perspective, iron is essential to organisms since it’s needed for the creation of hemoglobin. Humans need hemoglobin since these red pigments supply oxygen to our body through red blood cells. To increase intake of iron, foods like eggs, groundnuts and green leafy vegetables should be consumed. Without iron, our bodies would not be able to form hemoglobin and an organism would fail to survive.
Cobalt
Cobalt (Co) is a transition metal with an atomic weight of 58.93 and an atomic number of 27, right in between iron and nickel. Cobalt, however, is not as abundant as iron and only makes up of about 0.0020% of the Earth’s crust. This makes cobalt a little more rare and valuable than the other members of the triad because it is still used heavily in international industry. It is for this reason that this hard gray element is mined for and traded around the world.
Like iron, this element is commonly combined with other metals to create alloys. Superalloys that contain cobalt are used for parts of gas turbine engines that benefit both commercial and military devices. In current times, however, the rise of rechargeable batteries has made demand for the element rise to new heights since it is commonly used for electrodes in this relatively new technology. While these are its most popular applications, cobalt is also utilized in the production of petroleum, dyes, magnets, and electronics due to its ferromagnetic properties.
Cobalt and its alloys have recently made strides in the biomedical sphere by playing a large role in neurological, dental, orthopedic, and cardiovascular implant devices. This is due to the fact that the alloys function well with their corrosion resistance, mechanical properties, and biocompatibility.
Nickel
Nickel (Ni), a transition metal with an atomic weight of 58.69 and an atomic number of 28, is usually recognized by its silvery shine with a hint of gold color. It was first discovered by Axel Fredrik Cronstedt in 1751 and named nickel from the ore "kupfernickel". Although nickel is not extremely common, it still ranks as #24 on the list of the most abundant elements on earth.
This element is most abundant in Canada and about 300 million pounds are used in America every year. Nickel is most commonly used as a key ingredient in low alloy steels, stainless steels, and cast irons. In fact, four out of five of these 300 million pounds of Nickel go to this type of alloy making. Like cobalt, it shares the unique characteristic of corrosion resistance and is ideal for corrosion and heat resistant coatings, magnetic alloys, and controlled-expansion alloys.
Over 2000 years ago, the Bactrian civilization in Western Asia used a 75:25 alloy of copper and nickel for its coins. A modern US nickel has the same composition, but a modern Canadian nickel is nickel-plated steel and contains only 2.5% nickel by mass.
New nickel-cadmium rechargeable batteries have also brought the use of nickel into a new light. Like cobalt, the turning of the millennium and the increased use of rechargeable batteries has given nickel a new major use that will likely surpass others in the future. Nickel-cadmium batteries have improved the performance of cordless power tools, portable computers, and other portable electronic devices.
Table \(1\): Chemical and Physical Properties of Iron Triad Elements
Iron
Cobalt
Nickel
Atomic Number
26
27
28
Mass
55.85
58.93
58.69
Electron Configuration
3d64s2
3d74s2
3d84s1
Metallic radius, pm
124
125
125
Ionization Energy, kJ mol-1
762.5
760.4
737.1
• First
759
758
737
• Second
1561
1646
1753
• Third
2957
3232
3393
E°, Vb
-0.440
-0.277
-0.257
Common (+) Oxidation States
2, 3, 6
2, 3
2, 3
Melting Point (°C)
1530
1495
1455
Boiling Point (°C)
2862
2927
2732
Density, g cm-3
7.87
8.90
8.91
Electrical Conductivitye
16
25
23 | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.5%3A_The_Iron_Triad%3A_Iron_Cobalt_and_Nickel.txt |
The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times.
Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe2). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer.
Some properties of the coinage metals are listed in Table \(1\). The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns1(n − 1)d10 valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s25d10 valence electron configuration.
Table \(1\): Some Properties of the Elements of Groups 11
Group Element Z Valence Electron Configuration Electronegativity Metallic Radius (pm) Melting Point (°C) Density (g/cm3)
11 Cu 29 4s13d10 1.90 128 1085 8.96
Ag 47 5s14d10 1.93 144 962 10.50
Au 79 6s15d104f14 2.40 144 1064 19.30 | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.6%3A_Group_11%3A_Copper_Silver_and_Gold.txt |
We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries.
As shown in Table \(4\), the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns2(n − 1)d10 valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg22+.
Note
The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble.
All the possible group 12 dihalides (MX2) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl2). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d10 subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO3 and H2SO4. All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur.
Summary
The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.7%3A_Group_12%3A_Zinc_Cadmium_and_Mercury.txt |
The Lanthanides consist of the elements in the f-block of period six in the periodic table. While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements.
Introduction
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.
Like any other series in the periodic table, such as the Alkali metals or the Halogens, the Lanthanides share many similar characteristics. These characteristics include the following:
• Similarity in physical properties throughout the series
• Adoption mainly of the +3 oxidation state. Usually found in crystalline compounds)
• They can also have an oxidation state of +2 or +4, though some lanthanides are most stable in the +3 oxidation state.
• Adoption of coordination numbers greater than 6 (usually 8-9) in compounds
• Tendency to decreasing coordination number across the series
• A preference for more electronegative elements (such as O or F) binding
• Very small crystal-field effects
• Little dependence on ligands
• Ionic complexes undergo rapid ligand-exchange
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 \(1\)). 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.
Electron Configuration
Similarly, the Lanthanides have similarities in their electron configuration, which explains most of the physical similarities. These elements are different from the main group elements in the fact that they have electrons in the f orbital. After Lanthanum, the energy of the 4f sub-shell falls below that of the 5d sub-shell. This means that the electron start to fill the 4f sub-shell before the 5d sub-shell.
The electron configurations of these elements were primarily established through experiments (Table \(1\)). The technique used is based on the fact that each line in an emission spectrum reveals the energy change involved in the transition of an electron from one energy level to another. However, the problem with this technique with respect to the Lanthanide elements is the fact that the 4f and 5d sub-shells have very similar energy levels, which can make it hard to tell the difference between the two.
Table \(1\): Electron Configurations of the Lanthanide Elements
Symbol Idealized Observed Symbol Idealized Observed
La 5d16s2 5d16s2 Tb 4f85d16s2 4f9 6s2 or 4f85d16s2
Ce 4f15d16s2 4f15d16s2 Dy 4f95d16s2 4f10 6s2
Pr 4f25d16s2 4f3 6s2 Ho 4f105d16s2 4f11 6s2
Nd 4f35d16s2 4f4 6s2 Er 4f115d16s2 4f12 6s2
Pm 4f45d16s2 4f5 6s2 Tm 4f125d16s2 4f13 6s2
Sm 4f55d16s2 4f6 6s2 Yb 4f135d16s2 4f14 6s2
Eu 4f65d16s2 4f7 6s2 Lu 4f145d16s2 4f145d16s2
Gd 4f75d16s2 4f75d16s2
Another important feature of the Lanthanides is the Lanthanide Contraction, in which the 5s and 5p orbitals penetrate the 4f sub-shell. This means that the 4f orbital is not shielded from the increasing nuclear change, which causes the atomic radius of the atom to decrease that continues throughout the series.
Properties and Chemical Reactions
One property of the Lanthanides that affect how they will react with other elements is called the basicity. Basicity is a measure of the ease at which an atom will lose electrons. In another words, it would be the lack of attraction that a cation has for electrons or anions. For the Lanthanides, the basicity series is the following:
La3+ > Ce3+ > Pr3+ > Nd3+ > Pm3+ > Sm3+ > Eu3+ > Gd3+ > Tb3+ > Dy3+ > Ho3+ > Er3+ > Tm3+ > Yb3+ > Lu3+
In other words, the basicity decreases as the atomic number increases. Basicity differences are shown in the solubility of the salts and the formation of the complex species. Another property of the Lanthanides is their magnetic characteristics. The major magnetic properties of any chemical species are a result of the fact that each moving electron is a micromagnet. The species are either diamagnetic, meaning they have no unpaired electrons, or paramagnetic, meaning that they do have some unpaired electrons. The diamagnetic ions are: La3+, Lu3+, Yb2+ and Ce4+. The rest of the elements are paramagnetic.
Table \(2\): Properties of the Lathanides
Symbol Ionization Energy (kJ/mol) Melting Point (°C)
Boiling Point (°C)
La 538 920 3469
Ce 527 795 3468
Pr 523 935 3127
Nd 529 1024 3027
Pm 536
Sm 543 1072 1900
Eu 546 826 1429
Gd 593 1312 3000
Tb 564 1356 2800
Dy 572 1407 2600
Ho 581 1461 2600
Er 589 1497 2900
Tm 597 1545 1727
Yb 603 824 1427
Lu 523 1652 3327
Metals and their Alloys
The metals have a silvery shine when freshly cut. However, they can tarnish quickly in air, especially Ce, La and Eu. These elements react with water slowly in cold, though that reaction can happen quickly when heated. This is due to their electropositive nature. The Lanthanides have the following reactions:
• oxidize rapidly in moist air
• dissolve quickly in acids
• reaction with oxygen is slow at room temperature, but they can ignite around 150-200 °C
• react with halogens upon heating
• upon heating, react with S, H, C and N
Periodic Trends: Size
The size of the atomic and ionic radii is determined by both the nuclear charge and by the number of electrons that are in the electronic shells. Within those shells, the degree of occupancy will also affect the size. In the Lanthanides, there is a decrease in atomic size from La to Lu. This decrease is known as the Lanthanide Contraction. The trend for the entire periodic table states that the atomic radius decreases as you travel from left to right. Therefore, the Lanthanides share this trend with the rest of the elements.
Table \(3\): Periodic Trends
Element Atomic Radius (pm) Ionic Radius (3+) Element Atomic Radius (pm) Ionic Radius (3+)
La 187.7 106.1 Tb 178.2 92.3
Ce 182 103.4 Dy 177.3 90.8
Pr 182.8 101.3 Ho 176.6 89.4
Nd 182.1 99.5 Er 175.7 88.1
Pm 181 97.9 Tm 174.6 89.4
Sm 180.2 96.4 Yb 194.0 85.8
Eu 204.2 95.0 Lu 173.4 84.8
Gd 180.2 93.8
Color and Light Absorbance
The color that a substance appears is the color that is reflected by the substance. This means that if a substance appears green, the green light is being reflected. The wavelength of the light determines if the light with be reflected or absorbed. Similarly, the splitting of the orbitals can affect the wavelength that can be absorbed (Table \(4\)). This is turn would be affected by the amount of unpaired electrons.
Table \(4\): : Unpaired Electrons and Color
Ion Unpaired Electrons Color Ion Unpaired Electrons Color
La3+ 0 Colorless Tb3+ 6 Pale Pink
Ce3+ 1 Colorless Dy3+ 5 Yellow
Pr3+ 2 Green Ho3+ 4 Pink; yellow
Nd3+ 3 Reddish Er3+ 3 Reddish
Pm3+ 4 Pink; yellow Tm3+ 2 Green
Sm3+ 5 Yellow Yb3+ 1 Colorless
Eu3+ 6 Pale Pink Lu3+ 0 Colorless
Gd3+ 7 Colorless
Occurrence in Nature
Each known Lanthanide mineral contains all the members of the series. However, each mineral contains different concentrations of the individual Lanthanides. The three main mineral sources are the following:
• Monazite: contains mostly the lighter Lanthanides. The commercial mining of monazite sands in the United States is centered in Florida and the Carolinas
• Xenotime: contains mostly the heavier Lanthanides
• Euxenite: contains a fairly even distribution of the Lanthanides
In all the ores, the atoms with a even atomic number are more abundant. This allows for more nuclear stability, as explained in the Oddo-Harkins rule. The Oddo-Harkins rule simply states that the abundance of elements with an even atomic number is greater than the abundance of elements with an odd atomic number. In order to obtain these elements, the minerals must go through a separating process, known as separation chemistry. This can be done with selective reduction or oxidation. Another possibility is an ion-exchange method.
Note: The Oddo-Harkins Rule
The abundance of elements with an even atomic number is greater than the abundance of elements with an odd atomic number.
Applications
The pure metals of the Lanthanides have little use. However, the alloys of the metals can be very useful. For example, the alloys of Cerium have been used for metallurgical applications due to their strong reducing abilities. The Lanthanides can also be used for ceramic purposes. The almost glass-like covering of a ceramic dish can be created with the lanthanides. They are also used to improve the intensity and color balance of arc lights.
Like the Actinides, the Lanthanides can be used for nuclear purposes. The hydrides can be used as hydrogen-moderator carriers. The oxides can be used as diluents in nuclear fields. The metals are good for being used as structural components. The can also be used for structural-alloy-modifying components of reactors. It is also possible for some elements, such as Tm, to be used as portable x-ray sources. Other elements, such as Eu, can be used as radiation sources. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.8%3A_Lanthanides.txt |
Learning Objectives
• To become familiar with the properties of superconductors.
Contributors
• Anonymous
Video from Wisconsin MRSEC @ YouTube
23.E: Exercises
23.4: First-Row Transition Metal Elements: Scandium to Manganese
Conceptual Problems
1. The valence electron configuration of Sc is 4s23d1, yet it does not lose the d1 electron to form 1+ ion. Why?
2. Give the ground-state electron configuration for Mn, Mn2+, Au, Au3+, Mo, and Mo5+.
3. A great deal of research is being conducted on the use of titanium alloys as materials for transportation applications (airplanes, ships, automobiles, etc.). Why is Ti particularly suited to this purpose? What is the primary disadvantage that needs to be overcome?
4. Both Ti and Ta are used for bioimplants because they are highly resistant to corrosion. Their uses also extend to other applications where corrosion must be avoided. Why are these metals so corrosion resistant?
5. Give two reasons why Zr is used to make the casing for UO2 fuel in water-cooled nuclear reactors.
6. Why is chromium added to steel to form stainless steel? What other elements might also be effective additives for this purpose? Why did you select these elements?
7. Tungsten is commonly used as the filament in electric light bulbs. Why is tungsten particularly suited to this purpose?
8. Palladium metal is used to purify H2 by removing other gases. Why is Pd so permeable to H2?
9. Give the valence electron configuration for Sc, Fe, Re, Ag, Zr, Co, V, Pr, Hg, Cr, Ni, Ce, Cu, and Tb.
10. The Hg–Hg bond is much stronger than the Cd–Cd bond, reversing the trend found among the other transition-metal groups. Explain this anomaly.
11. Which of the transition metals are most likely to form compounds in the +6 oxidation state? Why?
Structure and Reactivity
1. Do you expect TiCl4, TiCl3, TiCl2, and Ti to be oxidized, reduced, or hydrolyzed by water? Explain your reasoning.
2. The atomic radii of vanadium, niobium, and tantalum are 134 pm, 146 pm, and 146 pm, respectively. Why does the radius increase from vanadium to niobium but not from niobium to tantalum?
3. The most stable oxidation state for the metals of groups 3, 4, and 5 is the highest oxidation state possible. In contrast, for nearly all the metals of groups 8, 9, and 10, intermediate oxidation states are most stable. Why?
4. Most of the transition metals can form compounds in multiple oxidation states. Ru, for example, can form compounds in the +8, +6, +4, +3, +2, and −2 oxidation states. Give the valence electron configuration of Ru in each oxidation state. Why does Ru exhibit so many oxidation states? Which ones are the most stable? Why?
5. Predict the maximum oxidation states of Cu, Cr, Mo, Rh, Zr, Y, Ir, Hg, and Fe.
6. In the +4 oxidation state all three group 7 metals form the dioxides (MO2). Which of the three metals do you expect to form the most stable dioxide? Why?
7. Of [Fe(H2O)6]+, OsBr7, CoF4, PtF6, FeI3, [Ni(H2O)6]2+, OsO4, IrO4, NiO, RhS2, and PtH, which do not exist? Why?
8. The chemistry of gold is somewhat anomalous for a metal. With which elements does it form the Au ion? Does it form a stable sulfide?
9. Of Os4+, Pt10+, Cr6+, Ir9+, Ru8+, Re7+, and Ni10+, which are not likely to exist? Why?
10. Of Ag2S, Cu2S, AuI3, CuF, AuF, AgN, and AuO, which are not likely to exist?
11. There is evidence that the Au ion exists. What would be its electron configuration? The compound CsAu has been isolated; it does not exhibit a metallic luster and does not conduct electricity. Is this compound an alloy? What type of bonding is involved? Explain your answers.
12. Of Hg2Cl2, ZnO, HgF2, Cs2[ZnCl5], and HgNa, which are not likely to exist?
13. Mercurous oxide (Hg2O) and mercurous hydroxide [Hg2(OH)2] have never been prepared. Why not? What products are formed if a solution of aqueous sodium hydroxide is added to an aqueous solution of mercurous nitrate [Hg2(NO3)2]?
14. Arrange Fe2O3, TiO2, V2O5, MoO3, Mn2O7, and OsO4 in order of increasing basicity.
15. Mercurous sulfide has never been prepared. What products are formed when H2S gas is bubbled through an aqueous solution of mercurous nitrate?
16. Arrange Sc2O3, VO, V2O5, Cr2O3, Fe2O3, Fe3O4, and ZnO in order of increasing acidity.
17. Arrange Sc2O3, V2O5, CrO3, Mn2O7, MnO2, and VO2 in order of increasing basicity.
18. Predict the products of each reaction and then balance each chemical equation.
1. Ti + excess Cl2, heated
2. V2O5 in aqueous base
3. K2Cr2O7 + H2SO4
4. RuBr2 + O2, in water
5. [CrO4]2− in aqueous acid
6. Hg2+ + Hg, in aqueous acid
1. Predict the products of each reaction and then balance each chemical equation.
1. AgBr + hν
2. W + excess Cl2, heated
3. CuO + H2, heated
4. Fe2O3 in aqueous acid
5. RhCl3 + NH3, in water
6. Fe2+ + [MnO4], in water
1. What do you predict to be the coordination number of Pt2+, Au+, Fe3+, and Os2+?
2. Of La, Sc, Cr, and Hf, which is most likely to form stable compounds in the +4 oxidation state? Why?
3. Give the most common oxidation state for Y, W, Ru, Ag, Hg, Zn, Cr, Nb, and Ti.
4. Give the most common oxidation state for Os, Cd, Hf, V, Ac, Ni, Mn, Pt, and Fe.
5. Give the highest oxidation state observed for Zr, Fe, Re, Hg, Ni, La, and Mo.
6. Give the highest oxidation state observed for Ag, Co, Os, Au, W, and Mn.
7. Arrange La, Cs, Y, Pt, Cd, Mo, Fe, Co, and Ir in order of increasing first ionization energy.
8. Briefly explain the following trends within the transition metals.
1. Transition-metal fluorides usually have higher oxidation states than their iodides.
2. For a given metal, the lowest-oxidation-state oxide is basic and the highest-oxidation-state oxide is acidic.
3. Halides of the transition metals become more covalent with increasing oxidation state and are more prone to hydrolysis.
1. Propose a method to prepare each of the following compounds: TiCl4[(CH3)2O]2, Na2TiO3, V2O5, and Na2Cr2O7.
2. Of the group 5 elements, which
1. has the greatest tendency to form ions in the lower oxidation states?
2. has the greatest tendency to form a polymeric fluoride?
3. does not form an MX2 species?
4. forms the most basic oxide?
5. has the greatest tendency to form complexes with coordination numbers higher than 6?
Answers
1. Pt10+, Ir9+, and Ni10+. Because ionization energies increase from left to right across the d block, by the time you reach group 9, it is impossible to form compounds in the oxidation state that corresponds to loss of all the valence electrons.
1. Hg22+(aq) + H2S(g) → Hg(l) + HgS(s) + 2H+(aq)
1. Mn2O7 < CrO3 < V2O5 < MnO2 ≈ VO2 < Sc2O3
1. 2AgBr(s) $\xrightarrow{\mathrm{light}}$ 2Ag(s) + Br2(l)
2. W(s) + excess Cl2(g) $\xrightarrow{\Delta}$ WCl6(s)
3. CuO(s) + H2(g) $\xrightarrow{\Delta}$ Cu(s) + H2O(g)
4. Fe2O3(s) + 6H+(aq) → 2Fe3+(aq) + 3H2O(l)
5. RhCl3(s) + 6NH3(aq) → [Rh(NH3)6]Cl3(aq)
6. 3Fe2+(aq) + MnO4(aq) + 7H2O(aq) → Fe3+(aq) + MnO2(s) + 5H+(l)
1. Os, +4; Cd, +2; Hf, +4; V, +5; Ac, +3; Ni, +2; Mn, +2; Pt, +2 & +4; Fe, +2 & +3
1. Ag, +3; Co, +4; Os, +8; Au, +5; W, +6; Mn, +7
23.5: The Iron Triad: Iron, Cobalt, and Nickel
Practice Problems
1. Fe2O3(s) + 3CO(g) --> 2Fe (l) + 2CO2(g)
2. 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) --> 5Fe3+(aq) + Mn2+(aq) +4H2O(l)
3. [Fe(H2O)6]3+ +SCN-(aq) --> [FeSCN(H2O)5]2+ + H2O(l)
4. 4FeS + 7O2 --> 2Fe2O3 + 4SO2
5. Ni + H2O --> No Reaction
23.8: Lanthanides
Practice Problems
1. Which elements are considered to be Lanthanides?
2. How do the Lanthanides react with oxygen?
3. What causes the Lanthanide Contraction?
4. Why do Lanthanides exhibit strong electromagnetic and light properties?
5. What do the Lanthanides have in common with the Noble Gases?
Answers
1. Elements Lanthanum (57) through Lutetium (71) on the periodic table are considered to be Lanthanides.
2. Lanthanides tend to react with oxygen to form oxides. The reaction at room temperature can be slow while heat can cause the reaction to happen rapidly.
3. The Lanthanide Contraction refers to the decrease in atomic size of the elements in which electrons fill the f-subshell. Since the f sub-shell is not shielded, the atomic size will decrease as the nuclear charge still increases.
4. Lanthanides exhibit strong electromagnetic and light properties because of the presence of unpaired electrons in the f-orbitals. The majority of the Lanthanides are paramagnetic, which means that they have strong magnetic fields.
5. Both the Lanthanides and Noble Gases tend to bind with more electronegative atoms, such as Oxygen or Fluorine.
23.9: High Temperature Superconductors
1. Why does the BCS theory predict that superconductivity is not possible at temperatures above approximately 30 K?
2. How does the formation of Cooper pairs lead to superconductivity?
Solution
1. According to BCS theory, the interactions that lead to formation of Cooper pairs of electrons are so weak that they should be disrupted by thermal vibrations of lattice atoms above about 30 K. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.9%3A_High_Temperature_Superconductors.txt |
• 24.1: Werner’s Theory of Coordination Compounds
A metal complex consists of a central metal atom or ion that is bonded to one or more ligands, which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, positively charged, or negatively charged. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
• 24.2: Ligands
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Ligands are ions or neutral molecules that bond to a central metal atom or ion. Ligands act as Lewis bases and the central atom acts as a Lewis acid.
• 24.3: Complex Ion Nomenclature
Coordination complexes have their own classes of isomers, different magnetic properties and colors, and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal Anions Neutrals]±Charge, while names are written Prefix Ligands Metal(Oxidation State).
• 24.4: Isomerism in Coordination Complexes
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 dimens
• 24.5: Bonding in Complex Ions: Crystal Field Theory
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting depends on the nature of the ligands bonded to the metal.
• 24.6: Magnetic Properties of Coordination Compounds and Crystal Field Theory
The magnetic properties of a compound can be determined from its electron configuration and the size of its atoms. Because magnetism is generated by electronic spin, the number of unpaired electrons in a specific compound indicates how magnetic the compound is. In this section, the magnetism of the d-block elements (or transition metals) are evaluated. These compounds tend to have a large number of unpaired electrons.
• 24.7: Color and the Colors of Complexes
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed.
• 24.8: Aspects of Complex-Ion Equilibria
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
• 24.9: Acid-Base Reactions of Complex Ions
• 24.10: Some Kinetic Considerations
Transition metal complexes which undergo rapid substitution of one ligand for another are labile, whereas complexes in which substitution proceed slowly or not at all are inert. For an inert complex, it is a large activation energy which prevents ligand substitution. Inert complexes are therefore kinetically stable compounds.
• 24.E: Exercises
24: Complex Ions and Coordination Compounds
Learning Objectives
• To know the most common structures observed for metal complexes.
• To predict the relative stabilities of metal complexes with different ligands
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more ligands (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions.
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: $\ce{KFe2(CN)6}$. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as $\ce{AlF3·3KF}$, $\ce{Fe(CN)2·4KCN}$, and $\ce{ZnCl2·2CsCl}$, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should $\ce{AlF3·3KF}$ exist but not $\ce{AlF3·4KF}$ or $\ce{AlF3·2KF}$? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, $\ce{CoCl3·6NH3}$, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, $\ce{CoCl3·6NH3}$, $\ce{CoCl3·5NH3}$, $\ce{CoCl3·4NH3}$, and $\ce{CoCl3·3NH3}$ were all known and had very different properties, but despite all attempts, chemists could not prepare $\ce{CoCl3·2NH3}$ or $\ce{CoCl3·NH3}$.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in Table $1$. The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in Table $1$ showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Alfred Werner (1866–1919)
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table $1$: Werner’s Data on Complexes of Ammonia with $PtCl_4$
Complex Conductivity (ohm−1) Number of Ions per Formula Unit Number of Cl Ions Precipitated by Ag+
PtCl4·6NH3 523 5 4
PtCl4·5NH3 404 4 3
PtCl4·4NH3 299 3 2
PtCl4·3NH3 97 2 1
PtCl4·2NH3 0 0 0
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligand-metal bonds bound to the metal ion. If $\ce{Pt}$ had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the $\ce{PtCl4·NH3}$ adducts by the following reactions, where the metal complex is enclosed in square brackets:
\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \[4pt] \mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}
Further work showed that the two missing members of the series—[Pt(NH3)Cl5] and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (Figure $\PageIndex{1a}$). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure $\PageIndex{1b}$. His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Example $1$
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA4B2 complex were known tell you about the three-dimensional structures of these complexes?
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Strategy:
Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand.
Solution
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise $1$
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer
square planar, 2; tetrahedral, 1
Structures of Metal Complexes
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Coordination Number 2
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Coordination Number 3
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3 ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Coordination Number 4
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4] and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
Coordination Number 5
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
Coordination Number 6
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
Coordination Number 7
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
Coordination Number 8
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
Coordination Number 9
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
Key Takeaways
• Coordination compounds are a major feature of the chemistry of over half the elements.
• Coordination compounds have important roles as industrial catalysts in controlling reactivity, and they are essential in biochemical processes.
Summary
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.01%3A_Werners_Theory_of_Coordination_Compounds.txt |
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Ligands are ions or neutral molecules that bond to a central metal atom or ion. Ligands act as Lewis bases (electron pair donors), and the central atom acts as a Lewis acid (electron pair acceptor). Ligands have at least one donor atom with an electron pair used to form covalent bonds with the central atom.
The term ligand come from the latin word ligare (which meaning to bind) was first used by Alfred Stock in 1916 in relation to silicon chemistry. Ligands can be anions, cations, or neutral molecules. Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced, hence the idea of bite angle etc. A monodentate ligand has only one donor atom used to bond to the central metal atom or ion.
Monodentate Ligands
The term "monodentate" can be translated as "one tooth," referring to the ligand binding to the center through only one atom. Some examples of monodentate ligands are: chloride ions (referred to as chloro when it is a ligand), water (referred to as aqua when it is a ligand), hydroxide ions (referred to as hydroxo when it is a ligand), and ammonia (referred to as ammine when it is a ligand).
Bidentate Ligands
Bidentate ligands have two donor atoms which allow them to bind to a central metal atom or ion at two points. Common examples of bidentate ligands are ethylenediamine (en), and the oxalate ion (ox). Shown below is a diagram of ethylenediamine: the nitrogen (blue) atoms on the edges each have two free electrons that can be used to bond to a central metal atom or ion.
Polydentate Ligands
Polydentate ligands range in the number of atoms used to bond to a central metal atom or ion. EDTA, a hexadentate ligand, is an example of a polydentate ligand that has six donor atoms with electron pairs that can be used to bond to a central metal atom or ion.
Unlike polydentate ligands, ambidentate ligands can attach to the central atom in two places. A good example of this is thiocyanate, \(SCN^−\), which can attach at either the sulfur atom or the nitrogen atom.
Example \(1\)
Draw metal complexes using the ligands below and metal ions of your choice.
Chelation
Chelation is a process in which a polydentate ligand bonds to a metal ion, forming a ring. The complex produced by this process is called a chelate, and the polydentate ligand is referred to as a chelating agent. The term chelate was first applied in 1920 by Sir Gilbert T. Morgan and H.D.K. Drew, who stated: "The adjective chelate, derived from the great claw or chela (chely- Greek) of the lobster or other crustaceans, is suggested for the caliperlike groups which function as two associating units and fasten to the central atom so as to produce heterocyclic rings." As the name implies, chelating ligands have high affinity for metal ions relative to ligands with only one binding group (which are called monodentate = "single tooth") ligands. Both Ethylenediamine (Figure \(2\)) and Ethylenediaminetetraaceticacid acid (Figure \(3\)) are examples of chelating agents, but many others are commonly found in the inorganic laboratory.
The Chelate Effect
The chelate effect is the enhanced affinity of chelating ligands for a metal ion compared to the affinity of a collection of similar nonchelating (monodentate) ligands for the same metal.
The macrocyclic effect follows the same principle as the chelate effect, but the effect is further enhanced by the cyclic conformation of the ligand. Macrocyclic ligands are not only multi-dentate, but because they are covalently constrained to their cyclic form, they allow less conformational freedom. The ligand is said to be "pre-organized" for binding, and there is little entropy penalty for wrapping it around the metal ion. For example heme b is a tetradentate cyclic ligand that strongly complexes transition metal ions, including Fe+2 in hemogloben (Figure \(4\)).
Some other common chelating and cyclic ligands are shown below:
• Acetylacetonate (acac-, top) is an anionic bidentate ligand that coordinates metal ions through two oxygen atoms. Acac- is a hard base so it prefers hard acid cations. With divalent metal ions, acac-forms neutral, volatile complexes such as Cu(acac)2 and Mo(acac)2 that are useful for chemical vapor deposition (CVD) of metal thin films.
• 2,2'-Bipyridine 2,2'-Bipyridine (Figure \(5\): left) and related bidentate ligands such as 1,10-phenanthroline form propeller-shaped complexes with metals such as Ru2+. The [Ru(bpy)3]2+ complex is photoluminescent and can also undergo photoredox reactions, making it an interesting compound for both photocatalysis and artificial photosynthesis.
• Crown ethers such as 18-crown-6 2,2'-Bipyridine (Figure \(5\): center) are cyclic hard bases that can complex alkali metal cations. Crowns can selectively bind Li+, Na+, or K+ depending on the number of ethylene oxide units in the ring.
• The chelating properties of crown ethers are mimetic of the natural antibiotic valinomycin ( 2,2'-Bipyridine (Figure \(5\): right), which selectively transports K+ ions across bacterial cell membranes, killing the bacterium by dissipating its membrane potential. Like crown ethers, valinomycin is a cyclic hard base. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.02%3A_Ligands.txt |
Learning Objectives
• To learn the basis for complex ion and compound nomenclature
Coordination complexes have their own classes of isomers, different magnetic properties and colors, and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal ligands]±Charge, while names are written Prefix Ligands Metal (Oxidation State).
Introduction
According to the Lewis base theory, ligands are Lewis bases since they can donate electrons to the central metal atom. The metals, in turn, are Lewis acids since they accept electrons. Coordination complexes consist of a ligand and a metal center cation. The overall charge can be positive, negative, or neutral. Coordination compounds are complex or contain complex ions, for example:
• Complex Cation: \(\ce{[Co(NH3)6]^{3+}}\)
• Complex Anion: \(\ce{[CoCl4(NH3)2]^{-}}\)
• Neutral Complex: \(\ce{[CoCl3(NH3)3]}\)
• Coordination Compound: \(\ce{K4[Fe(CN)6]}\)
A ligand can be an anion or a neutral molecule that donates an electron pair to the complex (NH3, H2O, Cl-). The number of ligands that attach to a metal depends on whether the ligand is monodentate or polydentate. To begin naming coordination complexes, here are some things to keep in mind.
1. Ligands are named first in alphabetical order.
2. The name of the metal comes next.
3. The oxidation state of the metal follows, noted by a Roman numeral in parentheses (II, IV).
Rule 1: Anionic Ligands
Ligands that act as anions which end in "-ide" are replaced with an ending "-o" (e.g., Chloride → Chloro). Anions ending with "-ite" and "-ate" are replaced with endings "-ito" and "-ato" respectively (e.g., Nitrite → Nitrito, Nitrate → Nitrato).
Table \(1\): Anionic Monodentate Ligands
Molecular Formula Ligand Name Molecular Formula Ligand Name
F- Fluoro OH- Hydroxo
Cl- Chloro SO42- Sulfato
Br- Bromo S2O32- Thiosulfato
I- Iodo NO2- Nitrito-N-; Nitro
O2- Oxo ONO- Nitrito-O-; Nitrito
CN- Cyano SCN- Thiocyanato-S-; Thiocyanato
NC- Isocyano NCS- Thiocyanato-N-; Isothiocyanato
Rule 2: Neutral Ligands
Most neutral molecules that are ligands carry their normal name. The few exceptions are the first four on the chart: ammine, aqua, carbonyl, and nitrosyl.
Table \(2\): Select Neutral Monodentate Ligands. Note: Ammine is spelled with two m's when referring to a ligand. Amines are a class of organic nitrogen-containing compounds.
Molecular Formula of Ligand Ligand Name
NH3 Ammine
H2O Aqua
CO Carbonyl
NO Nitrosyl
CH3NH2 Methylamine
C5H5N Pyridine
Polydentate ligands follow the same rules for anions and neutral molecules.
Table \(3\): Select Polydentate ligands
Short name Extended name
en Ethylenediamine
ox2- Oxalato
EDTA4- Ethylenediaminetetraacetato
Rule 3: Ligand Multiplicity
The number of ligands present in the complex is indicated with the prefixes di, tri, etc. The exceptions are polydentates that have a prefix already in their name (en and EDTA4- are the most common). When indicating how many of these are present in a coordination complex, put the ligand's name in parentheses and use bis (for two ligands), tris (for three ligands), and tetrakis (for four ligands).
Table \(4\): Prefixes for indicating number of ligands in a complex.
Number of Ligands Monodentate Ligands Polydentate Ligands
1 mono -
2 di bis
3 tri tris
4 tetra tetrakis
5 penta pentakis
6 hexa hexakis
Prefixes always go before the ligand name; they are not taken into account when putting ligands in alphabetical order. Note that "mono" often is not used. For example, \(\ce{[FeCl(CO)2(NH3)3]^{2+}}\) would be called triamminedicarbonylchloroiron(III) ion. Remember that ligands are always named first, before the metal is.
Example \(1\)
What is the name of this complex ion: \(\ce{[CrCl2(H2O)4]^{+}}\)?
Solution
Let's start by identifying the ligands. The ligands here are Cl and H2O. Therefore, we will use the monodentate ligand names of "chloro" and "aqua". Alphabetically, aqua comes before chloro, so this will be their order in the complex's name. There are 4 aqua's and 2 chloro's, so we will add the number prefixes before the names. Since both are monodentate ligands, we will say "tetra[aqua]di[chloro]".
Now that the ligands are named, we will name the metal itself. The metal is Cr, which is chromium. Therefore, this coordination complex is called tetraaquadichlorochromium(III) ion. See the next section for an explanation of the (III).
Exercise \(1\)
What is the name of this complex ion: \(\ce{[CoCl_2(en)_2]^{+}}\)?
Answer
We take the same approach. There are two chloro and ethylenediamine ligands. The metal is Co, cobalt. We follow the same steps, except that \(en\) is a polydentate ligand with a prefix in its name (ethylenediamine), so "bis" is used instead of "di", and parentheses are added. Therefore, this coordination complex is called dichlorobis(ethylenediamine)cobalt(III) ion.
Rule 4: The Metals
When naming the metal center, you must know the formal metal name and the oxidation state. To show the oxidation state, we use Roman numerals inside parenthesis. For example, in the problems above, chromium and cobalt have the oxidation state of +3, so that is why they have (III) after them. Copper, with an oxidation state of +2, is denoted as copper(II). If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. Copper +2 will change into cuprate(II). The following change to their Latin names when part of an anion complex:
Table \(5\): Latin terms for Select Metal Ion
Transition Metal Latin
Iron Ferrate
Copper Cuprate
Tin Stannate
Silver Argentate
Lead Plumbate
Gold Aurate
The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).
Finally, when a complex has an overall charge, "ion" is written after it. This is not necessary if it is neutral or part of a coordination compound (Example \(3\)). Here are some examples with determining oxidation states, naming a metal in an anion complex, and naming coordination compounds.
Example \(2\)
What is the name of [Cr(OH)4]- ?
Solution
Immediately we know that this complex is an anion. There is only one monodentate ligand, hydroxide. There are four of them, so we will use the name "tetrahydroxo". The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". The oxidation state of the metal is 3 (x+(-1)4=-1). Write this with Roman numerals and parentheses (III) and place it after the metal to get tetrahydroxochromate(III) ion.
Exercise \(2\)
What is the name of \(\ce{[CuCl4]^{2-}}\)?
Answer
tetrachlorocuprate(II) ion
A last little side note: when naming a coordination compound, it is important that you name the cation first, then the anion. You base this on the charge of the ligand. Think of NaCl. Na, the positive cation, comes first and Cl, the negative anion, follows.
Example \(3\)
What is the name of \([\ce{Pt(NH3)4}][\ce{Pt(Cl)4}]\)?
Solution
NH3 is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH3 first, followed by the one with Cl (the same order as the formula). This coordination compound is called tetraammineplatinum(II) tetrachloroplatinate(II).
Exercise \(3\): The Nitro/Nitrito Ambidentate Ligand
What is the name of \(\ce{[CoCl(NO2)(NH3)4]^{+}}\) ?
Answer
This coordination complex is called tetraamminechloronitrito-N-cobalt(III). N comes before the O in the symbol for the nitrite ligand, so it is called nitrito-N. If an O came first, as in [CoCl(ONO)(NH3)4]+, the ligand would be called nitrito-O, yielding the name tetraamminechloronitrito-O-cobalt(III).
Nitro (for NO2) and nitrito (for ONO) can also be used to describe the nitrite ligand, yielding the names tetraamminechloronitrocobalt(III) and tetraamminechloronitritocobalt(III).
Writing Formulas of Coordination Complexes
While chemistry typically follow the nomenclature rules for naming complexes and compounds, there is disagreement with the rules for constructing formulas of inorganic complex. The order of ligand names in their formula has been ambiguous with different conventions being used (charged vs neutral, number of each ligand, etc.). In 2005, IUPAC adopted the recommendation that all ligand names in formulas be listed alphabetically (in the same way as in the naming convention) irrespective of the charge or number of each ligand type.
However, this rule is not adhered to in many chemistry laboratories. For practice, the order of the ligands in chemical formulas does not matter as long as you write the transition metal first, which is the stance taken here.
Examples \(4\)
Write the chemical formulas for:
1. Amminetetraaquachromium(II) ion
2. Amminesulfatochromium(II)
Solution
1. Amminetetraaquachromium(II) ion could be written as \(\ce{[Cr(H2O)4(NH3)]^{+2}}\) or \(\ce{[Cr(NH3)(H2O)4]^{+2}}\).
2. Amminesulfatochromium (II) could be written as \(\ce{[Cr(SO4)(NH3)]}\) or \(\ce{[Cr(NH3)(SO4)]}\).
Exercise \(4\)
Write the chemical formulas for
1. Amminetetraaquachromium (II) sulfate
2. Potassium hexacyanoferrate (III)
Answer
1. Amminetetraaquachromium (II) sulfate can be written as \(\ce{[Cr(H2O)4(NH3)]SO4}\). Although \(\ce{[Cr(NH3)(H2O)4]SO4}\) is also acceptable.
2. Potassium hexacyanoferrate (III) is be written as \(\ce{K3[Fe(CN)6]}\)
Contributors and Attributions
• Justin Hosung Lee (UCD), Sophia Muller (UCD) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.03%3A_Nomenclature.txt |
Learning Objectives
• Explain the differences of Structural and Geometric isomerization in a coordination complexes or complex ions.
• Define Ionization, Linkage, and Coordination Isomerization (structural isomer classes)
• Define cis/tran and mer/fac isomerization (geometric isomer classes)
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. 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. Isomers are compounds with the same molecular formula but different structural formulas and do not necessarily share similar properties. There are many different classes of isomers, like stereoisomers, enantiomers, and geometrical isomers. There are two main forms of isomerism: structural isomerism and stereoisomerism (spatial isomerism).
Class I: Structural Isomers
Isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another are called structural isomers, which differ in structure or bond type. For inorganic complexes, there are three types of structural isomers: ionization, coordination and linkage. Structural isomers, as their name implies, differ in their structure or bonding, which are separate from stereoisomers that differ in the spatial arrangement of the ligands are attached, but still have the bonding properties. The different chemical formulas in structural isomers are caused either by a difference in what ligands are bonded to the central atoms or how the individual ligands are bonded to the central atoms. When determining a structural isomer, you look at (1) the ligands that are bonded to the central metal and (2) which atom of the ligands attach to the central metal.
Ionization Isomerism
Ionization isomers occur when a ligand that is bound to the metal center exchanges places with an anion or neutral molecule that was originally outside the coordination complex. The geometry of the central metal ion and the identity of other ligands are identical. For example, an octahedral isomer will have five ligands that are identical, but the sixth will differ. The non-matching ligand in one compound will be outside of the coordination sphere of the other compound. Because the anion or molecule outside the coordination sphere is different, the chemical properties of these isomers is different. A hydrate isomer is a specific kind of ionization isomer where a water molecule is one of the molecules that exchanges places.
The difference between the ionization isomers can be view within the context of the ions generated when each are dissolved in solution. For example, when pentaamminebromocobalt(II) chloride is dissolved in water, $\ce{Cl^{-}}$ ions are generated:
$\ce{CoBr(NH_3)_5Cl {(s)} \rightarrow CoBr(NH_3)^{+}5 (aq) + Cl^{-} (aq)}$
whereas when pentaamminechlorocobalt(II) bromide is dissolved, $\ce{Br^{-}}$ ions are generated:
$\ce{CoCl(NH_3)_5Br {(s)} \rightarrow CoCl(NH_3)^{+}_{5} (aq) + Br^{-} (aq)}$
Notice that both anions are necessary to balance the charge of the complex, and that they differ in that one ion is directly attached to the central metal, but the other is not.
Solvate and Hydrate Isomerism: A special Type of Ionization Isomers
A very similar type of isomerism results from replacement of a coordinated group by a solvent molecule (Solvate Isomerism), which in the case of water is called Hydrate Isomerism. The best known example of this occurs for chromium chloride ($\ce{CrCl_3 \cdot 6H_2O}$) which may contain 4, 5, or 6 coordinated water molecules (assuming a coordination number of 6). The dot here is used essentially as an expression of ignorance to indicate that, though the parts of the molecule separated by the dot are bonded to one another in some fashion, the exact structural details of that interaction are not fully expressed in the resulting formula. Alfred Werner’s coordination theory indicates that several of the water molecules are actually bonded directly (via coordinate covalent bonds) to the central chromium ion. In fact, there are several possible compounds that use the brackets to signify bonding in the complex and the dots to signify "water molecules that are not bound to the central metal, but are part of the lattice:
• $\ce{[CrCl_2(H_2O)_4]Cl \cdot 2H_2O}$: bright-green colored
• $\ce{[CrCl(H_2O)_5]Cl_2 \cdot H_2O}$: grey-green colored
• $\ce{[Cr(H_2O)_6]Cl_3}$: violet colored
These isomers have very different chemical properties and on reaction with $\ce{AgNO_3}$ to test for $\ce{Cl^{-}}$ ions, would find 1, 2, and 3 $\ce{Cl^{-}}$ ions in solution, respectively.
Upon crystallization from water, many compounds incorporate water molecules in their crystalline frameworks. These "waters of crystallization" refers to water that is found in the crystalline framework of a metal complex or a salt, which is not directly bonded to the metal cation. In the first two hydrate isomers, there are water molecules that are artifacts of the crystallization and occur inside crystals. These waters of crystallization contribute to the total weight of water in a substance and are mostly present in a definite (stoichiometric) ratio.
What are "Waters of Crystallization"?
A compound with associated water of crystallization is known as a hydrate. The structure of hydrates can be quite elaborate, because of the existence of hydrogen bonds that define polymeric structures. For example, consider the aquo complex $\ce{NiCl_2 \cdot 6H_2O}$ that consists of separated trans-[NiCl2(H2O)4] molecules linked more weakly to adjacent water molecules. Only four of the six water molecules in the formula are bound to the nickel (II) cation, and the remaining two are waters of crystallization as the crystal structure resolves (Figure $3$).
Water is particularly common solvent to be found in crystals because it is small and polar. But all solvents can be found in some host crystals. Water is noteworthy because it is reactive, whereas other solvents such as benzene are considered to be chemically innocuous.
Coordination Isomerism
Coordination isomerism occurs in compounds containing complex anionic and complex cationic parts and can be viewed as an interchange of some ligands from the cation to the anion. Hence, there are two complex compounds bound together, one with a negative charge and the other with a positive charge. In coordination isomers, the anion and cation complexes of a coordination compound exchange one or more ligands. For example, the $\ce{[Zn(NH3)4] [Cu(Cl4)]}$ and $\ce{[Cu(NH3)4] [Zn(Cl4)]}$ compounds are coordination isomers (Figure $4$).
Example $2$: Coordination Isomers
What is the coordination isomer for the $\ce{[Co(NH3)6] [Cr(CN)6]}$ compound?
Solution
Coordination isomerism involves switching the metals between the cation and anion spheres.
Hence the $\ce{[Cr(NH3)6][Co(CN)6] }$ compound is a coordination isomer of $\ce{[Co(NH3)6] [Cr(CN)6]}$.
Linkage Isomerism
Linkage isomerism occurs with ambidentate ligands that are capable of coordinating in more than one way. The best known cases involve the monodentate ligands: $\ce{SCN^{-} / NCS^{-}}$ and $\ce{NO_2^{-} / ONO^{-}}$. The only difference is what atoms the molecular ligands bind to the central ion. The ligand(s) must have more than one donor atom, but bind to ion in only one place. For example, the ($\ce{NO2^{-}}$) ion is a ligand that can bind to the central atom through the nitrogen or the oxygen atom, but cannot bind to the central atom with both oxygen and nitrogen at once, in which case it would be called a polydentate ligand rather than an ambidentate ligand.
As with all structural isomers, the formula of the complex is unchanged for each isomer, but the properties may differ. The names used to specify the changed ligands are changed as well. For example, the $\ce{NO2^{-}}$ ion is called nitro when it binds with the $\ce{N}$ atom and is called nitrito when it binds with the $\ce{O}$ atom.
Example $3$: Linkage Isomerization
The cationic cobalt complex [Co(NH3)5(NO2)]Cl2 exists in two separable linkage isomers of the complex ion: (NH3)5(NO2)]2+.
When donation is from nitrogen to a metal center, the complex is known as a nitro- complex and when donation is from one oxygen to a metal center, the complex is known as a nitrito- complex. An alternative formula structure to emphasize the different coordinate covalent bond for the two isomers
• $\ce{[Co(ONO)(NH_3)_5]Cl}$: the nitrito isomer -O attached
• $\ce{[Co(NO_2)(NH_3)_5]Cl}$: the nitro isomer - N attached.
Another example of an ambidentate ligand is thiocyanate, $\ce{SCN^{−}}$, which can attach to the transition metal at either the sulfur atom or the nitrogen atom. Such compounds give rise to linkage isomerism. Other ligands that give rise to linkage isomers include selenocyanate, $\ce{SeCN^{−}}$ – isoselenocyanate, $\ce{NCSe^{−}}$ and sulfite, $\ce{SO3^{2−}}$.
Exercise $3$
Are $\ce{[FeCl5(NO2)]^{3–}}$ and $\ce{[FeCl5(ONO)]^{3–}}$ complex ions linkage isomers of each other?
Answer
Here, the difference is in how the ligand bonds to the metal. In the first isomer, the ligand bonds to the metal through an electron pair on the nitrogen. In the second isomer, the ligand bonds to the metal through an electron pair on one of the oxygen atoms. It's easier to see it:
Class 2: 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:
Because there is no way to convert the cis structure to the trans by rotating or flipping the molecule in space, they are fundamentally different arrangements of atoms in space. Probably the best-known examples of cis and trans isomers of an MA2B2 square planar complex are cis-Pt(NH3)2Cl2, also known as cisplatin, and trans-Pt(NH3)2Cl2, which is actually toxic rather than therapeutic.
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, 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:
Exercise $4$
Draw the cis and trans isomers of the following compounds:
1. (NH3)2IrCl(CO)
2. (H3P)2PtHBr
3. (AsH3)2PtH(CO)
Exercise $5$
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.
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 $4$
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Solution
Given: formula of complex
Asked for: structures of geometrical isomers
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 $6$
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
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. 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.
Table $1$ emphasizes the key differences of the three classes of structural isomers discussed below. The highlighted ions are the ions that switch or change somehow to make the type of structural isomer it is.
Table $1$: Overview of Structural Isomer Classes
Structural Isomer Examples
Ionization [CoBr(H2O)5]+Cl- and [CoCl(H2O)5]+Br-
Coordination [Zn(NH3)4]+[CuCl4]2- and [Cu(NH3)4]+[ZnCl4]2-
Linkage [Co(NO2)6]3- and [Co(ONO)6]3-
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.04%3A_Isomerism.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
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0.$
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 $2$, 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).
Factor 1: 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.
Factor 2: 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.
Factor 3: 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.
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−
Answer a
octahedral; high spin; five
Answer b
square planar; low spin; no unpaired electrons
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)
t2g eg t2g eg
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.
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_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.05%3A_Bonding_in_Complex_Ions%3A_Crystal_Field_Theory.txt |
The magnetic properties of a compound can be determined from its electron configuration and the size of its atoms. Because magnetism is generated by electronic spin, the number of unpaired electrons in a specific compound indicates how magnetic the compound is. In this section, the magnetism of the d-block elements (or transition metals) are evaluated. These compounds tend to have a large number of unpaired electrons.
Introduction
An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number \(m_s\) as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic.
Bulk Magnetism: Ferromagnetism
Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets. This means the compound shows permanent magnetic properties rather than exhibiting them only in the presence of a magnetic field (Figure \(1\)). In a ferromagnetic element, electrons of atoms are grouped into domains in which each domain has the same charge. In the presence of a magnetic field, these domains line up so that charges are parallel throughout the entire compound. Whether a compound can be ferromagnetic or not depends on its number of unpaired electrons and on its atomic size.
Ferromagnetism, the permanent magnetism associated with nickel, cobalt, and iron, is a common occurrence in everyday life. Examples of the knowledge and application of ferromagnetism include Aristotle's discussion in 625 BC, the use of the compass in 1187, and the modern-day refrigerator. Einstein declared that electricity and magnetism are inextricably linked in his theory of special relativity.
Magnetic Moments of Molecules and Ions
Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N2 and ions such as Na+ and [Fe(CN)6]4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.
When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d6 [Fe(CN)6]4− confirms that iron is diamagnetic, whereas high-spin d6 [Fe(H2O)6]2+ has four unpaired electrons with a magnetic moment that confirms this arrangement.
Hunds' Rule states that electrons fill all available orbitals with single electrons before pairing up, while maintaining parallel spins (paired electrons have opposing spins). For a set of five degenerate d-orbitals in an uncomplexed metal atom, electrons fill all orbitals before pairing to conserve pairing energy. With the addition of ligands, the situation is more complicated. The splitting energy between the d-orbitals increases the energy required to place single electrons into the higher-energy orbitals. Once the lower-energy orbitals have been half-filled (one electron per orbital), an electron can either be placed in a higher-energy orbital or paired with an electron in a lower-energy orbital. The strength of the ligands determine which option is chosen. If the splitting energy is greater than the pairing energy, the electrons will pair up; if the pairing energy is greater, unpaired electrons will occupy higher energy orbitals. In other words, with a strong-field ligand, low-spin complexes are usually formed; with a weak-field ligand, a high-spin complex is formed.
Low-spin complexes contain more paired electrons because the splitting energy is larger than the pairing energy. These complexes, such as [Fe(CN)6]3-, are more often diamagnetic or weakly paramagnetic. Likewise, high-spin complexes usually contain more unpaired electrons because the pairing energy is larger than the splitting energy. With more unpaired electrons, high-spin complexes are often paramagnetic. The unpaired electrons in paramagnetic compounds create tiny magnetic fields, similar to the domains in ferromagnetic materials. The strength of the paramagnetism of a coordination complex increases with the number of unpaired electrons; a higher-spin complex is more paramagnetic. The occurrence and relative strength of paramagnetism can be predicted by determining whether the compound is coordinated to a weak field ligand or a strong field ligand.
Example \(1\)
Which ligand generates a stronger magnetic complex ion when bound to \(Fe^{+2}\): EDTA or \(CN^-\)?
Solution
Since \([Fe(EDTA)_3^{-2}]\) has more unpaired electrons than \([FeCN_6^{-3}]\), it is more paramagnetic.
Example \(2\)
\(ZnI_4\) have eight valence electrons. If it is found to be diamagnetic, then does it occupy a tetrahedral or square plan geometry?
Solution
The splitting pattern for the two geometries differ and hence the electron configuration from adding the eight electrons also differ.
The tetrahedral geometry has two unpaired electrons and the square planer geometry has zero. Since \(ZnI_4\) is diamagnetic, it must have a square planar geometry.
Exercise \(1\)
For each of the following coordination complexes, identify if it is paramagnetic or diamagnetic?
1. octahedral, low spin, \(d^4\),
2. octahedral, low spin, \(d^6\),
3. tetrahedral, low spin, \(d^4\)
4. square planar, low spin, \(d^8\)
Measuring the Strength of Magnetism
The Gouy balance is used to measure paramagnetism by suspending the complex in question against an equivalent weight with access to a magnetic field. In principle, a magnetic measurement can be done very simply. All it takes is a balance and a magnetic field. We all know how a balance works. In a simple model from an earlier time, we place the sample in one pan. The balance tips over. We find the appropriate weight - a little piece of carefully prepared metal, certified by some bureau of standards. The balance balances. The masses in the two pans must be equal.
With a Gouy balance, the same idea applies, but we throw in a magnetic field, too. Even when it should be balanced, the balance tips, because of an attraction to the magnetic field. We need to find another weight that will get the balance even again. The weight needed to balance the scale is proportional to the attraction of the material to the magnetic field.
If the compound is paramagnetic, it will be pulled visibly towards the electromagnet, which is the distance proportional to the magnitude of the compound's paramagnetism. If the compound, however, is diamagnetic, it will not be pulled towards the electromagnet, instead, it might even slightly be repelled by it. This will be proven by the decreased weight or the no change in weight. The change in weight directly corresponds to the amount of unpaired electrons in the compound. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.06%3A_Magnetic_Properties_of_Coordination_Compounds_and_Crystal_Field_Theory.txt |
The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow (Figure $1$).
The blue color of the $\ce{[Cu(NH3)4]^{2+}}$ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure $2$). If white light (ordinary sunlight, for example) passes through $\ce{[Cu(NH3)4]SO4}$ solution, some wavelengths in the light are absorbed by the solution. The $\ce{[Cu(NH3)4]^{2+}}$ ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through a $\ce{[Cu(NH_3)_4]SO_4}$ solution.
Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You would not have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
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 complementary color wheel.
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 $4$). 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.
Example $1$: Colors of Complexes
The octahedral complex [Ti(H2O)6]3+ has a single d electron. To excite this electron from the ground state t2g orbital to the eg orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δo and occurs at 499 nm. Calculate the value of Δo in Joules and predict what color the solution will appear.
Solution
We can convert wavelength to frequency:
\begin{align*} \nu &=\dfrac{c}{λ} \[4pt] &= \dfrac{3.00 \times 10^8 \, m/s} { (499\, \cancel{nm}) \times \left( \dfrac{1 \,m}{10^9\, \cancel{ nm}} \right)} \[4pt] &=6.01 \times 10^{14} s^{-1} =6.01 \times 10^{14} \,Hz \end{align*}
And then using Planck's equation that related the frequency of light to energy
$E=h \nu \nonumber$
so
\begin{align*} E &= (6.63 \times 10^{−34} \textrm{J⋅s} ) \times (6.01 \times 10^{14} \,Hz) \[4pt] &=3.99 \times 10^{−19} \,J \end{align*}
Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.
Note: This is the energy for one transition (i.e., in one complex). If you want to calculate the energy in J/mol, then you have to multiply this by Avogadro's number ($N_A$).
Exercise $1$
A complex that appears green, absorbs photons of what wavelengths?
Answer
red, 620–800 nm
Color Depends on Oxidation State
Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure $4$, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.
Watch this video of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.
Color Depends on Ligand Field
The specific ligands coordinated to the metal center also influence the color of coordination complexes. 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. For example, the iron(II) complex [Fe(H2O)6]SO4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure $5$). In contrast, the low-spin iron(II) complex K4[Fe(CN)6] appears pale yellow because it absorbs higher-energy violet photons.
In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δoct). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.
Strong-field ligands cause a large split in the energies of d orbitals of the central metal atom and transition metal coordination compounds with these ligands are typically yellow, orange, or red because they absorb higher-energy violet or blue light. Coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.
Example $2$: Matching Colors to Ligand Fields
Identify the color (either blue, green, yellow, or orange) for the following complex ions formed with $Co^{3+}$:
1. $[Co(CN)_6]_3^{-2}$
2. $[Co(NH_3)_6]_3^{3+}$
3. $[CoF_6]_3^{-4}$
4. $[Co(H_2O)_6]^{3+}$
Solution
Each of these complex ions has the same metal with the same oxidation state, so the ligand field is the relevant factors. Each of the complex ions also has an octahedral ligand field, so we only need to compare the strength of the ligands in determining $\Delta_o$, which is determined by the spectrochemical series.
The ligands for each complex ions are: (a) $CN^-$, (b) $NH_3$, (c) $F^-$ and (d) $H_2O$, which are ranked in increasing $\Delta _o$ magnitude:
$F^- < H_2O < NH_3 < CN^- \label{eq0}$
The relationship between the $\Delta_o$ and the energy of the photons are absorbed in the d-d transition of $\ce{Co^{3+}}$ is given by equating Planck's equation to the crystal field splitting parameter:
$E= h \nu= \dfrac{hc}{\lambda} = \Delta_o \label{eq1}$
Now, we need to get a relative correlation between observed color (to the eye) and the wavelength of the light that is absorbed. From the complementary color wheel in Figure $1$ we get the following relationships (arranged from highest energy absorbed to lowest):
• 400-nm Violet light absorbed → Green-yellow observed
• 430-nm Blue light absorbed → Orange observed
• 450-nm Blue light absorbed → Yellow observed
• 490-nm Blue-green light absorbed → Red observed
• 570-nm Yellow-green light absorbed → Violet observed
• 580-nm Yellow light absorbed → Dark blue observed
• 600-nm Orange light absorbed → Blue observed
• 650-nm Red light absorbed → Green observed
Of the four possible colors given in the problem (blue, green, yellow, and orange), the corresponding colors that are absorbed are (600 nm, 650 nm, 450 nm, and 430, respectively). From Equation $\ref{eq1}$, the smaller $\lambda$ of absorbed light corresponds to the higher energy photons, so we would correlate the four wavelengths of absorbing photons in terms of increasing energy to observed color:
$\underbrace{650 \,nm }_{green} < \underbrace{600 \,nm}_{blue} < \underbrace{450 \, nm}_{yellow} < \underbrace{430 \, nm}_{orange} \label{eq3}$
Now, just correlate the ranking in Equation $\ref{eq3}$ to the ranking in Equation $\ref{eq0}$:
$\underbrace{F^- }_{green} < \underbrace{ H_2O}_{blue} < \underbrace{NH_3}_{yellow} < \underbrace{CN^-}_{orange} \nonumber$
and more specifically in terms of the original question
$\underbrace{[CoF_6]_3^{-4} }_{green} < \underbrace{ [Co(H_2O)_6]^{3+}}_{blue} < \underbrace{[Co(NH_3)_6]_3^{3+}}_{yellow} < \underbrace{[Co(CN)_6]_3^{-2}}_{orange} \nonumber$
Exercise $1$
If a complex ion $[M(NH_3)_6]^{2+}$ is red in solution, which of the following ligands, after a ligand exchange reaction to substitute for the ammine ligands, may shift the solution to be orange? More than one answer is possible.
$Cl^-$, $CN^-$, $CO$, $F^-$ $H_2O$, $I^-$, $en$, $NO_2^-$, $OH^-$, $SCN^-$
Answer
Shifting from red to orange is an increase in $\Delta_o$, so those ligands that are stronger field ligands than $NH_3$ in the list (deterimined via the spectrochemical series) are $CN^-$, $CO$, $en$, and $NO_2^-$.
A coordination compound of the Cu+ ion has a d10 configuration, and all the eg orbitals are filled. To excite an electron to a higher level, such as the 4p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN)2], for example, is colorless. On the other hand, octahedral Cu2+ complexes have a vacancy in the eg orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure $6$). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes.
Gems
A ruby is a pink to blood-red colored gemstone that consist of trace amounts of chromium in the mineral corundum $Al_2O_3$. In contrast, emeralds are colored green by trace amounts of chromium within a Be3Al2Si6O18 matrix. 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.
The absorbance spectrum of a ruby is shown Figure $\PageIndex{7; left}$. The number and positions of the peaks in the spectrum is determined by the electronic structure of the compound, which in this case depends upon the identity of the metal and the identities, number, and geometry of the surrounding ions. Crystal field theory may be used to predict the electronic structure and thus the absorbance spectrum. If white light is shown on the gem, the absorbance spectrum indicates which wavelengths of light are removed. In this case, there are strong bands centered at 414 and 561 nm. These wavelengths correspond with blue and yellow-green light, respectively. For the most part, these colors are not present in the light reaching ones eyes. An alternative way to express this concept is to recognize that the spectrum of light reaching the eye is the product of the spectrum of the incident light (white light) and the transmittance spectrum. For this ruby, the transmittance spectrum has a peak at 481 nm and a broad plateau past 620 nm. (Note that there is significant attenuation of the light across the entire spectrum.) Thus only light with wavelengths near 481 nm (cyan) and greater than 620 nm (red) reach the eye.
A similar analysis of the spectrum of an emerald is possible. The absorbance spectrum (Figure $\PageIndex{8; left}$) shows strong bands at 438 and 606 nm, which remove blue and orange light, respectively. The light that is not absorbed is shown by the transmittance spectrum, which indicates the dominant band of light reaching the eye is centered at 512 nm (green light) with smaller contributions from the far blue and red portions of the spectrum. This combination of wavelengths produces a deep green color.
Summary
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.07%3A_Color_and_the_Colors_of_Complexes.txt |
Learning Objectives
• To be introduced to complex ions, including ligands.
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
The Formation Constant
The replacement of water molecules from $\ce{[Cu(H2O)6]^{2+}}$ by ammonia occurs in sequential steps. Omitting the water molecules bound to $\ce{Cu^{2+}}$ for simplicity, we can write the equilibrium reactions as follows:
$\ce{Cu^{2+}(aq) + NH3(aq) <=> [Cu(NH3)]^{2+}(aq)} \tag{step 1}$
with $K_1 = \dfrac{\ce{[Cu(NH3)]^{2+}(aq)}}{[\ce{Cu^{2+}(aq)}][\ce{NH3(aq)}]} \nonumber$
$\ce{[Cu(NH_3)]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)2]^{2+}(aq)} \tag{step 2}$
with $K_2= \dfrac{\ce{[Cu(NH3)2]^{2+}(aq)}}{[\ce{[Cu(NH3)]^{2+}(aq)}][\ce{NH3(aq)}]} \nonumber$
$\ce{ [Cu(NH3)2]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)3]^{2+}(aq)} \tag{step 3}$
with $K_3= \dfrac{\ce{[Cu(NH3)3]^{2+}(aq)}}{[\ce{[Cu(NH3)2]^{2+}(aq)}][\ce{NH3(aq)}]} \nonumber$
$\ce{[Cu(NH3)3]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \tag{step 4}$
with $K_4= \dfrac{\ce{[Cu(NH3)4]^{2+}(aq)}}{[\ce{[Cu(NH3)3]^{2+}(aq)}][\ce{NH3(aq)}]} \nonumber$
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $NH_3$ ligands, not six.
$\ce{Cu^{2+}(aq) + 4NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \label{2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant ($K_f$). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
\begin{align*} K_\textrm f &=\dfrac{\left[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4} \[4pt] &=2.1\times10^{13} \[4pt] &=K_1 \times K_2 \times K_3 \times K_4 \label{3} \end{align*}
The formation constant ($K_f$) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion Equilibrium Equation $K_f$
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32− ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42− ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of $\ce{Cu(NO3)2•6H2O}$ is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of $\ce{Cu^{2+}(aq)}$?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of $\ce{Cu^{2+}(aq)}$ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant ($K_f$) to calculate the equilibrium concentration of $\ce{Cu^{2+}(aq)}$.
Solution
Adding an ionic compound that contains $\ce{Cu^{2+}(aq)}$ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation \ref{2} We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of $\ce{Cu^{2+}(aq)}$ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M} \nonumber$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
$\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber$
ICE [Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant ($K_f$) and assuming that $x \ll 0.0846$, which allows us to remove $x$ from the sum and difference,
\begin{align*} K_\textrm f &=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4} \[4pt] &=\dfrac{0.0846-x}{x(0.66+4x)^4} \[4pt] &\approx \dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \[4pt] x &=2.1\times10^{-14}\end{align*}
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise $1$
The ferrocyanide ion ($\ce{[Fe(CN)6]^{4−}}$) is very stable, with a $K_f$ of $1 × 10^{35}$. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of $\ce{K4[Fe(CN)6]}$.
Answer
2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$\ce{AgBr(s) <=> Ag^+(aq) + Br^{−}(aq)} \label{4a}$
with
$K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{4b}$
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{5a}$
with
$K_f = 2.9 \times 10^{13} \label{5b}$
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of $\ce{AgBr}$ by writing the appropriate reactions and adding them together:
\begin{align*}\mathrm{AgBr(s)} &\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)} \quad K_{\textrm{sp}} =5.35\times10^{-13} \[4pt] \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)} &\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)} \quad K_\textrm f =2.9\times10^{13} \[4pt] \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)} &\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)} \quad K =K_{\textrm{sp}}K_{\textrm f}=15\end{align*}
Comparing $K$ with $K_{sp}$ shows that the formation of the complex ion increases the solubility of $\ce{AgBr}$ by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$: Common Ion Effect in Complexation
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. If we let x equal the solubility of AgCl, then at equilibrium $\ce{[Ag^{+}] = [Cl^{−}] = x \,M}. \nonumber$ Substituting this value into the solubility product expression, \begin{align*} K_{sp} &= [\ce{Ag^{+}}][\ce{Cl^{−}}] \[4pt] &= (x)(x) \[4pt] &= x^2 \[4pt] &=1.77 \times 10^{−10} \[4pt] x &= 1.33 \times 10^{−5} \end{align*} Thus the solubility of $\ce{AgCl}$ in pure water at 25°C is 1.33 × 10−5 M.
2. If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that $x \le 1.0$: \begin{align*} K_{sp} &= [\ce{Ag^{+}}][\ce{Cl^{−}}] \[4pt] &= (x)(1.0 + x) \[4pt] & \approx x(1.0) \[4pt] &= 1.77 \times 10^{−10} = x \end{align*} If the common ion effect were the only important factor, we would predict that $\ce{AgCl}$ is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
3. To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf: \begin{align*}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}} &=1.77 \times 10^{-10} \[4pt] \mathrm{Ag^+ aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \[4pt] \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align*}
4. If we let $x$ equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that $x \le 1.0$, \begin{align*} K &=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]} \[4pt] &=\dfrac{x}{1.0-x}\approx1.9\times 10^{-5} \[4pt] &=x \end{align*}
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide ($\ce{HgI2}$) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer a
1.9 × 10−10 M
Answer b
1.4 M
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing.
Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$\ce{Ca^{2+}(aq) + O_3POPO^{4−}4(aq) <=> [Ca(O3POPO3)]^{2−}(aq)} \nonumber$
with
$K_f = 4\times 10^4 \nonumber$
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Complexing Agents in MRIs
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $2$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown in Figure $2$.
Summary
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.08%3A_Aspects_of_Complex-Ion_Equilibria.txt |
The pH's of solutions containing hexaaqua ions vary a lot from one metal to another (assuming you are comparing solutions of equal concentrations). However, the underlying explanation is the same for all of them. Consider the hexaaquairon(III) ion, $[Fe(H_2O)_6]^{3+}$ with six water molecules are attached to the central iron(III) ion via a co-ordinate bond using one of the lone pairs on the oxygen. Imagine for the moment that the 3+ charge is located entirely on the iron.
When the lone pairs on the oxygen atoms form co-ordinate bonds with the iron, there is obviously a movement of electrons towards the iron. That has an effect on the electrons in the O-H bonds. These electrons, in turn, get pulled towards the oxygen even more than usual. That leaves the hydrogen nuclei more exposed than normal. The overall effect is that each of the hydrogen atoms is more positive than it is in ordinary water molecules. The 3+ charge is no longer located entirely on the iron, but spreads out over the whole ion - much of it on the hydrogen atoms.
The 3+ charge is no longer located entirely on the iron, but spreads out over the whole ion - much of it on the hydrogen atoms.
The hydrogen atoms attached to the water ligands are sufficiently positive that they can be pulled off in a reaction involving water molecules in the solution. The first stage of this process is:
$Fe(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Fe(H_2O)_5(OH)]^{2+} _{(aq)} + H_3O^+_{(aq)} \label{Eqa1}$
The complex ion is acting as an acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a base by accepting the hydrogen ion. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify Equation $\ref{Eqa1}$:
$[Fe(H_2O)_6]^{3+} _{(aq)} \rightleftharpoons [Fe(H_2O)_5(OH)]^{2+} _{(aq)} + H^+ _{(aq)}$
However, if you write it like this, remember that the hydrogen ion is not just falling off the complex ion. It is being pulled off by a water molecule in the solution. The hexaaquairon(III) ion is quite strongly acidic giving solutions with pH's around 1.5, depending on concentration. You can get further loss of hydrogen ions as well, from a second and a third water molecule.
Losing a second hydrogen ion:
$[ Fe(H_2O)_5(OH)]^{2+} _{(aq)} \rightleftharpoons [ Fe(H_2O)_4(OH)_2]^{+} _{(aq)} + H^+ _{(aq)}$
. . . and a third one:
$[ Fe(H_2O)_4(OH)_2]^{+} _{(aq)} \rightleftharpoons [ Fe(H_2O)_3(OH)_4] _{(s)} + H^+ _{(aq)}$
This time you end up with a neutral $[ Fe(H_2O)_3(OH)_4]_{(s)}$ complex t hat is weakly soluble in water and precipitates.
Video $1$: Combining solutions of sodium hydroxide and iron(III) nitrate produces a precipitate of iron(III) hydroxide. The precipitate can be dissolved by adding a small volume of hydrochloric acid. The precipitate reappears when the sodium hydroxide solution is added again. https://www.youtube.com/watch?v=asdDyWmE9KQ
Exerimental Perspective
Looking at the equilibrium showing the loss of the first hydrogen ion (Equation $\ref{Eqa1}$):
The color of the new complex ion on the right-hand side is so strong that it completely masks the color of the hexaaqua ion. In concentrated solutions, the equilibrium position will be even further to the right-hand side (Le Chatelier's Principle), and so the color darkens. You will also get significant loss of other hydrogen ions leading to some formation of the neutral complex - and so you get some precipitate. The position of this equilibrium can be shifted by adding extra hydrogen ions from a concentrated acid (e.g., by increasing pH by adding concentrated acid to the solution (Video $1$). The new hydrogen ions push the position of the equilibrium to the left so that you can see the color of the hexaaqua ion:
Solutions containing 3+ hexaaqua ions tend to have pH's in the range from 1 to 3. Solutions containing 2+ ions have higher pH's - typically around 5 - 6, although they can go down to about 3. Remember that the reason that these ions are acidic is because of the pull of the electrons towards the positive central ion. An ion with 3+ charges on it is going to pull the electrons more strongly than one with only 2+ charges.
In 3+ ions, the electrons in the O-H bonds will be pulled further away from the hydrogens than in 2+ ions. That means that the hydrogen atoms in the ligand water molecules will have a greater positive charge in a 3+ ion, and so will be more attracted to water molecules in the solution. If they are more attracted, they will be more readily lost - and so the 3+ ions are more acidic.
If you have ions of the same charge, it seems reasonable that the smaller the volume this charge is packed into, the greater the distorting effect on the electrons in the O-H bonds. Ions with the same charge but in a smaller volume (a higher charge density) would be expected to be more acidic. You would therefore expect to find that the smaller the radius of the metal ion, the stronger the acid. Unfortunately, it's not that simple!
Summary
There probably is a relationship between ionic radius and acid strength, but that it is nothing like as simple and straightforward as most books at this level pretend. The problem is that there are other more important effects operating as well (quite apart from differences in charge) and these can completely swamp the effect of the changes in ionic radius. You have to look in far more detail at the bonding in the hexaaqua ions and the product ions | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.09%3A_Acid-Base_Reactions_of_Complex_Ions.txt |
Transition metal complexes which undergo rapid substitution of one ligand for another are labile, whereas complexes in which substitution proceed slowly or not at all are inert. For an inert complex, it is a large activation energy which prevents ligand substitution. Inert complexes are therefore kinetically stable compounds. As a classic example, the substitution reaction:
$[Co(NH_3)_6]^{3+}_{(aq)} + 6H_3O^+_{(aq)} \rightarrow [Co(H_2O)_6]^{3+}_{(aq)} + 6NH^+_{4(aq)} \label{Eq1}$
has an equilibrium constant of $10^{64}$. The large equilibrium constant suggests that the complex ion $[Co(NH_3)_6]^{3+}$ is thermodynamically unstable. This reaction is highly thermodynamically favored, yet the inert $[Co(NH_3)_6]^{3+}$ complex ion lasts for weeks in acidic solutions because the rate of the reaction is very slow. Thus, the large activation energy acts as an efficient barrier for ligand substitution rendering the $[Co(NH_3)_6]^{3+}$ ion thermodynamically metastable.
The allotropes of carbon
Although diamond is thermodynamically less stable than graphite, diamonds last a long time (~100 million years) since the conversion of diamond into graphite occurs extremely slowly.
Here is an example of a labile complex (notice difference in oxidation state for $Co$ in Equation $\ref{Eq1}$:
$[Co(NH_3)_6]^{2+}_{(aq)} + 6H_3O^+_{(aq)} \rightarrow [Co(H_2O)_6]^{2+}_{(aq)} + 6NH^+_{4(aq)}$
This reaction is virtually complete in a few seconds. The $[Co(NH_3)_6]^{2+}$ complex is thermodynamically unstable and also labile. Notice that the only difference between the two ammine cobalt complexes shown in the examples above is the oxidation number of the cobalt atom. The inert complex has Co(III) while the labile one has Co(II). Both ammine complexes are octahedral and in the case of Co(III), a d6 species, the t2g levels are filled. Co(II), on the other hand, has partially filled eg orbitals. It is straightforward to demonstrate lability since changes will occur and ligand substitution in complexes is normally accompanied by a color change. Inertness, however, is somewhat dull since nothing happens. In this experiment the inertness of ligand substitution by chromium (III) ions is compared with other reactions that do proceed at reasonably fast rates.
How to Predict if a Complex is Labile or Inert
Henry Taube (Nobel Prize, 1983) tried to understand lability by comparing the factors that govern bond strengths in ionic complexes to observations about the rates of reaction of coordination complexes. He saw some things that were unsurprising. Taube observed that many M+1 ions (M = metal) are more labile than many M+3 ions, in general. That is not too surprising, since metal ions function Lewis acids and ligands function Lewis bases in forming coordination complexes. In other words, metals with higher charges ought to be stronger Lewis acids, and so they should bind ligands more tightly. However, there were exceptions to that general rule. For example, Taube also observed that Mo(V) compounds are more labile than Mo(III) compounds. That means there is more going on here than just charge effects. Another factor that governs ionic bond strengths is the size of the ion. Typically, ions with smaller atomic radii form stronger bonds than ions with larger radii. Taube observed that Al3+, V3+, Fe3+ and Ga3+ ions are all about the same size. All these ions exchange ligands at about the same rate. However, other factors are in play that are outside the scope of this Module.
Labile and Inert octahedral Complexes 1. https://www.youtube.com/watch?v=g4fJSM_cE-s
Example $1$
In which compound from each pair would you expect the strongest ionic bonds? Why?
1. LiF vs KBr
2. CaCl2 vs. KCl
Contributors and Attributions
• Angel C. de Dios (Georgetown University) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.10%3A_Some_Kinetic_Considerations.txt |
24.1: Werner’s Theory of Coordination Compounds
Conceptual Problems
1. Give two reasons a metal can bind to only a finite number of ligands. Based on this reasoning, what do you predict is the maximum coordination number of Ti? of Ac?
2. Can a tetrahedral MA2B2 complex form cis and trans isomers? Explain your answer.
3. The group 12 elements are never found in their native (free) form but always in combination with one other element. What element is this? Why? Which of the group 12 elements has the highest affinity for the element you selected?
Answer
1. The group 12 metals are rather soft and prefer to bind to a soft anion such as sulfide rather than to a hard anion like oxide; hence they are usually found in nature as sulfide ores. Because it is the softest of these metals, mercury has the highest affinity for sulfide.
Structure and Reactivity
1. Complexes of metals in the +6 oxidation state usually contain bonds to which two Lewis bases? Why are these bonds best described as covalent rather than ionic? Do Ca, Sr, and Ba also form covalent bonds with these two Lewis bases, or is their bonding best described as ionic?
2. Cr, Mn, Fe, Co, and Ni form stable CO complexes. In contrast, the earlier transition metals do not form similar stable complexes. Why?
3. The transition metals Cr through Ni form very stable cyanide complexes. Why are these complexes so much more stable than similar compounds formed from the early transition metals?
4. Of Co(en)33+, CoF63, Co(NH3)63+, and Co(dien)23+, which species do you expect to be the most stable? Why?
5. Of Ca2+, Ti2+, V2+, Mn2+, Fe2+, Co2+, Ni2+, and Zn2+, which divalent metal ions forms the most stable complexes with ligands such as NH3? Why?
6. Match each Lewis base with the metal ions with which it is most likely to form a stable complex:
Lewis bases: NH3, F, RS, OH, and Cl
Metals: Sc3+, Cu+, W6+, Mg2+, V3+, Fe3+, Zr4+, Co2+, Ti4+, Au+, Al3+, and Mn7+
1. Of ReF2, ReCl5, MnF6, Mn2O7, and ReO, which are not likely to exist?
2. Of WF2, CrF6, MoBr6, WI6, CrO3, MoS2, W2S3, and MoH, which are not likely to exist?
Answers
1. Metals in the +6 oxidation state are stabilized by oxide (O2−) and fluoride (F). The M−F and M−O bonds are polar covalent due to extreme polarization of the anions by the highly charged metal. Ca, Sr, and Ba can be oxidized only to the dications (M2+), which form ionic oxides and fluorides.
1. Cyanide is a relatively soft base, and the early transition-metal cations are harder acids than the later transition metals.
1. The formation of complexes between NH3 and a divalent cation is largely due to electrostatic interactions between the negative end of the ammonia dipole moment and the positively charged cation. Thus the smallest divalent cations (Ni2+, Zn2+, and Cu2+) will form the most stable complexes with ammonia.
1. Re2+ is a very soft cation, and F and O2− are very hard bases, so ReO and ReF2 are unlikely to exist. MnF6 is also unlikely to exist: although fluoride should stabilize high oxidation states, in this case Mn6+ is probably too small to accommodate six F ions.
24.2: Ligands
Problems
1. Do ligands act like Lewis acids or Lewis bases? Why?
2. Do ligands form ionic bonds with the central metal atom?
3. What are chelating agents?
4. What is a monodentate ligand?
5. Describe polydentate ligands and provide an example.
6. What are hexadentate ligands?
Answers
1. Ligands act like Lewis bases because they share their electron pairs (electron donors) with the central metal atom.
2. No, ligands do not form ionic bonds the with the central metal atom. Rather, they form covalent bonds with the central metal atom because they share electron pairs.
3. Chelating agents are ligands that have two or more atoms with donating electron pairs that are able to attach a metal ion at the same time. These chelating ligands are monodentate and tridentate ligands
4. A monodentate ligand is a ligand that uses only one pair of electrons to bond to the central metal atom or ion.
5. Polydentate ligands are ligands which are able to donate more than two electron pairs to the central metal they bond to. EDTA is an example of a polydentate ligand.
6. Hexadentate ligands are ligands which have six lone pairs of electrons which can all bond to the central metal atom.
7. tetraamminecopper(II) sulfate
8. Tris(ethylenediamine)cobalt(II) nitrate
Problems
Write the name of the following complexes (Chapter 24 / Custom Edition Chapter 21 Exercises):
1. [CoCl3(NH3)3]
2. [Co(ONO)3(NH3)3]
3. [Fe(ox)2(H2O)2]-
4. Ag2[HgI4]
Answers
1. triamminetrichlorocobalt(III)
2. triamminetrinitrito-O-cobalt(III); or triamminetrinitritocobalt(III)
3. diaquadioxalatoferrate(III) ion
4. silver(I) tetraiodomercurate(II)
24.4: Isomerism
Problems
1. Write the Coordination Isomer for: [Co(NH3)6][Cr(CN)6]
2. Write the corresponding linkage isomer as well as names of the two complexes for: [CoCl(NO2)(NH3)4Cl]
3. What is the coordination isomer of: [Cr(NH3)6][Fe(CN)6]
4. Write the Ionization isomer for: [CoBr(NH3)5]SO4
5. Explain a polydentate ligand.
Answers
1. [Cr(NH3)6][Co(CN)6]
2. [CoCl(ONO)(NH3)4Cl]
3. [Fe(NH3)6][Cr(CN)6]
4. [CoSO4(NH3)5]Br
5. A polydentate ligand is a ligand that can bind to the central atom of a complex compound at many places at one time.
24.5: Bonding in Complex Ions: Crystal Field Theory
Conceptual Problems
1. Describe crystal field theory in terms of its
1. assumptions regarding metal–ligand interactions.
2. weaknesses and strengths compared with valence bond theory.
1. In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?
2. Will the value of Δo increase or decrease if I ligands are replaced by NO2 ligands? Why?
3. For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?
4. How can CFT explain the color of a transition-metal complex?
Structure and Reactivity
1. Do strong-field ligands favor a tetrahedral or a square planar structure? Why?
2. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [TiCl6]3−
2. [CoCl4]2−
1. For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Cu(NH3)4]2+
2. [Ni(CN)4]2−
1. The ionic radii of V2+, Fe2+, and Zn2+ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.
Answers
1. d9, square planar, neither high nor low spin, single unpaired electron
2. d8, square planar, low spin, no unpaired electrons
24.8: Aspects of Complex-Ion Equilibria
Conceptual Problems
1. What is the difference between Keq and Kf?
2. Which would you expect to have the greater tendency to form a complex ion: Mg2+ or Ba2+? Why?
3. How can a ligand be used to affect the concentration of hydrated metal ions in solution? How is Ksp affected? Explain your answer.
4. Co(II) forms a complex ion with pyridine (C5H5N). Which is the Lewis acid, and which is the Lewis base? Use Lewis electron structures to justify your answer.
Numerical Problems
1. Fe(II) forms the complex ion [Fe(OH)4]2− through equilibrium reactions in which hydroxide replaces water in a stepwise manner. If log K1 = 5.56, log K2 = 4.21, log K3 = −0.10, and log K4 = −1.09, what is Kf? Write the equilibrium equation that corresponds to each stepwise equilibrium constant. Do you expect the [Fe(OH)4]2− complex to be stable? Explain your reasoning.
2. Zn(II) forms the complex ion [Zn(NH3)4]2+ through equilibrium reactions in which ammonia replaces coordinated water molecules in a stepwise manner. If log K1 = 2.37, log K2 = 2.44, log K3 = 2.50, and log K4 = 2.15, what is the overall Kf? Write the equilibrium equation that corresponds to each stepwise equilibrium constant. Do you expect the [Zn(NH3)4]2+ complex to be stable? Explain your reasoning.
3. Although thallium has limited commercial applications because it is toxic to humans (10 mg/kg body weight is fatal to children), it is used as a substitute for mercury in industrial switches. The complex ion [TlBr6]3− is highly stable, with log Kf = 31.6. What is the concentration of Tl(III)(aq) in equilibrium with a 1.12 M solution of Na3[TlBr6]?
Answer
[Fe(H2O)6]2+(aq) + OH(aq) ⇌ [Fe(H2O(aq))5(OH)] + (aq) + H2O log K1 = 5.56
[Fe(H2O)5(OH)] + (aq) + OH(aq) ⇌ [Fe(H2O)4(OH)2](aq) + H2O(aq) log K2 = 9.77
[Fe(H2O)4(OH)2](aq) + OH(aq) ⇌ [Fe(H2O)3(OH)3](aq) + H2O(aq) log K3 = 9.67
[Fe(H2O)3(OH)3](aq) + OH(aq) ⇌ [Fe(OH)4]2−(aq) + 3H2O(l) log K4 = 8.58
[Fe(H2O)6]2+(aq) + 4OH(aq) ⇌ [Fe(OH)4]2−(aq) + 6H2O(l)
log Kf = log K1 + log K2 + log K2 + log K4
= 33.58
Thus, Kf = 3.8 × 1033. Because [Fe(OH)4]2− has a very large value of Kf, it should be stable in the presence of excess OH. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.E%3A_Exercises.txt |
Learning Objectives
• Write and balance nuclear equations
• To know the different kinds of radioactive decay.
• To balance a nuclear reaction.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
$\large \ce{^{A}_{Z}X} \label{Eq1a}$
where
• $X$ is the symbol for the element,
• $A$ is the mass number, and
• $Z$ is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger).
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Nuclear Equations
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an α particle with magnesium-25 $(\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber$
where
• $\ce A$ is the mass number and
• $\ce Z$ is the atomic number of the new nuclide, $\ce X$.
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$\mathrm{25+4=A+1} \nonumber$
so
$\mathrm{A=28} \nonumber$
Similarly, the charges must balance, so:
$\mathrm{12+2=Z+1} \nonumber$
so
$\mathrm{Z=13} \nonumber$
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$.
Exercise $1$
The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber$
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction, a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the $^A _Z \textrm{X}$ notation for nuclides to include radioactive emissions. Table $1$ lists the name and symbol for each type of emitted radiation. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.
Table $1$: Nuclear Decay Emissions and Their Symbols
Identity Symbol Charge Mass (amu)
helium nucleus $^4_2\alpha$ +2 4.001506
electron $^0_{-1}\beta$ or $\beta ^-$ −1 0.000549
photon $_0^0\gamma$
neutron $^1_0\textrm n$ 0 1.008665
proton $^1_1\textrm p$ +1 1.007276
positron $^0_{+1}\beta$ or $\beta ^+$ +1 0.000549
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol $^0_{-1}\textrm e$, which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol $^0_{-1}\beta$, which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, $^4_{2}\textrm{He}^{2+}$ refers to the nucleus of a helium atom, and $^4_{2}\alpha$ denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure $1$. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha $\alpha$ Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay, which results in the emission of a helium-4 nucleus as an alpha (α) particle, $^4_{2}\alpha$. The general reaction is as follows:
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\ \textrm{particle}}{^4_2 \alpha}\label{Eq1}$
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
$^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2}$
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta $\beta^-$ Decay
Nuclei that contain too many neutrons often undergo beta (β) decay, in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
$\underset{\textrm{unstable} \ \textrm{neutron in} \ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \ \textrm{retained} \ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \ \textrm{emitted by} \ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3}$
The general reaction for beta decay is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4}$
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
$^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber$
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron $\beta^+$ Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
$^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6}$
The general reaction for positron emission is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber$
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
$^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber$
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC), in which an electron in an inner shell reacts with a proton to produce a neutron:
$^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9}$
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
$\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber$
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
$^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
$^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma $\gamma$ Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma ($\gamma$) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
$^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \ \textrm{nuclear} \ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber$
If we disregard the decay event that created the excited nucleus, then
$^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber$
or more generally,
$^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber$
Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a $γ$ ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $2$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$: $^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: $^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows: $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber$
Exercise $2$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer a
$^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
Answer d
$^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
Answer c
$^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer a
beta decay
Answer d
positron emission or electron capture
Answer c
alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure $2$). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
$^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}$
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
$^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18}$
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure $3$). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
$^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19}$
Example $4$
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, $^{30}_{15}\textrm{P}$.
B The balanced nuclear equation for the reaction is as follows:
$^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber$
Exercise $4$
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei $(^{2}_{1}\textrm{H})$ produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
Answer
neutron, $^{1}_{0}\textrm{n}$ ; $^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}$ :
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as $^{235}_{92}\textrm{U}$ and $^{239}_{94}\textrm{Pu}$ undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure $4$:
$^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20}$
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure $4$. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure $5$, a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $2$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
Table $2$: Some Reactions Used to Synthesize Transuranium Elements
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}$
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}$
$^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}$
$^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}$
$^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}$
$^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}$
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $6$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary and Key Takeaway
• Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, $^4 _2 \alpha$, and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Equations
alpha decay
$^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber$
beta decay
$^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber$
positron emission
$^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber$
electron capture
$^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber$
gamma emission
$^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.01%3A_Radioactivity.txt |
Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
Relative Abundances of the Elements on Earth and in the Known Universe
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure $1$. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure $1$ illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table $1$ for some common elements.
Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Table $1$: Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Synthesis of the Elements in Stars
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure $2$). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
$_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta \_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma \_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}$
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two $\gamma$ rays, and a great deal of energy:
$4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2}$
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
$2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3}$
Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
$_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4}$
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
$_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5}$
$_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6}$
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
$_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7}$
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
The Formation of Heavier Elements in Supernovas
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure $2$). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure $3$). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
$_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}$
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
Example $1$: Carbon Burning Stars
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution
1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{12}^{24}\textrm{Mg}$.
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{10}^{20}\textrm{Ne}+\,_2^4\alpha$.
Exercise $1$
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer
19 neutrons; $_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta$
Summary
Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.02%3A_Naturally_Occurring_Radioactive_Isotopes.txt |
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles, gamma rays and conversion electrons Although radioactivity is observed as a natural occurring process, it can also be artificially induced typically via the bombarding atoms of a specific element by radiating particles, thus creating new atoms.
Introduction
Ernest Rutherford was a prominent New Zealand scientist, and a winner of the Nobel Prize in chemistry in 1908. Amongst his vast list of discoveries, Rutherford was also the first to discover artificially induced radioactivity. Through the bombardment of alpha particles against nuclei of $\ce{^{14}N}$ with 7 protons/electrons, Rutherford produced $\ce{^{17}O}$ (8 protons/electrons) and protons (Figure $1$). Through this observation, Rutherford concluded that atoms of one specific element can be made into atoms of another element. If the resulting element is radioactive, then this process is called artificially induced radioactivity
Rutherford was the first researcher to create protons outside of the atomic nuclei and the $\ce{^{17}O}$ isotope of oxygen, which is nonradioactive. Similarly, other nuclei when bombarded with alpha particles will generate new elements (Figure $2$) that may be radioactive and decay naturally or that may be stable and persist like $\ce{^{17}O}$.
Before this discovery of artificial induction of radioactivity, it was a common belief that atoms of matter are unchangeable and indivisible. After the very first discoveries made by Ernest Rutherford, Irene Joliot-Curie and her husband, Frederic Joliot, a new point of view was developed. The point of view that although atoms appear to be stable, they can be transformed into new atoms with different chemical properties. Today over one thousand artificially created radioactive nuclides exist, which considerably outnumber the nonradioactive ones created.
Note: Irene Joliet-Curie and Frederic Joliot
Irene Joliet-Curie and her husband Frédéric both were French scientists who shared winning the Nobel Prize award in chemistry in 1935 for artificially synthesizing a radioactive isotope of phosphorus by bombarding aluminum with alpha particles. $\ce{^{30}P}$ with 15 protons was the first radioactive nuclide obtained through this method of artificially inducing radioactivity.
$\ce{^27_13Al + ^4_2He \rightarrow ^30_15P + ^1_0n}$
$\ce{^30_15P \rightarrow ^30_14Si + ^0_{-1}\beta}$
Activation (or radioactivation) involves making a radioactive isotope by neutron capture, e.g. the addition of a neutron to a nuclide resulting in an increase in isotope number by 1 while retaining the same atomic number (Figure $3$). Activation is often an inadvertent process occurring inside or near a nuclear reactor, where there are many neutrons flying around. For example, Coba in or near a nuclear reactor will capture a neutron forming the radioactive isotope Co-60.
$\ce{ ^1_0n + ^{59}Co \rightarrow ^{60}Co }$
The $\ce{ ^{60}Co}$ isotope is unstable (half life of 5.272 years) and disintegrates into $\ce{ ^{60}Ni }$ via the emission of $\beta$ particle and $\gamma$ radiation Figure $4$.
Example $1$: Neutron Bombardment
Write a nuclear equation for the creation of 56Mn through the bombardment of 59Co with neutrons.
Solution
A unknown particle is produced with 56Mn, in order to find the mass number (A) of the unknown we must subtract the mass number of the Manganese atom from the mass number of the Cobalt atom plus the neutron being thrown. In simpler terms,
Now, by referring to a periodic table to find the atomic numbers of Mn and Co, and then subtracting the atomic number of Mn from Co, we will receive the atomic number of the unknown particle
Thus, the unknown particle has A = 4, and Z = 2, which would make it a Helium particle, and the nuclear formula would be as follows:
$\ce{^{50}_{27}Co + ^1_0n \rightarrow ^{56}_{25}Mn + ^{4}_{2}\alpha } \nonumber$
Example $2$: Calcium Bombardment
Write a nuclear equation for the production of $\ce{^{147}Eu}$ by bombardment of $\ce{^{139}La}$ with $\ce{^{12}Ca}$.
Solution
Like the above example, you must first find the mass number of the unknown particle.
Thus, the mass number of the unknown particle is 4. Again by referring to a periodic table and finding the atomic numbers of Lanthanum, Carbon and Europium, we are able to calculate the atomic number of the unknown particle,
The atomic number for the unknown particle equals to zero, therefore 4 neutrons are emitted, and the nuclear equation is written as follows:
$\ce{^{139}_{57}La + ^{12}_6C \rightarrow ^{147}_{63}Eu + 4 ^{0}_{1}n } \nonumber$
Summary
Induced radioactivity occurs when a previously stable material has been made radioactive by exposure to specific radiation. Most radioactivity does not induce other material to become radioactive. This Induced radioactivity was discovered by Irène Curie and F. Joliot in 1934. This is also known as man-made radioactivity. The phenomenon by which even light elements are made radioactive by artificial or induced methods is called artificial radioactivity. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.03%3A_Nuclear_Reactions_and_Artificially_Induced_Radioactivity.txt |
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta \label{25.4.1}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta \label{25.4.2}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $1$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability." As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval.
Table $1$: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95 $\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}$
curium Cm 96 $\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}$
californium Cf 98 $\ce{^{242}_{96}Cm + ^4_2He⟶ ^{243}_{97}Bk + 2^1_0n}$
einsteinium Es 99 $\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}$
mendelevium Md 101 $\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}$
nobelium No 102 $\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}$
rutherfordium Rf 104 $\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}$
seaborgium
Sg
106
$\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}$
$\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}$
meitnerium Mt 107 $\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}$
Particle Accelerators
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $2$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.04%3A_Transuranium_Elements.txt |
Learning Objectives
• Recognize common modes of radioactive decay
• Identify common particles and energies involved in nuclear decay reactions
• Write and balance nuclear decay equations
• Calculate kinetic parameters for decay processes, including half-life
• Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure $1$).
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $2$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, $γ$ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and $γ$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Alpha ($α$) decay is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay:
$\ce{^{210}_{84}Po⟶ ^4_2He + ^{206}_{82}Pb} \hspace{40px}\ce{or}\hspace{40px} \ce{^{210}_{84}Po ⟶ ^4_2α + ^{206}_{82}Pb}\nonumber$
Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). Because the loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing α decay lies below the band of stability, the daughter nuclide will lie closer to the band.
Beta (β) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay:
$\ce{^{131}_{53}I ⟶ ^0_{-1}e + ^{131}_{54}X} \hspace{40px}\ce{or}\hspace{40px} \ce{^{131}_{53}I ⟶ ^0_{-1}β + ^{131}_{54}Xe}\nonumber$
Beta decay, which can be thought of as the conversion of a neutron into a proton and a β particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission (γ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a γ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits γ radiation and is used in many applications including cancer treatment:
$\mathrm{^{60}_{27}Co^* ⟶\, ^0_0γ +\, ^{60}_{27}Co}\nonumber$
There is no change in mass number or atomic number during the emission of a γ ray unless the γ emission accompanies one of the other modes of decay.
Positron emission ($β^+$ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
$\ce{^{15}_8O ⟶ ^0_{+1}e + ^{15}_7N} \hspace{40px}\ce{or}\hspace{40px} \ce{^{15}_8O ⟶ ^0_{+1}β + ^{15}_7N}\nonumber$
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
$\ce{^{40}_{19}K + ^0_{-1}e ⟶ ^{40}_{18}Ar}\nonumber$
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur. Figure $3$ summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure $4$). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
For example, F-18 is produced by proton bombardment of 18O $(\ce{^{18}_8O + ^1_1p⟶ ^{18}_9F + ^1_0n})$ and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The 18F emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure $5$). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure $6$). In a given cobalt-60 source, since half of the $\ce{^{60}_{27}Co}$ nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, N, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
$\text{decay rate} = \lambda N\nonumber$
with $\lambda$ is the decay constant for the particular radioisotope.
The decay constant, $\lambda$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, t1/2:
$λ=\dfrac{\ln 2}{t_{1/2}}=\dfrac{0.693}{t_{1/2}} \hspace{40px}\ce{or}\hspace{40px} t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}\nonumber$
The first-order equations relating amount, N, and time are:
$N_t=N_0e^{−kt} \hspace{40px}\ce{or}\hspace{40px} t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
where N0 is the initial number of nuclei or moles of the isotope, and Nt is the number of nuclei/moles remaining at time t. Example $1$ applies these calculations to find the rates of radioactive decay for specific nuclides.
Example $1$: Rates of Radioactive Decay
$\ce{^{60}_{27}Co}$ decays with a half-life of 5.27 years to produce $\ce{^{60}_{28}Ni}$.
1. What is the decay constant for the radioactive disintegration of cobalt-60?
2. Calculate the fraction of a sample of the $\ce{^{60}_{27}Co}$ isotope that will remain after 15 years.
3. How long does it take for a sample of $\ce{^{60}_{27}Co}$ to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5.27\:y}=0.132\:y^{−1}} \nonumber$
(b) The fraction of $\ce{^{60}_{27}Co}$ that is left after time t is given by $\dfrac{N_t}{N_0}$. Rearranging the first-order relationship Nt = N0eλt to solve for this ratio yields:
$\dfrac{N_t}{N_0}=e^{-λt}=e^\mathrm{-(0.132/y)(15.0/y)}=0.138 \nonumber$
The fraction of $\ce{^{60}_{27}Co}$ that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the $\ce{^{60}_{27}Co}$ originally present will remain after 15 years.
(c) 2.00% of the original amount of $\ce{^{60}_{27}Co}$ is equal to 0.0200 × N0. Substituting this into the equation for time for first-order kinetics, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)=−\dfrac{1}{0.132\:\ce y^{−1}}\ln\left(\dfrac{0.0200×N_0}{N_0}\right)=29.6\:\ce y \nonumber$
Exercise $1$
Radon-222, $\ce{^{222}_{86}Rn}$, has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of $\ce{^{209}_{83}Bi}$ is 1.9 × 1019 years; $\ce{^{239}_{94}Ra}$ is 24,000 years; $\ce{^{222}_{86}Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 × 10–3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table $1$, and others are listed in Appendix N1.
Table $1$: Half-lives of Radioactive Isotopes Important to Medicine
Type Decay Mode Half-Life Uses
F-18 β+ decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m1 γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit $γ$ radiation to rid themselves of excess energy and become (more) stable.
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: $\ce{^{12}_6C}$, which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
$\ce{^{14}_7N + ^1_0n⟶ ^{14}_6C + ^1_1H}\nonumber$
All isotopes of carbon react with oxygen to produce CO2 molecules. The ratio of $\ce{^{14}_6CO2}$ to $\ce{^{12}_6CO2}$ depends on the ratio of $\ce{^{14}_6CO}$ to $\ce{^{12}_6CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_6CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_6C ^{14}_6CO2}$ and $\ce{^{12}_6CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_6C: ^{12}_6C}$ ratio found in a living plant is the same as the $\ce{^{14}_6C: ^{12}_6C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_6C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years:
$\ce{^{14}_6C⟶ ^{14}_7N + ^0_{-1}e}\nonumber$
Thus, the $\ce{^{14}_6C: ^{12}_6C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $7$ visually depicts this process.
For example, with the half-life of $\ce{^{14}_6C}$ being 5730 years, if the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of $\ce{^{14}_6C : ^{12}_6C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Example $2$: Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, N, in the relationship:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)⟶t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right) \nonumber$
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{5730\: y}=1.21×10^{−4}\:y^{−1}} \nonumber$
Substituting and solving, we have:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{\ce{Rate}_t}{\ce{Rate}_0}\right)=\mathrm{−\dfrac{1}{1.21×10^{−4}\:y^{−1}}\ln\left(\dfrac{10.8\:dis/min/g\: C}{13.6\:dis/min/g\: C}\right)=1910\: y}\nonumber$
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure $8$).
Exercise $2$
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the $\ce{^{14}_6C : ^{12}_6C}$ ratio. The accuracy of a straightforward application of this technique depends on the $\ce{^{14}_6C : ^{12}_6C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of CO2 molecules (largely $\ce{^{12}_6CO2}$) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the $\ce{^{14}_6C}$ has decayed), the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the atmosphere may be changing. This manmade increase in $\ce{^{12}_6CO2}$ in the atmosphere causes the $\ce{^{14}_6C : ^{12}_6C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example $3$: Radioactive Dating of Rocks
An igneous rock contains 9.58 × 10–5 g of U-238 and 2.51 × 10–5 g of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
$\mathrm{9.58×10^{−5}\cancel{g\: U}×\left( \dfrac{1\: mol\: U}{238\cancel{g\: U}}\right )=4.03×10^{−7}\:mol\: U}\nonumber$
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
$\mathrm{2.51×10^{-5}\cancel{g\: Pb}×\left( \dfrac{1\cancel{mol\: Pb}}{206\cancel{g\: Pb}}\right )×\left(\dfrac{1\: mol\: U}{1\cancel{mol\: Pb}}\right)=1.22×10^{-7}\:mol\: U}\nonumber$
The total amount of U-238 originally present in the rock is therefore:
$\mathrm{4.03×10^{−7}\:mol+1.22×10^{−7}\:mol=5.25×10^{−7}\:mol\: U}\nonumber$
The amount of time that has passed since the formation of the rock is given by:
$t=−\dfrac{1}{λ}\ln\left(\dfrac{N_t}{N_0}\right)\nonumber$
with N0 representing the original amount of U-238 and Nt representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of 4.5 × 109 y, so the decay constant λ is:
$λ=\dfrac{\ln 2}{t_{1/2}}=\mathrm{\dfrac{0.693}{4.5×10^9\:y}=1.54×10^{−10}\:y^{−1}}\nonumber$
Substituting and solving, we have:
$t=\mathrm{−\dfrac{1}{1.54×10^{−10}\:y^{−1}}\ln\left(\dfrac{4.03×10^{−7}\cancel{mol\: U}}{5.25×10^{−7}\cancel{mol\: U}}\right)=1.7×10^9\:y}\nonumber$
Therefore, the rock is approximately 1.7 billion years old.
Exercise $3$
A sample of rock contains 6.14 × 10–4 g of Rb-87 and 3.51 × 10–5 g of Sr-87. Calculate the age of the rock. (The half-life of the β decay of Rb-87 is 4.7 × 1010 y.)
Answer
3.7 × 109 y
Summary
Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.
Key Equations
• decay rate = λN
• $t_{1/2}=\dfrac{\ln 2}{λ}=\dfrac{0.693}{λ}$
Glossary
alpha (α) decay
loss of an alpha particle during radioactive decay
beta (β) decay
breakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle
daughter nuclide
nuclide produced by the radioactive decay of another nuclide; may be stable or may decay further
electron capture
combination of a core electron with a proton to yield a neutron within the nucleus
gamma (γ) emission
decay of an excited-state nuclide accompanied by emission of a gamma ray
half-life (t1/2)
time required for half of the atoms in a radioactive sample to decay
parent nuclide
unstable nuclide that changes spontaneously into another (daughter) nuclide
positron emission
(also, β+ decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted
radioactive decay
spontaneous decay of an unstable nuclide into another nuclide
radioactive decay series
chains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product
radiocarbon dating
highly accurate means of dating objects 30,000–50,000 years old that were derived from once-living matter; achieved by calculating the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the object vs. the ratio of $\ce{^{14}_6C : ^{12}_6C}$ in the present-day atmosphere
radiometric dating
use of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.05%3A_Rate_of_Radioactive_Decay.txt |
Learning Objectives
• To calculate a mass-energy balance and a nuclear binding energy.
• To understand the differences between nuclear fission and fusion.
Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy.
Mass–Energy Balance
The relationship between mass (m) and energy (E) is expressed in the following equation:
$E = mc^2 \label{Eq1}$
where
• $c$ is the speed of light ($2.998 \times 10^8\; m/s$), and
• $E$ and $m$ are expressed in units of joules and kilograms, respectively.
Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to Equation $\ref{Eq1}$, every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected. As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows:
$\textrm{C(graphite)} + \frac{1}{2}\textrm O_2(\textrm g)\rightarrow \mathrm{CO_2}(\textrm g)\hspace{5mm}\Delta H^\circ=-393.5\textrm{ kJ/mol} \label{Eq2}$
Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. When a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as
$\Delta{E}=(\Delta m)c^2 \label{Eq3}$
we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy:
$\Delta m=\dfrac{\Delta E}{c^2} \label{Eq4}$
Because 1 J = 1 (kg•m2)/s2, the change in mass is as follows:
$\Delta m=\dfrac{-393.5\textrm{ kJ/mol}}{(2.998\times10^8\textrm{ m/s})^2}=\dfrac{-3.935\times10^5(\mathrm{kg\cdot m^2})/(\mathrm{s^2\cdot mol})}{(2.998\times10^8\textrm{ m/s})^2}=-4.38\times10^{-12}\textrm{ kg/mol} \label{Eq5}$
This is a mass change of about 3.6 × 10−10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible.
In contrast, for a typical nuclear reaction, such as the radioactive decay of 14C to 14N and an electron (a β particle), there is a much larger change in mass:
$^{14}\textrm C\rightarrow \,^{14}\textrm N+\,^0_{-1}\beta \label{Eq6}$
We can use the experimentally measured masses of subatomic particles and common isotopes given in Table 20.1 to calculate the change in mass directly. The reaction involves the conversion of a neutral 14C atom to a positively charged 14N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral 14N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral 14N atom (14.003074 amu) and the mass of a 14C atom (14.003242 amu):
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}} \&=14.003074\textrm{ amu} - 14.003242\textrm{ amu} = - 0.000168\textrm{ amu}\end{align} \label{Eq7}
The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of 14C is −0.000168 g = −1.68 × 10−4 g = −1.68 × 10−7 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows:
\begin{align}\Delta E &=(\Delta m)c^2=(-1.68\times10^{-7}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \ &=-1.51\times10^{10}(\mathrm{kg\cdot m^2})/\textrm s^2=-1.51\times10^{10}\textrm{ J}=-1.51\times10^7\textrm{ kJ}\end{align} \label{Eq8}
The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of 14C is a relatively low-energy nuclear reaction.
Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 103 eV), megaelectronvolts (1 MeV = 106 eV), and even gigaelectronvolts (1 GeV = 109 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of 14C is thus
$\mathrm{(-1.68\times10^{-4}\, amu) \left(\dfrac{931\, MeV}{amu}\right) = -0.156\, MeV = -156\, keV}\label{Eq9}$
Example $1$
Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of 238U to 234Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom.
Given: nuclear decay reaction
Asked for: changes in mass and energy
Strategy:
A Use the mass values in Table 20.1 to calculate the change in mass for the decay reaction in atomic mass units.
B Use Equation $\ref{Eq4}$ to calculate the change in energy in joules per mole.
C Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom.
Solution
A Using particle and isotope masses from Table 20.1, we can calculate the change in mass as follows:
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}}=(\mathrm{mass \;^{234}Th+mass\;^4_2He})-\mathrm{mass\;^{238}U} \&=(234.043601\textrm{ amu}+4.002603\textrm{ amu}) - 238.050788\textrm{ amu} = - 0.004584\textrm{ amu}\end{align}
B Thus the change in mass for 1 mol of 238U is −0.004584 g or −4.584 × 10−6 kg. The change in energy in joules per mole is as follows:
ΔE = (Δm)c2 = (−4.584 × 10−6 kg)(2.998 × 108 m/s)2 = −4.120 × 1011 J/mol
C The change in energy in electronvolts per atom is as follows:
$\Delta E = -4.584\times10^{-3}\textrm{ amu}\times\dfrac{\textrm{931 MeV}}{\textrm{amu}}\times\dfrac{1\times10^6\textrm{ eV}}{\textrm{1 MeV}}=-4.27\times10^6\textrm{ eV/atom}$
Exercise $1$
Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium (3H) to 3He and a β particle.
Answer
Δm = −2.0 × 10−5 amu; ΔE = −1.9 × 106 kJ/mol = −19 keV/atom
Nuclear Binding Energies
We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 (1H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium (2H) is 2.014102 amu, although its calculated mass is 2.016490 amu:
\begin{align}m_{^2\textrm H}&=m_{\textrm{neutron}}+m_{\textrm{proton}}+m_{\textrm{electron}} \&=1.008665\textrm{ amu}+1.007276\textrm{ amu}+0.000549\textrm{ amu}=2.016490\textrm{ amu} \end{align}\label{Eq10}
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation $\ref{Eq4}$, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The amount of energy released when a nucleus forms from its component nucleons is the nuclear binding energy (Figure $1$). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus.
Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components.
Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in Figure $2$, the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., 4He, 12C, and 16O). As mentioned earlier, these are particularly stable combinations.
Because the maximum binding energy per nucleon is reached at 56Fe, all other nuclei are thermodynamically unstable with regard to the formation of 56Fe. Consequently, heavier nuclei (toward the right in Figure $2$) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in Figure $2$) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern.
Heavier nuclei spontaneously undergo nuclear reactions that decrease their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that increase their atomic number.
Example $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 56Fe. The experimental mass of the nuclide is given in Table A4.
Given: nuclide and mass
Asked for: nuclear binding energy and binding energy per nucleon
Strategy:
A Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of 1H atoms (because its mass is the same as the mass of one electron and one proton).
B Calculate the mass defect by subtracting the experimental mass from the calculated mass.
C Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon.
Solution
A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 1H atoms, we can calculate the sum of the masses more quickly as follows:
\begin{align*}\textrm{calculated mass}&=26(\textrm{mass }^1_1\textrm H)+30(\textrm{mass }^1_0 \textrm n)\[4pt] &=26(1.007825)\textrm{amu}+30(1.008665)\textrm{amu}=56.463400\textrm{ amu}\ \textrm{experimental mass} &=55.934938 \end{align*} \nonumber
B We subtract to find the mass defect:
\begin{align*}\textrm{mass defect}&=\textrm{calculated mass}-\textrm{experimental mass} \&=56.463400\textrm{ amu}-55.934938\textrm{ amu}=0.528462\textrm{ amu}\end{align*} \nonumber
C The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon.
Exercise $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 238U.
Answer
1800 MeV/238U; 7.57 MeV/nucleon
Summary
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: ΔE = (Δm)c2. Large changes in energy are usually reported in kiloelectronvolts or megaelectronvolts (thousands or millions of electronvolts). With the exception of 1H, the experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect of the nucleus. The energy corresponding to the mass defect is the nuclear binding energy, the amount of energy released when a nucleus forms from its component particles. In nuclear fission, nuclei split into lighter nuclei with an accompanying release of multiple neutrons and large amounts of energy. The critical mass is the minimum mass required to support a self-sustaining nuclear chain reaction. Nuclear fusion is a process in which two light nuclei combine to produce a heavier nucleus plus a great deal of energy. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.06%3A_Energetics_of_Nuclear_Reactions.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force, an extremely powerful but very short-range attractive force between nucleons (Figure \(1\)). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(2\). The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure \(3\), more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Note
Most stable nuclei contain even numbers of both neutrons and protons
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(2\), the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(_{15}^{30} \textrm P\)
2. \(_{43}^{98} \textrm{Tc}\)
3. tin-118
4. \(_{94}^{239} \textrm{Pu}\)
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(2\), its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is.
b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is.
c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable.
d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive.
Exercise \(1\)
Classify each nuclide as stable or radioactive.
1. \(_{90}^{232} \textrm{Th}\)
2. \(_{20}^{40} \textrm{Ca}\)
3. \(_8^{15} \textrm{O}\)
4. \(_{57}^{139} \textrm{La}\)
Answer:
1. radioactive
2. stable
3. radioactive
4. stable
Superheavy Elements
In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements, with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.07%3A_Nuclear_Stability.txt |
Learning Objectives
• Explain nuclear fission
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(1\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(2\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure \(3\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(4\)).
An atomic bomb (Figure \(5\)) contains several pounds of fissionable material, \(\ce{^{235}_{92}U}\) or \(\ce{^{239}_{94}Pu}\), a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(6\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (Figure \(4\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water \(\ce{( ^2_1H2O)}\) or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
\[\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He}\]
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Video \(1\): Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
\[\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g)\]
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(8\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(1\)0).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.08%3A_Nuclear_Fission.txt |
Learning Objectives
• Describe the nuclear reactions in a nuclear fusion reaction
• Quantify the energy released or absorbed in a fusion reaction
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1H$ and a triton, $^3_1H$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $2$).
Example $1$
Calculate the energy released in each of the following hypothetical processes.
1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$
2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$
3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$
Solution
1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$
2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$
3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$
Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.
Nuclear Reactors
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $3$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.09%3A_Nuclear_Fusion.txt |
Learning Objectives
• Describe the biological impact of ionizing radiation.
• Define units for measuring radiation exposure.
• Explain the operation of common tools for detecting radioactivity.
• List common sources of radiation exposure in the US.
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure $1$).
Ionizing vs. Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure $2$).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure $4$). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure $5$).
Radon is found in buildings across the country, with amounts dependent on location. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the level found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure $6$). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Table $1$). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = $3.7 \times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose.
Table $1$: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes $\mathrm{3.7 \times 10^{10}\; decays/second}$
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE × Gy
roentgen equivalent for man (rem) Rem = RBE × rad
The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy), along with a biological factor referred to as the RBE (for relative biological effectiveness), that is an approximate measure of the relative damage done by the radiation. These are related by:
$\text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2}$
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Example $1$: Amount of Radiation
Cobalt-60 (t1/2 = 5.26 y) is used in cancer therapy since the $\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
$\textrm{Activity}=λN=\left( \dfrac{\ln 2}{t_{1/2} } \right) N=\mathrm{\left( \dfrac{\ln 2}{5.26\ y} \right) \times 5.00 \ g=0.659\ \dfrac{g}{y} \ of\ \ce{^{60}Co} \text{ that decay}} \nonumber$
And to convert this to decays per second:
$\mathrm{0.659\; \frac{g}{y} \times \dfrac{y}{365 \;day} \times \dfrac{1\; day}{ 24\; hours} \times \dfrac{1\; h}{3,600 \;s} \times \dfrac{1\; mol}{59.9\; g} \times \dfrac{6.02 \times 10^{23} \;atoms}{1 \;mol} \times \dfrac{1\; decay}{1\; atom}} \nonumber$
$\mathrm{=2.10 \times 10^{14} \; \frac{decay}{s}} \nonumber$
(a) Since $\mathrm{1\; Bq = 1\; \frac{ decay}{s}}$, the activity in Becquerel (Bq) is:
$\mathrm{2.10 \times 10^{14} \dfrac{decay}{s} \times \left(\dfrac{1\ Bq}{1 \; \frac{decay}{s}} \right)=2.10 \times 10^{14} \; Bq} \nonumber$
(b) Since $\mathrm{1\ Ci = 3.7 \times 10^{11}\; \frac{decay}{s}}$, the activity in curie (Ci) is:
$\mathrm{2.10 \times 10^{14} \frac{decay}{s} \times \left( \dfrac{1\ Ci}{3.7 \times 10^{11} \frac{decay}{s}} \right) =5.7 \times 10^2\;Ci} \nonumber$
Exercise $1$
Tritium is a radioactive isotope of hydrogen ($t_{1/2} = \mathrm{12.32\; years}$) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer a
$\mathrm{3.56 \times 10^{11} Bq}$
Answer b
$\mathrm{0.962\; Ci}$
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure $8$, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table $2$.
Table $2$: Health Effects of Radiation
Exposure (rem) Health Effect Time to Onset (Without Treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Summary
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source of radiation, and limiting time of exposure.
Footnotes
1. 1 Source: US Environmental Protection Agency
Glossary
becquerel (Bq)
SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s.
curie (Ci)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
Geiger counter
Instrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube.
gray (Gy)
SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue.
ionizing radiation
Radiation that can cause a molecule to lose an electron and form an ion.
millicurie (mCi)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
nonionizing radiation
Radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules.
radiation absorbed dose (rad)
SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy.
radiation dosimeter
Device that measures ionizing radiation and is used to determine personal radiation exposure.
relative biological effectiveness (RBE)
Measure of the relative damage done by radiation.
roentgen equivalent man (rem)
Unit for radiation damage, frequently used in medicine; 1 rem = 1 Sv.
scintillation counter
Instrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation.
sievert (Sv)
SI unit measuring tissue damage caused by radiation; takes energy and biological effects of radiation into account. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.10%3A_Effect_of_Radiation_on_Matter.txt |
Learning Objectives
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\ce{(^{99}_{43}Tc)}$, thallium-201 $\ce{(^{201}_{81}Tl)}$, iodine-131 $\ce{(^{131}_{53}I)}$, and sodium-24 $\ce{(^{24}_{11}Na)}$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure $1$) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure $2$). The parent nuclide Mo-99 is part of a molybdate ion, $\ce{MoO4^2-}$; when it decays, it forms the pertechnetate ion, $\ce{TcO4-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure $3$). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
$\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber$
The overall decay scheme for this is shown graphically in Figure $4$.
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
$\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber$
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of $\ce{^{14}_6C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure $5$). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure $6$). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.
Summary
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
Glossary
chemotherapy
similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
external beam radiation therapy
radiation delivered by a machine outside the body
internal radiation therapy
(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
radiation therapy
use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive tracer
(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.11%3A_Applications_of_Radioisotopes.txt |
Learning Objectives
• To recognize the composition and properties typical of organic and inorganic compounds.
• To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
• Write condensed structural formulas for alkanes given complete structural formulas.
• Draw line-angle formulas given structural formulas.
• To name alkanes by the IUPAC system and write formulas for alkanes given IUPAC names
• To describe functional groups and explain why they are useful in the study of organic chemistry.
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH4Cl), expecting to get ammonium cyanate (NH4OCN). What he expected is described by the following equation.
$AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{Eq1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
Note
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$.
Hydrocarbons
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
Note
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) and they are shown again in Figure $1$.
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure $2$).
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $2$.
Table $2$: The First 10 Straight-Chain Alkanes
Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers
methane CH4 CH4
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3 2
pentane C5H12 CH3CH2CH2CH2CH3 3
hexane C6H14 CH3CH2CH2CH2CH2CH3 5
heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9
octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18
nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35
decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75
Consider the series in Figure $3$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
We use several kinds of formulas to describe organic compounds. A molecular formula shows only the kinds and numbers of atoms in a molecule. For example, the molecular formula C4H10 tells us there are 4 carbon atoms and 10 hydrogen atoms in a molecule, but it doesn’t distinguish between butane and isobutane. A structural formula shows all the carbon and hydrogen atoms and the bonds attaching them. Thus, structural formulas identify the specific isomers by showing the order of attachment of the various atoms. Unfortunately, structural formulas are difficult to type/write and take up a lot of space. Chemists often use condensed structural formulas to alleviate these problems. The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated for butane:
The ultimate condensed formula is a line-angle formula, in which carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. For example, we can represent pentane (CH3CH2CH2CH2CH3) and isopentane [(CH3)2CHCH2CH3] as follows:
Note
Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom.
Nomenclature
As noted in Table $2$:, the number of isomers increases rapidly as the number of carbon atoms increases. There are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes. It would be difficult to assign unique individual names that we could remember. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. (Some of the names we used earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called common names.) A stem name (Table $3$) indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called substituents, are then named, with their positions indicated by numbers. For now, we will consider only those substituents called alkyl groups.
Table $3$: Stems That Indicate the Number of Carbon Atoms in Organic Molecules
Stem Number
meth- 1
eth- 2
prop- 3
but- 4
pent- 5
hex- 6
hept- 7
oct- 8
non- 9
dec- 10
An alkyl group is a group of atoms that results when one hydrogen atom is removed from an alkane. The group is named by replacing the -ane suffix of the parent hydrocarbon with -yl. For example, the CH3 group derived from methane (CH4) results from subtracting one hydrogen atom and is called a methyl group. The alkyl groups we will use most frequently are listed in Table $4$. Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically.
Table $4$: Common Alkyl Groups
Parent Alkane Alkyl Group Condensed Structural Formula
methane methyl CH3
ethane ethyl CH3CH2
propane propyl CH3CH2CH2
isopropyl (CH3)2CH–
butane butyl* CH3CH2CH2CH2
*There are four butyl groups, two derived from butane and two from isobutane. We will introduce the other three where appropriate.
Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example $1$).
1. Name alkanes according to the LCC of carbon atoms in the molecule (rather than the total number of carbon atoms). This LCC, considered the parent chain, determines the base name, to which we add the suffix -ane to indicate that the molecule is an alkane.
2. If the hydrocarbon is branched, number the carbon atoms of the LCC. Numbers are assigned in the direction that gives the lowest numbers to the carbon atoms with attached substituents. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. (The LCC need not be written in a straight line; for example, the LCC in the following has five carbon atoms.)
1. Place the names of the substituent groups in alphabetical order before the name of the parent compound. If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di-, tri-, tetra-, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word.
When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to put it to work, not just memorize the rules. It’s easier than it looks.
Example $1$
Name each compound.
Solution
1. The LCC has five carbon atoms, and so the parent compound is pentane (rule 1). There is a methyl group (rule 2) attached to the second carbon atom of the pentane chain. The name is therefore 2-methylpentane.
2. The LCC has six carbon atoms, so the parent compound is hexane (rule 1). Methyl groups (rule 2) are attached to the second and fifth carbon atoms. The name is 2,5-dimethylhexane.
3. The LCC has eight carbon atoms, so the parent compound is octane (rule 1). There are methyl and ethyl groups (rule 2), both attached to the fourth carbon atom (counting from the right gives this carbon atom a lower number; rule 3). The correct name is thus 4-ethyl-4-methyloctane.
Exercise $1$
Name each compound.
Example $2$
Draw the structure for each compound.
1. 2,3-dimethylbutane
2. 4-ethyl-2-methylheptane
Solution
In drawing structures, always start with the parent chain.
1. The parent chain is butane, indicating four carbon atoms in the LCC.
Then add the groups at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (CH3) groups, one on the second carbon atom and one on the third.
Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds.
2. Adding the groups at their proper positions gives
Filling in all the hydrogen atoms gives the following condensed structural formulas:
Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing.
Exercise $2$
Draw the structure for each compound.
1. 4-ethyloctane
2. 3-ethyl-2-methylpentane
3. 3,3,5-trimethylheptane
Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups. In our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups tend to behave in organic reactions.
Previously, we considered several kinds of hydrocarbons. Now we examine some of the many organic compounds that contain functional groups. We first introduced the idea of the functional group, a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to the molecule. If you understand the behavior of a particular functional group, you will know a great deal about the general properties of that class of compounds. In this chapter, we make a brief yet systematic study of some of organic compound families. Each family is based on a common, simple functional group that contains an oxygen atom or a nitrogen atom. Some common functional groups are listed in Table $5$ and a more comprehensive list if found here.
Table $5$: Selected Organic Functional Groups
Name of Family General Formula Functional Group Suffix*
alkane RH none -ane
alkene R2C=CR2 -ene
alkyne RC≡CR –C≡C– -yne
alcohol ROH –OH -ol
thiol RSH –SH -thiol
ether ROR –O– ether
aldehyde -al
ketone -one
carboxylic acid -oic acid
*Ethers do not have a suffix in their common name; all ethers end with the word ether.
Summary
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit.
• Condensed chemical formulas show the hydrogen atoms (or other atoms or groups) right next to the carbon atoms to which they are attached.
• Line-angle formulas imply a carbon atom at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.
• Alkanes have both common names and systematic names, specified by IUPAC.
• The functional group, a structural arrangement of atoms and/or bonds, is largely responsible for the properties of organic compound families. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.1%3A_Organic_Compounds_and_Structures%3A_An_Overview.txt |
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula \(C_nH_{2n+2}\) and can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons.
Conformation
When we smell something, the information travels to us via molecules, and almost always these are organic molecules. These molecules can be detected by a variety of organs, including noses in dogs and antennae in crickets, but no matter what organ is sensing the smell, one of the crucial factors in determining how an organism reacts to a compound is the shape of the molecule. The sense of smell depends on thousands of different receptors in the organ, working conceptually on a lock-and-key basis: a molecule with a given shape can fit into a given receptor, and when it does, a signal is sent telling the nervous system that the organism has encountered that particular type of molecule, and the organism reacts appropriately.
What gives a molecule its shape? Given a structural formula, could you determine the shape of the corresponding molecule in three dimensions? Could you predict its biological activity, including not only its smell, but also a host of other behaviors linked to the shape of molecular messengers, such as anti-cancer activity or narcotic properties? These questions are at the cutting edge of biological chemistry. Although they are best answered through computer modeling, we can develop some of the qualitative ideas used in these models.
A conformation of an acyclic molecule is a specific disposition of atoms in the molecule in space within the molecule due to free rotation around bonds (e.g., ethane (CH3CH3)). Due to free rotation around the carbon-carbon bond, the ethane molecule could assume an infinite number of conformations, two of which are shown below as saw-horse formulas.
In the eclipsed conformation of ethane the smallest dihedral angle is 0º, whereas in the staggered conformation it is 60º, which can be seen more clearly in the respective Newman projection.
Conformations of Butane
In the case of ethane, conformational changes are very subtle, but in others they are more obvious. Butane (CH3CH2CH2CH3) has four tetrahedral carbons and three carbon-carbon bonds connecting them together. Let's number the carbons along the chain C1, C2, C3 and C4. Rotating around the C-C bonds connected to the terminal carbons -- C1-C2 and C3-C4 -- only subtle changes in shape would be apparent. However, rotating about C2-C3 produces some pretty obvious shape changes.
Pay attention to where the two methyl groups are with respect to each other. If we call C1-C2-C3-C4 the dihedral angle, then at 0 degrees the molecule is in an eclipsed conformation, apparent when looking at the Newman projection. Looked at from other vantage points, the molecule is curled up into the shape of a letter C (Figure \(2\)).
At 60 degrees, the molecule is no longer eclipsed, and just as in ethane the energy is a little bit lower, but the overall shape when viewed from the side is still a sort of twisted C (Figure \(3\)).
At 120 degrees the molecule is eclipsed again, but from the side it has now twisted almost into a shape like the letter Z ( Figure \(4\)).
At 180 degrees, the molecule is staggered again and has settled into a regular, zig-zag, letter Z shape.
These conformations of butane are really pretty different. Which shape would the molecule prefer? From what we learned about ethane, we could probably rule out the eclipsed conformations. Each of those would have 3 kcal/mol of torsional strain.
However, there is another factor that destabilizes the initial conformation at 0 degrees, in favor of the Z-shaped one with the methyl groups 180 degrees apart from each other. This factor is called "sterics" and it refers to the idea that molecules, or parts of molecules, take up space, and so two parts of the butane can't occupy the same place at the same time. Put more simply, sterics refers to crowdedness. When the two methyl groups in butane are too close together, they are too crowded, and they are at higher energy. When they get farther apart, crowding subsides and the energy in the molecule goes down.
• Torsional strain forces bonds on neighboring carbons to be staggered.
• Steric strain forces groups away from each other to relieve crowding.
• Torsion and sterics both contribute to strain energy.
• Strain can be released by getting the molecule in a lower energy conformation.
For butane, that means getting those two methyl groups away from each other and keeping the bonds staggered. There is some additional jargon that is used to describe these butane conformations:
• In an anti conformer, the largest groups are 180 degrees from each other.
• In a gauche conformer, the bonds are staggered but the largest groups are 600 from each other.
Preparation of Alkanes
An important way to generate alkanes is the hydrogenation of an alkene (discussed later) The double bond of an alkene consists of a sigma (σ) bond and a pi (π) bond. Because the carbon-carbon π bond is relatively weak, it is quite reactive and can be easily broken and reagents can be added to carbon. Reagents are added through the formation of single bonds to carbon in an addition reaction.
An example of an alkene addition reaction is a process called hydrogenation.In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated alkane. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability.
Although the hydrogenation of an alkene is a thermodynamically favorable reaction, it will not proceed without the addition of a catalyst. Common catalysts used are insoluble metals such as palladium and platinum. Hydrogenation reactions are extensively used to create commercial goods.Hydrogenation is used in the food industry to make a large variety of manufactured goods, like spreads and shortenings, from liquid oils. This process also increases the chemical stability of products and yields semi-solid products like margarine. Hydrogenation is also used in coal processing. Solid coal is converted to a liquid through the addition of hydrogen. Liquefying coal makes it available to be used as fuel.
Alkanes from Petroleum
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be: | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.2%3A_Alkanes.txt |
Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360 degrees, going through many different conformations. That complete rotation isn't possible in a cyclic system, because the parts that you would be trying to twist away from each other would still be connected together. Cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation". Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
Cyclopropane
A three membered ring has no rotational freedom whatsoever. A plane is defined by three points, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane.
Furthermore, if you look at a model you will find that the neighboring C-H bonds (C-C bonds, too) are all held in eclipsed conformations. Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a line drawing of cyclopropane as shown from the side. In this oblique view, the dark lines mean that those sides of the ring are closer to you.
However, the ring isn't big enough to introduce any steric strain, which does not become a factor until we reach six membered rings. Until that point, rings are not flexible enough for two atoms to reach around and bump into each other. The really big problem with cyclopropane is that the C-C-C bond angles are all too small.
• All the carbon atoms in cyclopropane appear to be tetrahedral.
• These bond angles ought to be 109 degrees.
• The angles in an equilateral triangle are actually 60 degrees, about half as large as the optimum angle.
• This factor introduces a huge amount of strain in the molecule, called ring strain.
Cyclobutane
Cyclobutane is a four membered ring. In two dimensions, it is a square, with 90 degree angles at each corner.
However, in three dimensions, cyclobutane is flexible enough to buckle into a "butterfly" shape, relieving torsional strain a little bit. When it does that, the bond angles get a little worse, going from 90 degrees to 88 degrees. In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
• With bond angles of 88 rather than 109 degrees, cyclobutane has a lot of ring strain, but less than in cyclopropane.
• Torsional strain is still present, but the neighboring bonds are not exactly eclipsed in the butterfly.
• Cyclobutane is still not large enough that the molecule can reach around to cause crowding. Steric strain is very low.
• Cyclobutanes are a little more stable than cyclopropanes and are also a little more common in nature.
Cyclopentanes
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common paraffinic ring in nature, after cyclohexanes. In two dimensions, a cyclopentane appears to be a regular pentagon.
In three dimensions, there is enough freedom of rotation to allow a slight twist out of this planar shape. In a line drawing, this three-dimensional shape is drawn from an oblique view, just like cyclobutane.
The ideal angle in a regular pentagon is about 107 degrees, very close to a tetrahedral bond angle. Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four, and as a result, ring strain is entirely removed. The envelope removes torsional strain along the sides and flap of the envelope. However, the neighbouring carbons are eclipsed along the "bottom" of the envelope, away from the flap. There is still some torsional strain in cyclopentane. Again, there is no steric strain in this system.
Cyclohexane
Cyclohexane is one of the most important cycloalkanes and so is the focus discussion of the conformations of cycloalkanes. If the carbons of a cyclohexane ring were placed at the corners of a regular planar hexagon, all the C-C-C bond angles would have to be 120°. Because the expected normal C-C-C bond angle should be near the tetrahedral value of 109.5°, the suggested planar configuration of cyclohexane would have angle strain at each of the carbons, and would correspond to less stable cyclohexane molecules than those with more normal bond angles. The actual normal value for the C-C-C bond angle of an open-chain \(\ce{-CH_2-CH_2-CH_2}\) unit appears to be about 112.5°, which is 3° greater than the tetrahedral value. From this we can conclude that the angle strain at each carbon of a planar cyclohexane would be (120° — 112.5°) = 1.5°. Angle strain is not the whole story with regard to the instability of the planar form, because in addition to having C-C-C bond angles different from their normal values, the planar structure also has its carbons and hydrogens in the unfavorable eclipsed arrangement, as shown in Figure 12-2. How both of these factors can be taken into account is illustrated in Exercises 12-2 and 12-3.
If the carbon valence angles are kept near the tetrahedral value, you will find that you can construct ball-and-stick models of the cyclohexane six-carbon ring with two quite different conformations. These are known as the “chair” and “boat” conformations (Figure \(1\)). It has not been possible to separate cyclohexane at room temperature into pure isomeric forms that correspond to these conformations, and actually the two forms appear to be rapidly interconverted. The chair conformation is considerably more stable and comprises more than 99.9% of the equilibrium mixture at room temperature.1
Why is the boat form less stable than the chair form, if both have normal C—C—C bond angles? The answer is that the boat form has unfavorable nonbonded interactions between the hydrogen atoms around the ring. If we make all of the bond angles normal and orient the carbons to give the “extreme boat” conformation of Figure 12-4, a pair of 1,4 hydrogens (the so-called “flagpole” hydrogens) have to be very close together (1.83 A).
Table \(1\): Strain in Cycloalkane Rings and Heats of Combustion of Cycloalkanes
Compound \(n\) Angle Strain at each \(CH_2\) Heat of Combustion \(\Delta{H}^o\) (kcal/mol) Heat of Combustion \(\Delta{H}^o\) per \(CH_2/N\) (kcal/mol) Total Strain (kcal/mol)
ethene 2 109.5 337.2 168.6 22.4
cyclopropane 3 49.5 499.9 166.6 27.7
cyclobutane 4 19.5 655.9 164.0 26.3
cyclopentane 5 1.5 793.4 158.7 6.5
cyclohexane 6 10.5 944.8 157.5 0.4
cycloheptane 7 19.1 1108.1 158.4 6.3
cyclooctane 8 25.5 1268.9 158.6 9.7
cyclononane 9 30.5 1429.5 158.8 12.9
cyclodecane 10 34.5 1586.1 158.6 12.1
cyclopentadecane 15 46.5 2362.5 157.5 1.5
open chain alkane 157.4 -
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.3%3A_Cycloalkanes.txt |
In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women.
Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects.
Baby born to a mother who had taken thalidomide while pregnant.
Researchers later realized the that problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms.
One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp3-hybridized carbon. These two forms of thalidomide are stereoisomers.
Note that the carbon in question has four different substituents (two of these just happen to be connected by a ring structure). Tetrahedral carbons with four different substituent groups are called stereocenters.
Example \(1\)
Locate all of the carbon stereocenters in the molecules below.
Solution
Looking at the structures of what we are referring to as the two isomers of thalidomide, you may not be entirely convinced that they are actually two different molecules. In order to convince ourselves that they are indeed different, let’s create a generalized picture of a tetrahedral carbon stereocenter, with the four substituents designated R1-R4. The two stereoisomers of our simplified model look like this:
If you look carefully at the figure above, you will notice that molecule A and molecule B are mirror images of each other (the line labeled 's' represents a mirror plane). Furthermore, they are not superimposable: if we pick up molecule A, flip it around, and place it next to molecule B, we see that the two structures cannot be superimposed on each other. They are different molecules!
See a figure of the two enantiomers of the amino acid alanine
If you make models of the two stereoisomers of thalidomide and do the same thing, you will see that they too are mirror images, and cannot be superimposed (it well help to look at a color version of the figure below).
Thalidomide is a chiral molecule. Something is considered to be chiral if it cannot be superimposed on its own mirror image – in other words, if it is asymmetric (lacking in symmetry). The term ‘chiral’ is derived from the Greek word for ‘handedness’ – ie. right-handedness or left-handedness. Your hands are chiral: your right hand is a mirror image of your left hand, but if you place one hand on top of the other, both palms down, you see that they are not superimposable.
A pair of stereoisomers that are non-superimposable mirror images of one another are considered to have a specific type of stereoisomeric relationship – they are a pair of enantiomers. Thalidomide exists as a pair of enantiomers. On the macro level, your left and right hands are also a pair of enantiomers.
Naming Enantiomers: The R,S, System
Chemists need to have a convenient way to distinguish between stereoisomers. The Cahn-Ingold-Prelog system is a set of rules that allows us to unambiguously define the stereochemical configuration of any stereocenter, using the designations ‘R’ (from the Latin rectus, meaning right-handed) or ‘S’ (from the Latin sinister, meaning left-handed).
The rules for this system of stereochemical nomenclature are, on the surface, fairly simple. Because thalidomide is a somewhat complicated molecule, we’ll use the simple 3-carbon sugar glyceraldehyde as our first example. You will want to make a model of the stereoisomer of glyceraldehyde shown below (be sure that you are making the correct enantiomer!).
The first thing that we must do is to assign a priority to each of the four substituents bound to the chiral carbon. In this nomenclature system, the priorities are based on atomic number, with higher atomic numbers having a higher priority. We first look at the atoms that are directly bonded to the chiral carbon: these are H, O (in the hydroxyl), C (in the aldehyde), and C (in the CH2OH group).
Two priorities are easy: hydrogen, with an atomic number of 1, is the lowest (#4) priority, and the hydroxyl oxygen, with atomic number 8, is priority #1. Carbon has an atomic number of 6. Which of the two ‘C’ groups is priority #2, the aldehyde or the CH2OH? To determine this, we move one more bond away from the stereocenter: for the aldehyde we have a double bond to an oxygen, while on the CH2OH group we have a single bond to an oxygen. If the atom is the same, double bonds have a higher priority than single bonds. Therefore, the aldehyde group is assigned #2 priority and the CH2OH group the #3 priority. With our priorities assigned, we next make sure that the #4 priority group (the hydrogen) is pointed back away from ourselves, into the plane of the page (it is already).
Then, we trace a circle defined by the #1, #2, and #3 priority groups, in increasing order. For our glyceraldehyde example, this circle is clockwise, which tells us that this carbon has the ‘R’ configuration, and that this molecule is (R)-glyceraldehyde. For (S)-glyceraldehyde, the circle described by the #1, #2, and #3 priority groups is counter-clockwise (but first, we must flip the molecule over so that the H is pointing into the plane of the page).
We can follow a similar procedure for a molecule such as lactate. Clearly, H is the #4 substituent and OH is #1. Owing to its three bonds to oxygen, the CO2 group takes priority #2.
The structure shown is the S enantiomer of lactate.
Now, let’s try to assign a stereochemical configuration to the enantiomer of thalidomide that is the effective medication. The #4 priority is again hydrogen (in fact, the #4 group will be hydrogen for most of the stereocenters we encounter). The nitrogen group is #1, the carbonyl side of the ring is #2, and the –CH2 side of the ring is #3.
The hydrogen is shown pointing away from us, and the prioritized substituents trace a clockwise circle: this is the R enantiomer of thalidomide.
Example \(2\)
Label stereochemistry for the figures of glyceraldehyde, alanine, lactate and amphetamine shown earlier in the chapter (section 3.3).
Example \(3\)
Using solid or dashed wedges to show stereochemistry, draw the (R)-enantiomers of malate and ibuprofin (structures in exercise 3.6, this section), and mevalonate, 2-methylerithritol-4-phosphate, and dihydroorotate (structures in section 3.3).
Solutions
Summary
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existance of these molecules are determined by concept known as chirality. The word "chiral" was derived from the Greek word for hand, because our hands display a good example of chirality since they are non-superimposable mirror images of each other. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.4%3A_Stereoisomerism_in_Organic_Compounds.txt |
As noed before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Alkenes
Some representative alkenes—their names, structures, and physical properties—are given in Table \(1\).
Table \(1\): Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
We used only condensed structural formulas in Table \(1\). Thus, CH2=CH2 stands for
The double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
The first two alkenes in Table \(1\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(1\)). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Note
Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8.
Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC):
1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene.
2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene).
3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus,
is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering.
Example \(1\)
Name each compound.
Solution
1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.
2. The longest chain containing the double bond has four carbon atoms, so the parent compound is a butene (rule 1). (The longest chain overall has five carbon atoms, but it does not contain the double bond, so the parent name is not pentene.) To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 1-butene. There is an ethyl group on the second carbon atom (rule 3), so the compound’s name is 2-ethyl-1-butene.
Exercise \(1\)
Name each compound.
1. CH3CH2CH2CH2CH2CH=CHCH3
Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name.
Example \(2\)
Draw the structure for each compound.
1. 3-methyl-2-pentene
2. cyclohexene
Solution
1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:
Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.
2. First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond.
Exercise \(2\)
Draw the structure for each compound.
1. 2-ethyl-1-hexene
2. cyclopentene
Steroisomerism in Alkenes
There is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. Look at the two chlorinated hydrocarbons in Figure \(2\).
Table \(2\): Rotation about Bonds. In 1,2-dichloroethane (a), free rotation about the C–C bond allows the two structures to be interconverted by a twist of one end relative to the other. In 1,2-dichloroethene (b), restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond are significant.
In 1,2-dichloroethane (part (a) of Figure \(2\)), there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule:
In 1,2-dichloroethene (part (b) of Figure \(2\)), however, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule.
Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism (Figure \(3\)).
Table \(3\): Ball-and-Spring Models of (a) Cis-2-Butene and (b) Trans-2-Butene. Cis-trans isomers have different physical, chemical, and physiological properties.
Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule. Their structural formulas are as follows:
Note, however, that the presence of a double bond does not necessarily lead to cis-trans isomerism. We can draw two seemingly different propenes:
However, these two structures are not really different from each other. If you could pick up either molecule from the page and flip it over top to bottom, you would see that the two formulas are identical. Thus there are two requirements for cis-trans isomerism:
1. Rotation must be restricted in the molecule.
2. There must be two nonidentical groups on each doubly bonded carbon atom.
In these propene structures, the second requirement for cis-trans isomerism is not fulfilled. One of the doubly bonded carbon atoms does have two different groups attached, but the rules require that both carbon atoms have two different groups. In general, the following statements hold true in cis-trans isomerism:
• Alkenes with a C=CH2 unit do not exist as cis-trans isomers.
• Alkenes with a C=CR2 unit, where the two R groups are the same, do not exist as cis-trans isomers.
• Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.
Cis-trans isomerism also occurs in cyclic compounds. In ring structures, groups are unable to rotate about any of the ring carbon–carbon bonds. Therefore, groups can be either on the same side of the ring (cis) or on opposite sides of the ring (trans). For our purposes here, we represent all cycloalkanes as planar structures, and we indicate the positions of the groups, either above or below the plane of the ring.
Example \(3\)
Which compounds can exist as cis-trans (geometric) isomers? Draw them.
1. CHCl=CHBr
2. CH2=CBrCH3
3. (CH3)2C=CHCH2CH3
4. CH3CH=CHCH2CH3
Solution
All four structures have a double bond and thus meet rule 1 for cis-trans isomerism.
1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers:
2. This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers.
3. This compound has two methyl (CH3) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers.
4. This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:
Exercise \(3\)
Which compounds can exist as cis-trans isomers? Draw them.
1. CH2=CHCH2CH2CH3
2. CH3CH=CHCH2CH3
3. CH3CH2CH=CHCH2CH3
E,Z Convention
E,Z Convention is a method used to specify relative configuration at carbon atoms in alkene groups. According to E,Z convention, if the pair of carbon atoms in an alkene group could exist in two relative configurations, one is designated using the label E and the other the label Z.
To determine whether an alkene group is E or Z, use the following two-step procedure.
Step 1: Assign priority numbers to the two ligands on each carbon atom in the alkene group. (See R,S convention for the procedure.)
Step 2:
Alkynes
The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C2H2 and is known by its common name—acetylene (Figure \(1\)). Its structure is H–C≡C–H.
Note
Acetylene is used in oxyacetylene torches for cutting and welding metals. The flame from such a torch can be very hot. Most acetylene, however, is converted to chemical intermediates that are used to make vinyl and acrylic plastics, fibers, resins, and a variety of other products.
Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is -yne rather than -ene. The IUPAC name for acetylene is ethyne. The names of other alkynes are illustrated in the following exercises. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.5%3A_Alkenes_and_Alkynes.txt |
Next we consider a class of hydrocarbons with molecular formulas like those of unsaturated hydrocarbons, but which, unlike the alkenes, do not readily undergo addition reactions. These compounds comprise a distinct class, called aromatic hydrocarbons, with unique structures and properties. We start with the simplest of these compounds. Benzene (C6H6) is of great commercial importance. The formula C6H6 seems to indicate that benzene has a high degree of unsaturation. (Hexane, the saturated hydrocarbon with six carbon atoms has the formula C6H14—eight more hydrogen atoms than benzene.) However, despite the seeming low level of saturation, benzene is rather unreactive. It does not, for example, react readily with bromine, which, is a test for unsaturation.
Note
Benzene is a liquid that smells like gasoline, boils at 80°C, and freezes at 5.5°C. It is the aromatic hydrocarbon produced in the largest volume. It was formerly used to decaffeinate coffee and was a significant component of many consumer products, such as paint strippers, rubber cements, and home dry-cleaning spot removers. It was removed from many product formulations in the 1950s, but others continued to use benzene in products until the 1970s when it was associated with leukemia deaths. Benzene is still important in industry as a precursor in the production of plastics (such as Styrofoam and nylon), drugs, detergents, synthetic rubber, pesticides, and dyes. It is used as a solvent for such things as cleaning and maintaining printing equipment and for adhesives such as those used to attach soles to shoes. Benzene is a natural constituent of petroleum products, but because it is a known carcinogen, its use as an additive in gasoline is now limited.
To explain the surprising properties of benzene, chemists suppose the molecule has a cyclic, hexagonal, planar structure of six carbon atoms with one hydrogen atom bonded to each. We can write a structure with alternate single and double bonds, either as a full structural formula or as a line-angle formula:
However, these structures do not explain the unique properties of benzene. Furthermore, experimental evidence indicates that all the carbon-to-carbon bonds in benzene are equivalent, and the molecule is unusually stable.
Chemists often represent benzene as a hexagon with an inscribed circle:
The inner circle indicates that the valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized, or spread out, over all the carbon atoms). It is understood that each corner of the hexagon is occupied by one carbon atom, and each carbon atom has one hydrogen atom attached to it. Any other atom or groups of atoms substituted for a hydrogen atom must be shown bonded to a particular corner of the hexagon. We use this modern symbolism, but many scientists still use the earlier structure with alternate double and single bonds.
Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons.
The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms. This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers.
Aromaticity
Aromaticity is a property of conjugated cycloalkenes in which the stabilization of the molecule is enhanced due to the ability of the electrons in the orbitals to delocalize. This act as a framework to create a planar molecule. Why do we care if a compound is aromatic or not? Because we encounter aromatics every single day of our lives. Without aromatic compounds, we would not only be lacking many material necessities, our bodies would also not be able to function. Aromatic compounds are essential in industry; about 35 million tons of aromatic compounds are produced in the world every year to produce important chemicals and polymers, such as polyester and nylon. Aromatic compounds are also vital to the biochemistry of all living things. Three of the twenty amino acids used to form proteins ("the building blocks of life") are aromatic compounds and all five of the nucleotides that make up DNA and RNA sequences are all aromatic compounds. Needless to say, aromatic compounds are vital to us in many aspects.
The three general requirements for a compound to be aromatic are:
1. The compound must be cyclic
2. Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.
3. The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).
Among the many distinctive features of benzene, its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene.
The delocalization of the p-orbital carbons on the sp2 hybridized carbons is what gives the aromatic qualities of benzene.
Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring.
Aromatic Hydrocarbon Nomenclature
In the International Union of Pure and Applied Chemistry (IUPAC) system, aromatic hydrocarbons are named as derivatives of benzene. Figure \(1\) shows four examples. In these structures, it is immaterial whether the single substituent is written at the top, side, or bottom of the ring: a hexagon is symmetrical, and therefore all positions are equivalent.
Although some compounds are referred to exclusively by IUPAC names, some are more frequently denoted by common names, as is indicated in Table \(1\).
When there is more than one substituent, the corners of the hexagon are no longer equivalent, so we must designate the relative positions. There are three possible disubstituted benzenes, and we can use numbers to distinguish them (Figure \(2\)). We start numbering at the carbon atom to which one of the groups is attached and count toward the carbon atom that bears the other substituent group by the shortest path.
In Figure \(2\), common names are also used: the prefix ortho (o-) for 1,2-disubstitution, meta (m-) for 1,3-disubstitution, and para (p-) for 1,4-disubstitution. The substituent names are listed in alphabetical order. The first substituent is given the lowest number. When a common name is used, the carbon atom that bears the group responsible for the name is given the number 1:
Example \(2\)
Name each compound using both the common name and the IUPAC name.
Solution
1. The benzene ring has two chlorine atoms (dichloro) in the first and second positions. The compound is o-dichlorobenzene or 1,2-dichlorobenzene.
2. The benzene ring has a methyl (CH3) group. The compound is therefore named as a derivative of toluene. The bromine atom is on the fourth carbon atom, counting from the methyl group. The compound is p-bromotoluene or 4-bromotoluene.
3. The benzene ring has two nitro (NO2) groups in the first and third positions. It is m-dinitrobenzene or 1,3-dinitrobenzene.
Note: The nitro (NO2) group is a common substituent in aromatic compounds. Many nitro compounds are explosive, most notably 2,4,6-trinitrotoluene (TNT).
Exercise \(2\)
Name each compound using both the common name and the IUPAC name.
Sometimes an aromatic group is found as a substituent bonded to a nonaromatic entity or to another aromatic ring. The group of atoms remaining when a hydrogen atom is removed from an aromatic compound is called an aryl group. The most common aryl group is derived from benzene (C6H6) by removing one hydrogen atom (C6H5) and is called a phenyl group, from pheno, an old name for benzene. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.6%3A_Aromatic_Hydrocarbons.txt |
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups tend to behave in organic reactions.
Alcohols and Phenols
We have already seen the simplest possible example of an alcohol functional group in methanol. In the alcohol functional group, a carbon is single-bonded to an OH group (this OH group, by itself, is referred to as a hydroxyl). If the central carbon in an alcohol is bonded to only one other carbon, we call the group a primary alcohol. In secondary alcohols and tertiary alcohols, the central carbon is bonded to two and three carbons, respectively. Methanol, of course, is in class by itself in this respect.
Compounds in which an OH group is attached directly to an aromatic ring are designated ArOH and called phenols. Phenols differ from alcohols in that they are slightly acidic in water. They react with aqueous sodium hydroxide (NaOH) to form salts.
$ArOH_{(aq)} + NaOH_{(aq)} \rightarrow ArONa_{(aq)} + H_2O$
The parent compound, C6H5OH, is itself called phenol. (An old name, emphasizing its slight acidity, was carbolic acid.) Phenol is a white crystalline compound that has a distinctive (“hospital smell”) odor.
(Left) Structure of Phenol (right) Approximately two grams of phenol in glass vial. Image used with permisison from Wikipedia
Preparation and Uses of Alcohols
There are two primary methods to make alcohols in the laboratory: Hydration of an alkene and hydrolysis of an alkyl halide.
Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. The reaction is reversible.
Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH3CH=CH2. In principle, there are two different alcohols which might be formed:
You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol. If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH2 rather than the CH. The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around.
The other common method to make alcholes is a substitution reaction, the halogen atom is replaced by an -OH group to give an alcohol. For example:
Or, as an ionic equation:
In this example, 2-bromopropane is converted into propan-2-ol. The halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.
The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met.
Ethanol is usually sold as industrial methylated spirits, which is ethanol with a small quantity of methanol and possibly some color added. Because methanol is poisonous, industrial methylated spirits are unfit to drink, allowing purchasers to avoid the high taxes levied on alcoholic drinks. Ethanol burns to produce carbon dioxide and water, as shown in the equation below, and can be used as a fuel in its own right or in mixtures with petrol (gasoline). "Gasohol" is a petrol/ethanol mixture containing approximately 10–20% ethanol. Because ethanol can be produced by fermentation, this is a useful method for countries without an oil industry to reduce the amount of petrol imports.
$CH_3CH_2OH +3O_2 \rightarrow 2CO_2 + 3H_2O$
Ethanol is widely used as a solvent. It is relatively safe and can be used to dissolve many organic compounds that are insoluble in water. It is used, for example, in many perfumes and cosmetics.
Methanol also burns to form carbon dioxide and water:
$2CH_3OH +3O_2 \rightarrow 2CO_2 + 4H_2O$
It can be used a a petrol additive to improve combustion, and its use as a fuel in its own right is under investigation. Furthremore, most methanol is used to make other compounds, for example, methanal (formaldehyde), ethanoic acid, and methyl esters of various acids. In most cases, these are then converted into further products.
Phenols are widely used as antiseptics (substances that kill microorganisms on living tissue) and as disinfectants (substances intended to kill microorganisms on inanimate objects such as furniture or floors). The first widely used antiseptic was phenol. Joseph Lister used it for antiseptic surgery in 1867. Phenol is toxic to humans, however, and can cause severe burns when applied to the skin. In the bloodstream, it is a systemic poison—that is, one that is carried to and affects all parts of the body. Its severe side effects led to searches for safer antiseptics, a number of which have been found.
Ethers
In an ether functional group, a central oxygen is bonded to two carbons. Below are the line and Lewis structures of diethyl ether, a common laboratory solvent and also one of the first medical anaesthesia agents.
Preparation and Uses of Ethers
Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized method of preparing symmetrical ethers. As shown in the following two equations, the success of this procedure depends on the temperature. At 110º to 130 ºC an SN2 reaction of the alcohol conjugate acid leads to an ether product. At higher temperatures (over 150 ºC) an E2 elimination takes place.
2 CH3CH2-OH + H2SO4 130 ºC
CH3CH2-O-CH2CH3 + H2O
CH3CH2-OH + H2SO4 150 ºC
CH2=CH2 + H2O
In this reaction alcohol has to be used in excess and the temperature has to be maintained around 413 K. If alcohol is not used in excess or the temperature is higher, the alcohol will preferably undergo dehydration to yield alkene.
If ethanol is dehydrated to ethene in presence of sulfuric acid at 433 K but as 410 K, ethoxyethane is the main product. The dehydration of secondary and tertiary alcohols to get corresponding ethers is unsuccessful as alkenes are formed easily in these reactions.
This reaction cannot be employed to prepare unsymmetrical ethers. It is because a mixture of products is likely to be obtained.
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Preparation and Uses of Aldehydes and Ketones
Aldehydes and ketones can be prepared using a wide variety of reactions. Although these reactions are discussed in greater detail in other sections, they are listed here as a summary and to help with planning multistep synthetic pathways. A common way to synthesize aldehydes is the ixidation of 1o alcohols to form aldehydes
Hydration of an alkyne to form aldehydes via an addition reaction of a hydroxyl group to an alkyne forms an aldehyde. The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl.
Carboxylic acids
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in the acid-base chapter!).
Preparation and Uses of Carboxylic Acids
The oxidation of aldehydes or primary alcohols forms carboxylic acids:
In the presence of an oxidizing agent, ethanol is oxidized to acetaldehyde, which is then oxidized to acetic acid.
This process also occurs in the liver, where enzymes catalyze the oxidation of ethanol to acetic acid.
$\mathrm{CH_3CH_2OH \underset{oxidizing\: agent}{\xrightarrow{alcohol\: dehydrogenase}} CH_3CHO \underset{oxidizing\: agent}{\xrightarrow{alcohol\: dehydrogenase}} CH_3COOH\%}$
Acetic acid can be further oxidized to carbon dioxide and water.
Esters
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
Preparation and Uses of Esters
Some esters can be prepared by esterification, a reaction in which a carboxylic acid and an alcohol, heated in the presence of a mineral acid catalyst, form an ester and water:
The reaction is reversible. As a specific example of an esterification reaction, butyl acetate can be made from acetic acid and 1-butanol.
Esters are common solvents. Ethyl acetate is used to extract organic solutes from aqueous solutions—for example, to remove caffeine from coffee. It also is used to remove nail polish and paint. Cellulose nitrate is dissolved in ethyl acetate and butyl acetate to form lacquers. The solvent evaporates as the lacquer “dries,” leaving a thin film on the surface. High boiling esters are used as softeners (plasticizers) for brittle plastics.
Amides
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
Preparation and Uses of Amides
The addition of ammonia (NH3) to a carboxylic acid forms an amide, but the reaction is very slow in the laboratory at room temperature. Water molecules are split out, and a bond is formed between the nitrogen atom and the carbonyl carbon atom.
In living cells, amide formation is catalyzed by enzymes. Proteins are polyamides; they are formed by joining amino acids into long chains. In proteins, the amide functional group is called a peptide bond.
With the exception of formamide (HCONH2), which is a liquid, all simple amides are solids (Table $1$). The lower members of the series are soluble in water, with borderline solubility occurring in those that have five or six carbon atoms. Like the esters, solutions of amides in water usually are neutral—neither acidic nor basic.
Table $1$: Physical Constants of Some Unsubstituted Amides
Condensed Structural Formula Name Melting Point (°C) Boiling Point (°C) Solubility in Water
HCONH2 formamide 2 193 soluble
CH3CONH2 acetamide 82 222 soluble
CH3CH2CONH2 propionamide 81 213 soluble
CH3CH2CH2CONH2 butyramide 115 216 soluble
C6H5CONH2 benzamide 132 290 slightly soluble
The amides generally have high boiling points and melting points. These characteristics and their solubility in water result from the polar nature of the amide group and hydrogen bonding (Figure $1$). (Similar hydrogen bonding plays a critical role in determining the structure and properties of proteins, deoxyribonucleic acid [DNA], ribonucleic acid [RNA], and other giant molecules so important to life processes.
Amines
Ammonia is the simplest example of a functional group called amines. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations.
Preparation and Uses of Amines
The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical halogenoalkane. You get a mixture of amines formed together with their salts. The reactions happen one after another.
The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.
$CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+Br^-$
There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
$CH_3CH_2NH_3^+Br^- + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4^+ Br^-$
The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored.
The reaction does not stop at a primary amine. The ethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get a salt formed - this time, diethylammonium bromide. Think of this as ammonium bromide with two hydrogens replaced by ethyl groups.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the diethylammonium ion to leave a secondary amine - diethylamine. A secondary amine is one which has two alkyl groups attached to the nitrogen. The reaction does not stop! The diethylamine also reacts with bromoethane - in the same two stages as before. In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
The ammonia removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine. A tertiary amine is one which has three alkyl groups attached to the nitrogen.
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Molecules with Multiple Functional Groups
A single compound often contains several functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table on the inside back cover provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
Example $1$
Identify the functional groups in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Example $2$
Draw one example each (there are many possible correct answers) of compounds fitting the descriptions below, using line structures. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule).
1. a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
2. an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
3. A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.7%3A_Organic_Compounds_Containing_Functional_Groups.txt |
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $\pi$ bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DoU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
Problems
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Contributors
• Kim Quach (UCD) | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/26%3A_Structure_of_Organic_Compounds/26.8%3A_From_Molecular_Formula_to_Molecular_Structure.txt |
These are the four "prototypical" organic chemistry reactions, though several others which can be categorized as one of these are generally referred to by other names. Look at these reactions and ask yourself this question for each: what bonds are broken, and what bonds are formed.
At least 80% of the reactions students in organic chemistry fall into one of these four categories. The sooner you can get into the habit of recognizing bond formation and breakage the better off you will be. A fifth reaction is also discussed: rearrangement reactions.
Acid/Base Reactions
Another important category of organic reactions are straight-forward Brønsted–Lowry acid-base reactions (Figure $2$.
You’ll notice that they might look very different on the surface – all those different structures! – but the basic plot of each reaction is the same. we are breaking an H-(atom) bond and forming an H-(atom) bond. At the same time we are also connecting the two “leftover” partners to form a salt, composed of two oppositely-charged ions. Let’s look at that last reaction in more detail. Here, we are breaking a C-H bond and an (ionic) Na-NH2 bond, and forming an N-H bond as well as an (ionic) C-Na bond.
There are four “actors” in this reaction – as there are in every acid-base reaction – and we have names for all of them.
• The reactant where the bond to H is breaking is the acid.
• The reactant where the bond to H is forming is the base
• The product formed when the bond to H is broken is called the conjugate base.
• The product formed when the bond to H is formed is called the conjugate acid.
We can also draw the reverse of the previous reaction. Look at this carefully. we are still breaking a bond to H and forming a bond to H, but we’ve swapped everything. we are breaking N-H and C-Na, and forming N-Na and C-H. It’s still an acid-base reaction.
However, experiment tells us that this reaction does not happen to any appreciable extent.
Exercise $1$
In this acid/base reaction, which bonds are broken and with are formed, if any?
$\ce{(CH3-CH2)3N} + \ce{HCl} \rightarrow \ce{(CH3-CH2)3NH^+} + \ce{Cl^-} \nonumber$
Addition Reactions
When you take an alkene (or alkyne) and add certain types of reagents to them, you get results like this. See if you can recognize the bonds formed and broken.
Here’s the basic pattern: break a C-C multiple bond (also called a π bond) and form two new single bonds (“σ-bonds” to carbon). This reaction is called an “addition reaction” and it’s an extremely common type of reaction. Note that the reaction occurs only at the carbons that are a part of a multiple bond – nothing else on the molecule is affected.
Exercise $2$
In this addition reaction, which bonds are broken and with are formed, if any?
$\ce{CH3-C(=O)-CH3} \ce{->[\ce{ LiAlH4 }][\ce{ H^+/H2O }]} \ce{CH3-CH(OH)-CH3} \nonumber$
Note: This reaction could be described as a (nucleophilic) addition reaction or a reduction.
Substitution Reaction
Here are three examples of nucleophilic substitution reactions.
In each case, we are breaking a bond at carbon, and forming a new bond at carbon. This is an extremely common pattern for organic chemistry reactions. Here’s some interesting results that experiments tell us. We don’t get this information by thinking about what happens and predicting – we have to interrogate nature in order to get her to give up her secrets.
Exercise $3$
In this substitution reaction , which bonds are broken and with are formed, if any?
$\ce{CH3-C(=O)-OH + CH3-OH} \ce{->[\ce{H2SO4}][\text{ Δ and reflux }]} \ce{CH3-C(=O)-O-CH3} + \ce{H2O} \nonumber$
Note: This reaction would usually be called a condensation reaction.
Elimination Reactions
Let’s look at two simple cases. One important thing to keep in mind: although you see “Na OCH3” written here, the “Na+” (sodium) is not important for our purposes. It could alternatively be K+ (potassium) or Li+ (lithium). It’s just balancing the negative charge on the oxygen. When you take an alkyl halide and add a strong base (such as $\ce{NaOCH3}$ or $\ce{NaOCH2CH3}$) a reaction occurs. See if you can recognize the bonds broken and formed.
This reaction results in the forming of a new C-C double bond (π bond) and breaking two single bonds to carbon (in these cases, one of them is H and the other is a halide such as Cl or Br). We are also forming a new bond between H and the O. So in this respect, this reaction incorporates a pattern we have seen before – an acid-base reaction. Finally, we also form a salt in this reaction. Since we are primarily interested in the organic product (that is, the one containing carbon), you might find that the salt byproduct is not written in some reaction schemes, but that doesn’t mean that it’s not there.
Note
The three events in the elimination reactyio of Figure $3$ (formation of C-C π bond, breakage of two adjacent carbon sigma bonds) are the exact opposite of the addition reactions discussed above.
Another example of an elimination is when 2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.
Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask.
Exercise $4$
In this elimination reaction, which bonds are broken and with are formed, if any?
$\ce{CH3-CH(OH)-CH3} \ce{->[H_2SO_4][K_2Cr_2O_7 ]} \ce{CH3-C(=O)-CH3} \nonumber$
Note: This reaction could also be described an oxidation reaction.
Rearrangement Reactions
Rearrangement reactions can accompany many of the reactions we’ve previously covered such as substitution, addition, and elimination reactions. These reactions are comparatively rare. In fact, if you don’t look closely, sometimes you can miss the fact that a rearrangement reaction has occurred. Let’s look at a substitution reaction first.
On the top is a “typical” substitution reaction: we are taking an alkyl halide and adding water. The C-Br bond is broken and a C-OH bond is formed. If you look at the table on the right you’ll see this follows the typical pattern of substitution reactions. However if we change one thing about this alkyl halide – move the bromine to C-3 instead of C-2 – now when we run this reaction we see a different product emerge. It is also a substitution reaction (we are replacing Br with OH) but it’s on a different carbon. That’s because if you look closely, you can see there are actually 3 bonds broken and 3 bonds formed. The C2-H bond broke and the C3-H bond formed.
Exercise $5$
In this Rearrangement reaction, which bonds are broken and with are formed, if any?
$\ce{CH3-CH2-CH2-C(OH)=CH2} → \ce{CH3-CH2-CH2-C(=O)-CH3} \nonumber$
Note: This reaction could also be described an oxidation reaction. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.01%3A_Organic_Reactions%3A_An_Introduction.txt |
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. In the same way, we will see later that nucleophiles and leaving groups are sometimes protonated and sometimes not, so for now, for the sake of simplicity, we will not include protons on 'Nu' or 'X'. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
Associative nucleophilic substitution: the SN2 reaction
There are two mechanistic models for how a nucleophilic substitution reaction can proceed. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously.
This is called an 'associative', or 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product.
Dissociative nucleophilic substitution: the SN1 reaction
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybriduzed carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
While most nonenzymatic SN1 reactions are not stereoselective, we will see later that enzyme-catalyzed nucleophilic substitution reactions - whether SN1 or SN2 - almost always are stereoselective. The direct result of an enzymatic nucleophilic substitution reaction is more often than not inversion of configuration - this is because the leaving group usually remains bound in the enzyme's active site long enough to block a nucleophilic attack from that side. This does not mean, however, that enzymes can only catalyze substitution reactions with inversion of configuration: we will see in the next chapter (section 9.2) an example of an enzymatic nucleophilic substitution reaction in which the overall result is 100% retention of configuration.
In the SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was first introduced in Chapter 6. Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Recall (section 6.2) that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Example
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Solution (8.5)
Many SN1 reactions are of a class that are referred to as solvolysis, where a solvent molecule participates in the reaction as a nucleophile. The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
Because water and alcohols are relatively weak nucleophiles, they are unlikely to react in an SN2 fashion.
Nucleophilic substitutions occur at sp3-hybridized carbons
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons (section 1.5C), meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in section 8.4B).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this chapter focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
What makes a nucleophile?
As a general rule, nucleophile substitution reactions that involve powerful nucleophiles tend to occur with SN2 mechanisms, while reactions with weaker nucleophiles tend to be SN1. But what makes a group a strong or weak nucleophile? Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Protonation states and nucleophilicity
The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an SN2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile.
A neutral amine is nucleophilic, whereas a protonated ammonium cation is not. This is why enzymes which have evolved to catalyze nucleophilic reactions often have a basic amino acid side chain poised in position to accept a proton from the nucleophilic atom as the nucleophilic attack occurs.
Depending on the specific reaction being discussed, deprotonation of the nucleophile might occur before, during, or after the actual nucleophilic attack. In general for enzymatic reactions, however, it is most accurate to depict the proton abstraction and nucleophilic attack occurring simultaneously.
The basic enzymatic group could be a histidine, a neutral (deprotonated) arginine or lysine, or a negatively-charged (deprotonated) aspartate, glutamate, or tyrosine. For example, a more complete picture of the DNA methylation reaction we saw in section 8.1 shows an aspartate from the enzyme's active site accepting a proton from the nucleophilic amine as it attacks the carbon electrophile.
As it is deprotonated by the aspartate, the amine nitrogen becomes more electron-rich, and therefore more nucleophilic.
Periodic trends and solvent effects in nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall from section 7.3A that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic (you may want to review the reasoning for this trend in section 7.3A). What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity (see section 7.3C). If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
Steric effects on nucleophilicity
Steric hindrance is an important consideration when evaluating nucleophility. For example, tert-butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!).
It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.02%3A_Introduction_to_Substitution_Reactions.txt |
Elimination reactions are also possible at positions that are isolated from carbonyls or any other electron-withdrawing groups. This type of elimination can be described by two model mechanisms: it can occur in a single concerted step (proton abstraction at Cα occurring at the same time as Cβ-X bond cleavage), or in two steps (Cβ-X bond cleavage occurring first to form a carbocation intermediate, which is then 'quenched' by proton abstraction at the alpha-carbon).
These mechanisms, termed E2 and E1, respectively, are important in laboratory organic chemistry, but are less common in biological chemistry. As explained below, which mechanism actually occurs in a laboratory reaction will depend on the identity of the R groups (ie., whether the alkyl halide is primary, secondary, tertiary, etc.) as well as on the characteristics of the base.
E1 and E2 reactions in the laboratory
E2 elimination reactions in the laboratory are carried out with relatively strong bases, such as alkoxides (deprotonated alcohols). 2-bromopropane will react with ethoxide, for example, to give propene.
Propene is not the only product of this reaction, however - the ethoxide will also to some extent act as a nucleophile in an SN2 reaction.
Chemists carrying out nonenzymatic nucleophilic substitution or elimination reactions always have to be aware of the competition between the two mechanisms, because bases can also be nucleophiles, and vice-versa. However, a chemist can tip the scales in one direction or another by carefully choosing reagents. Primary carbon electrophiles like 1-bromopropane, for example, are much more likely to undergo substitution (by the SN2 mechanism) than elimination (by the E2 mechanism) – this is because the electrophilic carbon is unhindered and a good target for a nucleophile.
SN1 and E1 mechanisms are unlikely with such compounds because of the relative instability of primary carbocations.
The nature of the electron-rich species is also critical. Acetate, for example, is a weak base but a reasonably good nucleophile, and will react with 2-bromopropane mainly as a nucleophile.
In order to direct the reaction towards elimination, a strong hindered base such as tert-butoxide can be used. The bulkiness of tert-butoxide makes it difficult for the oxygen to reach the carbon (in other words, to act as a nucleophile). It is more likely to pluck off a proton, which is much more accessible than the electrophilic carbon).
E1 reactions occur by the same kinds of carbocation-favoring conditions that have already been described for SN1 reactions (section 9.4): a secondary or tertiary substrate, a protic solvent, and a relatively weak base/nucleophile. In fact, E1 and SN1 reactions generally occur simultaneously, giving a mixture of substitution and elimination products after formation of a common carbocation intermediate. When tert-butyl chloride is stirred in a mixture of ethanol and water, for example, a mixture of SN1 products (tert-butyl alcohol and tert-butyl ethyl ether) and E1 product (2-methylpropene) results.
Exercise
A straightforward functional group conversion that is often carried out in the undergraduate organic lab is the phosphoric acid-catalyzed dehydration of cyclohexanol to form cyclohexene. No solvent is necessary in this reaction - pure liquid cyclohexanol is simply stirred together with a few drops of concentrated phosphoric acid. In order to drive the equilibrium of this reversible reaction towards the desired product, cyclohexene is distilled out of the reaction mixture as it forms (the boiling point of cyclohexene is 83 oC, significantly lower than that of anything else in the reaction solution). Any cyclohexyl phosphate that might form from the competing SN1 reaction remains in the flask, and is eventually converted to cyclohexene over time. Draw a mechanism for the cyclohexene synthesis reaction described above. Also, draw a mechanism showing how the undesired cyclohexyl phosphate could form.
Solutions to exercises
Next, let’s put aside the issue of competition between nucleophilic substitution and elimination, and focus on the regioselectivity of elimination reactions. In many cases an elimination reaction can result in more than one constitutional isomer or stereoisomer. The elimination products of 2-chloropentane provide a good example:
This reaction is both regiospecific and stereospecific. In general, more substituted alkenes are more stable, and as a result, the product mixture will contain less 1-butene than 2-butene (this is the regiochemical aspect of the outcome, and is often referred to as Zaitsev's rule). In addition, we already know that trans (E) alkenes are generally more stable than cis (Z) alkenes (section 3.7C), so we can predict that more of the E product will form compared to the Z product.
The Hoffman elimination is a well-studied E2 elimination in which the leaving group is a quaternary amine - note that there is no proton on the quaternary amine that could protonate the base in the reaction:
In practice, the quaternary amine is made by treating a primary or secondary amine with excess methyl iodide and weak base (this, of course, is an SN2 methylation reaction, covered in section 9.1B). Silver oxide in water generates the necessary hydroxide ion.
There is very little competing substitution under these conditions.
The Hoffman elimination is somewhat unique in that, unlike other elimination reactions, it is usually the least substituted alkene that is the predominant product. This is due to steric factors: the large size of the quaternary ammonium leaving group results in the most accessible (least hindered) proton being abstracted - meaning the proton from the least substituted carbon.
The Cope elimination is another well-known laboratory E2 reaction involving an amine oxide:
Notice that the leaving group is also the base. The transition state is thought to be planar (or close to it). | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.03%3A_Introduction_to_Elimination_Reactions.txt |
Learning Objectives
1. Give two major types of reactions of alcohols.
2. Describe the result of the oxidation of a primary alcohol.
3. Describe the result of the oxidation of a secondary alcohol.
Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH-bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure \(1\), two—dehydration and oxidation—are considered here. The third reaction type—esterification—is covered elsewhere.
Table \(1\): Reactions of Alcohols. Oxidation and dehydration of alcohols are considered here.
Dehydration of Alcohols
As noted in Figure \(1\), an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule:
Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule.
(Ethers are discussed in elsewhere) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols.
Note
Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway
Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds.
Oxidation of Acohols
Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows:
We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones. The oxidation of isopropyl alcohol by potassium dichromate (K2Cr2O7) gives acetone, the simplest ketone:
Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH2OH) and secondary (R2CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group.
Note
These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows:
Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group:
The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone.
Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized.
Example \(1\)
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
1. CH3CH2CH2CH2CH2OH
Solution
The first step is to recognize the class of each alcohol as primary, secondary, or tertiary.
1. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid.
2. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs.
3. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone.
Exercise \(1\)
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
Reactions of Alcohols with Hydrogen Halides (HX)
When alcohols react with a hydrogen halide, a substitution occurs, producing an alkyl halide and water:
• The order of reactivity of alcohols is 3° > 2° > 1° methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation in an \(S_N1\) reaction with the protonated alcohol acting as the substrate. The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group (\)H_2O\_), which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we perform the reaction in the presence of acid and in the presence of halide ions and not at elevated temperatures. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and because halide ions are present in a high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an \(S_n1\) reaction.
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism. In these reactions, the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
Acid protonates the alcohol hydroxyl group, making it a good leaving group. However, other strong Lewis acids can be used instead of hydrohalic acids. Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride does not react with primary or secondary alcohols unless zinc chloride or a similar Lewis acid is added to the reaction mixture as well. Zinc chloride, a good Lewis acid, forms a complex with the alcohol through association with an unshared pair of electrons on the oxygen atom. This enhances the hydroxyl’s leaving group potential sufficiently so that chloride can displace it.
As we might expect, many reactions of alcohols with hydrogen halides, particularly those in which carbocations are formed, are accompanied by rearrangements. The general rule is that if rearrangement CAN OCCUR (to form more stable or equally stable cations), it will! In these reactions, mixtures of products can be formed. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.04%3A_Reactions_of_Alcohols.txt |
The most common chemical transformation of a carbon-carbon double bond is the addition reaction. A large number of reagents, both inorganic and organic, have been found to add to this functional group, and in this section we shall review many of these reactions.
• Addition of Lewis Acids (Electrophilic Reagents)
• Addition of Strong Brønsted Acids
• Rearrangement of Carbocations
A large number of reagents, both inorganic and organic, have been found to add to this functional group, and in this section we shall review many of these reactions. A majority of these reactions are exothermic, due to the fact that the C-C pi-bond is relatively weak (ca. 63 kcal/mole) compared to the sigma-bonds formed with the atoms or groups of the reagent. Remember, the bond energies of a molecule are the energies required to break (homolytically) all the covalent bonds in the molecule. Consequently, if the bond energies of the product molecules are greater than the bond energies of the reactants, the reaction will be exothermic. The following calculations for the addition of H-Br are typical. Note that by convention exothermic reactions have a negative heat of reaction.
Addition of Hydrogen: Hydrogenation of Alkenes
Addition of hydrogen to a carbon-carbon double bond is called hydrogenation. The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst,
Ethene reacts with hydrogen in the presence of a finely divided palladium catalyst at a temperature of about 150°C. Ethane is produced.
This is a fairly pointless reaction because ethene is a far more useful compound than ethane! However, what is true of the reaction of the carbon-carbon double bond in ethene is equally true of it in much more complicated cases.
Addition of Hydrogen Halides to Alkenes
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other. For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions. When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be. Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the $pi$ bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this. Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes. The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Addition to unsymmetrical alkenes
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond. If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.
This is in line with Markovnikov's Rule.
Definition: Markovnikov's Rule
When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant.
Addition of Halogens: Halogenation of Alkenes
As halogen molecule, for example Br2, approaches a double bond of the alkene, electrons in the double bond repel electrons in bromine molecule causing polarization of the halogen bond. This creates a dipolar moment in the halogen molecule bond. Heterolytic bond cleavage occurs and one of the halogens obtains positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but stereoselective.Stereochemistry of this addition can be explained by the mechanism of the reaction.In the first step electrophilic halogen with a positive charge approaches the double carbon bond and 2 p orbitals of the halogen, bond with two carbon atoms and create a cyclic ion with a halogen as the intermediate step. In the second step, halogen with the negative charge attacks any of the two carbons in the cyclic ion from the back side of the cycle as in the SN2 reaction. Therefore stereochemistry of the product is vicinial dihalides through anti addition.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X}$
Halogens that are commonly used in this type of the reaction are: $Br$ and $Cl$. In thermodynamical terms $I$ is too slow for this reaction because of the size of its atom, and $F$ is too vigorous and explosive. Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl4) can be used in this reaction. Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.
Electrophilic addition mechanism consists of two steps. Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene.
Nucleophile Double bond in alkene
Electrophile Br2, Cl2
Regiochemistry not relevant
Stereochemistry
ANTI
Step 1: In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond.
Step 2: In the second step, bromide anion attacks any carbon of the bridged bromonium ion from the back side of the cycle. Cycle opens up and two halogens are in the position anti. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.05%3A_Introduction_to_Addition_Reactions%3A_Reactions_of_Alkenes.txt |
Aromatics, or arenes, are derivatives of benzene or other compounds with aromatic ring systems. That is, they are cyclic, planar, fully conjugated and have an odd number of π-electron pairs. Like alkenes, aromatics have π-electrons that are loosely held and are easily attracted to electrophiles. However, aromatics don't undergo the typical reactions of alkenes.
For example, bromine will not add across the double bond of benzene.
Instead, a bromine atom replaces one of the hydrogen atoms on the benzene. This reaction is greatly accelerated in the presence of Lewis acids, such as ferric chloride.
A similar reaction happens with chlorine. If treated with chlorine gas and a metal catalyst, a chlorine atom from chlorine gas can replace a hydrogen atom on benzene. However, the same thing doesn't work as smoothly with the other halogens, iodine and fluorine.
The reactions of chlorine and bromine with benzene and other aromatics can be catalysed by a variety of Lewis acidic metal catalysts. So can the reactions of alkyl halides and acyl halides, which we don't normally think of as electrophiles for alkene addition.
There are some limitations on what kind of groups can be added in this way. The carbon attached to the halide should be tetrahedral. Typically, it is much easier to add secondary or tertiary alkyls than primary ones. That is, the carbon attached to the halogen had best be attached to two or three other carbons as well. Methyls are very, very difficult to add in this way.
There is an exception - the carbon attached to the halogen need not be tetrahedral, provided it is a carbonyl carbon. That reaction is called an acylation.
In these cases, it is the alkyl or acyl, rather than the halogen, that replaces a hydrogen atom on the benzene. Remember, benzene is most likely acting as a nucleophile in this reaction, even though it is following a different pathway than an alkene would. It is reacting with the most electrophilic part of the alkyl halide or acyl halide.
Aromatics have a limited repertoire of electrophiles with which they commonly undergo reaction. In addition to these Lewis acid-catalysed reactions, there are also reactions strong acidic media, such as a mixture of nitric and sulfuric acid.
Another acidic medium, referred to as "fuming sulfuric acid," is really a mixture of sulfuric acid and sulfur trioxide.
Just as with the acid-catalyzed reactions, the nitro group and the sulfonate group just replace a hydrogen atom on the benzene ring. The overall reaction involves bond formation between a benzene carbon and the electrophile, and bond cleavage between the same carbon and a proton.
Activation and Deactivation and Directing Effects
Because the benzene acts as a nucleophile in electrophilic aromatic substitution, substituents that make the benzene more electron-rich can accelerate the reaction. Substituents that make the benzene moor electron-poor can retard the reaction. In the mid-twentieth century, physical organic chemists including Christopher Ingold conducted a number of kinetic studies on electrophilic aromatic substitution reactions. In table 1, you can see that some substituents confer a rate of reaction that is much higher than that of benzene (R = H). Phenol, C6H5OH, undergoes nitration a thousand times faster than benzene does. Nitrobenzene, C6H5NO2, undergoes the reaction millions of times more slowly.
Table \(1\): Rate of nitration in benzene derivatives
R in C6H5R Relative rate
OH 1,000
CH3 25
H 1
CH2Cl 0.71
I 0.18
F 0.15
Cl 0.033
Br 0.030
CO2Et 0.0037
NO2 6 x 10-8
NMe3+ 1.2 x 10-8
These observations are consistent with the role of the aromatic as a nucleophile in this reaction. Substituents that draw electron density away from the aromatic ring slow the reaction down. These groups are called deactivating groups in this reaction. Substituents that readily donate electron desnity to the ring, or that effectively stabilize the cationic intermediate, promote the reaction. These groups are called activating groups in this reaction.
The roles of these groups are related to their electronic interactions with the electrons in the ring. Some groups might be π-donors, providing additional electron density to the benzene ring via conjugation.
Other groups may be π-acceptors, drawing electron density away from the ring via conjugation.
Still others may be σ-acceptors, drawing electron density away from the ring via a simple inductive effect which arises from the electronegativity of a substituent.
In some cases, there may be multiple effects, and the overall influence of the substituents is determined by the balance of the effects. One effect may be stronger in one case than the other, so it wins out in one case and loses in another.
In addition to exerting an effect on the speed of reaction, substituents on the benzene ring also influence the regiochemistry of the reaction. That is, they control where the new substituent appears in the product.
Remember, there are three different position on the bezene ring where a new substituent can attach, relative to the original substituent. Substitution could actually occur on five positions around the ring, but two pairs are related by symmetry. Isomerism in disubstituted benzenes can be described by numbering the substituents (1,2- etc) or by the relationships ortho-, meta- and para-. There are two positions ortho- to the initial substituent and two positions meta- to it.
Ingold and colleagues investigated the question of regiochemistry in nitration. They reported the following observations:
Table \(2\) : Substitution patterns during nitration of benzene derivatives
R in C6H5R % o- product % m- product % p- product
CH3 56 3 41
Cl 30 0 70
Br 38 0 62
OH 10 0 90
CHO 19 72 9
CO2Et 28 68 3
CN 17 81 2
NO2 6 94 0
In looking at the table, you might see that there are two groups of substituents. One group reacts to make mixtures of ortho- and para- products. There may be different ratios of ortho- to para- and there may be small amounts of meta-, but don't get bogged down in the details. Focus on the bigger picture. Some groups are "ortho-/para-directors".
The other group reacts to makemostly meta-substituted products. Here may be small amounts ofortho- and para- products, but don't worry about that. Focus on the bigger picture. Some groups are "meta-directors". These regiochemical effects are very closely related to the activating and directing effects we have already seen. If we want to understand this data, we need to think about things like π-donation, π-acceptance, inductive effects and cation stability. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.06%3A_Electrophilic_Aromatic_Substitution.txt |
Alkanes (the most basic of all organic compounds) undergo very few reactions. The two reactions of more importaces is combustion and halogenation, (i.e., substitution of a single hydrogen on the alkane for a single halogen) to form a haloalkane. The halogen reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!
Example $1$: Propane Combustion
For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:
$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$
Counting the oxygens leads directly to the final version:
$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
Example $2$: Butane Combustion
With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.
$C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$
Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O2 molecules on the left.
$C_4H_{10} + 6\dfrac{1}{2}\, O_2 \rightarrow 4CO_2 + 5H_2O$
If that offends you, double everything:
$2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O$
The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.
Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.
Note: Why carbon monoxide is poisonous
Oxygen is carried around the blood by hemoglobin, which unfortunately binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.
Halogenation of Alkanes
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane, shown below, provides a simple example of this reaction.
CH4 + Cl2 + energy → CH3Cl + HCl
Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride.
CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine. The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions that happen at room temperature. We'll look at the reactions with chlorine, although the reactions with bromine are similar, but evolve more slowly.
Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
The original mixture of a colorless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the exception of the chloromethane which is a gas.
If you were using bromine, you could either mix methane with bromine vapor , or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green. One would not choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate. The mechanisms for the reactions are explained on separate pages.
Larger alkanes and chlorine
You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem! For example, with propane, you could get one of two isomers:
If chance was the only factor, you would expect to get three times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the center carbon atom.
Cycloalkanes
The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane. In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.
This can still happen in the presence of light - but you will get substitution reactions as well. The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds. The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.07%3A_Reactions_of_Alkanes.txt |
Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by Hermann Staudinger, a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of macromolecules composed of 10,000 or more atoms. He formulated a polymeric structure for rubber, based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms polymer and monomer were derived from the Greek roots poly (many), mono (one) and meros (part).
Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections.
There are two general types of polymerization reactions: addition polymerization and condensation polymerization. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers. Ethylene molecules are joined together in long chains.
Note
Many natural materials—such as proteins, cellulose and starch, and complex silicate minerals—are polymers. Artificial fibers, films, plastics, semisolid resins, and rubbers are also polymers. More than half the compounds produced by the chemical industry are synthetic polymers.
Chain-Reaction (Addition) Polymerization
The polymerization can be represented by the reaction of a few monomer units:
The bond lines extending at the ends in the formula of the product indicate that the structure extends for many units in each direction. Notice that all the atoms—two carbon atoms and four hydrogen atoms—of each monomer molecule are incorporated into the polymer structure. Because displays such as the one above are cumbersome, the polymerization is often abbreviated as follows:
nCH2=CH2 [ CH2CH2 ] n
During the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator.
Some common addition polymers are listed in Table $1$. Note that all the monomers have carbon-to-carbon double bonds. Many polymers are mundane (e.g., plastic bags, food wrap, toys, and tableware), but there are also polymers that conduct electricity, have amazing adhesive properties, or are stronger than steel but much lighter in weight.
Table $1$ : Some Addition Polymers
Monomer Polymer Polymer Name Some Uses
CH2=CH2 ~CH2CH2CH2CH2CH2CH2~ polyethylene plastic bags, bottles, toys, electrical insulation
CH2=CHCH3 polypropylene carpeting, bottles, luggage, exercise clothing
CH2=CHCl polyvinyl chloride bags for intravenous solutions, pipes, tubing, floor coverings
CF2=CF2 ~CF2CF2CF2CF2CF2CF2~ polytetrafluoroethylene nonstick coatings, electrical insulation
Step 1: Chain Initiation
The oxygen reacts with some of the ethene to give an organic peroxide. Organic peroxides are very reactive molecules containing oxygen-oxygen single bonds which are quite weak and which break easily to give free radicals. You can short-cut the process by adding other organic peroxides directly to the ethene instead of using oxygen if you want to. The type of the free radicals that start the reaction off vary depending on their source. For simplicity we give them a general formula: $Ra ^{\bullet}$
Step 2: Chain Propagation
In an ethene molecule, CH2=CH2, the two pairs of electrons which make up the double bond aren't the same. One pair is held securely on the line between the two carbon nuclei in a bond called a sigma bond. The other pair is more loosely held in an orbital above and below the plane of the molecule known as a $\pi$ bond.
Note
It would be helpful - but not essential - if you read about the structure of ethene before you went on. If the diagram above is unfamiliar to you, then you certainly ought to read this background material.
Imagine what happens if a free radical approaches the $\pi$ bond in ethene.
Note
Don't worry that we've gone back to a simpler diagram. As long as you realise that the pair of electrons shown between the two carbon atoms is in a $\pi$ bond - and therefore vulnerable - that's all that really matters for this mechanism.
The sigma bond between the carbon atoms isn't affected by any of this. The free radical, Ra, uses one of the electrons in the $\pi$ bond to help to form a new bond between itself and the left hand carbon atom. The other electron returns to the right hand carbon. You can show this using "curly arrow" notation if you want to:
Note
If you aren't sure about about curly arrow notation you can follow this link.
This is energetically worth doing because the new bond between the radical and the carbon is stronger than the $\pi$ bond which is broken. You would get more energy out when the new bond is made than was used to break the old one. The more energy that is given out, the more stable the system becomes. What we've now got is a bigger free radical - lengthened by CH2CH2. That can react with another ethene molecule in the same way:
So now the radical is even bigger. That can react with another ethene - and so on and so on. The polymer chain gets longer and longer.
Step 3: Chain Termination
The chain does not, however, grow indefinitely. Sooner or later two free radicals will collide together.
That immediately stops the growth of two chains and produces one of the final molecules in the poly(ethene). It is important to realise that the poly(ethene) is going to be a mixture of molecules of different sizes, made in this sort of random way. Because chain termination is a random process, poly(ethene) will be made up of chains of different lengths.
Step-Reaction (Condensation) Polymerization
A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety.
Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram.
Characteristics of Condensation Polymers
Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative.
Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers
The difference in Tg and Tm between the first polyester (completely aliphatic) and the two nylon polyamides (5th & 6th entries) shows the effect of intra-chain hydrogen bonding on crystallinity. The replacement of flexible alkylidene links with rigid benzene rings also stiffens the polymer chain, leading to increased crystalline character, as demonstrated for polyesters (entries 1, 2 &3) and polyamides (entries 5, 6, 7 & 8). The high Tg and Tm values for the amorphous polymer Lexan are consistent with its brilliant transparency and glass-like rigidity. Kevlar and Nomex are extremely tough and resistant materials, which find use in bullet-proof vests and fire resistant clothing.
Many polymers, both addition and condensation, are used as fibers The chief methods of spinning synthetic polymers into fibers are from melts or viscous solutions. Polyesters, polyamides and polyolefins are usually spun from melts, provided the Tm is not too high. Polyacrylates suffer thermal degradation and are therefore spun from solution in a volatile solvent. Cold-drawing is an important physical treatment that improves the strength and appearance of these polymer fibers. At temperatures above Tg, a thicker than desired fiber can be forcibly stretched to many times its length; and in so doing the polymer chains become untangled, and tend to align in a parallel fashion. This cold-drawing procedure organizes randomly oriented crystalline domains, and also aligns amorphous domains so they become more crystalline. In these cases, the physically oriented morphology is stabilized and retained in the final product. This contrasts with elastomeric polymers, for which the stretched or aligned morphology is unstable relative to the amorphous random coil morphology.
This cold-drawing treatment may also be used to treat polymer films (e.g. Mylar & Saran) as well as fibers.
Step-growth polymerization is also used for preparing a class of adhesives and amorphous solids called epoxy resins. Here the covalent bonding occurs by an SN2 reaction between a nucleophile, usually an amine, and a terminal epoxide. In the following example, the same bisphenol A intermediate used as a monomer for Lexan serves as a difunctional scaffold to which the epoxide rings are attached. Bisphenol A is prepared by the acid-catalyzed condensation of acetone with phenol. | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.08%3A_Polymers_and_Polymerization_Reactions.txt |
Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules.
The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does he cut, tailor and glue the components on a molecule that he cannot see? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of Chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. In this task, he is further guided by several ‘thumb rules’ that chemists have evolved over the past two centuries. The discussions on the topics Name Reactions, Reagents for synthesis, Spectroscopy and Chromatography are beyond the scope of this write-up. Let us begin with a brief look at some of the important ‘Rules’ in organic chemistry that guide us in planning organic synthesis. We would then discuss Protection and Deprotection of some important functional groups. We could then move on to the Logic of planning Organic Synthesis.
Modern Synthesis
A multi-step synthesis of any organic compound requires the chemist to accomplish three related tasks:
1. Constructing the carbon framework or skeleton of the desired molecule.
2. Introducing, removing or transforming functional groups in a fashion that achieves the functionality of the desired compound.
3. Exercising selective stereocontrol at all stages in which centers of stereoisomerism are created or influenced. These are not discrete independent tasks to be attacked and solved in turn, but must be integrated and correlated in an overall plan. Thus, the assembly of the molecular framework will depend in part on the structure and functionality of available starting materials, the selectivity (regio and stereo) of the various reactions that may be used to stitch them together, and the loss or relocation of functional groups in the intermediate compounds formed on the way to the final product.
Other factors must also be considered, always in context with those listed above:
1. Since a successful synthesis must produce the desired product in reasonable amount, it should be as short and efficient as possible. A two or three-step sequence is usually better than a six or seven step procedure, even if the individual step yields are better in the longer route. Most reactions do not proceed in 100% yield, and the losses are multiplied with each additional step. Furthermore, a long multi-step synthesis requires many hours of effort by the chemists conducting the reactions.
2. The format of a synthesis is important. Assuming a constant yield for each step, a linear sequence of reactions gives a poorer overall yield than the same number of convergent reactions, as shown in the following diagram.
3. If numerous functional groups are present at intermediate stages, some of these may require protection from unwanted reaction. Since the use of a protective group requires its introduction and later removal, such operations can add many steps to a synthesis. A similar cost is attached to the use of blocking and activating groups.
4. The starting compounds and reagents for a synthesis must be purchased, so economic considerations are often significant. Recovery or recycling of expensive reagents and catalysts is often desirable. Also, if large amounts of useless by-products are formed in the synthesis, the expense of their disposal is a factor. The term atom efficiency has been coined to reflect the latter point.
5. For a given target molecule, several different reaction sequences may serve to accomplish its synthesis. Indeed, new synthetic routes to known compounds continue to be reported, particularly as new reactions are developed and applied to difficult transformations. Evaluating the quality and efficacy of these diverse procedures involves consideration of all the above. Over the course of the past hundred years, a very large number of syntheses for a wide variety of compounds have been recorded. For all but the simplest of these, a majority of the reactions in the synthesis involve functional group modification, preceding or following a smaller number of carbon-carbon bond forming reactions. Because functional group chemistry consists of such a vast number of addition, elimination and substitution interconversions, it is not possible to identify a general pattern in their application to synthesis. Instead, an example from the synthesis of reserpine by the R. B. Woodward group (Harvard), displayed in the following diagram, will serve to illustrate the importance of regio and stereo-control in the course of functional group modification.
28.1: Chemical Structure of Living Matter: An Overview
Learning Objectives
• To identify the common structural units of important biological molecules.
Contributors
• Anonymous
Modified by Joshua B. Halpern | textbooks/chem/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.09%3A_Synthesis_of_Organic_Compounds.txt |
Learning Objectives
• To get an overview of hydrocarbons molecules and their four primary classifications
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors.
The classifications for hydrocarbons, defined by IUPAC nomenclature of organic chemistry are as follows:
1. Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. The general formula for saturated hydrocarbons is \(C_nH_{2n+2}\) (assuming non-cyclic structures). Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
2. Unsaturated hydrocarbons have one or more double or triple bonds between carbon atoms. Those with double bond are called alkenes and those with one double bond have the formula \(C_nH_{2n}\) (assuming non-cyclic structures). Those containing triple bonds are called alkynes, with general formula \(C_nH_{2n-2}\). The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The smallest alkyne is ethyne, which is also known as acetylene:
1. Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The general formula for a saturated hydrocarbon containing one ring is \(C_nH_{2n}\).
2. Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds.
The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that the bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30).
• Wikipedia
7.2: The Alkanes
Learning Objectives
• To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) and they are shown again in Figure $1$.
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure $2$).
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $1$.
Table $1$: The First 10 Straight-Chain Alkanes
Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers
methane CH4 CH4
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3 2
pentane C5H12 CH3CH2CH2CH2CH3 3
hexane C6H14 CH3CH2CH2CH2CH2CH3 5
heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9
octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18
nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35
decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75
Consider the series in Figure $3$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Key Takeaway
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.1%3A_Petroleum_Refining_and_the_Hydrocarbons.txt |
Learning Objectives
• To name alkenes given formulas and write formulas for alkenes given names.
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Some representative alkenes—their names, structures, and physical properties—are given in Table \(1\).
Table \(1\): Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
We used only condensed structural formulas in Table \(1\). Thus, CH2=CH2 stands for
The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
The first two alkenes in Table \(1\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(1\)). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8.
Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC):
1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene.
2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene).
3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus, the structure below is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering.
Example \(1\)
Name each compound.
Solution
1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.
2. The longest chain containing the double bond has five carbon atoms, so the parent compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the third carbon atom (rule 3), so the compound’s name is 3-methyl-2-pentene.
Exercise \(1\)
Name each compound.
1. CH3CH2CH2CH2CH2CH=CHCH3
Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name.
Example \(2\)
Draw the structure for each compound.
1. 3-methyl-2-pentene
2. cyclohexene
Solution
1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:
Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.
• First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond.
Exercise \(2\)
Draw the structure for each compound.
1. 2-ethyl-1-hexene
2. cyclopentene
Key Takeaway
• Alkenes are hydrocarbons with a carbon-to-carbon double bond. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.3%3A_The_Alkenes_and_Alkynes.txt |
Learning Objectives
• To describe the bonding in benzene and the way typical reactions of benzene differ from those of the alkenes.
Next we consider a class of hydrocarbons with molecular formulas like those of unsaturated hydrocarbons, but which, unlike the alkenes, do not readily undergo addition reactions. These compounds comprise a distinct class, called aromatic hydrocarbons, with unique structures and properties. We start with the simplest of these compounds. Benzene (C6H6) is of great commercial importance, but it also has noteworthy health effects.
The formula C6H6 seems to indicate that benzene has a high degree of unsaturation. (Hexane, the saturated hydrocarbon with six carbon atoms has the formula C6H14—eight more hydrogen atoms than benzene.) However, despite the seeming low level of saturation, benzene is rather unreactive. It does not, for example, react readily with bromine, which, is a test for unsaturation.
Benzene is a liquid that smells like gasoline, boils at 80°C, and freezes at 5.5°C. It is the aromatic hydrocarbon produced in the largest volume. It was formerly used to decaffeinate coffee and was a significant component of many consumer products, such as paint strippers, rubber cements, and home dry-cleaning spot removers. It was removed from many product formulations in the 1950s, but others continued to use benzene in products until the 1970s when it was associated with leukemia deaths. Benzene is still important in industry as a precursor in the production of plastics (such as Styrofoam and nylon), drugs, detergents, synthetic rubber, pesticides, and dyes. It is used as a solvent for such things as cleaning and maintaining printing equipment and for adhesives such as those used to attach soles to shoes. Benzene is a natural constituent of petroleum products, but because it is a known carcinogen, its use as an additive in gasoline is now limited.
To explain the surprising properties of benzene, chemists suppose the molecule has a cyclic, hexagonal, planar structure of six carbon atoms with one hydrogen atom bonded to each. We can write a structure with alternate single and double bonds, either as a full structural formula or as a line-angle formula:
However, these structures do not explain the unique properties of benzene. Furthermore, experimental evidence indicates that all the carbon-to-carbon bonds in benzene are equivalent, and the molecule is unusually stable. Chemists often represent benzene as a hexagon with an inscribed circle:
The inner circle indicates that the valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized, or spread out, over all the carbon atoms). It is understood that each corner of the hexagon is occupied by one carbon atom, and each carbon atom has one hydrogen atom attached to it. Any other atom or groups of atoms substituted for a hydrogen atom must be shown bonded to a particular corner of the hexagon. We use this modern symbolism, but many scientists still use the earlier structure with alternate double and single bonds.
To Your Health: Benzene and Us
Most of the benzene used commercially comes from petroleum. It is employed as a starting material for the production of detergents, drugs, dyes, insecticides, and plastics. Once widely used as an organic solvent, benzene is now known to have both short- and long-term toxic effects. The inhalation of large concentrations can cause nausea and even death due to respiratory or heart failure, while repeated exposure leads to a progressive disease in which the ability of the bone marrow to make new blood cells is eventually destroyed. This results in a condition called aplastic anemia, in which there is a decrease in the numbers of both the red and white blood cells.
Key Takeaway
• Aromatic hydrocarbons appear to be unsaturated, but they have a special type of bonding and do not undergo addition reactions.
7.5: Fullerenes
If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C60) has been named buckminsterfullerene because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called fullerenes. These materials represent a third class of carbon allotropes. Alternating views of the C60 fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor. By clicking on this graphic, a model of C60 will be displayed.
Although C60 is composed of fused benzene rings its chemical reactivity resembles that of the cycloalkenes more than benzene. Indeed, exposure to light and oxygen slowly degrade fullerenes to cage opened products. Most of the reactions thus far reported for C60 involve addition to, rather than substitution of, the core structure. These reactions include hydrogenation, bromination and hydroxylation. Strain introduced by the curvature of the surface may be responsible for the enhanced reactivity of C60.
. Larger fullerenes, such as C70, C76, C82 & C84have ellipsoidal or distorted spherical structures, and fullerene-like assemblies up to C240 have been detected. A fascinating aspect of these structures is that the space within the carbon cage may hold atoms, ions or small molecules. Such species are called endohedral fullerenes. The cavity of C60 is relatively small, but encapsulated helium, lithium and atomic nitrogen compounds have been observed. Larger fullerenes are found to encapsulate lanthanide metal atoms.
Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.4%3A_Aromatic_Hydrocarbons.txt |
Learning Objectives
• To know the major classes of organic compounds and identify important functional groups.
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure \(1\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
The first family listed in Figure \(1\) is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(2\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Summary
Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Conceptual Problems
1. Can two substances have the same systematic name and be different compounds?
2. Is a carbon–carbon multiple bond considered a functional group? | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.6%3A_Functional_Groups_and_Organic_Reactions.txt |
This page explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number), complex ion formation, colored ions, and catalytic activity.
What is a transition metal?
The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms. We'll explore d block elements first:
You will remember that when you are building the Periodic Table and working out where to put the electrons using the Aufbau Principle, something odd happens after argon. At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium. Only after that do the 3d levels fill. The elements in the Periodic Table which correspond to the d levels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below.
The electronic structures of the d block elements shown are:
Sc [Ar] 3d14s2
Ti [Ar] 3d24s2
V [Ar] 3d34s2
Cr [Ar] 3d54s1
Mn [Ar] 3d54s2
Fe [Ar] 3d64s2
Co [Ar] 3d74s2
Ni [Ar] 3d84s2
Cu [Ar] 3d104s1
Zn [Ar] 3d104s2
You will notice that the pattern of filling is not entirely tidy! It is broken at both chromium and copper.Transition metals
Not all d block elements count as transition metals!
A transition metal is one that forms one or more stable ions which have incompletely filled d orbitals. On the basis of this definition, scandium and zinc do not count as transition metals - even though they are members of the d block.
• Scandium has the electronic structure [Ar] 3d14s2. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The Sc3+ ion has no d electrons and so does not meet the definition.
• Zinc has the electronic structure [Ar] 3d104s2. When it forms ions, it always loses the two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d10. The zinc ion has full d levels and does not meet the definition either.
By contrast, copper, [Ar] 3d104s1, forms two ions. In the Cu+ ion the electronic structure is [Ar] 3d10. However, the more common Cu2+ ion has the structure [Ar] 3d9. Copper is definitely a transition metal because the Cu2+ ion has an incomplete d level.
Transition metal ions
Here you are faced with one of the most irritating facts in chemistry at this level! When you work out the electronic structures of the first transition series (from scandium to zinc) using the Aufbau Principle, you do it on the basis that the 3d orbitals have higher energies than the 4s orbitals.
That means that you work on the assumption that the 3d electrons are added after the 4s ones. However, in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first.
The 4s electrons are lost first in forming d-block ions
To write the electronic structure for Co2+:
Co [Ar] 3d74s2
Co2+ [Ar] 3d7
The 2+ ion is formed by the loss of the two 4s electrons.
To write the electronic structure for V3+:
V [Ar] 3d34s2
V3+ [Ar] 3d2
The 4s electrons are lost first followed by one of the 3d electrons.
Variable oxidation state (number)
One of the key features of transition metal chemistry is the wide range of oxidation states (oxidation numbers) that the metals can show. It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like Sulfur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals. However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent.
Examples of variable oxidation states in the transition metals
• Iron: Iron has two common oxidation states (+2 and +3) in, for example, Fe2+ and Fe3+. It also has a less common +6 oxidation state in the ferrate(VI) ion, FeO42-.
• Manganese: Manganese has a very wide range of oxidation states in its compounds. For example:
+2 in Mn2+
+3 in Mn2O3
+4 in MnO2
+6 in MnO42-
+7 in MnO4-
You will find the above examples and others looked at in detail if you explore the chemistry of individual metals from the transition metal menu. There is a link to this menu at the bottom of the page.
Explaining the variable oxidation states in the transition metals
We'll look at the formation of simple ions like Fe2+ and Fe3+. When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound. There are several energy terms to think about, but the key ones are:
• The amount of energy needed to ionize the metal (the sum of the various ionization energies)
• The amount of energy released when the compound forms. This will either be lattice enthalpy if you are thinking about solids, or the hydration enthalpies of the ions if you are thinking about solutions.
The more highly charged the ion, the more electrons you have to remove and the more ionization energy you will have to provide. But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion.
Thinking about a typical non-transition metal (calcium)
The formula for Calcium chloride is CaCl2. Why is that? If you tried to make CaCl, (containing a Ca+ ion), the overall process is slightly exothermic. By making a Ca2+ ion instead, you have to supply more ionization energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca2+ ions than there is if you only have a 1+ ion. The overall process is very exothermic. Because the formation of CaCl2 releases much more energy than making CaCl, then CaCl2 is more stable - and so forms instead.
What about CaCl3? This time you have to remove yet another electron from calcium. The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionization energy between the second and third electron removed. Although there will be a gain in lattice enthalpy, it is not anything like enough to compensate for the extra ionization energy, and the overall process is very endothermic. It definitely is not energetically sensible to make CaCl3!
Thinking about a typical transition metal (iron)
Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion.
Fe [Ar] 3d64s2
Fe2+ [Ar] 3d6
Fe3+ [Ar] 3d5
The 4s orbital and the 3d orbitals have very similar energies. There is not a huge jump in the amount of energy you need to remove the third electron compared with the first and second. The figures for the first three ionization energies (in kJ mol-1) for iron compared with those of calcium are:
metal 1st IE 2nd IE 3rd IE
Ca 590 1150 4940
Fe 762 1560 2960
There is an increase in ionization energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. However, there is much less increase when you take the third electron from iron than from calcium.
In the iron case, the extra ionization energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made. The net effect of all this is that the overall enthalpy change is not vastly different whether you make, say, FeCl2 or FeCl3. That means that it is not too difficult to convert between the two compounds.
The formation of complex ions
What is a complex ion?
A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate (dative covalent) bonds (in some cases, the bonding is actually more complicated). The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions.
What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion.
Some examples of complex ions formed by transition metals
[Fe(H2O)6]2+
[Co(NH3)6]2+
[Cr(OH)6]3-
[CuCl4]2-
Other metals also form complex ions - it is not something that only transition metals do. Transition metals do, however, form a very wide range of complex ions.
The formation of colored compounds
The diagrams show approximate colors for some common transition metal complex ions.
You will find these and others discussed if you follow links to individual metals from the transition metal menu (link at the bottom of the page). Alternatively, you could explore the complex ions menu (follow the link in the help box which has just disappeared off the top of the screen).
The origin of color in the transition metal ions
When white light passes through a solution of one of these ions, or is reflected off it, some colors in the light are absorbed. The color you see is how your eye perceives what is left. Attaching ligands to a metal ion has an effect on the energies of the d orbitals. Light is absorbed as electrons move between one d orbital and another. This is explained in detail on another page.
Catalytic Activity
Transition metals and their compounds are often good catalysts. A few of the more obvious cases are mentioned below, but you will find catalysis explored in detail elsewhere on the site (follow the link after the examples). Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is explored in the main catalysis section.
Iron in the Haber Process
The Haber Process combines hydrogen and nitrogen to make ammonia using an iron catalyst.
Nickel in the hydrogenation of C=C bonds
This reaction is at the heart of the manufacture of margarine from vegetable oils. However, the simplest example is the reaction between ethene and hydrogen in the presence of a nickel catalyst.
Transition metal compounds as catalysts
Vanadium(V) oxide in the Contact Process
At the heart of the Contact Process is a reaction which converts Sulfur dioxide into Sulfur trioxide. Sulfur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst.
Iron ions in the reaction between persulfate ions and iodide ions
Persulphate ions (peroxodisulphate ions), S2O82-, are very powerful oxidizing agents. Iodide ions are very easily oxidized to iodine. And yet the reaction between them in solution in water is very slow. The reaction is catalyzed by the presence of either iron(II) or iron(III) ions.
$S_2O_8^{2-} +2I^- \rightarrow 2SO_4^{2-} + I_2 \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.1%3A_Chemistry_of_the_Transition_Meta.txt |
The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Almost all of the transition metals have multiple oxidation states experimentally observed.
Introduction
Filling atomic orbitals requires a set number of electrons. The s-block is composed of elements of Groups I and II, the alkali and alkaline earth metals (sodium and calcium belong to this block). Groups XIII through XVIII comprise of the p-block, which contains the nonmetals, halogens, and noble gases (carbon, nitrogen, oxygen, fluorine, and chlorine are common members). Transition metals reside in the d-block, between Groups III and XII. If the following table appears strange, or if the orientations are unclear, please review the section on atomic orbitals.
Table $1$
s Orbital p Orbitals d Orbitals
1 orbital, 2 electrons 3 orbitals: px, py, pz; 6 electrons 5 orbitals: dx2-y2, dz2, dxy, dyz, dxz; 10 electrons
Highest energy orbital for a given quantum number n Degenerate with s-orbital of quantum number n+1
The key thing to remember about electronic configuration is that the most stable noble gas configuration is ideal for any atom. Forming bonds are a way to approach that configuration. In particular, the transition metals form more lenient bonds with anions, cations, and neutral complexes in comparison to other elements. This is because the d orbital is rather diffused (the f orbital of the lanthanide and actinide series more so).
Neutral-Atom Electron Configurations
Counting through the periodic table is an easy way to determine which electrons exist in which orbitals. As mentioned before, by counting protons (atomic number), you can tell the number of electrons in a neutral atom. Organizing by block quickens this process. For example, if we were interested in determining the electronic organization of Vanadium (atomic number 23), we would start from hydrogen and make our way down the the Periodic Table).
1s (H, He), 2s (Li, Be), 2p (B, C, N, O, F, Ne), 3s (Na, Mg), 3p (Al, Si, P, S, Cl, Ar), 4s (K, Ca), 3d (Sc, Ti, V).
If you do not feel confident about this counting system and how electron orbitals are filled, please see the section on electron configuration.
Referring to the periodic table below confirms this organization. We have three elements in the 3d orbital. Therefore, we write in the order the orbitals were filled.
1s2 2s2 2p6 3s2 3p6 4s2 3d3
or
[Ar] 4s2 3d3.
The neutral atom configurations of the fourth period transition metals are in Table $2$.
Table $2$
Sc Ti V Cr Mn Fe Co Ni Cu Zn
[Ar] 4s23d1 [Ar] 4s23d2 [Ar] 4s23d3 [Ar] 4s23d4 [Ar] 4s23d5 [Ar] 4s23d6 [Ar] 4s23d7 [Ar] 4s23d8 [Ar] 4s23d9 [Ar] 4s23d10
[Ar] 4s13d5 [Ar] 4s13d10
Chromium and copper appear anomalous. Take a brief look at where the element Chromium (atomic number 24) lies on the Periodic Table (Figure $1$). The electronic configuration for chromium is not [Ar] 4s23d4 but instead it is [Ar] 4s13d5. This is because the half-filled 3d manifold (with one 4s electron) is more stable than a partially filled d-manifold (and a filled 4s manifold). You will notice from Table $2$ that the copper exhibits a similar phenomenon, although with a fully filled d-manifold.
Oxidation States of Transition Metal Ions
When considering ions, we add or subtract negative charges from an atom. Keeping the atomic orbitals when assigning oxidation numbers in mind helps in recognizing that transition metals pose a special case, but not an exception to this convenient method. An atom that accepts an electron to achieve a more stable configuration is assigned an oxidation number of -1. The donation of an electron is then +1. When a transition metal loses electrons, it tends to lose it's s orbital electrons before any of its d orbital electrons. For more discussion of these compounds form, see formation of coordination complexes.
Example $1$
Write the electronic configurations of:
1. neutral iron,
2. iron(II) ion, and
3. iron(III) ion.
Answer
The atomic number of iron is 26 so there are 26 protons in the species.
1. Fe: [Ar] 4s2 3d6
2. Fe2+: [Ar] 3d6
3. Fe3+: [Ar] 3d5
Note that the s-orbital electrons are lost first, then the d-orbital electrons.
Example $2$
Determine the more stable configuration between the following pair:
1. [Kr] 5s2 4d6 vs. [Kr] 5s1 4d7
2. Ag1+ vs. Ag2+
Answer
1. This describes Ruthenium. There is only one 5s electron.
2. Once-oxidized silver ([Kr] 4d10) is more stable than twice- ([Kr] 4d9).
Multiple Oxidation States
Most transition metals have multiple oxidation states, since it is relatively easy to lose electron(s) for transition metals compared to the alkali metals and alkaline earth metals. Alkali metals have one electron in their valence s-orbital and their ions almost always have oxidation states of +1 (from losing a single electron). Similarly, alkaline earth metals have two electrons in their valences s-orbitals, resulting in ions with a +2 oxidation state (from losing both). However, transitions metals are more complex and exhibit a range of observable oxidation states due primarily to the removal of d-orbital electrons. The following chart describes the most common oxidation states of the period 3 elements.
Scandium is one of the two elements in the first transition metal period which has only one oxidation state (zinc is the other, with an oxidation state of +2). All the other elements have at least two different oxidation states. Manganese, which is in the middle of the period, has the highest number of oxidation states, and indeed the highest oxidation state in the whole period since it has five unpaired electrons (see table below).
To help remember the stability of higher oxidation states for transition metals it is important to know the trend: the stability of the higher oxidation states progressively increases down a group. For example, in group 6, (chromium) Cr is most stable at a +3 oxidation state, meaning that you will not find many stable forms of Cr in the +4 and +5 oxidation states. By contrast, there are many stable forms of molybdenum (Mo) and tungsten (W) at +4 and +5 oxidation states.
Example $3$
What makes zinc stable as Zn2+? What makes scandium stable as Sc3+?
Answer
Zinc has the neutral configuration [Ar]4s23d10. Losing 2 electrons does not alter the complete d orbital. Neutral scandium is written as [Ar]4s23d1. Losing 3 electrons brings the configuration to the noble state with valence 3p6.
Example $4$
Why is iron almost always Fe2+ or Fe3+?
Answer
Iron is written as [Ar]4s23d6. Losing 2 electrons from the s-orbital (3d6) or 2 s- and 1 d-orbital (3d5) electron are fairly stable oxidation states.
Example $5$
Write manganese oxides in a few different oxidation states. Which ones are possible and/or reasonable?
Answer
Although Mn+2 is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn(s), Mn(II) as MnO, Mn(II,III) as Mn3O4, Mn(IV) as MnO2, or manganese dioxide, Mn(VII) in the permanganate ion MnO4-, and so on.
Oxidation State of Transition Metals in Compounds
When given an ionic compound such as $\ce{AgCl}$, you can easily determine the oxidation state of the transition metal. In this case, you would be asked to determine the oxidation state of silver (Ag). Since we know that chlorine (Cl) is in the halogen group of the periodic table, we then know that it has a charge of -1, or simply Cl-. In addition, by seeing that there is no overall charge for $\ce{AgCl}$, (which is determined by looking at the top right of the compound, i.e., AgCl#, where # represents the overall charge of the compound) we can conclude that silver ($\ce{Ag}$) has an oxidation state of +1. This gives us Ag+ and Cl-, in which the positive and negative charge cancels each other out, resulting with an overall neutral charge; therefore +1 is verified as the oxidation state of silver (Ag).
Example $6$
Determine the oxidation state of cobalt in $\ce{CoBr2}$.
Answer
Similar to chlorine, bromine ($\ce{Br}$) is also a halogen with an oxidation charge of -1 ($\ce{Br^{-}}$). Since there are two bromines each with a charge of -1. In addition, we know that $\ce{CoBr2}$ has an overall neutral charge, therefore we can conclude that the cation (cobalt), $\ce{Co}$ must have an oxidation state of +2 to neutralize the -2 charge from the two bromine anions.
Example $7$
What is the oxidation state of zinc in $\ce{ZnCO3}$. (Note: the $\ce{CO3}$ anion has a charge state of -2)
Answer
Knowing that $\ce{CO3}$ has a charge of -2 and knowing that the overall charge of this compound is neutral, we can conclude that zinc has an oxidation state of +2. This gives us $\ce{Zn^{2+}}$ and $\ce{CO3^{-2}}$, in which the positive and negative charges from zinc and carbonate will cancel with each other, resulting in an overall neutral charge expected of a compound.
Polyatomic Transition Metal Ions
Consider the manganese ($\ce{Mn}$) atom in the permanganate ($\ce{MnO4^{-}}$) ion. Since oxygen has an oxidation state of -2 and we know there are four oxygen atoms. In addition, this compound has an overall charge of -1; therefore the overall charge is not neutral in this example. Thus, since the oxygen atoms in the ion contribute a total oxidation state of -8, and since the overall charge of the ion is -1, the sole manganese atom must have an oxidation state of +7. This gives us $\ce{Mn^{7+}}$ and $\ce{4 O^{2-}}$, which will result as $\ce{MnO4^{-}}$.
This example also shows that manganese atoms can have an oxidation state of +7, which is the highest possible oxidation state for the fourth period transition metals.
Manganese: A Case Study
Manganese is widely studied because it is an important reducing agent in chemical analysis and is also studied in biochemistry for catalysis and in metallurgy in fortifying alloys. In plants, manganese is required in trace amounts; stronger doses begin to react with enzymes and inhibit some cellular function. Due to manganese's flexibility in accepting many oxidation states, it becomes a good example to describe general trends and concepts behind electron configurations.
Electron configurations of unpaired electrons are said to be paramagnetic and respond to the proximity of magnets. Fully paired electrons are diamagnetic and do not feel this influence. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is.
$\ce{Mn2O3}$ is manganese(III) oxide with manganese in the +3 state. 4 unpaired electrons means this complex is paramagnetic.
$\ce{[Ar]} 4s^{0} 3d^{4}\nonumber$
$\ce{MnO2}$ is manganese(IV) oxide, where manganese is in the +4 state. 3 unpaired electrons means this complex is less paramagnetic than Mn3+.
$\ce{[Ar]} 4s^{0} 3d^{3}\nonumber$
$\ce{KMnO4}$ is potassium permanganate, where manganese is in the +7 state with no electrons in the 4s and 3d orbitals.
$\ce{[Ar]} 4s^{0} 3d^{0}\nonumber$
Since the 3p orbitals are all paired, this complex is diamagnetic.
Summary
Oxidation states of transition metals follow the general rules for most other ions, except for the fact that the d orbital is degenerated with the s orbital of the higher quantum number. Transition metals achieve stability by arranging their electrons accordingly and are oxidized, or they lose electrons to other atoms and ions. These resulting cations participate in the formation of coordination complexes or synthesis of other compounds.
Questions
Determine the oxidation states of the transition metals found in these neutral compounds. Note: The transition metal is underlined in the following compounds.
(A) Copper(I) Chloride: CuCl (B) Copper(II) Nitrate: Cu(NO3)2 (C) Gold(V) Fluoride: AuF5
(D) Iron(II) Oxide: FeO (E) Iron(III) Oxide: Fe2O3 (F) Lead(II) Chloride: PbCl2
(G) Lead(II) Nitrate: Pb(NO3)2 (H) Manganese(II) Chloride: MnCl2 (I) Molybdenum trioxide: MoO3
(J) Nickel(II) Hydroxide: Ni(OH)2 (K) Platinum(IV) Chloride: PtCl4 (L) Silver Sulfide: Ag2S
(M) Tungsten(VI) Fluoride: WF6 (N) Vanadium(III) Nitride: VN (O) Zirconium Hydroxide: Zr(OH)4
1. Determine the oxidation state of the transition metal for an overall non-neutral compound: Manganate (MnO42-)
2. Why do transition metals have a greater number of oxidation states than main group metals (i.e. alkali metals and alkaline earth metals)?
3. Which transition metal has the most number of oxidation states?
4. Why does the number of oxidation states for transition metals increase in the middle of the group?
5. What two transition metals have only one oxidation state?
Contributors and Attributions
• Margaux Kreitman (UCD), Joslyn Wood, Liza Chu (UCD) | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.2%3A_Bonding_in_Simple_Molecules_That.txt |
Learning Objectives
• To introduce complex ions and the basic principles of metal-ligand bonding
A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate (dative covalent) bonds (in some cases, the bonding is actually more complicated than that. The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions.
What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. All ligands are lone pair donors. In other words, all ligands function as Lewis bases.
Bonding in Simple Complex Ions
We are going to look in detail at the bonding in the complex ion formed when water molecules attach themselves to an aluminum ion to give Al(H2O)63+. Start by thinking about the structure of a naked aluminum ion before the water molecules bond to it.
Example \(1\): \(Al(H_2O)_6^{3+}\)
Aluminum has the electronic structure
1s22s22p63s23px1
When it forms an Al3+ ion it loses the 3-level electrons to leave
1s22s22p6
That means that all the 3-level orbitals are now empty. The aluminium uses all six of these empty 3-level orbitals to accept lone pairs from six water molecules. It re-organizes (hybridizes) the 3s, the three 3p, and two of the 3d orbitals to produce six new orbitals all with the same energy.
You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules that will fit around an aluminum ion (and most other metal ions) due to steric constraints. By making the maximum number of bonds, it releases most energy and so becomes most energetically stable.
Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminum and so isn't involved in the bonding. The resulting ion looks like this:
Because of the movement of electrons towards the center of the ion, the 3+ charge is no longer located entirely on the aluminum, but is now spread over the whole of the ion. Because the aluminum is forming 6 bonds, the co-ordination number of the aluminum is said to be 6. The co-ordination number of a complex ion counts the number of co-ordinate bonds being formed by the metal ion at its center.
In a simple case like this, that obviously also counts the number of ligands - but that is not necessarily so, as you will see later. Some ligands can form more than one co-ordinate bond with the metal ion.
Example \(2\): \(Fe(H_2O)_6^{3+}\)
Iron has the electronic structure
1s22s22p63s23p63d64s2
When it forms an Fe3+ ion it loses the 4s electrons and one of the 3d electrons to leave
1s22s22p63s23p63d5
Looking at this as electrons-in-boxes, at the bonding level:
The single electrons in the 3d level are NOT involved in the bonding in any way. Instead, the ion uses 6 orbitals from the 4s, 4p and 4d levels to accept lone pairs from the water molecules. Before they are used, the orbitals are re-organized (hybridized) to produce 6 orbitals of equal energy.
Once the co-ordinate bonds have been formed, the ion looks exactly the same as the equivalent aluminium ion.
Because the iron is forming 6 bonds, the co-ordination number of the iron is 6.
Example \(3\): \(CuCl_4^{2-}\)
This is a simple example of the formation of a complex ion with a negative charge.
Copper has the electronic structure
1s22s22p63s23p63d104s1
When it forms a Cu2+ ion it loses the 4s electron and one of the 3d electrons to leave
1s22s22p63s23p63d9
To bond the four chloride ions as ligands, the empty 4s and 4p orbitals are used (in a hybridized form) to accept a lone pair of electrons from each chloride ion. Because chloride ions are bigger than water molecules, you can't fit 6 of them around the central ion - that's why you only use 4.
Only one of the 4 lone pairs on each chloride ion is shown. The other three are pointing away from the copper ion, and aren't involved in the bonding. That gives you the complex ion:
The ion carries 2 negative charges overall. That comes from a combination of the 2 positive charges on the copper ion and the 4 negative charges from the 4 chloride ions. In this case, the co-ordination number of the copper is 4. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.3%3A_Introduction_to_Coordination_Che.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_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.4%3A_Structures_of_Coordination_Compl.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_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.5%3A_Crystal_Field_Theory%3A_Optical_.txt |
Learning Objectives
• To get a simple overview of the origin of color and magnetism in complex ions.
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. 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 and have certain characteristics, including amplitude, wavelength, and frequency (Figure $1$).
General properties of all electromagnetic radiation include:
1. Electromagnetic radiation can travel through empty space, while 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 to be heard.
2. The speed of light ($c$) is always a constant (2.99792458 x 108 m s-1).
3. Wavelengths ($\lambda$) are measured between the distances of either crests or troughs.
The energy of a photon is expressed by Planck's law in terms of the frequency ($u$) of the photon
$E=h u \label{24.5.1}$
since $\lambda u =c$ for all light Plancks law can be also expressed in terms of the wavelength of the photon
$E = h u = \dfrac{hc}{\lambda} \label{24.5.2}$
If white light is passed through a prism, it splits into all the colors of the rainbow (Figure $2$). Visible light is simply a small part of an electromagnetic spectrum most of which we cannot see - gamma rays, X-rays, infra-red, radio waves and so on. Each of these has a particular wavelength, ranging from 10-16 meters for gamma rays to several hundred meters for radio waves. Visible light has wavelengths from about 400 to 750 nm (1 nanometer = 10-9 meters).
Example $1$: Blue Color of Copper (II) Sulfate in Solution
If white light (ordinary sunlight, for example) passes through copper(II) sulfate solution, some wavelengths in the light are absorbed by the solution. Copper(II) ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through copper(II) sulfate solution.
Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You would not have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
Origin of Colors
The process of absorption involves the excitation of the valence electrons in the molecule typically from the low lying level called the Highest Occupied Molecular Orbital (HOMO) into a higher lying state called the the Lowest Unoccupied Molecular Orbital (LUMO). When this HOMO and LUMO transition (Figure $3$) involves the absorption of visible light, the sample is colored.
The HOMO-LUMO energy difference
$\Delta E = E_{HOMO} - E_{LUMO} \label{24.5.3A}$
depends on the nature of the molecule and can be connected to the wavelength of the light absorbed
$\Delta E = h u = \dfrac{hc}{\lambda} \label{24.5.3B}$
Equation $\ref{24.5.3B}$ is the most important equation in the field of light-matter interactions (spectroscopy).
As Example $1$ demonstrated, when white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other (Figure $5$). Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absorption close to 400 nm as well as absorption near 800 nm.
Colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. Copper(II) sulfate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum and cyan is the complementary color of red (Table $1$).
Table $1$: The Visible Spectrum
Color Wavelength (nm) ΔE HOMO - LUMO gap (eV)
UV 100 - 400 12.4 - 3.10
Violet 400 - 425 3.10 - 2.92
Blue 425 - 492 2.92 - 2.52
Green 492 - 575 2.52 - 2.15
Yellow 575 - 585 2.15 - 2.12
Orange 585 - 647 2.12 - 1.92
Red 647 - 700 1.92 - 1.77
Near IR 700 - 10,000 1.77 - 0.12
If the compound absorbs in one region of the spectra, it appears with the opposite (complementary) color, since all of the absorbed color has been removed. For example:
• the material absorbs violet light ⇒ color is yellow
• the material absorbs blue light ⇒ color is orange
• the material absorbs yellow-green light ⇒ color is red-violet
The Origin of Color in Complex Ions
We often casually talk about the transition metals as being those in the middle of the Periodic Table where d orbitals are being filled, but these should really be called d block elements rather than transition elements (or metals). The definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. Zinc with the electronic structure [Ar] 3d104s2 does not count as a transition metal whichever definition you use. In the metal, it has a full 3d level. When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. At the other end of the row, scandium ([Ar] 3d14s2) does not really counts as a transition metal either. Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. The Sc3+ ion does not count as a transition metal ion because its 3d level is empty.
Example $3$: Hexaaqua Metal Ions
The diagrams show the approximate colors of some typical hexaaqua metal ions, with the formula [ M(H2O)6 ]n+. The charge on these ions is typically 2+ or 3+.
• Non-transition metal ions
• Transition metal ions
The corresponding transition metal ions are colored. Some, like the hexaaquamanganese(II) ion (not shown) and the hexaaquairon(II) ion, are quite faintly colored - but they are colored.
So, what causes transition metal ions to absorb wavelengths from visible light (causing color) whereas non-transition metal ions do not? And why does the color vary so much from ion to ion? This is discussed in the next sections.
Magnetism
The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only the first two are introduced below.
Paramagnetism
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, $O_2$ is a good example of paramagnetism (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet:
A chemical demonstration of the paramagnetism of oxygen, as shown by the attraction of liquid oxygen to a magnet. Carleton University, Ottawa, Canada.
As shown in the video, molecular oxygen ($O_2$ is paramagnetic and is attracted to the magnet. Incontrast, Molecular nitrogen, $N_2$, however, has no unpaired electrons and it is diamagnetic (this concept is discussed below); it is therefore unaffected by the magnet.
There are some exceptions to the paramagnetism rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include $Sc^{3+}$, $Ti^{4+}$, $Zn^{2+}$, and $Cu^+$. These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here.
Diamagnetism
Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the Pauli Exclusion Principle which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure $6$.
Figure $6$: Levitating pyrolytic carbon: A small (~6 mm) piece of pyrolytic graphite levitating over a permanent neodymium magnet array (5 mm cubes on a piece of steel). Note that the poles of the magnets are aligned vertically and alternate (two with north facing up, and two with south facing up, diagonally). from Wikipedia.
How to Tell if a Substance is Paramagnetic or Diamagnetic
The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps:
1. Find the electron configuration
2. Draw the valence orbitals
3. Look for unpaired electrons
4. Determine whether the substance is paramagnetic (one or more unpaired electrons) or diamagnetic (all electrons paired)
Example $4$: Chlorine atoms
Are chlorine atoms paramagnetic or diamagnetic?
Step 1: Find the electron configuration
For Cl atoms, the electron configuration is 3s23p5
Step 2: Draw the valence orbitals
Ignore the core electrons and focus on the valence electrons only.
Step 3: Look for unpaired electrons
There is one unpaired electron.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Since there is an unpaired electron, Cl atoms are paramagnetic (but is quite weak).
Example 2: Zinc Atoms
Step 1: Find the electron configuration
For Zn atoms, the electron configuration is 4s23d10
Step 2: Draw the valence orbitals
Step 3: Look for unpaired electrons
There are no unpaired electrons.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Because there are no unpaired electrons, Zn atoms are diamagnetic. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.6%3A_Optical_Properties_and_the_Spect.txt |
Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced. Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed." The halides, phosphines, ammonia and amines seen previously are monodentate ligands. Bidentate ligands bind through two donor sites. Bidentate means "two-toothed." An example of a bidentate ligand is ethylenediamine. It can bind to a metal via two donor atoms at once.
Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the "chelate effect." This term comes from the Greek chelos, meaning "crab." A crab does not have any teeth at all, but it does have two claws for tightly holding onto something for a couple of reasons. A very simple analogy is that, if you are holding something with two hands rather than one, you are not as likely to drop it.
Complex metal ions containing more complicated ligands
In the examples previously disccussed, each ligand only forms one bond with the central metal ion to give the complex ion. Such a ligand is said to be unidentate. That means literally that it only has one tooth! It only has one pair of electrons that it can use to bond to the metal - any other lone pairs are pointing in the wrong direction. Some ligands, however, have rather more teeth! These are known generally as multidentate or polydentate ligands, but can be broken down into a number of different types.
Bidentate ligands
Bidentate ligands have two lone pairs, both of which can bond to the central metal ion. The two commonly used examples are 1,2-diaminoethane (old name: ethylenediamine - often given the abbreviation "en"), and the ethanedioate ion (old name: oxalate).
In the ethanedioate ion, there are lots more lone pairs than the two shown, but these are the only ones we are interested in. You can think of these bidentate ligands rather as if they were a pair of headphones, carrying lone pairs on each of the "ear pieces". These will then fit snuggly around a metal ion.
Example $1$: $Ni (NH_2CH_2CH_2NH_2)_3^{2+}$
You might find this abbreviated to $[Ni(en)_3]^{2+}$. The structure of the ion looks like this:
In this case, the "ear pieces" are the nitrogen atoms of the NH2 groups - and the "bit that goes over your head " is the $-CH_2CH_2-$ group. If you were going to draw this in an exam, you would obviously want to draw it properly - but for learning purposes, drawing all the atoms makes the diagram look unduly complicated!
Notice that the arrangement of the bonds around the central metal ion is exactly the same as it was with the ions with 6 water molecules attached. The only difference is that this time each ligand uses up two of the positions - at right angles to each other.
Because the nickel is forming 6 co-ordinate bonds, the coordination number of this ion is 6, despite the fact that it is only joined to 3 ligands. Coordination number counts the number of bonds, not the number of ligands.
Example 5: $Cr(C_2O_4)_3^{3-}$
This is the complex ion formed by attaching 3 ethanedioate (oxalate) ions to a chromium(III) ion. The shape is exactly the same as the previous nickel complex. The only real difference is the number of charges. The original chromium ion carried 3+ charges, and each ethanedioate ion carried -2, i.e.,
$(+3) + (3 \times -2) = -3. \nonumber$
The structure of the ion looks like this:
Again, if you drew this in an exam, you would want to show all the atoms properly. If you need to be able to do this, practice drawing it so that it looks clear and tidy! Refer back to the diagram of the ethanedioate ion further up the page to help you.
A Quadridentate Ligand
A quadridentate ligand has four lone pairs, all of which can bond to the central metal ion. An example of this occurs in haemoglobin (American: hemoglobin). The functional part of this is an iron(II) ion surrounded by a complicated molecule called heme. This is a sort of hollow ring of carbon and hydrogen atoms, at the center of which are 4 nitrogen atoms with lone pairs on them. Heme is one of a group of similar compounds called porphyrins. They all have the same sort of ring system, but with different groups attached to the outside of the ring. You aren't going to need to know the exact structure of the haem at this level.
We could simplify the heme with the trapped iron ion as:
Each of the lone pairs on the nitrogen can form a co-ordinate bond with the iron(II) ion - holding it at the center of the complicated ring of atoms. The iron forms 4 co-ordinate bonds with the heme, but still has space to form two more - one above and one below the plane of the ring. The protein globin attaches to one of these positions using a lone pair on one of the nitrogens in one of its amino acids. The interesting bit is the other position.
The water molecule which is bonded to the bottom position in the diagram is easily replaced by an oxygen molecule (again via a lone pair on one of the oxygens in $O_2$) - and this is how oxygen gets carried around the blood by the haemoglobin. When the oxygen gets to where it is needed, it breaks away from the haemoglobin which returns to the lungs to get some more.
Carbon Monoxide Poisoning
You probably know that carbon monoxide is poisonous because it reacts with hemeoglobin. It bonds to the same site that would otherwise be used by the oxygen - but it forms a very stable complex. The carbon monoxide doesn't break away again, and that makes that hemeoglobin molecule useless for any further oxygen transfer.
A Hexadentate Ligand
A hexadentate ligand has 6 lone pairs of electrons - all of which can form co-ordinate bonds with the same metal ion. The best example is EDTA. The diagram shows the structure of the ion with the important atoms and lone pairs picked out.
The EDTA ion entirely wraps up a metal ion using all 6 of the positions that we have seen before. The co-ordination number is again 6 because of the 6 co-ordinate bonds being formed by the central metal ion. The diagram below shows this happening with a copper(II) ion. Here is a simplified version. Make sure that you can see how this relates to the full structure above.
The overall charge, of course, comes from the 2+ on the original copper(II) ion and the 4- on the $EDTA^{4-}$ ion. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/8%3A_Bonding_in_Transition_Metal_Compounds_and_Coordination_Complexes/8.7%3A_Bonding_in_Coordination_Complexe.txt |
Some of the first real breakthroughs in the study of chemistry happened in the study of the gaseous state. In gases, the volume of the actual gas particles is but a tiny fraction of the total volume that the gas occupies. This allowed early chemists to relate parameters such as volume and the number of gas particles, leading to the development of the mole concept. As we have seen in previous chapters, the notion of a chemical mole allows us to do quantitative chemistry and lead us to the point where we can routinely address reaction stoichiometry, etc. In this chapter, we will visit some of the early observations that lead to our current understand of gasses and how they behave. We will see how the relationships between pressure and volume; volume and temperature and volume and moles lead to the ideal gas laws and how these simple rules can allow us to do quantitative calculations in the gas phase.
Thumbnail: The kinetic theory of gases describes this state of matter as composed of tiny particles in constant motion with a lot of distance between the particles.
10.1: Bulk Properties of Liquids - Molecular Interpretation
Gas, liquid, and solid are known as the three states of matter or material, but each of solid and liquid states may exist in one or more forms. Thus, another term is required to describe the various forms, and the term phase is used. Each distinct form is called a phase, but the concept of phase defined as a homogeneous portion of a system, extends beyond a single material, because a phase may also involve several materials. A solid has a definite shape and volume. A liquid has a definite volume but it takes the shape of a container whereas a gas fills the entire volume of a container. You already know that diamond and graphite are solids made up of the element carbon. They are two phases of carbon, but both are solids.Solids are divided into subclasses of amorphous (or glassy) solids and crystalline solids. Arrangements of atoms or molecules in crystalline solids are repeated regularly over a very long range of millions of atoms, but their arrangements in amorphous solids are somewhat random or short range of say some tens or hundreds of atoms. In general, there is only one liquid phase of a material. However, there are two forms of liquid helium, each have some unique properties. Thus, the two forms are different (liquid) phases of helium. At a definite temperature and pressure, the two phases co-exist.
• 10.1: Bulk Properties of Liquids - Molecular Interpretation
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance and the intermolecular forces try to draw the particles together.
• 10.2: Intermolecular Forces - Origins in Molecular Structure
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold molecules and polyatomic ions together. 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.
• 10.3: Intermolecular Forces in Liquids
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid. Surfactants are molecules that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. The viscosity of a liquid is its resistance to flow.
• 10.4: Phase Equilibrium
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation.
• 10.5: Phase Transitions
Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic.
• 10.6: Phase Diagrams
The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a phase diagram, which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point.
• 10.E: Solids, Liquids, and Phase Transitions (Exercises)
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al.
10: Solids Liquids and Phase Transitions
Learning Objectives
• To describe the unique properties of liquids.
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to four unique properties of liquids that intimately depend on the nature of intermolecular interactions:
1. surface tension,
2. capillary action,
3. viscosity, and
4. boiling point.
Surface Tension
If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of forming a thin, continuous film? The answer lies in a property called surface tension, which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of $7.29 \times 10^{-2} J/m^2$ (at 20°C), while mercury with metallic (electrostatic) bonds has as surface tension that is 6.5-times greater: $4.86 \times 10^{-1} J/m^2$ (at 20°C).
Figure $1$ presents a microscopic view of a liquid droplet. A typical molecule in the interior of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no net force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the surface experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads.
The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends above the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is better demonstrated in the zero-gravity conditions of space (Figure $2$).
Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m2). The values of the surface tension of some representative liquids are listed in Table $2$. Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding.
Table $2$: Surface Tension, Viscosity, Vapor Pressure (at 25°C Unless Otherwise Indicated), and Normal Boiling Points of Common Liquids
Substance Surface Tension (× 10−3 J/m2) Viscosity (mPa•s) Vapor Pressure (mmHg) Normal Boiling Point (°C)
Organic Compounds
diethyl ether 17 0.22 531 34.6
n-hexane 18 0.30 149 68.7
acetone 23 0.31 227 56.5
ethanol 22 1.07 59 78.3
ethylene glycol 48 16.1 ~0.08 198.9
Liquid Elements
bromine 41 0.94 218 58.8
mercury 486 1.53 0.0020 357
Water
0°C 75.6 1.79 4.6
20°C 72.8 1.00 17.5
60°C 66.2 0.47 149
100°C 58.9 0.28 760
Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids.
Surface Energies and Surface Tension
Any material - solid, liquid or (non-ideal) gas - wants to bond to itself. This is why condensed phase materials "sticks" together in the first place. However, a surface disrupts this bonding, and so incurs an energy penalty. This is why liquids in zero gravity ball up into spherical drops (Figure $2$) - the sphere is the shape with the lowest surface area for a fixed volume. We can describe this surface energy with dimensions of energy per unit area, which is the amount of extra energy needed to create new surface or extend a surface (e.g., cracking a solid or parting a liquid). Hence, surface tension is typically given in J/m2 units (Table $2$).
Capillary Action
Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in Figure $3$. When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. The height of the water does not depend on the angle at which the capillary is tilted.
Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary.
• Cohesive forces bind molecules of the same type together
• Adhesive forces bind a substance to a surface
Water has both strong adhesion to glass, which contains polar $\ce{SiOH}$ groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (part (a) in Figure $4$). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure $\PageIndex{4b}$).
Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body.
Polar substances are drawn up a glass capillary and generally have concave meniscuses and nonpolar substances general avoid the capillary and exhibit convex meniscuses.
Viscosity
Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a low viscosity. Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. The two most common methods for evaluating the viscosity of a liquid are
1. to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and
2. to measure the time it takes steel balls to fall through a given volume of the liquid.
The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table $1$ and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH2CH2OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces.
There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous.
Viscosity increases as intermolecular interactions or molecular size increases.
Application: Motor Oils
Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing (Table $3$). As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures.
The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity. So-called single-grade oils can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are multigrade, with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils.
Table $3$: Dynamic Viscosity (N s/m2)
SAE Temperature (oC)
0 20 50 100
10 0.31 0.079 0.020 0.005
20 0.72 0.170 0.033 0.007
30 1.53 0.310 0.061 0.010
40 2.61 0.430 0.072 0.012
50 3.82 0.630 0.097 0.015
Example $1$
Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.)
Given: substance and composition of the glass surface
Asked for: behavior of oil and the shape of meniscus
Strategy:
1. Identify the cohesive forces in the motor oil.
2. Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus.
Solution:
A Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains.
B Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury.
Exercise $1$
Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)?
Answer
Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave.
Oddity of Science: Superfluid Helium-4
Superfluid helium-4 is the superfluid form of helium-4, an isotope of the element helium. A superfluid is a state of matter in which the matter behaves like a fluid with zero viscosity. The substance, which looks like a normal liquid, flows without friction past any surface, which allows it to continue to circulate over obstructions and through pores in containers which hold it, subject only to its own inertia.
Many ordinary fluids, like alcohol or petroleum, creep up solid walls, driven by their surface tension. However, in the case of superfluid helium-4, the flow of the liquid in the layer is not restricted by its viscosity but by a critical velocity which is about 20 cm/s. This is a fairly high velocity so superfluid helium can flow relatively easily up the wall of containers, over the top, and down to the same level as the surface of the liquid inside the container. In a container, lifted above the liquid level, it forms visible droplets as seen above.
Boiling Points
The vapor pressure of a liquid is defined as the pressure exerted by a vapor in equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system (discussed in more detail in next Sections of Chapter). As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $8$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm.
Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $4$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $4$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $8$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Mercury boils at 356 °C at room pressure. To see video go to www.youtube.com/watch?v=0iizsbXWYoo
Given: data in Figure $8$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $8$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethylene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $8$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Surface tension, capillary action, boiling points, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. Surfactants are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when cohesive forces, the intermolecular forces in the liquid, are weaker than adhesive forces, the attraction between a liquid and the surface of the capillary. The shape of the meniscus, the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The viscosity of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. The boiling point of a liquid is the temperature when the vapor pressure of the liquid equals the external pressure, or the atmospheric pressure in the case of an open container. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/09%3A_The_Gaseous_State.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.
Coulombic Interactions
Through various experiments, Charles Augustin de Coulomb found a way to explain the interactions between charged particles, which in turn helped to explain where the stabilities and instabilities of various particles come from. While the entities that hold atoms together within a molecule can be attributed to bonds, the forces that create these bonds can be explained by Coulomb Forces. Thus, the physical basis behind the bonding of two atoms can be explained.
Coulomb’s findings indicate that like charges repel each other and unlike charges attract one another. Thus electrons, which are negatively charged, repel each other but attract protons. Likewise, protons repel each other.
Each atom is made up of a nucleus in the center, which consists of a number of protons and neutrons, depending upon the element in question. Surrounding the nucleus are electrons that float around the nucleus in what can be thought of as a cloud. As two atoms approach one another, the protons of one atom attract the electrons of the other atom. Similarly, the protons of the other atom attract the electrons of the first atom. As a result, the simultaneous attraction of the components from one atom to another create a bond. This interaction can be summarized mathematically and is known as Coulombic forces:
$F = k \dfrac{q_{1}q_{2}}{r^{2}} \label{C}$
In this mathematical representation of Coulomb's observations,
• $F$ is the electrical force acting between two atoms
• with $q_1$ and $q_2$ representing the magnitude of the charges of each atom,
• $r$ is the distance between the two atoms.
• and $k$ is a constant.
From Equation \ref{C}, the electrostatic force between two charges is inversely proportional to the square of the distance separating the two atoms.
Ion–Ion Interactions
The interactions between ions (ion - ion interactions or charge-charge interactions) are the easiest to understand since such interactions are just a simple application of Coulombic forces (Equation \ref{C}). This specific interaction operates over relatively long distances in the gas phase and is responsible for the attraction of opposite charge ions and the repulsion of like charged ions.
Coulombic forces are also involved in all forms of chemical bonding; when they act between separate charged particles they are especially strong. Thus the energy required to pull a mole of $\ce{Na^{+}}$ and $\ce{F^{–}}$ ions apart in the sodium fluoride crystal is greater than that needed to break the a covalent bonds of a mole of $\ce{H2}$. The effects of ion-ion attraction are seen most directly in salts such as $\ce{NaF}$ and $\ce{NaCl}$ that consist of oppositely-charged ions arranged in inter-penetrating crystal lattices.
According to Coulomb's Law the force between two charged particles is given by
$\underbrace{F= \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}}_{\text{ion-ion Force}} \label{7.2.1}$
Instead of using SI units, chemists often prefer to express atomic-scale distances in picometers and charges as electron charge (±1, ±2, etc.) Using these units, the proportionality constant $1/4\pi\epsilon$ works out to $2.31 \times 10^{16}\; J\; pm$. The sign of $F$ determines whether the force will be attractive (–) or repulsive (+); notice that the latter is the case whenever the two q's have the same sign. Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them.
They form an ion-pair, a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature).
Equation \ref{7.2.1} is an example of an inverse square law; the force falls off as the square of the distance. A similar law governs the manner in which the illumination falls off as you move away from a point light source; recall this the next time you walk away from a street light at night, and you will have some feeling for what an inverse square law means.
The stronger the attractive force acting between two particles, the greater the amount of work required to separate them. Work represents a flow of energy, so the foregoing statement is another way of saying that when two particles move in response to a force, their potential energy is lowered. This work is found by integrating the negative of the force function with respect to distance over the distance moved. Thus the energy that must be supplied in order to completely separate two oppositely-charged particles initially at a distance $r_0$ is given by
$w= - \int _{r_o} ^{\infty} \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}dr = - \dfrac{q_1q_2}{4\pi\epsilon_0 r_o} \label{7.2.2}$
hence, the potential ($V_{ion-ion}$) responsible for the ion-ion force is
$\underbrace{V_{ion-ion} = \dfrac{q_1q_2}{4\pi\epsilon_0 r} }_{\text{ion-ion potential}} \label{7.2.3}$
Example $1$
When sodium chloride is melted, some of the ion pairs vaporize and form neutral $\ce{NaCl}$ dimers. How much energy would be released when one mole of $\ce{Na^{+}}$ and $\ce{Cl^{–}}$ ions are brought together to generate dimers in this way? The bondlength of $\ce{NaCl}$ is 237 pm.
Solution
The interactions involved in forming $\ce{NaCl}$ dimers is the ion-ion forces with a potential energy given by Equation \ref{7.2.3}. However, this is the energy of interaction for one pair of $\ce{Na^{+}}$ and $\ce{Cl^{–}}$ ion and needs to be scaled by a mole. So the energy released will be
\begin{align*}E &= N_a V(\ce{NaCl}) \[4pt] &= N_a\dfrac{q_1q_2}{4\pi\epsilon_0 r} \end{align*}
The $r$ in this equation is the distance between the two ions, which is the bondlength of 237 pm ($237 \times 10^{-12}m$).
\begin{align*}E &= (6.022 \times 10^{23} ) \underbrace{(8.987 \times 10^9 N m^2/C^2 )}_{1/4\pi\epsilon_o} \dfrac{(+1.6 \times 10^{-19}C) (-1.6 \times 10^{-19}C) }{ 237 \times 10^{-12} m} \[4pt] &= –584 \;kJ/mol \end{align*}
This is not the energy needed to separate one mole of NaCl since that is a lattice and has more than pairwise interactions and require addressing the geometric orientation of the lattice (see Madelung Constants for more details).
Intermolecular forces are electrostatic in nature; that is, they arise from the electrostatic 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 (i.e., real gases).
Ion-Dipole Interactions
A dipole that is close to a positive or negative ion will orient itself so that the end whose partial charge is opposite to the ion charge will point toward the ion. This kind of interaction is very important in aqueous solutions of ionic substances; H2O is a highly polar molecule, so that in a solution of sodium chloride, for example, the Na+ ions will be enveloped by a shell of water molecules with their oxygen-ends pointing toward these ions, while H2O molecules surrounding the Cl ions will have their hydrogen ends directed inward. As a consequence of ion-dipole interactions, all ionic species in aqueous solution are hydrated; this is what is denoted by the suffix in formulas such as K+(aq), etc.
The strength of ion-dipole attraction depends on the magnitude of the dipole moment and on the charge density of the ion. This latter quantity is just the charge of the ion divided by its volume. Owing to their smaller sizes, positive ions tend to have larger charge densities than negative ions, and they should be more strongly hydrated in aqueous solution. The hydrogen ion, being nothing more than a bare proton of extremely small volume, has the highest charge density of any ion; it is for this reason that it exists entirely in its hydrated form H3O+ in water.
Since there is now both attractive and repulsive interactions and they both get weaker as the ion and dipole distance increases while also approaching each other in strength, the net ion-dipole is an inverse square relationship as shown in Equation \ref{11.2.2}.
$\underbrace{ E\: \propto \: \dfrac{-|q_1|\mu_2}{r^2}}_{\text{ion-dipole potential}} \label{11.2.2}$
• $r$ is the distance of separation.
• $q$ is the charge of the ion ( only the magnitude of the charge is shown here.)
• $\mu$ is the permanent dipole moment of the polar molecule. $\vec{\mu} = q' \, \vec{d} \label{11.2.3}$ where $q'$ is the partial charge of each end of the dipole and $d$ is the separation between the charges within the dipole
Ion-Ion vs. Ion-Dipole potentials
There are several differences between ion-ion potential (Equation \ref{7.2.3}) and the ion-dipole potential (Equation \ref{11.2.2}) interactions.
• First, the potential of ion/dipole interactions are negative and net interaction will always be attractive, since the attraction of the opposite dipole to the ion will make it closer than the dipole with the like charge. By using the absolute value of the charge of the ion, and placing a negative sign in front of the equation, this results in a lowering of the potential energy ($\mu$ is positive).
• Second, the potential drops off quicker in Equation \ref{11.2.2}, where it is an inverse square relationship to the radius ($1/r^2$), while a simple charge-charge interaction (Equation \ref{7.2.3}) has a linear inverse relationship ($1/r$). This means the ion-dipole are a shorter range interaction and diminish more rapidly the father the polar molecule is from the ion.
• Third, note that the units of the two equations are the same, as $\mu$ has the units of charge X distance. However, the distance in $\mu$ is the distance between the dipoles of the polar molecule, while the distance denoted by the "r" is the distance between the ion and the dipole.
It needs to be understood that the molecules in a solution are rotating and vibrating and actual systems are quite complicated (Figure $4$). What is important to realize is that these interactions are Coulombic in nature and how the mathematical equations describe this in terms of the magnitude of the charges and their distances from each other. In this course we will not be calculating dipole moments or the magnitudes of them, but understanding how to read the equations, and developing qualitative understandings that allow us to predict trends.
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 part (a) in Figure $1$.
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 $\PageIndex{5b}$, are attractive intermolecular interactions, whereas those in Figure $\PageIndex{5d}$ 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/r6, so doubling the distance between the dipoles decreases the strength of the interaction by 26, or 64-fold:
$V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T} \label{5}$
Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas 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 $1$. Using what we learned about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules.
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.
Example $2$
Arrange ethyl methyl ether ($\ce{CH3OCH2CH3}$), 2-methylpropane [isobutane, $\ce{(CH3)2CHCH3}$], and acetone ($\ce{CH3COCH3}$) 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, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and 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.
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)
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/r6, 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. Like dipole–dipole interactions, 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. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.2%3A_Intermolecular_Forces_-_Origins_in_Molecular_Structure.txt |
Learning Objectives
• To describe the intermolecular forces in liquids.
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 \(1\). 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 \(2\). 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, cagelike 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.
It takes two to tango...
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.
Example \(1\)
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 \(1\)
Considering \(\ce{CH3CO2H}\), \(\ce{(CH3)3N}\), \(\ce{NH3}\), and \(\ce{CH3F}\), which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
\(\ce{CH3CO2H}\) and \(\ce{NH3}\);
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 \(\ce{HF}\) 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 \(2\)
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 \(2\)
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)
Special Properties of Water
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.
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 two 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 water's strong intermolecular interaction
The properties of water make it suitable for organisms to survive in during differing weather conditions. Water expands as it freezes, 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.
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/r6, 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. Like dipole–dipole interactions, 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, cagelike structure that is less dense than liquid water. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.3%3A_Intermolecular_Forces_in_Liquids.txt |
Learning Objectives
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
• To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.
Evaporation and Condensation
Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $1$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.
To understand the causes of vapor pressure, consider the apparatus shown in Figure $2$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $2$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.
As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $2$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $3$.
Equilibrium Vapor Pressure
Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid.
If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $4$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.
The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $4$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be]
The exponential rise in vapor pressure with increasing temperature in Figure $4$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• $ΔH_{vap}$ is the enthalpy of vaporization,
• $R$ is the universal gas constant [8.314 J/(mol•K)],
• $T$ is the temperature in kelvins, and
• $C$ is the y-intercept, which is a constant for any given line.
Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −ΔHvap/R. Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation, can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2}$
Conversely, if we know ΔHvap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $1$.
A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be]
Example $1$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
experimentally measured vapor pressures of liquid Hg at four temperatures
T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845
From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)
Given: vapor pressures at four temperatures
Asked for: ΔHvap of mercury and vapor pressure at 160°C
Strategy:
1. Use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.
2. Substitute the calculated value of ΔHvap into Equation $\ref{Eq2}$ to obtain the unknown pressure (P2).
Solution:
A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of ΔHvap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq2}$ gives
\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber$
Using the relationship $e^{\ln x} = x$, we have
\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber
At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.
Exercise $1$: Vapor Pressure of Nickel
The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?
Answer
1896°C
Boiling Points
As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $4$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $1$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $1$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $4$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Given: Data in Figure $4$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $4$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethlyene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $4$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.4%3A_Phase_Equilibrium.txt |
Learning Objectives
• To calculate the energy changes that accompany phase changes.
We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO2, as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure $1$.
Energy Changes That Accompany Phase Changes
Phase changes are always accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is endothermic. Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is exothermic. The energy change associated with each common phase change is shown in Figure $1$.
ΔH is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state.
Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar enthalpies of fusion ($ΔH_{fus}$), the energy required to convert from a solid to a liquid, a process known as fusion (or melting), as well as the normal boiling points and enthalpies of vaporization ($ΔH_{vap}$) of selected compounds are listed in Table $1$.
Table $1$: Melting and Boiling Points and Enthalpies of Fusion and Vaporization for Selected Substances. Values given under 1 atm. of external pressure.
Substance Melting Point (°C) ΔHfus (kJ/mol) Boiling Point (°C) ΔHvap (kJ/mol)
N2 −210.0 0.71 −195.8 5.6
HCl −114.2 2.00 −85.1 16.2
Br2 −7.2 10.6 58.8 30.0
CCl4 −22.6 2.56 76.8 29.8
CH3CH2OH (ethanol) −114.1 4.93 78.3 38.6
CH3(CH2)4CH3 (n-hexane) −95.4 13.1 68.7 28.9
H2O 0 6.01 100 40.7
Na 97.8 2.6 883 97.4
NaF 996 33.4 1704 176.1
The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid).
Less energy is needed to allow molecules to move past each other than to separate them totally.
The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (ΔHsub). Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO2 (dry ice); iodine (Figure $2$); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in Figure $1$, the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same T; this is an application of Hess’s law.
$ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1}$
Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the chlorofluorocarbons (CFCs) and the hydrofluorocarbons (HCFCs). The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C.
Temperature Curves
The processes on the right side of Figure $1$—freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.
Heating Curves
Figure $3$ shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat (Cs) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is greater than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water.
Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils.
The temperature of a sample does not change during a phase change.
If heat is added at a constant rate, as in Figure $3$, then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure $3$, the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion.
A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form.
Cooling Curves
The cooling curve, a plot of temperature versus cooling time, in Figure $4$ plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure $3$, the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system.
Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO2 (dry ice) into the cloud from an airplane. Solid CO2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO2 sublimes, it absorbs heat from the cloud, often with the desired results.
A Video Discussing the Thermodynamics of Phase Changes. Video Source: The Thermodynamics of Phase Changes, YouTube(opens in new window) [youtu.be]
Example $1$: Cooling Tea
If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol.
Given: mass, volume, initial temperature, density, specific heats, and $ΔH_{fus}$
Asked for: final temperature
Strategy
Substitute the given values into the general equation relating heat gained (by the ice) to heat lost (by the tea) to obtain the final temperature of the mixture.
Solution
When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by
$q=mC_sΔT \nonumber$
where $q$ is heat, $m$ is mass, $C_s$ is the specific heat, and $ΔT$ is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C.
The amount of heat gained by the ice cube as it melts is determined by its enthalpy of fusion in kJ/mol:
$q=nΔH_{fus} \nonumber$
For our 50.0 g ice cube:
\begin{align*} q_{ice} &= 50.0 g⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol \[4pt] &= 16.7\, kJ \end{align*} \nonumber
Thus, when the ice cube has just melted, it has absorbed 16.7 kJ of heat from the tea. We can then substitute this value into the first equation to determine the change in temperature of the tea:
$q_{tea} = - 16,700 J = 500 mL⋅\dfrac{1.00\: g}{1\: mL}⋅4.184 J/(g•°C) ΔT \nonumber$
$ΔT = - 7.98 °C = T_f - T_i \nonumber$
$T_f = 12.02 °C \nonumber$
This would be the temperature of the tea when the ice cube has just finished melting; however, this leaves the melted ice still at 0.0°C. We might more practically want to know what the final temperature of the mixture of tea will be once the melted ice has come to thermal equilibrium with the tea. To determine this, we can add one more step to the calculation by plugging in to the general equation relating heat gained and heat lost again:
\begin{align*} q_{ice} &= - q_{tea} \[4pt] q_{ice} &= m_{ice}C_sΔT = 50.0g⋅4.184 J/(g•°C)⋅(T_f - 0.0°C) \[4pt] &= 209.2 J/°C⋅T_f \end{align*} \nonumber
$q_{tea} = m_{tea}C_sΔT = 500g⋅4.184 J/(g•°C)⋅(T_f - 12.02°C) = 2092 J/°C⋅T_f - 25,150 J \nonumber$
$209.2 J/°C⋅T_f = - 2092 J/°C⋅T_f + 25,150 J \nonumber$
$2301.2 J/°C⋅T_f = 25,150 J \nonumber$
$T_f = 10.9 °C \nonumber$
The final temperature is in between the initial temperatures of the tea (12.02 °C) and the melted ice (0.0 °C), so this answer makes sense. In this example, the tea loses much more heat in melting the ice than in mixing with the cold water, showing the importance of accounting for the heat of phase changes!
Exercise $1$: Death by Freezing
Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example $1$
Answer
200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow.
Summary
Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes. Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic. The conversion of a solid to a liquid is called fusion (or melting). The energy required to melt 1 mol of a substance is its enthalpy of fusion (ΔHfus). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (ΔHvap). The direct conversion of a solid to a gas is sublimation. The amount of energy needed to sublime 1 mol of a substance is its enthalpy of sublimation (ΔHsub) and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called heating curves. Heating curves relate temperature changes to phase transitions. A superheated liquid, a liquid at a temperature and pressure at which it should be a gas, is not stable. A cooling curve is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a supercooled liquid, a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a seed crystal of the same or another substance can induce crystallization. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.5%3A_Phase_Transitions.txt |
Learning Objectives
• To understand the basics of a one-component phase diagram as a function of temperature and pressure in a closed system.
• To be able to identify the triple point, the critical point, and four regions: solid, liquid, gas, and a supercritical fluid.
The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system.
Introduction
A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure $1$). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure $1$) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid.
The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure.
The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure $1$, the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in Section 11.5. It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the only combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point.
Remember that a phase diagram, such as the one in Figure $1$, is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error.
The Phase Diagram of Water
Figure $2$ shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee).
The phase diagram for water illustrated in Figure $\PageIndex{2b}$ shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure $1$; that is, the melting point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is less dense than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.
In Figure $\PageIndex{2b}$ point A is located at P = 1 atm and T = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines.
Ice Skating: An Incorrect Hypothesis of Phase Transitions
Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation.
Recall that pressure (P) is the force (F) applied per unit area (A):
$P=\dfrac{F}{A} \nonumber$
To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is
$F = mg \nonumber$
where m is the mass and g is the acceleration due to Earth’s gravity (9.81 m/s2). Thus the force is
$F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber$
If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is
$A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber$
If the skater is gliding on one foot, the pressure exerted on the ice is
$P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber$
The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases.
Example $1$: Water
Referring to the phase diagram of water in Figure $2$:
1. predict the physical form of a sample of water at 400°C and 150 atm.
2. describe the changes that occur as the sample in part (a) is slowly allowed to cool to −50°C at a constant pressure of 150 atm.
Given: phase diagram, temperature, and pressure
Asked for: physical form and physical changes
Strategy:
1. Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region.
2. Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes.
Solution:
1. A Locate the starting point on the phase diagram in part (a) in Figure $2$. The initial conditions correspond to point A, which lies in the region of the phase diagram representing water vapor. Thus water at T = 400°C and P = 150 atm is a gas.
2. B Cooling the sample at constant pressure corresponds to moving left along the horizontal line in part (a) in Figure $2$. At about 340°C (point B), we cross the vapor pressure curve, at which point water vapor will begin to condense and the sample will consist of a mixture of vapor and liquid. When all of the vapor has condensed, the temperature drops further, and we enter the region corresponding to liquid water (indicated by point C). Further cooling brings us to the melting curve, the line that separates the liquid and solid phases at a little below 0°C (point D), at which point the sample will consist of a mixture of liquid and solid water (ice). When all of the water has frozen, cooling the sample to −50°C takes us along the horizontal line to point E, which lies within the region corresponding to solid water. At P = 150 atm and T = −50°C, therefore, the sample is solid ice.
Exercise $2$
Referring to the phase diagram of water in Figure $2$, predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm.
Answer
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
The Phase Diagram of Carbon Dioxide
In contrast to the phase diagram of water, the phase diagram of CO2 (Figure $3$) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed.
Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps.
The Critical Point
As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure (Pc), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure $1$.
Figure $1$: Critical Temperatures and Pressures of Some Simple Substances
Substance Tc (°C) Pc (atm)
NH3 132.4 113.5
CO2 31.0 73.8
CH3CH2OH (ethanol) 240.9 61.4
He −267.96 2.27
Hg 1477 1587
CH4 −82.6 46.0
N2 −146.9 33.9
H2O 374.0 217.7
High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.
Supercritical Fluids
A Video Discussing Phase Diagrams. Video Source: Phase Diagrams(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.6%3A_Phase_Diagrams.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q.1
If a substance has relatively strong intermolecular forces, a high melting point, and is not easily compressed, will this substance be a solid, liquid, or gas?
Solution
The substance is a solid because solids have relatively stronger intermolecular forces than liquids or gases. These strong intermolecular forces strongly hold the molecules in the solid together which makes it hard to compress a solid. Solids also have higher melting points because you need to put a lot more energy into solids to break apart the intermolecular forces to make the solid change phases.
Q.3
An unknown sample at $25^oC$ had a volume of $3.00\times 10^{14} \mu m^3$ and a mass of $3.20\times 10^3\ g$.
1. Is this sample most likely in a gaseous state or a condensed one? Why? (HINT: The density of water is $1.000\frac{g}{mL})$.
2. After completing mass spectroscopy the molar mass of the material was found to be 127 g mol-1, what is the molar volume at $25^oC$ from using the answer in part (a).
Solution
We need to compute the density of the liquid first:
$\left(\dfrac{3.2 \times 10^3 g\;}{3.0 \times 10^{14} \cancel{µm^{3}}} \right)\left(\dfrac{1.0 \times 10^{12}\cancel{µm^{3}}\;}{\textrm{1} \textrm{ cm}^{3}\;}\right )=1.07 \times 10^{1}\textrm{g cm}^{-3}\nonumber$
This material is known to be condensed because its density is higher than that of water.
Calculate the molar volume by:
$\left(\dfrac{\textrm{1} \textrm{ cm}^{3}\;}{\textrm{10.7} \cancel{g}\;}\right)\left(\dfrac{127\cancel{g}\;}{\textrm{1} \textrm{ mol}\;}\right )=1.19\times10^{1}\textrm{cm}^{3} {mol}^{-1}\nonumber$
Q.6A
If the volume of a substance changes from 300 $cm^3$ to $313\ cm^3$ when it is heated from $25^oC$ to $40^oC$, is the substance ideal gas, nearly ideal gas, or condensed?
Solution
Temperature change = $15^oC$
$\text{Volume change ratio:} \dfrac{(313-300)}{300} = \dfrac{13}{300} \nonumber$
Hence, change percentage per $^oC$ is
$\dfrac{13}{300} \times \dfrac{1}{15} \times 100\%= \dfrac{13}{1500} \% = 0.288\%\nonumber$
Since $0.288\%$ is close to the ideal gas percentage of $0.366\%$, it is a nearly ideal gas.
Q.6B
Cooling a sample of matter from 130° to 50° at a constant pressure causes its volume to increase from 769.1 to 930.1 cm3. Classify the material as nearly ideal gas, a non ideal gas, or condensed.
Solution
Charles' Law - Ideal Gases
$\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}\nonumber$
$0.00191 \dfrac{L}{K} = 0.00288 \dfrac{L}{K}\nonumber$
The material is a non-ideal gas.
Q.7
The molar volume of a substance is the volume occupied by one mole of the substance. The molar volume of a typical solid or liquid is $10\frac{cm^3}{mol}$ to $100\frac{cm^3}{mol}$, while the molar volume of a gas under the same conditions is about $24,000\frac{cm^3}{mol}$. What does the similarity of the molar volumes of solid and liquid forms of the same substance suggest? What does the great difference of the the molar volumes of solid/liquid and gas forms of the same substance suggest?
Solution
The similarity of the molar volumes of solid and liquid forms of the same substance suggests that the distance between neighboring molecules of the substance in solid phase is approximately the same as the distance between neighboring molecules in liquid phase. The great difference of the molar volumes of solid/liquid and gas forms of the same substance suggests that the distance between neighboring molecules of the substance in gas phase is greater the distance between neighboring molecules in solid/liquid phase. This would explain why solids and liquids are called condensed states of matter, because the particles in these two phases are very close together, hence, solids and liquids have definite volumes. While the particles in the gas phase are so far away from each other that they can move freely at high speed, allowing the gas substance to assume the shape and volume of its container.
Q.9
Suppose you have to choose between solid $\ce{I_2}$ and solid $\ce{NH_4Cl}$ to fill your cushion and you prefer softer cushions, which one would you choose and why?
Solution
Non-directional ion-ion interaction is the dominant intermolecular interaction solid $\ce{NH_4Cl}$ has, and weaker London dispersion forces are the main intermolecular interaction $\ce{I_2}$ has, while the feeling of softness (indentation that breaks bonds in the solid) of the two solids depends on their strength of intermolecular interactions; in this case $\ce{I_2}$ has weaker intermolecular interactions and will feel softer.
Q.11
The diffusion constant is defined as the amount of substance that when diffusing from a region of high concentration to that of a low concentration goes through each unit of cross section per unit time. What happens to the diffusion constant, at constant temperature, as density of a liquid decreases, as the density of a solid increases, and as the density of a gas decreases? Explain the phenomena.
Solution
As the density of a liquid decreases, the diffusion constant will increase. This is because as density of the liquid decreases, there is more space in between the liquid molecules, and hence more movement is possible for the liquid molecules.
As density of a solid increases, diffusion constant will decrease. Similarly, in concept for the liquid molecules, if the density of a solid increases, there is even less space between the solid molecules than there was before. This further restricts the movements of the solid molecules and decreases diffusion. Note: solids also undergo diffusion; however, it occurs at an extremely slow rate.
As the density of gas decreases, diffusion constant will increase. As the density of a gas decreases, collision between gas molecules will occur less frequently, thus increasing freedom of movement for gas molecules and increasing the diffusion constant.
Q.15
For the following chemicals list attractive intermolecular forces that you expect to see with each chemical. Of the forces you listed for each example, rank the intermolecular forces from strongest to weakest.
1. $\ce{He}$
2. $\ce{H_2O}$
3. $\ce{NaCl}$
4. $\ce{CH_4}$
5. $\ce{CO}$
6. $\ce{O_2}$
Solution
1. London dispersion forces
2. Hydrogen bonding > dispersion forces
3. Ionic bonding > dispersion forces
4. London dispersion forces
5. Dipole-dipole forces > dispersion forces
6. London dispersion forces
Q.17
Which of the the following will be most strongly attracted to a lithium ion:
1. fluoride ion
2. a molecule of hydrogen fluoride
3. atom of Argon
Solution
A fluoride ion will be most strongly attracted to a lithium ion. The attraction between unlike charges such as lithium and fluoride are much stronger than the ion dipole attraction between $\ce{Li^+}$ and $\ce{HF}$ and the ion-induced dipole attraction between $\ce{Li^+}$ and $\ce{Ar}$.
Q.19
Estimate the bond length of $\ce{He2}$, $\ce{Ar2}$, and $\ce{Xe2}$ from the potential energy curves below. For each interacting pair, identify the attractive and repulsive distances. What are the intermolecular forces exist in each of the pair of molecules? Order the potential interactions in term of strength of intermolecular forces.
Solution
• $\ce{He2} \approx 2.7Å$
• $\ce{Ar2} \approx 3.8Å$
• $\ce{Xe2} \approx 4.5Å$
Repulsion is when R less than R at equilibrium. Attraction is when R is bigger than R at equilibrium.
Increasing order of intermolecular forces: $\ce{He2}$ < $\ce{Ar2}$ < $\ce{Xe2}$. The intermolecular force that the molecules have is London Dispersion Force (Van der Waals force). Larger molecules tend to have greater polarizability because they have more electrons and their electrons are further away from the nucleus. Therefore, LDF tend to get stronger when the molecule becomes larger. Based on the graph, the higher the potential energy, the stronger the interaction is.
Q.23
Arrange the following substances in order of decreasing normal boiling points and explain the rationale.
1. $\ce{CH_3CH_2CH_2CH_2CH_3}$
2. $\ce{CH_3CH_2CH_2CH_2OH}$
3. $\ce{CH_4}$
4. $\ce{CH_3CH(CH_3)CH_2CH_3}$
Solution
$\ce{CH_3CH_2CH_2CH_2OH > CH_3CH_2CH_2CH_2CH_3 > CH_3CH(CH_3)CH_2CH_3 > CH_4 }\nonumber$
1-Butanol has the greatest boiling point due to strong hydrogen bonding with the alcohol group. Pentane has the second strongest boiling point due to its systematical structure and large London Dispersion force. Although 2-methylbutane has the same molecular mass as pentane, its geometrical structure hinders stacking which leads to lower intermolecular force. Therefore, 2-methylbutane has the lowest boiling point.
Q.24
You are given four beakers A, B, C, and D with four different chemicals. You forgot to label the beakers, so to identify them you boiled them to find their normal boiling points. The order of the boiling points came out to be B>D>A>C. If the chemicals in the beakers were $\ce{HF}$ (Hydrogen Fluoride), $\ce{Ar}$ (Argon), Cesium Chloride $\ce{CsCl}$ and Hydrochloric Acid $\ce{HCl}$, assign the chemicals to their respective beakers and give the reasoning behind your answer.
Solution
The primary intermolecular forces in the compounds are
• $\ce{HF}$ → Hydrogen Bonding
• $\ce{Ar}$ → Dispersion Forces
• $\ce{CsCl}$ → Ionic Bonding
• $\ce{HCl}$ → Dipole-Dipole Interactions
Ionic Bonding > Hydrogen Bonding > Dipole-Dipole Interactions > Dispersion Forces
The stronger the intermolecular forces, the lower the vapor pressure, hence the boiling point would be higher.
Hence the beakers contain
• A = $\ce{HCl}$
• B = $\ce{CsCl}$
• C = $\ce{Ar}$
• D = $\ce{HF}$
Q.25
Like a normal human being, Kris enjoys breathing. But unfortunately for Kris, she is clinically paranoid, and feels as though she's breathing in a little too much ethanol. To calm Kris down, you say that as a vapor, ethanol exists, to an extent, as a dimer, ($\ce{(CH3CH2OH)2}$), in which two $\ce{CH3CH2OH}$ molecules are held together by hydrogen bonds. Propose and draw a reasonable structure for this dimer to help Kris deal with her problems.
Solution
Hydrogen bonds are a type of intermolecular forces. It is the bond between a hydrogen atom and a high electronegative atom such as N, O, and F. The only hydrogen bonding which can take place between the ethanol molecules is between one's oxygen and the other's hydrogen bonded onto the other's oxygen. So any two membered structure of two ethanol molecukes with this criteria indicated is acceptable.
Q.25
Acetic acid ($\ce{CH3COOH}$) forms a dimer in the gas phase, where two acetic acid molecules are held together by hydrogen bonds. Draw a reasonable structure for this dimer.
Solution
The diagram above shows a reasonable structure of a acetic acid dimer, where the dash lines represent hydrogen bonds.
Q.26
Hypochlorous acid ($\ce{HOCl}$) is a similar compound to $\ce{HOF}$, which is the simplest possible compound that allows comparison between fluoride and oxygen in their abilities to form hydrogen bonds. Although $\ce{Cl}$ attracts electrons more strongly than $\ce{O}$, solid $\ce{HOCl}$ cannot form $\ce{H-Cl}$ bonds. Draw the proposed structure for chains of $\ce{HOCl}$ molecules in the crystalline state. The bond angle for $\ce{HOCl}$ is $103^o$.
Solution
Q.27
How do the boiling points of hydrogen halides differ from that of hydrogen fluoride ($\ce{HF}$). Explain your reasoning. Answer: $\ce{HF}$ has a much higher boiling point ($20^oC$) compared to the other hydrogen halides. This is because $\ce{HF}$ is capable of forming hydrogen bonding within its compounds.
Solution
Fluorine is capable of doing hydrogen bonding as opposed to the other halogens in the periodic table. Since fluorine is the element with the highest electronegativity, the hydrogen-fluorine bond in hydrogen fluoride is highly polarized, creating a partial positive charge (δ+) on the hydrogen and, a negative charge (δ-) on the fluorine atom. Furthermore, the lone pairs situated on the fluorine atom are in the second energy level, thus they are very close and the negative charge on the atom is very concentrated, thus it has a strong attraction force. Hydrogen bonds form between the positively charged H atoms and the lone pair of the F atom. The other halogens are not capable of hydrogen bonding because they are not as electronegative as fluorine and they are larger in size. Thus the bond within the molecule is not as polarized and the lone pairs on the halogen atom are not as concentrated. Thus, the stronger the intermolecular forcers of the liquid, the harder it is for a molecule to gain enough energy to overcome the intermolecular forces that bond it in a liquid. The more energy is required to enter the gas phase. The higher the boiling point. Fluorine has higher intermolecular forces, thus it has a higher boiling point compared to the other hydrogen halides.
Q.31
Oxygen is stored at a temperature of 105 K and a pressure of 3.356 atm. If the volume of the container is 2.5 L, calculate the number of moles of oxygen in the container. Compare this number of moles to the number of moles in the same container at standard pressure and 298 K. Is it fewer, more, or unchanged?
Solution
This is an ideal gas law problem, so the formula $PV=nRT$ should be used.
$P=3.356 \, \text{atm}$ $V=2.5 \, \text{L}$ $n=?$ $R=0.0821 \dfrac{\text{L} \ \text{atm}}{\text{mol} \ \text{K}}$ $T=105 K$
$n= \dfrac{3.356 \times 2.5}{0.0821 \times 105} \, \text{mol} \nonumber$
$n= 0.9733 \, \text{mol} \nonumber$
Standard pressure = 1 atm
$n=\dfrac{1\times 2.5}{0.0821\times 298}\nonumber$
$n= 0.102 \, \text{mol} < 0.9733 \, \text{mol} \nonumber$
Therefore, under standard conditions, there are fewer moles of oxygen in the container.
Q.35
Consider the reaction at 25ºC.
$\ce{CaC2(s) + 2H2O(l) -> C2H2(g) + Ca(OH)2(s)} \nonumber$
If the total pressure is 0.9124 atm and the vapor pressure of water at this temperature is 0.0313 atm. What is the mass of acetylene ($\ce{C2H2}$) per liter of "wet" acetylene collected by this reaction (i.e., collected over a pool of water)? Assume all gases behave ideally.
Solution
Partial pressures in gases are additive via Dalton's Law.
• $P_T= 0.9124 atm$
• $P_{\ce{H2O}}= 0.0313 atm$
$P_T-P_{\ce{H2O}}=P_{\ce{C2H2}} = 0.8811 atm \nonumber$
Using the Ideal Gas Law, moles per liter of acetylene can be found by rearranging the equation:
$\ce\dfrac{n}{V}=\dfrac{P_{\ce{C2H2}}}{RT}\nonumber$
$\dfrac {0.8811}{0.082057 \times 298.15} =\dfrac{0.036 mol}{1 L}\nonumber$
The question asks for grams per liter, so:
$\left (\dfrac{0.036 \: \cancel{mol \; C_2H_2}}{1 L} \right ) \left (\dfrac{26.036\; g C_2H_2}{1\: \cancel{mol\; C_2H_2}} \right ) = 0.9376\; \dfrac{g}{L}\nonumber$
Q.39
The highest value of sea-level pressure on Earth occurs in Siberia where the Siberian High often allows water to boil at 80oC. Using chart 10.16 on page 431 of the text book, what can the fraction of pressure be in this location?
Solution
According to the graph, a location where the boiling temperature is 80oC correlates approximately to a fraction of 0.5 atm pressure.
Q.41
Water ($\ce{H2O}$) has a melting point of $0.00\text{°C}$ and a boiling point of $100.0\text{°C}$. Benzene ($\ce{C6H6}$) has a melting point of $5.5\text{°C}$ and a boiling point of $80.1\text{°C}$. At $20.0\text{°C}$, which of the two substances would be expected to have a greater surface tension? Explain?
Solution
At $20.0\text{°C}$, water $\ce{H2O}$ has a greater surface tension ($72.86\dfrac{mN}{m}$) than benzene, ($\ce{C6H6}$), ($28.88\dfrac{mN}{m}$) because the intermolecular forces between water molecules are stronger than the intermolecular forces between benzene molecules. Water is capable of forming hydrogen-bonds, whereas the only forces that act in benzene are London dispersion forces.
is the energy/work required to increase the surface area of a liquid. Molecules at the surface of a liquid do not experience any on one interface with air and therefore experience a net attractive force towards the center of the liquid. For surface molecules, the stronger the intermolecular forces between the molecules of the liquid, the stronger the net attractive force towards the center of the liquid will be. Compared to benzene, water possesses stronger intermolecular forces, therefore its surface molecules exhibit a stronger net attractive force towards the center of the liquid. As a result, more energy/work is required to counteract the net attractive force in order to increase the surface area. This indicates that for any liquid, stronger intermolecular forces allow for stronger surface tensions.
Q.43
At gallium’s melting point of $302.91K$, its density is $6.095\frac{g}{cm^3}$. The density of solid gallium under standard conditions is $5.91\frac{g}{cm^3}$. When $Ga$ is at $301.91K$ and pressure is strongly increased, will $Ga$ undergo a phase change?
Solution
Compression favors the denser phase. Since $Ga$ is a solid at $301.9K$, then it will undergo a phase change to become a liquid.
Q.45
Given the information below, sketch the phase diagram of Nitrogen:
• Triple point: 0.1252 bar, -210°C
• Normal boiling point: 1 bar, -195.8°C
• Normal melting point: 1 bar, -210°C
Solution
Normally, phase diagrams show pressures below that of the critical point.
Q.47
Use the phase diagram of water to determine whether water is a solid, a liquid, or a gas at each of the following combinations of temperature and pressure.
1. $1\ atm$ and $20^oC$
2. $0.006\ atm$ and $100^oC$
3. $210\ atm$ and $100^oC$
4. $10\ atm$ and $50^oC$
Figure used with permission (CK-12 Foundation – Christopher Auyeung)
Solution
Using the phase diagram of water,
1. Liquid
2. Gaseous
3. Liquid
4. Liquid
Q.53
What phase does candle wax exist in? How about butter?
Solution
Matter can exist in different phases depending on the temperature of the substance. For example, candle wax and butter can exist as a solid at room temperature, but when enough heat is applied to these substances, they can exist in a liquid or a gaseous phase.
Q.59
The equilibrium vapor pressure of water at $19^oC$ is $0.02168\,atm$. A humidifier is placed in a room with a volume of $180m^3$ and operates until the the room becomes saturated with water vapor. Initially there is no water vapor in water vapor in the air. Assuming that the room is closed completely from the outside such that water vapor cannot escape, calculate the mass of water that has passed into the air.
Solution
The air in the room becomes saturated once the vapor pressure of water in the room is equivalent to the equilibrium vapor pressure of water. Plug in given values into the ideal gas law to find the number of moles of water in the room when it is saturated, and then convert to grams.
$PV=nRT\nonumber$
$(0.02168atm)(180m^{3})(\dfrac{1000L}{m^{3}})=(n)(0.08206\frac{L\ atm}{mol\ K})(292.15K)\nonumber$
$n=162.77mols\nonumber$
$m=162.77mols\times \dfrac{18.02g}{mol}=2930g\nonumber$
Q.65
A cooling bath is a mixture that is primarily liquid which is kept at a constant, low temperature. While in the lab, Po’lah accidently severed its hand after trying to make some mac and cheese. As a result it had to immediately transfer its hand to a cooling bath to sufficiently preserve it but not so that it could be of some use later on. In order to achieve this, the cooling bath had to be kept at around $0^{\circ}C$. The only liquids it had access to were water and methanol, both at room temperature, and the only cooling agents that it had were liquid nitrogen and dry ice (solid carbon dioxide). What combination did Po’lah use and why?
Relevant Information:
• Carbon Dioxide: sublimes at $-78.5^{\circ}C$
• Nitrogen: boils at $-195.8^{\circ}C$
• Water: freezes at $0^{\circ}C$
• Ethanol: freezes at $-143.7^{\circ}C$
Solution
This question is a bit tricky. The coolants Liquid Nitrogen and Dry ice, ($\ce{CO2 (s)}$), both turn into gasses at very low temperatures. After they turn into gaseous, they are no longer “useful”, since they cannot be added into the bath. Therefore, after adding sufficient amounts of either of the coolants to either of the liquids, the temperature of the bath would decrease way past the freezing point of either liquid, resulting in a cooling “block” rather than a bath. So what is the deal? Is this problem unsolvable?
Fortunately, it isn’t. Think about a glass of ice water. If enough ice is put into the water, then eventually the water and ice would equilibrate at $0^{\circ}C$ which is the equilibrium point of the water where both its solid and liquid states can exit. Unfortunately, while the ice is not available at this time, the same concept can be applied. Remember, if enough of either coolant is used, then the entire cup of water will freeze over. However, if just enough of either coolant is put in so that the cooling bath remains at $0^{\circ}C$, a cooling bath that is somewhat solid and somewhat liquid will form, giving an adequate place for Po’lah to put its hand. Water is perfect for this because it remains a liquid at $0^{\circ}C$, and cannot exist as such at any lower temperature. Since Ethanol freezes at $-143.7^{\circ}C$, adding dry ice or liquid nitrogen would bring the cooling bath much lower than $0^{\circ}C$, which would destroy Po’lah’s hand
Thus, it doesn’t matter what cooling agent is used, either can bring water down to $0^{\circ}C$. However, methane cannot be used.
Abstract: Add a little of either coolant to bring the water down to $0^{\circ}C$, and it will remain at both a liquid and a solid which creates a $0^{\circ}C$ cooling bath. Methanol cannot be used.
Q.71
The melting points of the chlorides of third period elements are: $\ce{NaCl} = 1074 \,K$, $\ce{MgCl_2} = 987\,K$, $\ce{AlCl_3} =465\,K$, $\ce{SiCl_4} = 204.4\, K$, $\ce{PCl_3} = 179\, K$. Based on the applicable intermolecular forces present in each compound, explain this trend in melting points.
Solution
Experimentally, as the chloride compounds make their way across the third period elements, the melting points gradually decrease. This may appear an easy question to address since the melting points are a also measure of the magnitude of the intermolecular forces as play in the specific system, however it a quite complicated and to answer properly require knowledge of the chloride compounds.
Both ($\ce{NaCl}$ and $\ce{MgCl_2}$) are both "ionic solids" with strong ion-ion bonding and hence have the highest melting temperatures in the comparative series. We would expect the $\ce{Na+}$ ions in ($\ce{NaCl}$ to have weaker Coulombic interactions than the $\ce{Mg^{2+}}$ ions in $\ce{MgCl_2}$ since the latter are smaller and double the charge (i.e., a higher charge density). That would make the lattice energy higher because it would generate stronger ionic bonds (lattice energy of $\ce{NaCl}$ and $\ce{MgCl2}$ are 876 and 2526 kJ/mol). However, this ionic perspective is insufficient to explain the observed melting points with ($\ce{NaCl} > \ce{MgCl_2}$), which we ascribe to bonding never being 100% ionic. As expected for teh difference in electronegativity of chlorine, magnesium, and sodium, the percent ionic character of the $\ce{Na-Cl}$ bond is 71% and for $\ce{Mg-Cl}$ is even lower, at 58%. Hence, the $\ce{Na-Cl}$ bond is 29% covalent and the $\ce{Mg-Cl}$ bond is 42% covalent. The covalent character reduces the melting point of $\ce{MgCl2}$, even though $\ce{Mg+}$ has a higher charge-to-size ratio than $\ce{Na+}$.
$\ce{AlCl_3}$ is a molecular framework solid which means the solid is an extended covalent molecule and not individual $\ce{AlCl_3}$ molecules bound by intermolecular forces.
Both $\ce{SiCl_4}$ and $\ce{PCl_4}$ solids are molecular in nature. To understand the strength of intermolecular forces as play this properly, we need to apply basic VESPR rules to identify the structures of these molecules.
$\ce{SiCl_4}$ is a tetrahedral molecule with no dipole moment (nor quadrupole moment either). $\ce{PCl_4}$ is trigonal pyramidal and since $\ce{P}$ has a different electronegativity than $\ce{Cl}$, it has a dipole moment (0.97D). From this argument, we would think that $\ce{PCl_4}$ would have a higher melting point, however both molecules have dispersive forces (London) too. Since the magnitude of dispersive forces increased wtih larger molecules (with more electrons to shift around), this is higher in $\ce{PCl_4}$, which beats out the weak dipole-dipole interactions in $\ce{PCl_4}$.
Q.71
Why are the melting points of $\ce{NaCl}$ (801 °C) and $\ce{MgCl_2}$ (714 °C) so much higher than the melting points of $\ce{SCl_2}$ (-122 °C) and $\ce{Cl_2}$ (-101.5 °C)? Explain answer in terms of intermolecular forces and identify any potential forces involved.
Solution
As with all bulk properties, the melting points are a measure of the magnitude of the intermolecular forces as play in the specific system (geometric structures also affects bulk properties like the nature of the crystal lattice, but that is a secondary issue). From inspection, it is clear that the $\ce{NaCl}$ and $\ce{MgCl2}$ are ionic held together the strongest permanent electrostatic interaction (charge-charge or monopole-monopole); these are ionic solids. For $\ce{SCl2}$ and $\ce{Cl_2}$, the solid is a molecular solid wtih constituent molecules that lack charge and even dipole (easy to tell for $\ce{Cl2}$ and requires a little VSEPR effort for $\ce{SCl2}$. Hence, the next higher permanent electrostatic interactions are active in these molecules (i.e. quadrupole-quadrupole) along with dispersion and repulsive interactions; all of which are much weaker than the charge-charge interactions in $\ce{NaCl}$ and $\ce{MgCl2}$. The weaker the interactions, the lower the boiling point since less thermal energy is needed to break those interactions.
Now, a more advanced question is to explain the origin of the differences in the melting points of $\ce{NaCl}$ vs. $\ce{MgCl2}$. Also to explain the origin of the differences in the melting points of $\ce{SCl2}$ vs. $\ce{Cl2}$. No answer given. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/10.E%3A_Solids_Liquids_and_Phase_Transitions_%28Exercises%29.txt |
A solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.
• 11.1: Composition of Solutions
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible.
• 11.2: Nature of Dissolved Species
The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are unstable.
• 11.3: Reaction Stoichiometry in Solutions: Acid-Base Titrations
A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.
• 11.4: Reaction Stoichiometry in Solutions: Oxidation-Reduction Titrations
Redox titration are here the titrant is an oxidizing or reducing agent. In contrast to acid/base titrations, it is convenient for redox titrations to monitor the titration reaction’s potential instead of the concentration of one species.
• 11.5: Phase Equilibrium in Solutions - Nonvolatile Solutes
Colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law.
• 11.6: Phase Equilibrium in Solutions - Volatile Solutes
In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. Hence, the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in
• 11.7: Colloidal Suspensions
A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect.
• 11.E: Solutions (Exercises)
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al.
11: Solutions
Learning Objectives
• To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
• To be familiar with the different units used to express the concentrations of a solution.
There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example $1$ reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.
Example $1$: Molarity and Mole Fraction
Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Given: mass of substance and mass and density of solution
Asked for: molarity and mole fraction
Strategy:
1. Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
2. Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.
Solution:
A: The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass.
\begin{align*} \text{moles } \ce{CH_3CO_2H} &=\dfrac{3.78\; \cancel{\ce{g}}\; \ce{CH_3CO_2H}}{60.05\; \cancel{\ce{g}}/\ce{mol}} \[4pt] &=0.0629 \; \ce{mol} \end{align*} \nonumber
The volume of the solution equals its mass divided by its density.
\begin{align*} \text{volume} &=\dfrac{\text{mass}}{\text{density}} \[4pt] &=\dfrac{100.0\; \cancel{\ce{g}}\; \text{solution}}{1.00\; \cancel{\ce{g}}/\ce{mL}}=100\; mL\nonumber \end{align*} \nonumber
Then calculate the molarity directly.
\begin{align*} \text{molarity of } \ce{CH_3CO_2H} &=\dfrac{\text{moles } \ce{CH3CO2H} }{\text{liter solution}} \[4pt] &=\dfrac{0.0629\; mol\; \ce{CH_3CO_2H}}{(100\; \cancel{\ce{mL}})(1\; L/1000\; \cancel{\ce{mL}})}=0.629\; M \; \ce{CH_3CO_2H} \end{align*} \nonumber
This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than $\ce{ 1/2}$ mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.
B: To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have
$moles\; \ce{H_2O}=\dfrac{96.2\; \cancel{\ce{g}}\; \ce{H_2O}}{18.02\; \cancel{\ce{g}}/mol}=5.34\; mol\; \ce{H_2O}\nonumber$
The mole fraction $\chi$ of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:
\begin{align*} \chi_{\ce{CH3CO2H}} &=\dfrac{moles\; \ce{CH_3CO_2H}}{moles \; \ce{CH_3CO_2H} + moles\; \ce{H_2O}} \[4pt] &=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol} \[4pt] &=0.0116=1.16 \times 10^{−2} \end{align*} \nonumber
This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.
Exercise $1$: Molarity and Mole Fraction
A solution of $\ce{HCl}$ gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of $\ce{HCl}$ per 100.0 g of solution, and its density is 1.10 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Answer a
6.10 M HCl
Answer b
$\chi_{HCl} = 0.111$
The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:
$\text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1}$
Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of $H_2O$ under these conditions is very close to 1.0 L, and a 0.50 M solution of $KBr$ in water, for example, has approximately the same concentration as a 0.50 m solution.
Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):
\begin{align} \text{mass percentage}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2} \[4pt] \text{parts per million (ppm)} &=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3} \[4pt] \text{parts per billion (ppb)}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4} \end{align}
In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.
The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of $\ce{H_2SO_4}$ per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
Example $2$: Molarity and Mass
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
1. What is the molarity of the solution?
2. What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
1. Use the concentration of the solute in parts per million to calculate the molarity.
2. Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
\begin{align*} \text{molarity}&=\dfrac{\text{moles}}{\text{liter solution}} \[4pt] &=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \[4pt] &=1.63 \times 10^{-4} M\end{align*} \nonumber
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
\begin{align*} \text{mass of benzene} &=\dfrac{(12.7\; mg\; \text{benzene})(250\; \cancel{mL})}{1000\; \cancel{mL}} \[4pt] &=3.18\; mg \[4pt] &=3.18 \times 10^{-3}\; g\; \text{benzene} \end{align*} \nonumber
Exercise $2$: Molarity of Lead Solution
The maximum allowable concentration of lead in drinking water is 9.0 ppb.
1. What is the molarity of $\ce{Pb^{2+}}$ in a 9.0 ppb aqueous solution?
2. Use your calculated concentration to determine how many grams of $\ce{Pb^{2+}}$ are in an 8 oz glass of water.
Answer a
4.3 × 10−8 M
Answer b
2 × 10−6 g
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.
Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.
Table $1$ summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example $3$.
Table $1$: Different Units for Expressing the Concentrations of Solutions*
Unit Definition Application
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction ($\chi$) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
Example $3$: Vodka
Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.
1. the mass percentage
2. the mole fraction
3. the molarity
4. the molality
Given: volume percent and density
Asked for: mass percentage, mole fraction, molarity, and molality
Strategy:
1. Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
2. Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
3. Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
4. Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.
Solution:
The key to this problem is to use the density of pure ethanol to determine the mass of ethanol ($CH_3CH_2OH$), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:
$mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber$
If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.
B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:
\begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \[4pt]&= 34.5\%\end{align*} \nonumber
C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:
\begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*} \nonumber
Similarly, the number of moles of water is
$moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber$
The mole fraction of ethanol is thus
$\chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber$
D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is
$M_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\nonumber$
The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:
$m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber$
Exercise $3$: Toluene/Benzene Solution
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
1. mass percentage
2. mole fraction
3. molarity
4. molality
Answer a
mass percentage toluene = 24.8%
Answer b
$\chi_{toluene} = 0.219$
Answer c
2.35 M toluene
Answer d
3.59 m toluene
A Video Discussing Different Measures of Concentration. Video Link: Measures of Concentration, YouTube (opens in new window) [youtu.be]
A Video Discussing how to Convert Measures of Concentration. Video Link: Converting Units of Concentration, YouTube(opens in new window) [youtu.be]
Summary
Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.1%3A_Composition_of_Solutions.txt |
Learning Objectives
• To understand how enthalpy and entropy changes affect solution formation.
• To use the magnitude of the changes in both enthalpy and entropy to predict whether a given solute–solvent combination will spontaneously form a solution.
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table $1$ lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.
Table $1$: Types of Solutions
Solution Solute Solvent Examples
gas gas gas air, natural gas
liquid gas liquid seltzer water ($CO_2$ gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid $H_2$ in Pd (used for $H_2$ storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)
solid solid solid alloys and other "solid solutions"
Forming a Solution
The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:
$\ce{Zn(NO3)2(s) + H2O(l) \rightarrow Zn^{2+}(aq) + 2NO^{-}3(aq)} \label{13.1.1}$
Because $Zn(NO_3)_2$ can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:
$\ce{ Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \rightarrow Zn^{2+}(aq) + 2Cl^{-}(aq) + H2(g)} \label{13.1.2}$
When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation (that it is a physical change).
Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere.
The Role of Enthalpy in Solution Formation
Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.
Because enthalpy is a state function, we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by $ΔH_1$, $ΔH_2$, and $ΔH_3$ in Figure $2$. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3}$
When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent ($\Delta H_1$) and the solute ($\Delta H_2$). Because $ΔH$ is positive for both steps 1 and 2, the solute–solvent interactions ($\Delta H_3$) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic ($\Delta H_{soln} < 0$). When the solute is an ionic solid, $ΔH_2$ corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as $MgO$ (−3791 kJ/mol), are generally insoluble in all solvents.
A positive value for $ΔH_{soln}$ does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high $ΔH_{soln}$ is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive $ΔH_{soln}$ of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure $3$).
Entropy and Solution Formation
The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy ($S$) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.
The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, $ΔH_{soln}$ should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative $ΔH_{soln}$ but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with $ΔH \ge 0$ are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.
All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.
Table $2$ summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.
Table $2$: Relative Changes in Enthalpies for Different Solute–Solvent Combinations*
$ΔH_1$ (separation of solvent molecules) $ΔH_2$ (separation of solute particles) $ΔH_3$ (solute–solvent interactions) $ΔH_{soln}$ ($ΔH_1$ + $ΔH_2$ +$ΔH_3$) Result of Mixing Solute and Solvent†
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
*$ΔH_1$, $ΔH_2$, and $ΔH_3$ refer to the processes indicated in the thermochemical cycle shown in Figure $2$.
In all four cases, entropy increases.
In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process ($ΔH_{soln} \approx 0$), and the entropic factor due to the increase in disorder is dominant (Figure $4$). Consequently, all gases dissolve readily in one another in all proportions to form solutions.
Example $1$
Considering $\ce{LiCl}$, benzoic acid ($\ce{C6H5CO2H}$), and naphthalene, which will be most soluble and which will be least soluble in water?
Given: three compounds
Asked for: relative solubilities in water
Strategy: Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure $2$. Then use Table $2$ to predict the solubility of each compound in water and arrange them in order of decreasing solubility.
Solution:
The first substance, $\ce{LiCl}$, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in Equation $\ref{13.1.1}$). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect $ΔH_3$ to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect $ΔH_2$ to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak $ΔH_3 \approx 0$. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than $\ce{LiCl}$. We thus predict $\ce{LiCl}$ to be the most soluble in water and naphthalene to be the least soluble.
Exercise $1$
Considering ammonium chloride, cyclohexane, and ethylene glycol ($HOCH_2CH_2OH$), which will be most soluble and which will be least soluble in benzene?
Answer
The most soluble is cyclohexane; the least soluble is ammonium chloride.
Summary
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, $ΔH_{soln}$, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic ($ΔH_{soln} < 0$) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.2%3A_Nature_of_Dissolved_Species.txt |
Learning Objectives
• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration
• Explain the function of acid-base indicators
As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration.
Titration Curves
Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization.
Example $1$: Calculating pH for Titration Solutions: Strong Acid/Strong Base
A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure $1$. Calculate the pH at these volumes of added base solution:
1. 0.00 mL
2. 12.50 mL
3. 25.00 mL
4. 37.50 mL
Solution
Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of H3O+ is $\ce{[H3O+]_0}=0.100\:M$. When the base solution is added, it also dissociates completely, providing OH ions. The H3O+ and OH ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic.
The total initial amount of the hydronium ions is:
$\mathrm{n(H^+)_0=[H_3O^+]_0×0.02500\: L=0.002500\: mol} \nonumber$
Once X mL of the 0.100-M base solution is added, the number of moles of the OH ions introduced is:
$\mathrm{n(OH^-)_0=0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The total volume becomes:
$V=\mathrm{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The number of moles of H3O+ becomes:
$\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0=0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)} \nonumber$
The concentration of H3O+ is:
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{V}=\dfrac{0.002500\: mol-0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber$
$\mathrm{=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×X\: mL}{25.00\: mL+X\: mL}} \nonumber$
with the definition of $\mathrm{pH}$:
$\mathrm{pH=−\log([H_3O^+])} \label{phdef}$
The preceding calculations work if $\mathrm{n(H^+)_0-n(OH^-)_0>0}$ and so n(H+) > 0. When $\mathrm{n(H^+)_0=n(OH^-)_0}$, the H3O+ ions from the acid and the OH ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH particles to neutralize them. Therefore, in this case:
$\ce{[H3O+]}=\ce{[OH- ]},\:\ce{[H3O+]}=K_\ce{w}=1.0\times 10^{-14};\:\ce{[H3O+]}=1.0\times 10^{-7} \nonumber$
$\mathrm{pH=-log(1.0\times 10^{-7})=7.00} \nonumber$
Finally, when $\mathrm{n(OH^-)_0>n(H^+)_0}$, there are not enough H3O+ ions to neutralize all the OH ions, and instead of $\mathrm{n(H^+)=n(H^+)_0-n(OH^-)_0}$, we calculate: $\mathrm{n(OH^-)=n(OH^-)_0-n(H^+)_0}$
In this case:
$\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×X\: mL×\left(\dfrac{1\: L}{1000\: mL}\right)-0.002500\: mol}{(25.00\: mL+X\: mL)\left(\dfrac{1\: L}{1000\: mL}\right)}} \nonumber$
$\mathrm{=\dfrac{0.100\:\mathit{M}×X\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+X\: mL}} \nonumber$
then using the definition of $pOH$ and its relationship to $pH$ in room temperature aqueous solutios (Equation \ref{phdef}):
\begin{align} pH &=14-pOH \nonumber \&=14+\log([OH^-]) \nonumber\end{align} \nonumber
Let us now consider the four specific cases presented in this problem:
(a) X = 0 mL
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL}=0.1\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align} pH &= −\log(0.100) \nonumber \ &= 1.000 \nonumber\end{align} \nonumber
(b) X = 12.50 mL
$\mathrm{[H_3O^+]=\dfrac{n(H^+)}{\mathit{V}}=\dfrac{0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)-0.100\:\mathit{M}×12.50\: mL}{25.00\: mL+12.50\: mL}=0.0333\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align} pH &= −\log(0.0333) \nonumber \ &= 1.477 \nonumber\end{align} \nonumber
(c) X = 25.00 mL
Since the volumes and concentrations of the acid and base solutions are the same:
$\mathrm{n(H^+)_0=n(OH^-)_0} \nonumber$
and
$pH = 7.000 \nonumber$
as described earlier.
(d) X = 37.50 mL
In this case:
$\mathrm{n(OH^-)_0>n(H^+)_0} \nonumber$
$\mathrm{[OH^-]=\dfrac{n(OH^-)}{\mathit{V}}=\dfrac{0.100\:\mathit{M}×35.70\: mL-0.002500\: mol×\left(\dfrac{1000\: mL}{1\: L}\right)}{25.00\: mL+37.50\: mL}=0.0200\:\mathit{M}} \nonumber$
then using the definition of $pH$ (Equation \ref{phdef}):
\begin{align}[pH = 14 − pOH \nonumber\ &= 14 + \log([OH^{−}]) \nonumber \ &= 14 + \log(0.0200) \nonumber \ &= 12.30 \nonumber \end{align} \nonumber
Exercise $1$
Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO3(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base:
1. 0.00 mL,
2. 15.0 mL,
3. 25.0 mL, and
4. 40.0 mL.
Answer a
0.00: 1.000
Answer b
15.0: 1.5111
Answer c
25.0: 7e. Do not delete this text first.
Answer d
40.0: 12.523
In Example $1$, we calculated pH at four points during a titration. Table $1$ shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH.
Table $1$: pH Values in the Titrations of a Strong Acid with a Strong Base and of a Weak Acid with a Strong Base
Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M $CH_3CO_2H$2
0.0 0.0 1.00 2.87
5.0 0.00050 1.18 4.14
10.0 0.00100 1.37 4.57
15.0 0.00150 1.60 4.92
20.0 0.00200 1.95 5.35
22.0 0.00220 2.20 5.61
24.0 0.00240 2.69 6.13
24.5 0.00245 3.00 6.44
24.9 0.00249 3.70 7.14
25.0 0.00250 7.00 8.72
25.1 0.00251 10.30 10.30
25.5 0.00255 11.00 11.00
26.0 0.00260 11.29 11.29
28.0 0.00280 11.75 11.75
30.0 0.00300 11.96 11.96
35.0 0.00350 12.22 12.22
40.0 0.00400 12.36 12.36
45.0 0.00450 12.46 12.46
50.0 0.00500 12.52 12.52
1. Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH.
2. Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH.
The simplest acid-base reactions are those of a strong acid with a strong base. Table $1$ shows data for the titration of a 25.0-mL sample of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. The values of the pH measured after successive additions of small amounts of NaOH are listed in the first column of this table, and are graphed in Figure $1$, in a form that is called a titration curve. The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The point of inflection (located at the midpoint of the vertical part of the curve) is the equivalence point for the titration. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.00 and the points on the titration curve can be calculated using solution stoichiometry (Table $1$ and Figure $1$).
The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. Let us consider the titration of 25.0 mL of 0.100 M acetic acid (a weak acid) with 0.100 M sodium hydroxide and compare the titration curve with that of the strong acid. Table $1$ gives the pH values during the titration, Figure $\PageIndex{1b}$ shows the titration curve.
Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(l)+\ce{OH-}(aq) \nonumber$
After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases.
Example $2$: Titration of a Weak Acid with a Strong Base
The titration curve shown in Figure $\PageIndex{1b}$ is for the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:
$\ce{CH3CO2H + OH- ⟶ CH3CO2- + H2O} \nonumber$
1. What is the initial pH before any amount of the NaOH solution has been added? Ka = 1.8 × 10−5 for CH3CO2H.
2. Find the pH after 25.00 mL of the NaOH solution have been added.
3. Find the pH after 12.50 mL of the NaOH solution has been added.
4. Find the pH after 37.50 mL of the NaOH solution has been added.
Solution
(a) Assuming that the dissociated amount is small compared to 0.100 M, we find that:
$K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}≈\ce{\dfrac{[H3O+]^2}{[CH3CO2H]_0}} \nonumber$
and
$\ce{[H3O+]}=\sqrt{K_\ce{a}×\ce{[CH3CO2H]}}=\sqrt{1.8\times 10^{-5}×0.100}=1.3\times 10^{-3} \nonumber$
$\mathrm{pH=-\log(1.3\times 10^{-3})=2.87} \nonumber$
(b) After 25.00 mL of NaOH are added, the number of moles of NaOH and CH3CO2H are equal because the amounts of the solutions and their concentrations are the same. All of the CH3CO2H has been converted to $\ce{CH3CO2-}$. The concentration of the $\ce{CH3CO2-}$ ion is:
$\mathrm{\dfrac{0.00250\: mol}{0.0500\: L}=0.0500\: \ce{MCH3CO2-}} \nonumber$
The equilibrium that must be focused on now is the basicity equilibrium for $\ce{CH3CO2-}$:
$\ce{CH3CO2-}(aq)+\ce{H2O}(l)⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) \nonumber$
so we must determine Kb for the base by using the ion product constant for water:
$K_\ce{b}=\ce{\dfrac{[CH3CO2H][OH- ]}{[CH3CO2- ]}} \nonumber$
$K_\ce{a}=\ce{\dfrac{[CH3CO2- ][H+]}{[CH3CO2H]}},\textrm{ so }\ce{\dfrac{[CH3CO2H]}{[CH3CO2- ]}}=\dfrac{\ce{[H+]}}{K_\ce{a}}. \nonumber$
Since Kw = [H+][OH]:
\begin{align} K_\ce{b} &=\dfrac{\ce{[H+][OH- ]}}{K_\ce{a}} \ &=\dfrac{K_\ce{w}}{K_\ce{a}} \ &=\dfrac{1.0\times 10^{-14}}{1.8\times 10^{-5}} \ &=5.6\times 10^{-10} \end{align} \nonumber
Let us denote the concentration of each of the products of this reaction, CH3CO2H and OH, as x. Using the assumption that x is small compared to 0.0500 M, $K_\ce{b}=\dfrac{x^2}{0.0500\:M}$, and then:
$x=\ce{[OH- ]}=5.3\times 10^{−6} \nonumber$
$\ce{pOH}=-\log(5.3\times 10^{-6})=5.28 \nonumber$
$\ce{pH}=14.00−5.28=8.72 \nonumber$
Note that the pH at the equivalence point of this titration is significantly greater than 7.
(c) In (a), 25.00 mL of the NaOH solution was added, and so practically all the CH3CO2H was converted into $\ce{CH3CO2-}$. In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH3CO2H is converted into $\ce{CH3CO2-}$. The total initial number of moles of CH3CO2H is 0.02500L × 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH3CO2H and $\ce{CH3CO2-}$ are both approximately equal to $\mathrm{\dfrac{0.00250\: mol}{2}=0.00125\: mol}$, and their concentrations are the same.
Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation:
$\ce{pH}=\ce p K_\ce{a}+\log\ce{\dfrac{[Base]}{[Acid]}}=-\log(\mathit{K}_\ce{a})+\log\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]}}=-\log(1.8\times 10^{-5})+\log(1)$
(as the concentrations of $\ce{CH3CO2-}$ and CH3CO2H are the same)
Thus:
$\ce{pH}=−\log(1.8\times 10^{−5})=4.74 \nonumber$
(the pH = the pKa at the halfway point in a titration of a weak acid)
(d) After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L × 0.100 M = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH:
$\mathrm{[OH^-]=\dfrac{(0.003750\: mol−0.00250\: mol)}{0.06250\: L}}=2.00\times 10^{−2}\:M$
So:
$\mathrm{pOH=−\log(2.00\times 10^{−2})=1.70,\: and\: pH=14.00−1.70=12.30}$
Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point.
Exercise $2$
Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base:
1. 0.00 mL,
2. 15.0 mL,
3. 25.0 mL, and
4. 30.0 mL.
Answer a
0.00 mL: 2.37
Answer b
15.0 mL: 3.92
Answer c
25.00 mL: 8.29
Answer d
30.0 mL: 12.097
Acid-Base Indicators
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10−9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10−9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid-base indicators are either weak organic acids or weak organic bases.
The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use $\ce{HIn}$ as a simple representation for the complex methyl orange molecule:
$\underbrace{\ce{HIn (aq)}}_{\ce{red}}+\ce{H2O (l)}⇌\ce{H3O^{+} (aq)}+\underbrace{\ce{In^{-} (aq)}}_{\ce{yellow}} \nonumber$
$K_\ce{a}=\ce{\dfrac{[H3O+][In- ]}{[HIn]}}=4.0\times 10^{−4} \nonumber$
The anion of methyl orange, $\ce{In^{-}}$, is yellow, and the nonionized form, $\ce{HIn}$, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
An indicator’s color is the visible result of the ratio of the concentrations of the two species In and $\ce{HIn}$. If most of the indicator (typically about 60−90% or more) is present as $\ce{In^{-}}$, then we see the color of the $\ce{In^{-}}$ ion, which would be yellow for methyl orange. If most is present as $\ce{HIn}$, then we see the color of the $\ce{HIn}$ molecule: red for methyl orange. For methyl orange, we can rearrange the equation for Ka and write:
$\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}} \label{ABeq2}$
Equation \ref{ABeq2} shows us how the ratio of $\ce{\dfrac{[In- ]}{[HIn]}}$ varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged:
\begin{align} \dfrac{[H_3O^+]}{\mathit{K}_a} &=\dfrac{[HIn]}{[In^- ]} \[8pt] \log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right) &= \log\left(\dfrac{[HIn]}{[In^- ]}\right) \[8pt] \log([H_3O^+])-\log(\mathit{K}_a) &=-\log\left(\dfrac{[In^-]}{[HIn]}\right) \[8pt] -pH+p\mathit{K}_a & =-\log\left(\dfrac{[In^-]}{[HIn]}\right) \[8pt] pH &=p\mathit{K}_a+\log\left(\dfrac{[In^-]}{[HIn]}\right) \end {align} \nonumber
or in general terms
$pH=p\mathit{K}_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{HHeq}$
Equation \ref{HHeq} is the same as the Henderson-Hasselbalch equation, which can be used to describe the equilibrium of indicators.
When [H3O+] has the same numerical value as Ka, the ratio of [In] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10−4 M (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10−5 M (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the color-change interval of methyl orange; the pronounced color change takes place between these pH values.
There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure $2$ presents several indicators, their colors, and their color-change intervals.
Titration curves help us pick an indicator that will provide a sharp color change at the equivalence point. The best selection would be an indicator that has a color change interval that brackets the pH at the equivalence point of the titration.
The color change intervals of three indicators are shown in Figure $3$. The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. We can use it for titrations of either strong acid with strong base or weak acid with strong base.
Litmus is a suitable indicator for the HCl titration because its color change brackets the equivalence point. However, we should not use litmus for the CH3CO2H titration because the pH is within the color-change interval of litmus when only about 12 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 13 mL of NaOH, making litmus useless as an indicator of the equivalence point.
We could use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color, as Figure $2$ demonstrates, during the addition of nearly 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein; and (3) it goes from yellow to orange to red, making detection of a precise endpoint much more challenging than the colorless to pink change of phenolphthalein. Figure $2$ shows us that methyl orange would be completely useless as an indicator for the CH3CO2H titration. Its color change begins after about 1 mL of NaOH has been added and ends when about 8 mL has been added. The color change is completed long before the equivalence point (which occurs when 25.0 mL of NaOH has been added) is reached and hence provides no indication of the equivalence point.
We base our choice of indicator on a calculated pH, the pH at the equivalence point. At the equivalence point, equimolar amounts of acid and base have been mixed, and the calculation becomes that of the pH of a solution of the salt resulting from the titration.
Summary
A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.
Glossary
acid-base indicator
organic acid or base whose color changes depending on the pH of the solution it is in
color-change interval
range in pH over which the color change of an indicator takes place
titration curve
plot of the pH of a solution of acid or base versus the volume of base or acid added during a titration | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.3%3A_Reaction_Stoichiometry_in_Solutions%3A_Acid-Base_Titrations.txt |
Analytical titrations using redox reactions were introduced shortly after the development of acid–base titrimetry. The earliest Redox titration took advantage of the oxidizing power of chlorine. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl2, HCl, and HOCl) based on its ability to oxidize indigo, a dye that is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developed a similar method for determining chlorine in bleaching powder. In both methods the end point is a change in color. Before the equivalence point the solution is colorless due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution.
The number of redox titrimetric methods increased in the mid-1800s with the introduction of MnO4, Cr2O72–, and I2 as oxidizing titrants, and of Fe2+ and S2O32– as reducing titrants. Even with the availability of these new titrants, redox titrimetry was slow to develop due to the lack of suitable indicators. A titrant can serve as its own indicator if its oxidized and reduced forms differ significantly in color. For example, the intensely purple MnO4 ion serves as its own indicator since its reduced form, Mn2+, is almost colorless. Other titrants require a separate indicator. The first such indicator, diphenylamine, was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry.
Redox Titration Curves
To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H3O+ (as pH) or Mn+ (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species.
You may recall from Chapter 6 that the Nernst equation relates a solution’s potential to the concentrations of reactants and products participating in the redox reaction. Consider, for example, a titration in which a titrand in a reduced state, Ared, reacts with a titrant in an oxidized state, Box.
$A_\textrm{red}+B_\textrm{ox} \rightleftharpoons B_\textrm{red}+A_\textrm{ox}$
where Aox is the titrand’s oxidized form, and Bred is the titrant’s reduced form. The reaction’s potential, Erxn, is the difference between the reduction potentials for each half-reaction.
$E_\textrm{rxn}=E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}$
After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical.
$E_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}=E_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}$
This is an important observation because we can use either half-reaction to monitor the titration’s progress.
Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction
$E_\textrm{rxn}= E^o_{A_\mathrm{\Large ox}/A_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[A_\textrm{red}]}{[A_\textrm{ox}]}$
Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent formal potential often is used in its place. See Appendix 13 for the standard state potentials and formal potentials for selected half-reactions.
After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction.
$E_\textrm{rxn}= E^o_{B_\mathrm{\Large ox}/B_\mathrm{\Large red}}-\dfrac{RT}{nF}\ln\dfrac{[B_\textrm{red}]}{[B_\textrm{ox}]}$
Calculating the Titration Curve
Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 M HClO4. The reaction in this case is
$\textrm{Fe}^{2+}(aq)+\textrm{Ce}^{4+}(aq)\rightleftharpoons \textrm{Ce}^{3+}(aq)+\textrm{Fe}^{3+}(aq)\tag{9.15}$
In 1 M HClO4, the formal potential for the reduction of Fe3+ to Fe2+ is +0.767 V, and the formal potential for the reduction of Ce4+ to Ce3+ is +1.70 V.
Because the equilibrium constant for reaction 9.15 is very large—it is approximately 6 × 1015—we may assume that the analyte and titrant react completely.
Step 1
Calculate the volume of titrant needed to reach the equivalence point.
The first task is to calculate the volume of Ce4+ needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that
$\textrm{moles Fe}^{2+}=\textrm{moles Ce}^{4+}$
$M_\textrm{Fe}\times V_\textrm{Fe} = M_\textrm{Ce}\times V_\textrm{Ce}$
Solving for the volume of Ce4+ gives the equivalence point volume as
$V_\textrm{eq} = V_\textrm{Ce} = \dfrac{M_\textrm{Fe}V_\textrm{Fe}}{M_\textrm{Ce}}=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{(0.100 M)}}=\textrm{50.0 mL}$
Step 2:
alculate the potential before the equivalence point by determining the concentrations of the titrand’s oxidized and reduced forms, and using the Nernst equation for the titrand’s reduction half-reaction.
Before the equivalence point, the concentration of unreacted Fe2+ and the concentration of Fe3+ are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe3+/Fe2+ half-reaction.
$E = E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} - \dfrac{RT}{nF}\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}=+0.767\textrm V - 0.05916\log\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}\tag{9.16}$
For example, the concentrations of Fe2+ and Fe3+ after adding 10.0 mL of titrant are
\begin{align} [\textrm{Fe}^{2+}]&=\dfrac{\textrm{initial moles Fe}^{2+} - \textrm{moles Ce}^{4+}\textrm{ added}}{\textrm{total volume}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe} - M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe}+V_\textrm{Ce}}\ &\mathrm{= \dfrac{(0.100\;M)(50.0\;mL)-(0.100\;M)(10.0\;mL)}{50.0\;mL+10.0\;mL} = 6.67\times10^{-2}\;M} \end{align}
\begin{align} [\mathrm{Fe^{3+}}]&=\mathrm{\dfrac{moles\;Ce^{4+}\;added}{total\;volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}}{V_\textrm{Fe} + V_\textrm{Ce}}\ &=\dfrac{\textrm{(0.100 M)(10.0 mL)}}{\textrm{50.0 mL + 10.0 mL}}=1.67\times10^{-2}\textrm{ M} \end{align}
Substituting these concentrations into equation 9.16 gives a potential of
$E = +0.767\textrm{ V} - 0.05916 \log\dfrac{6.67\times10^{-2}\textrm{ M}}{1.67\times10^{-2}\textrm{ M}}=+0.731\textrm{ V}$
Step 3:
Calculate the potential after the equivalence point by determining the concentrations of the titrant’s oxidized and reduced forms, and using the Nernst equation for the titrant’s reduction half-reaction.
After the equivalence point, the concentration of Ce3+ and the concentration of excess Ce4+ are easy to calculate. For this reason we find the potential using the Nernst equation for the Ce4+/Ce3+ half-reaction.
$E=E^o_\mathrm{\large{Ce^{4+}/Ce^{3+}}}-\dfrac{RT}{nF}\log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}=+ 1.70\textrm{ V} - 0.05916 \log\mathrm{\dfrac{[Ce^{3+}]}{[Ce^{4+}]}}\tag{9.17}$
For example, after adding 60.0 mL of titrant, the concentrations of Ce3+ and Ce4+ are
\begin{align} [\textrm{Ce}^{3+}]&={\dfrac{\textrm{initial moles Fe}^{2+}}{\textrm{total volume}}}=\dfrac{M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\ &=\dfrac{\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=4.55\times10^{-3}\textrm{ M} \end{align}
\begin{align} [\textrm{Ce}^{4+}]&=\dfrac{\textrm{moles Ce}^{4+}\textrm{ added} - \textrm{initial moles Fe}^{2+}}{\textrm{total volume}}=\dfrac{M_\textrm{Ce}V_\textrm{Ce}-M_\textrm{Fe}V_\textrm{Fe}}{V_\textrm{Fe}+V_\textrm{Ce}}\ &=\dfrac{\textrm{(0.100 M)(60.0 mL)}-\textrm{(0.100 M)(50.0 mL)}}{\textrm{50.0 mL + 60.0 mL}}=9.09\times10^{-3}\textrm{ M} \end{align}
Substituting these concentrations into Equation 9.17 gives a potential of
$E=+1.70\textrm{ V}-0.05916\log\dfrac{4.55\times10^{-2}\textrm{ M}}{9.09\times10^{-3}\textrm{ M}}=+1.66\textrm{ V}$
Step 4
Calculate the potential at the equivalence point.
At the titration’s equivalence point, the potential, Eeq, in equation 9.16 and equation 9.17 are identical. Adding the equations together to gives
$2E_\textrm{eq}= E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}-0.05916\log\dfrac{\mathrm{[{Fe}^{2+}][Ce^{3+}]}}{\mathrm{[Fe^{3+}][Ce^{4+}]}}$
Because [Fe2+] = [Ce4+] and [Ce3+] = [Fe3+] at the equivalence point, the log term has a value of zero and the equivalence point’s potential is
$E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} + E^o_\mathrm{\large Ce^{4+}/Ce^{3+}}}{2}=\dfrac{\textrm{0.767 V + 1.70 V}}{2}=1.23\textrm{ V}$
Additional results for this titration curve are shown in Table 9.15 and Figure 9.36.
Table 9.15: Data for the Titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+
Volume of Ce4+ (mL) E (V) Volume Ce4+ (mL) E (V)
10.0 0.731 60.0 1.66
20.0 0.757 70.0 1.68
30.0 0.777 80.0 1.69
40.0 0.803 90.0 1.69
50.0 1.23 100.0 1.70
Figure 9.36 Titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+. The red points correspond to the data in Table 9.15. The blue line shows the complete titration curve.
Exercise $1$
Calculate the titration curve for the titration of 50.0 mL of 0.0500 M Sn2+ with 0.100 M Tl3+. Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is
$\textrm{Sn}^{2+}(aq)+\textrm{Tl}^{3+}(aq)\rightarrow \textrm{Sn}^{4+}(aq)+\textrm{Tl}^+(aq)$
Click here to review your answer to this exercise.
Sketching a Redox Titration Curve
To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a redox titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 M HClO4.
This is the same example that we used in developing the calculations for a redox titration curve. You can review the results of that calculation in Table 9.15 and Figure 9.36.
We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 50.0 mL. Next, we draw our axes, placing the potential, E, on the y-axis and the titrant’s volume on the x-axis. To indicate the equivalence point’s volume, we draw a vertical line corresponding to 50.0 mL of Ce4+. Figure 9.37a shows the result of the first step in our sketch.
Before the equivalence point, the potential is determined by a redox buffer of Fe2+ and Fe3+. Although we can easily calculate the potential using the Nernst equation, we can avoid this calculation by making a simple assumption. You may recall from Chapter 6 that a redox buffer operates over a range of potentials that extends approximately ±(0.05916/n) unit on either side of EoFe3+/Fe2+. The potential is at the buffer’s lower limit
$\textrm E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}-0.05916$
when the concentration of Fe2+ is 10× greater than that of Fe3+. The buffer reaches its upper potential
$\textrm E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+0.05916$
when the concentration of Fe2+ is 10× smaller than that of Fe3+. The redox buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume.
Figure 9.37b shows the second step in our sketch. First, we superimpose a ladder diagram for Fe2+ on the y-axis, using its EoFe3+/Fe2+ value of 0.767 V and including the buffer’s range of potentials. Next, we add points representing the pH at 10% of the equivalence point volume (a potential of 0.708 V at 5.0 mL) and at 90% of the equivalence point volume (a potential of 0.826 V at 45.0 mL).
We used a similar approach when sketching the acid–base titration curve for the titration of acetic acid with NaOH.
The third step in sketching our titration curve is to add two points after the equivalence point. Here the potential is controlled by a redox buffer of Ce3+ and Ce4+. The redox buffer is at its lower limit of E = EoCe4+/Ce3+ – 0.05916 when the titrant reaches 110% of the equivalence point volume and the potential is EoCe4+/Ce3+ when the volume of Ce4+ is 2×Veq.
Figure 9.37c shows the third step in our sketch. First, we add a ladder diagram for Ce4+, including its buffer range, using its EoCe4+/Ce3+ value of 1.70 V. Next, we add points representing the potential at 110% of Veq (a value of 1.66 V at 55.0 mL) and at 200% of Veq (a value of 1.70 V at 100.0 mL).
We used a similar approach when sketching the complexation titration curve for the titration of Mg2+ with EDTA.
Next, we draw a straight line through each pair of points, extending the line through the vertical line representing the equivalence point’s volume (Figure 9.37d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.37e). A comparison of our sketch to the exact titration curve (Figure 9.37f) shows that they are in close agreement.
Figure 9.37: Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in 1 M HClO4: (a) locating the equivalence point volume; (b) plotting two points before the equivalence point; (c) plotting two points after the equivalence point; (d) preliminary approximation of titration curve using straight-lines; (e) final approximation of titration curve using a smooth curve; (f) comparison of approximate titration curve (solid black line) and exact titration curve (dashed red line). See the text for additional details.
Exercise $2$
Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn4+ with 0.100 M Tl+. Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is
$\textrm{Sn}^{2+}(aq)+\textrm{Tl}^{3+}(aq)\rightarrow\textrm{Sn}^{4+}(aq)+\textrm{Tl}^+(aq)$
Compare your sketch to your calculated titration curve from Practice Exercise 9.17.
Click here to review your answer to this exercise.
9.4.2 Selecting and Evaluating the End point
A redox titration’s equivalence point occurs when we react stoichiometrically equivalent amounts of titrand and titrant. As is the case with acid–base and complexation titrations, we estimate the equivalence point of a complexation titration using an experimental end point. A variety of methods are available for locating the end point, including indicators and sensors that respond to a change in the solution conditions.
Where is the Equivalence Point?
For an acid–base titration or a complexometric titration the equivalence point is almost identical to the inflection point on the steeping rising part of the titration curve. If you look back at Figure 9.7 and Figure 9.28, you will see that the inflection point is in the middle of this steep rise in the titration curve, which makes it relatively easy to find the equivalence point when you sketch these titration curves. We call this a symmetric equivalence point. If the stoichiometry of a redox titration is symmetric—one mole of titrant reacts with each mole of titrand—then the equivalence point is symmetric. If the titration reaction’s stoichiometry is not 1:1, then the equivalence point is closer to the top or to bottom of the titration curve’s sharp rise. In this case we have an asymmetric equivalence point.
Example $1$
Derive a general equation for the equivalence point’s potential when titrating Fe2+ with MnO4.
$5\textrm{Fe}^{2+}(aq)+\textrm{MnO}_4^-(aq)+8\textrm H^+(aq)\rightarrow 5\textrm{Fe}^{3+}(aq)+\textrm{Mn}^{2+}(aq)+\mathrm{4H_2O}$
(We often use H+ instead of H3O+ when writing a redox reaction.)
Solution
The half-reactions for Fe2+ and MnO4 are
$\textrm{Fe}^{2+}(aq)\rightarrow\textrm{Fe}^{3+}(aq)+e^-$
$\textrm{MnO}_4^-(aq)+8\textrm H^+(aq)+5e^-\rightarrow \textrm{Mn}^{2+}(aq)+4\mathrm{H_2O}(l)$
for which the Nernst equations are
$E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}-0.05916\log\dfrac{[\textrm{Fe}^{2+}]}{[\textrm{Fe}^{3+}]}$
$E=E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-\dfrac{0.05916}{5}\log\dfrac{[\textrm{Mn}^{2+}]}{\ce{[MnO_4^- ][H^+]^8}}$
Before adding these two equations together we must multiply the second equation by 5 so that we can combine the log terms; thus
$6E=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-0.05916\log\mathrm{\dfrac{[Fe^{2+}][Mn^{2+}]}{[Fe^{3+}][\ce{MnO_4^-}][H^+]^8}}$
At the equivalence point we know that
$[\textrm{Fe}^{2+}]=5\times[\textrm{MnO}_4^-]$
$[\textrm{Fe}^{3+}]=5\times[\textrm{Mn}^{2+}]$
Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point.
$6E_\textrm{eq}=E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}-0.05916\log\mathrm{\dfrac{5[\ce{MnO_4^-}][Mn^{2+}]}{5[Mn^{2+}][\ce{MnO_4^-}][H^+]^8}}$
$E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}} + 5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}-\dfrac{0.05916}{6}\log\dfrac{1}{[\textrm H^+]^8}$
$E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}+\dfrac{0.05916\times8}{6}\log[\textrm H^+]$
$E_\textrm{eq}=\dfrac{E^o_\mathrm{\large Fe^{3+}/Fe^{2+}}+5E^o_\mathrm{\large MnO_4^-/Mn^{2+}}}{6}-0.07888\textrm{pH}$
Our equation for the equivalence point has two terms. The first term is a weighted average of the titrand’s and the titrant’s standard state potentials, in which the weighting factors are the number of electrons in their respective half-reactions. (Instead of standard state potentials, you can use formal potentials.) The second term shows that Eeq for this titration is pH-dependent. At a pH of 1 (in H2SO4), for example, the equivalence point has a potential of
$E_\textrm{eq}=\dfrac{0.768+5\times1.51}{6}-0.07888\times1=1.31\textrm{ V}$
Figure 9.38 shows a typical titration curve for titration of Fe2+ with MnO4. Note that the titration’s equivalence point is asymmetrical.
Figure 9.38: Titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.0200 M MnO4 at a fixed pH of 1 (using H2SO4). The equivalence point is shown by the red dot.
Exercise $3$
Derive a general equation for the equivalence point’s potential for the titration of U4+ with Ce4+. The unbalanced reaction is
$\textrm{Ce}^{4+}(aq)+\textrm U^{4+}(aq)\rightarrow \textrm{UO}_2^{2+}(aq)+\textrm{Ce}^{3+}(aq)$
What is the equivalence point’s potential if the pH is 1?
Click here to review your answer to this exercise.
Finding the End point with an Indicator
Three types of indicators are used to signal a redox titration’s end point. The oxidized and reduced forms of some titrants, such as MnO4, have different colors. A solution of MnO4 is intensely purple. In an acidic solution, however, permanganate’s reduced form, Mn2+, is nearly colorless. When using MnO4 as a titrant, the titrand’s solution remains colorless until the equivalence point. The first drop of excess MnO4 produces a permanent tinge of purple, signaling the end point.
Some indicators form a colored compound with a specific oxidized or reduced form of the titrant or the titrand. Starch, for example, forms a dark blue complex with I3. We can use this distinct color to signal the presence of excess I3 as a titrant—a change in color from colorless to blue—or the completion of a reaction consuming I3 as the titrand—a change in color from blue to colorless. Another example of a specific indicator is thiocyanate, SCN, which forms a soluble red-colored complex of Fe(SCN)2+ with Fe3+.
The most important class of indicators are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. When we add a redox indicator to the titrand, the indicator imparts a color that depends on the solution’s potential. As the solution’s potential changes with the addition of titrant, the indicator changes oxidation state and changes color, signaling the end point.
To understand the relationship between potential and an indicator’s color, consider its reduction half-reaction
$\mathrm{In_{ox}}+ne^-\rightleftharpoons \mathrm{In_{red}}$
where Inox and Inred are, respectively, the indicator’s oxidized and reduced forms.
For simplicity, Inox and Inred are shown without specific charges. Because there is a change in oxidation state, Inox and Inred cannot both be neutral.
The Nernst equation for this half-reaction is
$E=E^o_\mathrm{In_{\large ox}/In_{\large red}}-\dfrac{0.05916}{n}\log\mathrm{\dfrac{[In_{red}]}{[In_{ox}]}}$
As shown in Figure 9.39, if we assume that the indicator’s color changes from that of Inox to that of Inred when the ratio [Inred]/[Inox] changes from 0.1 to 10, then the end point occurs when the solution’s potential is within the range
$E=E^o_\mathrm{In_{\large ox}/In_{\large red}}\pm\dfrac{0.05916}{n}$
This is the same approach we took in considering acid–base indicators and complexation indicators.
Figure 9.39 Diagram showing the relationship between E and an indicator’s color. The ladder diagram defines potentials where Inred and Inox are the predominate species. The indicator changes color when E is within the range
E = EoInox/Inred ± 0.05916/n
A partial list of redox indicators is shown in Table 9.16. Examples of appropriate and inappropriate indicators for the titration of Fe2+ with Ce4+ are shown in Figure 9.40.
Table 9.16 Selected Examples of Redox Indicators
Indicator Color of Inox Color of Inred EoInox/Inred
indigo tetrasulfate blue colorless 0.36
methylene blue blue colorless 0.53
diphenylamine violet colorless 0.75
diphenylamine sulfonic acid red-violet colorless 0.85
tris(2,2´-bipyridine)iron pale blue red 1.120
ferroin pale blue red 1.147
tris(5-nitro-1,10-phenanthroline)iron pale blue red-violet 1.25
Figure 9.40: Titration curve for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+. The end point transitions for the indicators diphenylamine sulfonic acid and ferroin are superimposed on the titration curve. Because the transition for ferroin is too small to see on the scale of the x-axis—it requires only 1–2 drops of titrant—the color change is expanded to the right.
Other Methods for Finding the End point
Another method for locating a redox titration’s end point is a potentiometric titration in which we monitor the change in potential while adding the titrant to the titrand. The end point is found by visually examining the titration curve. The simplest experimental design for a potentiometric titration consists of a Pt indicator electrode whose potential is governed by the titrand’s or titrant’s redox half-reaction, and a reference electrode that has a fixed potential. A further discussion of potentiometry is found in Chapter 11. Other methods for locating the titration’s end point include thermometric titrations and spectrophotometric titrations.
The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical redox titrimetric method. Although each method is unique, the following description of the determination of the total chlorine residual in water provides an instructive example of a typical procedure. The description here is based on Method 4500-Cl B as published in Standard Methods for the Examination of Water and Wastewater, 20th Ed., American Public Health Association: Washington, D. C., 1998.
Representative Method 9.3: Determination of Total Chlorine Residual
Description of the Method
The chlorination of public water supplies produces several chlorine-containing species, the combined concentration of which is called the total chlorine residual. Chlorine may be present in a variety of states, including the free residual chlorine, consisting of Cl2, HOCl and OCl, and the combined chlorine residual, consisting of NH2Cl, NHCl2, and NCl3. The total chlorine residual is determined by using the oxidizing power of chlorine to convert I to I3. The amount of I3 formed is then determined by titrating with Na2S2O3 using starch as an indicator. Regardless of its form, the total chlorine residual is reported as if Cl2 is the only source of chlorine, and is reported as mg Cl/L.
Procedure
Select a volume of sample requiring less than 20 mL of Na2S2O3 to reach the end point. Using glacial acetic acid, acidify the sample to a pH of 3–4, and add about 1 gram of KI. Titrate with Na2S2O3 until the yellow color of I3 begins to disappear. Add 1 mL of a starch indicator solution and continue titrating until the blue color of the starch–I3 complex disappears (Figure 9.41). Use a blank titration to correct the volume of titrant needed to reach the end point for reagent impurities.
Questions
1. Is this an example of a direct or an indirect analysis?
This is an indirect analysis because the chlorine-containing species do not react with the titrant. Instead, the total chlorine residual oxidizes I to I3, and the amount of I3 is determined by titrating with Na2S2O3.
2. Why does the procedure rely on an indirect analysis instead of directly titrating the chlorine-containing species using KI as a titrant?
Because the total chlorine residual consists of six different species, a titration with I does not have a single, well-defined equivalence point. By converting the chlorine residual to an equivalent amount of I3, the indirect titration with Na2S2O3 has a single, useful equivalence point.
Even if the total chlorine residual is from a single species, such as HOCl, a direct titration with KI is impractical. Because the product of the titration, I3, imparts a yellow color, the titrand’s color would change with each addition of titrant, making it difficult to find a suitable indicator.
3. Both oxidizing and reducing agents can interfere with this analysis. Explain the effect of each type of interferent has on the total chlorine residual.
An interferent that is an oxidizing agent converts additional I to I3. Because this extra I3 requires an additional volume of Na2S2O3 to reach the end point, we overestimate the total chlorine residual. If the interferent is a reducing agent, it reduces back to I some of the I3 produced by the reaction between the total chlorine residual and iodide. As a result. we underestimate the total chlorine residual.
Figure 9.41 Endpoint for the determination of the total chlorine residual. (a) Acidifying the sample and adding KI forms a brown solution of I3. (b) Titrating with Na2S2O3 converts I3 to I with the solution fading to a pale yellow color as we approach the end point. (c) Adding starch forms the deep purple starch–I3 complex. (d) As the titration continues, the end point is a sharp transition from a purple to a colorless solution. The change in color from (c) to (d) typically takes 1–2 drops of titrant.
9.4.3 Quantitative Applications
Although many quantitative applications of redox titrimetry have been replaced by other analytical methods, a few important applications continue to be relevant. In this section we review the general application of redox titrimetry with an emphasis on environmental, pharmaceutical, and industrial applications. We begin, however, with a brief discussion of selecting and characterizing redox titrants, and methods for controlling the titrand’s oxidation state.
Adjusting the Titrand’s Oxidation State
If a redox titration is to be used in a quantitative analysis, the titrand must initially be present in a single oxidation state. For example, iron can be determined by a redox titration in which Ce4+ oxidizes Fe2+ to Fe3+. Depending on the sample and the method of sample preparation, iron may initially be present in both the +2 and +3 oxidation states. Before titrating, we must reduce any Fe3+ to Fe2+. This type of pretreatment can be accomplished using an auxiliary reducing agent or oxidizing agent.
A metal that is easy to oxidize—such as Zn, Al, and Ag—can serve as an auxiliary reducing agent. The metal, as a coiled wire or powder, is added to the sample where it reduces the titrand. Because any unreacted auxiliary reducing agent will react with the titrant, it must be removed before beginning the titration. This can be accomplished by simply removing the coiled wire, or by filtering.
An alternative method for using an auxiliary reducing agent is to immobilize it in a column. To prepare a reduction column an aqueous slurry of the finally divided metal is packed in a glass tube equipped with a porous plug at the bottom. The sample is placed at the top of the column and moves through the column under the influence of gravity or vacuum suction. The length of the reduction column and the flow rate are selected to ensure the analyte’s complete reduction.
Two common reduction columns are used. In the Jones reductor the column is filled with amalgamated zinc, Zn(Hg), prepared by briefly placing Zn granules in a solution of HgCl2. Oxidation of zinc
$\textrm{Zn(Hg)}(s)\rightarrow \textrm{Zn}^{2+}(aq)+\textrm{Hg}(l)+2e^-$
provides the electrons for reducing the titrand. In the Walden reductor the column is filled with granular Ag metal. The solution containing the titrand is acidified with HCl and passed through the column where the oxidation of silver
$\textrm{Ag}(s)+\textrm{Cl}^-(aq)\rightarrow \textrm{AgCl}(s)+e^-$
provides the necessary electrons for reducing the titrand. Table 9.17 provides a summary of several applications of reduction columns.
Table 9.17 Examples of Reactions For Reducing a Titrand’s Oxidation State Using a Reduction Column
Oxidized Titrand Walden Reductor Jones Reductor
Cr3+ Cr3+(aq) + e → Cr2+(aq)
Cu2+ Cu2+(aq) +e → Cu+(aq) Cu2+(aq) + 2e → Cr(s)
Fe3+ Fe3+(aq) + e → Fe2+(aq) Fe3+(aq) + e → Fe2+(aq)
TiO2+
TiO2+(aq) + 2H+(aq) + e → Ti3+(aq) + H2O(l)
MoO22+ MoO22+(aq) + e→ MoO2+(aq) MoO22+(aq) + 4H+(aq) + 3e→ Mo3+(aq) + 2H2O(l)
VO2+ VO2+(aq) + 2H+(aq) + e → VO2+(aq) + H2O(l) VO2+(aq) + 4H+(aq) + 3e → V2+(aq) + 2H2O(l)
Several reagents are commonly used as auxiliary oxidizing agents, including ammonium peroxydisulfate, (NH4)2S2O8, and hydrogen peroxide, H2O2. Peroxydisulfate is a powerful oxidizing agent
$\mathrm{S_2O_8^{2-}}(aq)+2e^-\rightarrow\mathrm{2SO_4^{2-}}(aq)$
capable of oxidizing Mn2+ to MnO4, Cr3+ to Cr2O72–, and Ce3+ to Ce4+. Excess peroxydisulfate is easily destroyed by briefly boiling the solution. The reduction of hydrogen peroxide in acidic solution
$\mathrm{H_2O_2}(aq)+\mathrm{2H^+}(aq)+2e^-\rightarrow\mathrm{2H_2O}(l)$
provides another method for oxidizing a titrand. Excess H2O2 is destroyed by briefly boiling the solution.
Selecting and Standardizing a Titrant
If it is to be used quantitatively, the titrant’s concentration must remain stable during the analysis. Because a titrant in a reduced state is susceptible to air oxidation, most redox titrations use an oxidizing agent as the titrant. There are several common oxidizing titrants, including MnO4, Ce4+, Cr2O72–, and I3. Which titrant is used often depends on how easy it is to oxidize the titrand. A titrand that is a weak reducing agent needs a strong oxidizing titrant if the titration reaction is to have a suitable end point.
The two strongest oxidizing titrants are MnO4 and Ce4+, for which the reduction half-reactions are
$\ce{MnO_4^-}(aq)+\mathrm{8H^+}(aq)+5e^-\rightleftharpoons \mathrm{Mn^{2+}}(aq)+\mathrm{4H_2O}(l)$
$\textrm{Ce}^{4+}(aq)+e^-\rightleftharpoons \textrm{Ce}^{3+}(aq)$
Solutions of Ce4+ usually are prepared from the primary standard cerium ammonium nitrate, Ce(NO3)4•2NH4NO3, in 1 M H2SO4. When prepared using a reagent grade material, such as Ce(OH)4, the solution is standardized against a primary standard reducing agent such as Na2C2O4 or Fe2+ (prepared using iron wire) using ferroin as an indicator. Despite its availability as a primary standard and its ease of preparation, Ce4+ is not as frequently used as MnO4– because it is more expensive.
Note
The standardization reactions are
$\mathrm{Ce^{4+}}(aq)+\mathrm{Fe^{2+}}(aq)\rightarrow \mathrm{Ce^{3+}}(aq)+\mathrm{Fe^{3+}}(aq)$
$\mathrm{2Ce^{4+}}(aq)+\mathrm{H_2C_2O_4}(aq)\rightarrow \mathrm{2Ce^{3+}}(aq)+\mathrm{2CO_2}(g)+\mathrm{2H^+}(aq)$
Solutions of MnO4 are prepared from KMnO4, which is not available as a primary standard. Aqueous solutions of permanganate are thermodynamically unstable due to its ability to oxidize water.
$\ce{4MnO_4^-}(aq)+\mathrm{2H_2O}(l)\rightleftharpoons\mathrm{4MnO_2}(s)+\mathrm{3O_2}(g)+\mathrm{4OH^-}(aq)$
This reaction is catalyzed by the presence of MnO2, Mn2+, heat, light, and the presence of acids and bases. A moderately stable solution of permanganate can be prepared by boiling it for an hour and filtering through a sintered glass filter to remove any solid MnO2 that precipitates. Standardization is accomplished against a primary standard reducing agent such as Na2C2O4 or Fe2+ (prepared using iron wire), with the pink color of excess MnO4 signaling the end point. A solution of MnO4 prepared in this fashion is stable for 1–2 weeks, although the standardization should be rechecked periodically.
Note
The standardization reactions are
$\ce{MnO_4^-}(aq)+\mathrm{5Fe^{2+}}(aq)+\mathrm{8H^+}(aq)\rightarrow \mathrm{Mn^{2+}}(aq)+\mathrm{5Fe^{3+}}(aq)+\mathrm{4H_2O}(l)$
$\ce{2MnO_4^-}(aq)+\mathrm{5H_2C_2O_4}(aq)+\mathrm{6H^+}(aq)\rightarrow\mathrm{2Mn^{2+}}(aq)+\mathrm{10CO_2}(g)+\mathrm{8H_2O}(l)$
Potassium dichromate is a relatively strong oxidizing agent whose principal advantages are its availability as a primary standard and the long term stability of its solutions. It is not, however, as strong an oxidizing agent as MnO4 or Ce4+, which makes it less useful when the titrand is a weak reducing agent. Its reduction half-reaction is
$\mathrm{Cr_2O_7^{2-}}(aq)+\mathrm{14H^+}(aq)+6e^-\rightleftharpoons \mathrm{2Cr^{3+}}(aq)+\mathrm{7H_2O}(l)$
Although a solution of Cr2O72– is orange and a solution of Cr3+ is green, neither color is intense enough to serve as a useful indicator. Diphenylamine sulfonic acid, whose oxidized form is red-violet and reduced form is colorless, gives a very distinct end point signal with Cr2O72–.
Iodine is another important oxidizing titrant. Because it is a weaker oxidizing agent than MnO4, Ce4+, and Cr2O72–, it is useful only when the titrand is a stronger reducing agent. This apparent limitation, however, makes I2 a more selective titrant for the analysis of a strong reducing agent in the presence of a weaker reducing agent. The reduction half-reaction for I2 is
$\textrm I_2(aq) + 2e^-\rightleftharpoons 2\textrm I^-(aq)$
Because iodine is not very soluble in water, solutions are prepared by adding an excess of I. The complexation reaction
$\textrm I_2(aq)+\textrm I^-(aq)\rightleftharpoons\textrm I_3^-(aq)$
increases the solubility of I2 by forming the more soluble triiodide ion, I3. Even though iodine is present as I3 instead of I2, the number of electrons in the reduction half-reaction is unaffected.
$\textrm I_3^-(aq)+2e^-\rightleftharpoons 3\textrm I^-(aq)$
Solutions of I3 are normally standardized against Na2S2O3 using starch as a specific indicator for I3.
Note
The standardization reaction is
$\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow 3\textrm I^-(aq)+\mathrm{2S_4O_6^{2-}}(aq)$
An oxidizing titrant such as MnO4, Ce4+, Cr2O72–, and I3, is used when the titrand is in a reduced state. If the titrand is in an oxidized state, we can first reduce it with an auxiliary reducing agent and then complete the titration using an oxidizing titrant. Alternatively, we can titrate it using a reducing titrant. Iodide is a relatively strong reducing agent that could serve as a reducing titrant except that a solution of I– is susceptible to the air-oxidation of I to I3.
$3\textrm I^-(aq)\rightleftharpoons \mathrm I_3^-(aq)+2e^-$
Note
A freshly prepared solution of KI is clear, but after a few days it may show a faint yellow coloring due to the presence of I3.
Instead, adding an excess of KI reduces the titrand, releasing a stoichiometric amount of I3. The amount of I3 produced is then determined by a back titration using thiosulfate, S2O32–, as a reducing titrant.
$\mathrm{2S_2O_3^{2-}}(aq)\rightleftharpoons\mathrm{2S_4O_6^{2-}}(aq)+2e^-$
Solutions of S2O32– are prepared using Na2S2O3•5H2O, and must be standardized before use. Standardization is accomplished by dissolving a carefully weighed portion of the primary standard KIO3 in an acidic solution containing an excess of KI. The reaction between IO3 and I
$\textrm{IO}_3^-(aq)+8\textrm I^-(aq)+6\textrm H^+(aq)\rightarrow \ce{3I_3^-}(aq)+\mathrm{3H_2O}(l)$
liberates a stoichiometric amount of I3. By titrating this I3 with thiosulfate, using starch as a visual indicator, we can determine the concentration of S2O32– in the titrant.
Note
The standardization titration is
$\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow 3\textrm I^-(aq)+\mathrm{2S_4O_6^{2-}}(aq)$
which is the same reaction used to standardize solutions of I3. This approach to standardizing solutions of S2O32−. is similar to the determination of the total chlorine residual outlined in Representative Method 9.3.
Although thiosulfate is one of the few reducing titrants that is not readily oxidized by contact with air, it is subject to a slow decomposition to bisulfite and elemental sulfur. If used over a period of several weeks, a solution of thiosulfate should be restandardized periodically. Several forms of bacteria are able to metabolize thiosulfate, which also can lead to a change in its concentration. This problem can be minimized by adding a preservative such as HgI2 to the solution.
Another useful reducing titrant is ferrous ammonium sulfate, Fe(NH4)2(SO4)2•6H2O, in which iron is present in the +2 oxidation state. A solution of Fe2+ is susceptible to air-oxidation, but when prepared in 0.5 M H2SO4 it remains stable for as long as a month. Periodic restandardization with K2Cr2O7 is advisable. The titrant can be used to directly titrate the titrand by oxidizing Fe2+ to Fe3+. Alternatively, ferrous ammonium sulfate is added to the titrand in excess and the quantity of Fe3+ produced determined by back titrating with a standard solution of Ce4+ or Cr2O72–.
Inorganic Analysis
One of the most important applications of redox titrimetry is evaluating the chlorination of public water supplies. Representative Method 9.3, for example, describes an approach for determining the total chlorine residual by using the oxidizing power of chlorine to oxidize I to I3. The amount of I3 is determined by back titrating with S2O32–.
The efficiency of chlorination depends on the form of the chlorinating species. There are two contributions to the total chlorine residual—the free chlorine residual and the combined chlorine residual. The free chlorine residual includes forms of chlorine that are available for disinfecting the water supply. Examples of species contributing to the free chlorine residual include Cl2, HOCl and OCl. The combined chlorine residual includes those species in which chlorine is in its reduced form and, therefore, no longer capable of providing disinfection. Species contributing to the combined chlorine residual are NH2Cl, NHCl2 and NCl3.
When a sample of iodide-free chlorinated water is mixed with an excess of the indicator N,N-diethyl-p-phenylenediamine (DPD), the free chlorine oxidizes a stoichiometric portion of DPD to its red-colored form. The oxidized DPD is then back titrated to its colorless form using ferrous ammonium sulfate as the titrant. The volume of titrant is proportional to the free residual chlorine.
Having determined the free chlorine residual in the water sample, a small amount of KI is added, catalyzing the reduction monochloramine, NH2Cl, and oxidizing a portion of the DPD back to its red-colored form. Titrating the oxidized DPD with ferrous ammonium sulfate yields the amount of NH2Cl in the sample. The amount of dichloramine and trichloramine are determined in a similar fashion.
The methods described above for determining the total, free, or combined chlorine residual also are used to establish a water supply’s chlorine demand. Chlorine demand is defined as the quantity of chlorine needed to completely react with any substance that can be oxidized by chlorine, while also maintaining the desired chlorine residual. It is determined by adding progressively greater amounts of chlorine to a set of samples drawn from the water supply and determining the total, free, or combined chlorine residual.
Another important example of redox titrimetry, which finds applications in both public health and environmental analyses is the determination of dissolved oxygen. In natural waters, such as lakes and rivers, the level of dissolved O2 is important for two reasons: it is the most readily available oxidant for the biological oxidation of inorganic and organic pollutants; and it is necessary for the support of aquatic life. In a wastewater treatment plant dissolved O2 is essential for the aerobic oxidation of waste materials. If the concentration of dissolved O2 falls below a critical value, aerobic bacteria are replaced by anaerobic bacteria, and the oxidation of organic waste produces undesirable gases, such as CH4 and H2S.
One standard method for determining the dissolved O2 content of natural waters and wastewaters is the Winkler method. A sample of water is collected without exposing it to the atmosphere, which might change the concentration of dissolved O2. The sample is first treated with a solution of MnSO4, and then with a solution of NaOH and KI. Under these alkaline conditions the dissolved oxygen oxidizes Mn2+ to MnO2.
$\mathrm{2Mn^{2+}}(aq)+\mathrm{4OH^-}(aq)+\mathrm O_2(g)\rightarrow \mathrm{2MnO_2}(s)+\mathrm{2H_2O}(l)$
After the reaction is complete, the solution is acidified with H2SO4. Under the now acidic conditions I is oxidized to I3 by MnO2.
$\mathrm{MnO_2}(s)+\mathrm{3I^-}(aq)+\mathrm{4H^+}(aq)\rightarrow \mathrm{Mn^{2+}}+\ce{I_3^-}(aq)+\mathrm{2H_2O}(l)$
The amount of I3 formed is determined by titrating with S2O32– using starch as an indicator. The Winkler method is subject to a variety of interferences, and several modifications to the original procedure have been proposed. For example, NO2 interferes because it can reduce I3 to I under acidic conditions. This interference is eliminated by adding sodium azide, NaN3, reducing NO2 to N2. Other reducing agents, such as Fe2+, are eliminated by pretreating the sample with KMnO4, and destroying the excess permanganate with K2C2O4.
Another important example of redox titrimetry is the determination of water in nonaqueous solvents. The titrant for this analysis is known as the Karl Fischer reagent and consists of a mixture of iodine, sulfur dioxide, pyridine, and methanol. Because the concentration of pyridine is sufficiently large, I2 and SO2 react with pyridine (py) to form the complexes py•I2 and py•SO2. When added to a sample containing water, I2 is reduced to I and SO2 is oxidized to SO3.
$\textrm{py}\bullet\textrm I_2+\textrm{py}\bullet\mathrm{SO_2}+\textrm{py}+\mathrm{H_2O}\rightarrow 2\textrm{py}\bullet\textrm{HI}+\textrm{py}\bullet\mathrm{SO_3}$
Methanol is included to prevent the further reaction of py•SO3 with water. The titration’s end point is signaled when the solution changes from the product’s yellow color to the brown color of the Karl Fischer reagent.
Organic Analysis
Redox titrimetry also is used for the analysis of organic analytes. One important example is the determination of the chemical oxygen demand (COD) of natural waters and wastewaters. The COD provides a measure of the quantity of oxygen necessary to completely oxidize all the organic matter in a sample to CO2 and H2O. Because no attempt is made to correct for organic matter that can not be decomposed biologically, or for slow decomposition kinetics, the COD always overestimates a sample’s true oxygen demand. The determination of COD is particularly important in managing industrial wastewater treatment facilities where it is used to monitor the release of organic-rich wastes into municipal sewer systems or the environment.
A sample’s COD is determined by refluxing it in the presence of excess K2Cr2O7, which serves as the oxidizing agent. The solution is acidified with H2SO4 using Ag2SO4 to catalyze the oxidation of low molecular weight fatty acids. Mercuric sulfate, HgSO4, is added to complex any chloride that is present, preventing the precipitation of the Ag+ catalyst as AgCl. Under these conditions, the efficiency for oxidizing organic matter is 95–100%. After refluxing for two hours, the solution is cooled to room temperature and the excess Cr2O72– is determined by back titrating using ferrous ammonium sulfate as the titrant and ferroin as the indicator. Because it is difficult to completely remove all traces of organic matter from the reagents, a blank titration must be performed. The difference in the amount of ferrous ammonium sulfate needed to titrate the sample and the blank is proportional to the COD.
Iodine has been used as an oxidizing titrant for a number of compounds of pharmaceutical interest. Earlier we noted that the reaction of S2O32– with I3 produces the tetrathionate ion, S4O62–. The tetrathionate ion is actually a dimer consisting of two thiosulfate ions connected through a disulfide (–S–S–) linkage. In the same fashion, I3 can be used to titrate mercaptans of the general formula RSH, forming the dimer RSSR as a product. The amino acid cysteine also can be titrated with I3. The product of this titration is cystine, which is a dimer of cysteine. Triiodide also can be used for the analysis of ascorbic acid (vitamin C) by oxidizing the enediol functional group to an alpha diketone
and for the analysis of reducing sugars, such as glucose, by oxidizing the aldehyde functional group to a carboxylate ion in a basic solution.
An organic compound containing a hydroxyl, a carbonyl, or an amine functional group adjacent to an hydoxyl or a carbonyl group can be oxidized using metaperiodate, IO4, as an oxidizing titrant.
$\ce{IO_4^-}(aq)+\mathrm{H_2O}(l)+2e^-\rightleftharpoons \ce{IO_3^-}(aq)+\mathrm{2OH^-}(aq)$
A two-electron oxidation cleaves the C–C bond between the two functional groups, with hydroxyl groups being oxidized to aldehydes or ketones, carbonyl functional groups being oxidized to carboxylic acids, and amines being oxidized to an aldehyde and an amine (ammonia if a primary amine). The analysis is conducted by adding a known excess of IO4 to the solution containing the analyte, and allowing the oxidation to take place for approximately one hour at room temperature. When the oxidation is complete, an excess of KI is added, which converts any unreacted IO4 to IO3 and I3.
$\ce{IO_4^-}(aq)+3\mathrm I^-(aq)+\mathrm{H_2O}(l)\rightarrow \ce{IO_3^-}(aq)+\textrm I_3^-(aq)+\mathrm{2OH^-}(aq)$
The I3 is then determined by titrating with S2O32– using starch as an indicator.
Quantitative Calculations
The quantitative relationship between the titrand and the titrant is determined by the stoichiometry of the titration reaction. If you are unsure of the balanced reaction, you can deduce the stoichiometry by remembering that the electrons in a redox reaction must be conserved.
Example 9.11
The amount of Fe in a 0.4891-g sample of an ore was determined by titrating with K2Cr2O7. After dissolving the sample in HCl, the iron was brought into the +2 oxidation state using a Jones reductor. Titration to the diphenylamine sulfonic acid end point required 36.92 mL of 0.02153 M K2Cr2O7. Report the ore’s iron content as %w/w Fe2O3.
(Although we can deduce the stoichiometry between the titrant and the titrand without balancing the titration reaction, the balanced reaction
$\mathrm{K_2Cr_2O_7}(aq)+\mathrm{6Fe^{2+}}(aq)+\mathrm{14H^+}(aq)\rightarrow \mathrm{2Cr^{3+}}(aq)+\mathrm{2K^+}(aq)+\mathrm{6Fe^{3+}}(aq)+\mathrm{7H_2O}(l)$
does provide useful information. For example, the presence of H+ reminds us that the reactionfs feasibility is pH-dependent.)
Solution
Because we have not been provided with the titration reaction, let’s use a conservation of electrons to deduce the stoichiometry. During the titration the analyte is oxidized from Fe2+ to Fe3+, and the titrant is reduced from Cr2O72– to Cr3+. Oxidizing Fe2+ to Fe3+ requires only a single electron. Reducing Cr2O72–, in which each chromium is in the +6 oxidation state, to Cr3+ requires three electrons per chromium, for a total of six electrons. A conservation of electrons for the titration, therefore, requires that each mole of K2Cr2O7 reacts with six moles of Fe2+.
The moles of K2Cr2O7 used in reaching the end point is
$\mathrm{(0.02153\;M\;K_2Cr_2O_7)\times(0.03692\;L\;K_2Cr_2O_7)=7.949\times10^{-4}\;mol\;K_2Cr_2O_7}$
which means that the sample contains
$\mathrm{7.949\times10^{-4}\;mol\;K_2Cr_2O_7\times\dfrac{6\;mol\;Fe^{2+}}{mol\;K_2Cr_2O_7}=4.769\times10^{-3}\;mol\;Fe^{2+}}$
Thus, the %w/w Fe2O3 in the sample of ore is
$\mathrm{4.769\times10^{-3}\;mol\;Fe^{2+}\times\dfrac{1\;mol\;Fe_2O_3}{2\;mol\;Fe^{2+}}\times\dfrac{159.69\;g\;Fe_2O_3}{mol\;Fe_2O_3}=0.3808\;g\;Fe_2O_3}$
$\mathrm{\dfrac{0.3808\;g\;Fe_2O_3}{0.4891\;g\;sample}\times100=77.86\%\;w/w\;Fe_2O_3}$
Practice Exercise 9.20
The purity of a sample of sodium oxalate, Na2C2O4, is determined by titrating with a standard solution of KMnO4. If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO4 to reach the titration’s end point, what is the %w/w Na2C2O4 in the sample.
Click here to review your answer to this exercise.
As shown in the following two examples, we can easily extend this approach to an analysis that requires an indirect analysis or a back titration.
Example 9.12
A 25.00-mL sample of a liquid bleach was diluted to 1000 mL in a volumetric flask. A 25-mL portion of the diluted sample was transferred by pipet into an Erlenmeyer flask containing an excess of KI, reducing the OCl to Cl, and producing I3. The liberated I3 was determined by titrating with 0.09892 M Na2S2O3, requiring 8.96 mL to reach the starch indicator end point. Report the %w/v NaOCl in the sample of bleach.
Solution
To determine the stoichiometry between the analyte, NaOCl, and the titrant, Na2S2O3, we need to consider both the reaction between OCl and I, and the titration of I3 with Na2S2O3.
First, in reducing OCl to Cl, the oxidation state of chlorine changes from +1 to –1, requiring two electrons. The oxidation of three I to form I3 releases two electrons as the oxidation state of each iodine changes from –1 in I to –⅓ in I3. A conservation of electrons, therefore, requires that each mole of OCl produces one mole of I3.
Second, in the titration reaction, I3. is reduced to I and S2O32– is oxidized to S4O62–. Reducing I3 to 3I requires two elections as each iodine changes from an oxidation state of –⅓ to –1. In oxidizing S2O32– to S4O62–, each sulfur changes its oxidation state from +2 to +2.5, releasing one electron for each S2O32–. A conservation of electrons, therefore, requires that each mole of I3 reacts with two moles of S2O32–.
Finally, because each mole of OCl produces one mole of I3, and each mole of I3 reacts with two moles of S2O32–, we know that every mole of NaOCl in the sample ultimately results in the consumption of two moles of Na2S2O3.
The balanced reactions for this analysis are:
$\mathrm{OCl^-}(aq)+\mathrm{3I^-}(aq)+\mathrm{2H^+}(aq)\rightarrow \ce{I_3^-}(aq)+\mathrm{Cl^-}(aq)+\mathrm{H_2O}(l)$
$\mathrm I_3^-(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow \mathrm{S_4O_6^{2-}}(aq)+\mathrm{3I^-}(aq)$
The moles of Na2S2O3 used in reaching the titration’s end point is
$\mathrm{(0.09892\;M\;Na_2S_2O_3)\times(0.00896\;L\;Na_2S_2O_3)=8.86\times10^{-4}\;mol\;Na_2S_2O_3}$
which means the sample contains
$\mathrm{8.86\times10^{-4}\;mol\;Na_2S_2O_3\times\dfrac{1\;mol\;NaOCl}{2\;mol\;Na_2S_2O_3}\times\dfrac{74.44\;g\;NaOCl}{mol\;NaOCl}=0.03299\;g\;NaOCl}$
Thus, the %w/v NaOCl in the diluted sample is
$\mathrm{\dfrac{0.03299\;g\;NaOCl}{25.00\;mL}\times100=0.132\%\;w/v\;NaOCl}$
Because the bleach was diluted by a factor of 40 (25 mL to 1000 mL), the concentration of NaOCl in the bleach is 5.28% (w/v).
Example 9.13
The amount of ascorbic acid, C6H8O6, in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a known amount of I3, and back titrating the excess I3 with Na2S2O3. A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of 0.01023 M I3. After the oxidation was complete, 13.82 mL of 0.07203 M Na2S2O3 was needed to reach the starch indicator end point. Report the concentration ascorbic acid in mg/100 mL.
Solution
For a back titration we need to determine the stoichiometry between I3 and the analyte, C6H8O6, and between I3 and the titrant, Na2S2O3. The later is easy because we know from Example 9.12 that each mole of I3 reacts with two moles of Na2S2O3.
The balanced reactions for this analysis are:
$\mathrm{C_6H_8O_6}(aq)+\ce{I_3^-}(aq)\rightarrow \mathrm{3I^-}(aq)+\mathrm{C_6H_6O_6}(aq)+\mathrm{2H^+}(aq)$
$\ce{I_3^-}(aq)+\mathrm{2S_2O_3^{2-}}(aq)\rightarrow \mathrm{S_4O_6^{2-}}(aq)+\mathrm{3I^-}(aq)$
In oxidizing ascorbic acid to dehydroascorbic acid, the oxidation state of carbon changes from +⅔ in C6H8O6 to +1 in C6H6O6. Each carbon releases ⅓ of an electron, or a total of two electrons per ascorbic acid. As we learned in Example 9.12, reducing I3 requires two electrons; thus, a conservation of electrons requires that each mole of ascorbic acid consumes one mole of I3.
The total moles of I3 reacting with C6H8O6 and with Na2S2O3 is
$\mathrm{(0.01023\;M\;\ce{I_3^-})\times(0.05000\;L\;\ce{I_3^-})=5.115\times10^{-4}\;mol\;\ce{I_3^-}}$
The back titration consumes
$\mathrm{0.01382\;L\;Na_2S_2O_3\times\dfrac{0.07203\;mol\;Na_2S_2O_3}{L\;Na_2S_2O_3}\times\dfrac{1\;mol\;\ce{I_3^-}}{2\;mol\;Na_2S_2O_3}=4.977\times10^{-4}\;mol\;\ce{I_3^-}}$
Subtracting the moles of I3 reacting with Na2S2O3 from the total moles of I3 gives the moles reacting with ascorbic acid.
$\mathrm{5.115\times10^{-4}\;mol\;\ce{I_3^-} - 4.977\times10^{-4}\;mol\;\ce{I_3^-}=1.38\times10^{-5}\;mol\;\ce{I_3^-}}$
The grams of ascorbic acid in the 5.00-mL sample of orange juice is
$\mathrm{1.38\times10^{-5}\;mol\;\ce{I_3^-}\times\dfrac{1\;mol\;C_6H_8O_6}{mol\;\ce{I_3^-}}\times\dfrac{176.13\;g\;C_6H_8O_6}{mol\;C_6H_8O_6}=2.43\times10^{-3}\;g\;C_6H_8O_6}$
There are 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg per 100 mL of orange juice.
Practice Exercise 9.21
A quantitative analysis for ethanol, C2H6O, can be accomplished by a redox back titration. Ethanol is oxidized to acetic acid, C2H4O2, using excess dichromate, Cr2O72–, which is reduced to Cr3+. The excess dichromate is titrated with Fe2+, giving Cr3+ and Fe3+ as products. In a typical analysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mL sample is taken and the ethanol is removed by distillation and collected in 50.00 mL of an acidified solution of 0.0200 M K2Cr2O7. A back titration of the unreacted Cr2O72– requires 21.48 mL of 0.1014 M Fe2+. Calculate the %w/v ethanol in the brandy.
Click here to review your answer to this exercise.
9.4.4 Evaluation of Redox Titrimetry
The scale of operations, accuracy, precision, sensitivity, time, and cost of a redox titration are similar to those described earlier in this chapter for acid–base or a complexation titration. As with acid–base titrations, we can extend a redox titration to the analysis of a mixture of analytes if there is a significant difference in their oxidation or reduction potentials. Figure 9.42 shows an example of the titration curve for a mixture of Fe2+ and Sn2+ using Ce4+ as the titrant. A titration of a mixture of analytes is possible if their standard state potentials or formal potentials differ by at least 200 mV.
Figure 9.42 Titration curve for the titration of 50.0 mL of 0.0125 M Sn2+ and 0.0250 M Fe2+ with 0.050 M Ce4+. Both the titrand and the titrant are 1M in HCl. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.4%3A_Reaction_Stoichiometry_in_Solutions%3A_Oxidation-Reduction_Titrations.txt |
Learning Objectives
• To describe the relationship between solute concentration and the physical properties of a solution.
• To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.
Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
Counting Concentrations and Effective Concentrations
When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, $\ce{NaCl}$, and $\ce{CaCl_2}$. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both $\ce{NaCl}$ and $\ce{CaCl_2}$ are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of $\ce{NaCl}$ contains 0.01 M $\ce{Na^{+}}$ ions and 0.01 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.02 M. Similarly, the $\ce{CaCl_2}$ solution contains 0.01 M $\ce{Ca^{2+}}$ ions and 0.02 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete.
At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as $\ce{Mg^{2+}}$ or $\ce{Al^{3+}}$), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. A greater discussion of this is below.
Vapor Pressure of Solutions and Raoult’s Law For Nonvolatile Solutes
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure $1$, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
Figure $2$ shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.
If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore
$P_A=\chi_AP^0_A \label{13.5.1}$
where $P_A$ is the vapor pressure of component A of the solution (in this case the solvent), $\chi_A$ is the mole fraction of $A$ in solution, and $P^0_A$ is the vapor pressure of pure $A$. Equation $\ref{13.5.1}$ is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute ($B$), then we can relate the mole fraction of the solvent to the solute
$\chi_A + \chi_B = 1 \nonumber$
and we can substitute $\chi_A = 1 − \chi_B$ into Equation \ref{13.5.1} to obtain
\begin{align*} P_A &=(1−\chi_B)P^0_A \[4pt] &=P^0_A−\chi_BP^0_A \end{align*}
Rearranging and defining $ΔP_A=P^0_A−P_A$, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:
\begin{align} P^0_A−P_A &=ΔP_A \[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align}
We can solve vapor pressure problems in either of two ways: by using Equation $\ref{13.5.1}$ to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation $\ref{13.5.3}$ to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example $1$: Anti-Freeze
Ethylene glycol ($\ce{HOCH_2CH_2OH}$), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.
Given: identity of solute, percentage by mass, and vapor pressure of pure solvent
Asked for: vapor pressure of solution
Strategy:
1. Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.
2. Use Raoult’s law (Equation \ref{13.5.1}) to calculate the vapor pressure of the solution.
Solution:
A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:
$moles \; EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber$
$moles \; \ce{H2O} =(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber$
The mole fraction of water is thus
$\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber$
B From Raoult’s law (Equation $\ref{13.5.1}$), the vapor pressure of the solution is
\begin{align*} P_{\ce{H2O}} &=(\chi_{\ce{H2O}})(P^0_{\ce{H2O}}) \[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*}
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation $\ref{13.5.3}$ to calculate the resulting decrease in vapor pressure:
\begin{align*} \chi_{EG} &=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O} \[4pt] &=0.112 \end{align*}
\begin{align*} ΔP_{\ce{H2O}} &=(\chi_{EG})(P^0_{\ce{H2O}}) \[4pt] &=(0.112)(760\; mmHg)=85.1\; mmHg \end{align*}
$P_{\ce{H2O}}=P^0_{\ce{H2O}}−ΔP_{\ce{H2O}}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber$
The same result is obtained using either method.
Exercise $1$
Seawater is an approximately 3.0% aqueous solution of $\ce{NaCl}$ by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of $\ce{NaCl}$, remembering that 1 mol of $\ce{NaCl}$ produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.
Answer
0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.
Boiling Point Elevation
Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure $4$). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to $P = 1\, atm$ at a higher temperature than does the curve for pure water.
The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure $5$).
We can define the boiling point elevation ($ΔT_b$) as the difference between the boiling points of the solution and the pure solvent:
$ΔT_b=T_b−T^0_b \label{13.5.8}$
where $T_b$ is the boiling point of the solution and $T^0_b$ is the boiling point of the pure solvent. We can express the relationship between $ΔT_b$ and concentration as follows
$ΔT_b = mK_b \label{13.5.9}$
where m is the concentration of the solute expressed in molality, and $K_b$ is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table $1$ lists characteristic $K_b$ values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
Table $1$: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents
Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m)
acetic acid 117.90 3.22 16.64 3.63
benzene 80.09 2.64 5.49 5.07
d-(+)-camphor 207.4 4.91 178.8 37.8
carbon disulfide 46.2 2.42 −112.1 3.74
carbon tetrachloride 76.8 5.26 −22.62 31.4
chloroform 61.17 3.80 −63.41 4.60
nitrobenzene 210.8 5.24 5.70 6.87
water 100.00 0.51 0.00 1.86
The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.
According to Table $1$, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous $\ce{NaCl}$ solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved ions. Hence a 1.00 m $\ce{NaCl}$ solution will have a boiling point of about 101.02°C.
Example $3$
In Example $1$, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: boiling point
Strategy:
Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation $\ref{13.5.9}$ to calculate the increase in boiling point.
Solution:
From Example $1$, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus
\begin{align*} \text{molality of ethylene glycol} &= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right) \[4pt] &=6.98\, m \end{align*}
From Equation $\ref{13.5.9}$, the increase in boiling point is therefore
\begin{align*} ΔT_b &= m K_b \[4pt] &=(6.98 \cancel{m})(0.51°C/\cancel{m}) \[4pt] &=3.6°C \end{align*}
The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation $\ref{13.5.9}$ may not be valid.
Exercise $3$
Assume that a tablespoon (5.00 g) of $\ce{NaCl}$ is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?
Answer
100.04°C, or 100°C to three significant figures. (Recall that 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)
Freezing Point Depression
The phase diagram in Figure $4$ shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to $P = 1\,atm$ at a lower temperature than the curve for pure water.
We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression ($ΔT_f$) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
$ΔT_f=T^0_f−T_f \label{13.5.10}$
where $T^0_f$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution.
The order of the terms is reversed compared with Equation $\ref{13.5.8}$ to express the freezing point depression as a positive number. The relationship between $ΔT_f$ and the solute concentration is given by an equation analogous to Equation $\ref{13.5.9}$:
$ΔT_f = mK_f \label{13.5.11}$
where $m$ is the molality of the solution and $K_f$ is the molal freezing point depression constant for the solvent (in units of °C/m).
Like $K_b$, each solvent has a characteristic value of $K_f$ (see Table $1$). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous $\ce{NaCl}$ solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.
The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example $4$: Salting the Roads
In colder regions of the United States, $\ce{NaCl}$ or $\ce{CaCl_2}$ is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of $\ce{NaCl}$ and one of $\ce{CaCl_2}$, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.
Given: solubilities of two compounds
Asked for: concentrations and freezing points
Strategy:
1. Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities.
2. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation $\ref{13.5.11}$ to calculate the freezing point depressions of the solutions.
Solution:
A. From Figure above, we can estimate the solubilities of $\ce{NaCl}$ and $\ce{CaCl_2}$ to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are
$m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber$
$m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber$
The lower formula mass of $\ce{NaCl}$ more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than $\ce{CaCl2}$.
B. Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of $\ce{NaCl}$ and $\ce{CaCl2}$, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for $\ce{NaCl}$ and 3 × 5.4 m = 16 m for $\ce{CaCl2}$. The resulting freezing point depressions can be calculated using Equation $\ref{13.5.11}$:
$\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber$
$\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber$
Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that $\ce{CaCl2}$ is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, $\ce{CaCl2}$ is the salt usually sold for home use, and it is also often used on highways.
Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer
Exercise $4$
Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples $1$ and $3$.
Answer
−13.0°C
Example $5$
Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m $\ce{KCl}$, 0.1 m glucose, 0.1 m $\ce{SrCl2}$, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m $\ce{HCl}$.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
1. Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.
2. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.
Solution:
A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. $\ce{KCl}$, $\ce{SrCl_2}$, and $\ce{HCl}$ are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: $\ce{KCl}$ and $\ce{HCl}$, 0.2 m; $SrCl_2$, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > $\ce{HCl}$ = $\ce{KCl}$ > $\ce{SrCl_2}$.
Exercise $5$
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m $\ce{NaCl}$, 0.3 m acetic acid, 0.1 m $\ce{CaCl_2}$, and 0.2 m sucrose.
Answer
0.2 m $\ce{NaCl}$ (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m $\ce{CaCl_2}$ < 0.2 m sucrose (highest freezing point)
Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing $K_b$ and $K_f$ values in Table $1$, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of $K_f$ (37.8°C/m), d-(+)-camphor (Table $1$) is often used to determine the molar mass of organic compounds by this method.
Example $6$: Sulfur
A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of $CS_2$ to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved $\ce{S_n}$ molecules (i.e., what is the value of $n$?).
Given: masses of solute and solvent and freezing point
Asked for: molar mass and number of $\ce{S}$ atoms per molecule
Strategy:
1. Use Equation $\ref{13.5.10}$, the measured freezing point of the solution, and the freezing point of $CS_2$ from Table $1$ to calculate the freezing point depression. Then use Equation $\ref{13.5.11}$ and the value of $K_f$ from Table $1$ to calculate the molality of the solution.
2. From the calculated molality, determine the number of moles of solute present.
3. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain $n$, the number of sulfur atoms per mole of dissolved sulfur.
Solution:
A The first step is to calculate the freezing point depression using Equation $\ref{13.5.10}$:
$ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber$
Then Equation $\ref{13.5.11}$ gives
$m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber$
B The total number of moles of solute present in the solution is
$\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber$
C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus
$\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber$
The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of $\ce{S_8}$.
Exercise $6$
One of the byproducts formed during the synthesis of $C_{60}$ is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of $CCl_4$ has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?
Answer
847 g/mol; $\ce{C_{70}}$
Osmotic Pressure
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure $6$, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure ($\Pi$) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
$\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12}$
where $M$ is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), $R$ is the ideal gas constant, and $T$ is the absolute temperature.
As shown in Example $7$, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example $7$
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% $\ce{NaCl}$ by mass; the solution density is 1.02 g/mL at 25°C.
1. Calculate the osmotic pressure of a 4.0% aqueous $\ce{NaCl}$ solution at 25°C.
2. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of $\ce{NaCl}$ solution; internal osmotic pressure of cell
Asked for: osmotic pressure of $\ce{NaCl}$ solution and concentration of glycerol needed
Strategy:
1. Calculate the molarity of the $\ce{NaCl}$ solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.
2. Use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution.
3. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{13.5.12} to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of $\ce{NaCl}$ per 100 g of solution. Using the formula mass of $\ce{NaCl}$ (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
\begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*}
Because 1 mol of $\ce{NaCl}$ produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution:
\begin{align*} \Pi &=MRT \[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\[4pt] &=34 \;atm\end{align*}
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure:
\begin{align*} M&=\dfrac{\Pi}{RT}\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*}
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the $\ce{NaCl}$ solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise $7$
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer
24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure $7$, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure $8$, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
Colligative Properties of Electrolyte Solutions
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of $\ce{NaCl}$ in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of $\ce{NaCl}$ and $\ce{KCl}$ are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.
The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van ’t Hoff factor ($i$) and is defined as follows:
$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13}$
Named for Jacobus Hendricus van ’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.
As the solute concentration increases, the van ’t Hoff factor decreases.
The van ’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van ’t Hoff factor, the greater the deviation. As the data in Table $2$ show, the van ’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van ’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.
Table $2$: van ’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
$NaCl$ 1.9 2.0
$HCl$ 1.9 2.0
$MgCl_2$ 2.7 3.0
$FeCl_3$ 3.4 4.0
$Ca(NO_3)_2$ 2.5 3.0
$AlCl_3$ 3.2 4.0
$MgSO_4$ 1.4 2.0
Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure $9$). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $Mg^{2+}$, $Al^{3+}$, $SO_4^{2−}$, and $PO_4^{3−}$ have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.
Example $8$: Iron Chloride in Water
A 0.0500 M aqueous solution of $\ce{FeCl3}$ has an osmotic pressure of 4.15 atm at 25°C. Calculate the van ’t Hoff factor $i$ for the solution.
Given: solute concentration, osmotic pressure, and temperature
Asked for: van ’t Hoff factor
Strategy:
1. Use Equation $\ref{13.5.12}$ to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
2. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation $\ref{13.5.13}$ to calculate the van ’t Hoff factor.
Solution:
A If $\ce{FeCl_3}$ dissociated completely in aqueous solution, it would produce four ions per formula unit (one $\ce{Fe^{3+}(aq)}$ and three \ce{Cl^{-}(aq)} ions) for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \begin{align*} \Pi &=MRT \[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 moles of particles per mole of \(\ce{FeCl_3} dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
$4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber$
or after rearranging
$M = 0.170 mol \nonumber$
The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 mole of particles per mole of $\ce{FeCl_3}$ dissolved. From Equation $\ref{13.5.13}$, the van ’t Hoff factor for the solution is
$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber$
Exercise $8$: Magnesium Chloride in Water
Calculate the van ’t Hoff factor for a 0.050 m aqueous solution of $\ce{MgCl2}$ that has a measured freezing point of −0.25°C.
Answer
2.7 (versus an ideal value of 3).
Summary
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. The boiling point elevation ($ΔT_b$) and freezing point depression ($ΔT_f$) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van ’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.
• Henry’s law: $C = kP \nonumber$
• Raoult’s law: $P_A=\chi_AP^0_A \nonumber$
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=\chi_BP^0_A \nonumber$
• vapor pressure of a system containing two volatile components: $P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \nonumber$
• boiling point elevation: $ΔT_b = mK_b \nonumber$
• freezing point depression: $ΔT_f = mK_f \nonumber$
• osmotic pressure: $\Pi=\dfrac{n}{V}RT=MRT \nonumber$
• van ’t Hoff factor: $i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.5%3A_Phase_Equilibrium_in_Solutions_-_Nonvolatile_Solutes.txt |
Learning Objectives
• To understand the relationship among temperature, pressure, and solubility.
• The understand that the solubility of a solid may increase or decrease with increasing temperature,
• To understand that the solubility of a gas decreases with an increase in temperature and a decrease in pressure.
Effect of Pressure on the Solubility of Gases: Henry’s Law
External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure $1$, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.
The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):
$C = k_HP \label{11.61}$
where $C$ is the concentration of dissolved gas at equilibrium, $P$ is the partial pressure of the gas, and $k_H$ is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature.
Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table $1$.
As the data in Table $1$ demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements of Group 18, the Henry’s law constants increase smoothly from $\ce{He}$ to $\ce{Ne}$ to $\ce{Ar}$. The table also shows that $\ce{O_2}$ is almost twice as soluble as $\ce{N_2}$. Although London dispersion forces are too weak to explain such a large difference, $\ce{O_2}$ is paramagnetic and hence more polarizable than $\ce{N_2}$, which explains its high solubility.
Table $1$: Henry’s Law Constants for Selected Gases in Water at 20°C
Gas Henry’s Law Constant [mol/(L·atm)] × 10−4
$\ce{Ar}$ 15
$\ce{CO_2}$ 392
$\ce{H_2}$ 8.1
$\ce{He}$ 3.9
$\ce{N_2}$ 7.1
$\ce{Ne}$ 4.7
$\ce{O_2}$ 14
Gases that react with the solvent do not obey Henry’s law
Gases that react chemically with water, such as $\ce{HCl}$ and the other hydrogen halides, $\ce{H_2S}$, and $\ce{NH_3}$, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, $\ce{HCl}$ reacts with water to give $\ce{H^{+}(aq)}$ and $\ce{Cl^{-}(aq)}$, not dissolved $\ce{HCl}$ molecules,
$\ce{HCl(g) + H2O(l) \rightarrow H3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
The dissociation of $\ce{HCl}$ into ions results in a much higher effective "solubility" than expected for a neutral molecule.
Henry’s law has important applications. For example, bubbles of $\ce{CO_2}$ form as soon as a carbonated beverage is opened because the drink was bottled under $\ce{CO_2}$ at a pressure greater than 1 atm. When the bottle is opened, the pressure of $\ce{CO_2}$ above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more $\ce{N_2}$ from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of $\ce{N_2}$ to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.
Due to the low Henry’s law constant for $\ce{O_2}$ in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the $\ce{O_2}$ concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind $\ce{O_2}$ reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds $\ce{O_2}$ and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four $\ce{O_2}$ molecules. Although the concentration of dissolved $\ce{O_2}$ in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved $\ce{O_2}$ concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for $\ce{O_2}$ result in dissolved oxygen concentrations comparable to those in normal blood.
Example $1$: Oxygen in Water
The Henry’s law constant for $\ce{O_2}$ in water at 25°C is $1.27 \times 10^{-3} M/atm$, and the mole fraction of $\ce{O_2}$ in the atmosphere is 0.21. Calculate the solubility of $\ce{O_2}$ in water at 25°C at an atmospheric pressure of 1.00 atm.
Given: Henry’s law constant, mole fraction of $\ce{O_2}$, and pressure
Asked for: solubility
Strategy:
1. Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen.
2. Use Henry’s law (Equation $\ref{11.61}$) to calculate the solubility, expressed as the concentration of dissolved gas.
Solution:
A According to Dalton’s law, the partial pressure of $O_2$ is proportional to the mole fraction of $O_2$:
$P_A = \chi_A P_t = (0.21)(1.00\; atm) = 0.21\; atm \nonumber$
B From Henry’s law, the concentration of dissolved oxygen under these conditions is
$CO_2=k_H P_{O_2}=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) =2.7 \times 10^{-4}\; M \nonumber$
Exercise $1$: Carbon Dioxide in Water
To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved $CO_2$ in a soft drink:
1. bottled under a pressure of 5.0 atm of $CO_2$
2. in equilibrium with the normal partial pressure of $CO_2$ in the atmosphere (approximately $3 \times 10^{-4} \;atm$). The Henry’s law constant for $CO_2$ in water at 25°C is $3.4 \times 10^{-2}\; M/atm$.
Answer a
0.17 M
Answer b
$1 \times 10^{-5} M$
Summary
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.6%3A_Phase_Equilibrium_in_Solutions_-_Volatile_Solutes.txt |
Learning Objectives
• To distinguish between true solutions and solutions with aggregate particles.
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table \(1\).
Table \(1\): Properties of Liquid Solutions, Colloids, and Suspensions
Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples
solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water
colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese
suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint
Colloids and Suspensions
Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols, which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible.
Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure \(1\).
Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic.
Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid.
In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure \(2\)). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin.
Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure \(3\). Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels.
Emulsions
Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”:
Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate \(\ce{[CH_3(CH_2)_{16}CO_2−Na^{+}]}\), and detergents, such as sodium dodecyl sulfate \(\ce{[CH_3(CH_2)_{11}OSO_3−Na^{+}]}\), whose structures are as follows:
When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise.
A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats.
Micelles
Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles, which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (Figure \(\PageIndex{4a}\)).
A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers, extended sheets consisting of a double layer of molecules. As shown in Figure \(\PageIndex{4b}\), the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution.
A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks.
Summary
A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.7%3A_Colloidal_Suspensions.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q.1
A student goes to the hospital and has blood work done and the results state that he has a bromide count of $3.17\frac{mg}{dL}$.
1. What is the molar concentration of the students blood?
2. What is the molality of the student's blood if the density of his blood was found to be $1.06\frac{g}{mL}$?
3. How much of the students blood would be require to get $3g$ of bromide?
Solution
The molar mass of bromide is $79.90\frac{g}{mol}$. $1L=10dL$, $1g=1000mg$
$\left(\dfrac{\textrm{3.17} \cancel{mg}\;}{\textrm{1} \cancel{dL}\;}\right)\left(\dfrac{\textrm{1} \cancel{g}\;}{\textrm{1000} \cancel{mg}\;}\right)\left(\dfrac{10\cancel{dL}}{\textrm{1} \textrm{L}\;}\right )\left(\dfrac{\textrm{1} \textrm{mol}\;}{\textrm{79.90} \cancel{g}\;}\right)=3.97\times10^{-4}\ \frac{mol}{L}\nonumber$
Assume there is 1L of solution. Modality (m) is calculated by:
$\textrm{3.17 x 10}^{-2} \cancel{g}\;\left(\dfrac{\textrm{1} \textrm{ mol}\;}{\textrm{79.90} \cancel{g}\;}\right)\left(\dfrac{1\textrm{ }}{\textrm{1.06} \textrm{ kg}\;}\right )=3.74\times10^{-4}\frac{mol}{kg}\nonumber$
If we know that there is $3.17\times 10^{-2}g$ of bromide in $1L$ of blood, so all that needs to be done is divide $3g$ by $3.17\times 10^{-2}\ \frac{g}{L}$:
Volume of Blood = 94.64 L
Q.5
Given a 3.1416 M aqueous solution of sucrose, $\ce{C12H22O11}$ with a density of $1.5986\ \mathrm{\frac{g}{mL}}$. Calculate the molality of this solution, remember that the molar mass of sucrose is $342.297 \mathrm{\frac{g}{mol}}$.
Solution
Before we calculate the molality of this solution, don't forget that molarity is just mol/L so we will multiply the molarity of the solution by 1 L to isolate the moles.
$1.00\; \cancel{L}\; \text{Water} = \dfrac{3.1416\: mol\ \text{Sucrose}}{1.00\; \cancel{L}\; \text{Water}}= 3.1416\; mol\ \text{Sucrose} \nonumber$
Once we have the moles of Sucrose we will multiply it by the molar mass of sucrose in order to find the number of sucrose grams present in the solution:
$3.1416\; \cancel{mol}\; \text{Sucrose} =\dfrac{342.297\: g\ \text{Sucrose}}{1.00\; \cancel{mol}\; \text{Sucrose}}= 1075.4\; g\ \text{Sucrose}\nonumber$
We next will multiply the density by 1000 mL so as get the number of grams per liter, this way we can find the total grams of the solution and can subtract the sucrose grams from the solution grams to find the grams of water.
$1000\; \cancel{mL}\; =\dfrac{1.5986\, g \: \text{SucroseSoln}}{1.00\; \cancel{mL}\; \text{SucroseSoln}}= 1598.6\; g/L\ \text{SucroseSoln} \nonumber$
$\text{Water grams} =[{1598.6g\; \text{Sucrose Soln.}}]-[{1075.4g\; \text{Sucrose} }]={523.2g}\; \text{Water} \; ={0.5232\; kg\; \text{Water} \;}\nonumber$
Lastly we divide the moles of Sucrose by the kg of water in order to get molality.
$\dfrac{3.1416\: mol\ \text{Sucrose}}{0.5232\; kg\; \text{Water}}= 6.005\; molal\nonumber$
Q.9
A solution contains 75% of ethanol ($\ce{C2H6O}$) by mass and the rest is water.
1. What is the density of the solution if there is 15 mol of ethanol per liter of solution?
2. To prepare a 3.5L of 2M ethanol, how many milliliters of the solution is needed?
Solution
a) For easy calculation, assume that there is 1 L of the solution. Using:
$\text{mol} = \text{molarity} \times \text{volume}$
The mol of 15M ethanol in 1L is 15 mol. The molar mass of ethanol is $46.068\frac{g}{mol}$. Using the equation below:
$\text{ethanol mol} = \dfrac{\text{mass ethanol}}{\text{ethanol molar mass}}$
$\text{ethanol mass in the solution} = \text{ethanol mol} \times \text{molar mass} = (15mol)(46.068\dfrac{g}{mol})=691.02g$
Ethanol is 75% of the solution by mass, therefore:
$\text{mass of solution} \times 0.75= \text{mass of ethanol}$
$\text{mass of solution} \times 0.75= 691.02g$
$\text{mass of solution} = \dfrac{691.02g}{0.75}=921.36g$
The equation to calculate density is:
$\text{Density} = \dfrac{\text{mass}}{\text{volume}}$
$\text{Density} = \dfrac{921.36g}{1L} = 0.921 \dfrac{g}{ml}$
b)To prepare a 3.5L of 2M ethanol;
$\text{ethanol mol needed} = \text{molarity} \times \text{volume} = (2 \, \text{M}) (3.5 \, \text{L})=7 \, \text{mol}$
$V_{\text{ethanol needed}}=\dfrac{\text{mol needed}}{\text{molarity}}=\dfrac{7 \, \text{mol}}{15 \, \text{mol}} \times 1 \, \text{L} = 0.467 \, \text{L}$
0.467L of the solution is needed, which is 467mL.
Q.14
Rewrite the following balanced equations as net ionic equations.
1. $\ce{H2SO4(aq) + 2NaOH(aq) \rightarrow 2H2O(l) + Na2SO4(aq)}$
2. $\ce{2Na3PO4(aq) + 3CaCl2(aq) \rightarrow 6NaCl(aq) + Ca3(PO4)2(s)}$
3. $\ce{BaCl2(aq) + Na2SO4(aq) \rightarrow BaSO4(s) + 2NaCl(aq)}$
4. $\ce{2AgNO3(aq) + ZnCl2(aq) \rightarrow 2AgCl(s) + Zn(NO3)2(aq)}$
Solution
1. $\ce{2H^{+} (aq) + 2OH^{-} (aq) -> 2H_{2} O (l) }$
2. $\ce{2PO_4^{3-} (aq) + 3Ca^{2+} (aq) -> Ca_{3}(PO_{4})_2 (s)}$
3. $\ce{Ba^{2+} (aq) + SO_{4}^{2-} (aq) -> BaSO4 (s)}$
4. $\ce{2Ag^{2+} (aq) + Cl^{-} (aq) -> 2AgCl (s)}$
Q.15
Write the balanced chemical equation for the titration of $\ce{H2SO4}$ with $\ce{KOH}$, then find the volume of a 5.49 M $\ce{KOH}$ solution needed to neutralize 36.2 g of $\ce{H2SO4}$ .
Solution
The balanced equation of $\ce{H2SO4}$ with KOH is:
$\ce{H2SO4 (aq) + 2 KOH (aq) -> K2SO4 (aq) + 2 H2O (l)} \nonumber$
Calculate the amount of moles of $\ce{H2SO4}$ being neutralized:
$n_{\ce{H2SO4}} = 36.2 \; \text{g} \; \ce{H2SO4} \times \dfrac{1 \; \text{mol} \; \ce{H2SO4}}{98.079 \; \text{g} \; \ce{H2SO4}} = 0.369 \; \text{mol} \; \ce{H2SO4} \nonumber$
Using stoichiometry to find the amount of moles of $\ce{KOH}$ needed to neutralize the amount of $\ce{H2SO4}$ present:
$n_{\ce{KOH}} = 0.369 \; \text{mol} \; \ce{H2SO4} \times \dfrac{2 \; \text{mol} \; \ce{KOH}}{1 \; \text{mol} \; \ce{H2SO4}} = 0.738 \; \text{mol} \; \ce{KOH} \nonumber$
Calculate the volume of the 5.49 M $\ce{KOH}$ solution needed to neutralize the amount of $\ce{H2SO4}$ present:
$V_{\ce{KOH}} = 0.738 \; \text{mol} \; \ce{KOH} \times \dfrac{1 \; \text{L}}{5.49 \; \text{mol} \; \ce{KOH}} = 0.134 \; \text{L} = 134 \; \text{mL} \nonumber$
Therefore, approximately 134 mL of the 5.49 M $\ce{KOH}$ solution is needed to neutralize 36.2 g $\ce{H2SO4}$.
Q.16
Phosphoric acid is made industrially as a by-product of the extraction of calcium phosphate
$\ce{Ca_3(PO_4)_2 + 6HNO_3 + 12H_2O -> 2H_3PO_4 + 3Ca(NO_3)_2 + 12H_2O} \nonumber$
What volume of 8.5 M phosphoric acid is generated by the reaction of 4.5 metric tons (4500 kg) of calcium phosphate?
Solution
$4500\,\text{kg} \ \ce{Ca3(PO4)2} \times \left(\dfrac{1000\,\text{g}}{1\ \text{kg}} \right) \times \left( \dfrac{1\ \text{mol}}{310 \ \text{g}} \right) \times \left( \dfrac{2 \, \ce{H3PO4}}{1 \, \ce{Ca3(PO4)2}} \right) \times \left( \dfrac{1\ \text{L}}{8.5\ \text{moles}} \right) =3415.56 \, \text{L} \ \ce{H3PO4} \nonumber$
Q.19
Write a balanced equation for the acid-base reaction that leads to the production of each of the following salts.
1. $\ce{BaSO4}$
2. $\ce{MgSO3}$
3. $\ce{PbSO4}$
4. $\ce{AgCl}$
Solution
1. $\ce{Ba(OH)2 (aq) + H2SO4 (aq) <=> BaSO4 (s) + 2 H2O (l)}$
2. $\ce{Mg(OH)2 (s) + H2SO3 (aq) <=> MgSO3 (s) + 2 H2O (l)}$
3. $\ce{Pb(OH)2 (s) + H2SO4 (aq) <=> PbSO4 (s) + 2 H2O (l)}$
4. $\ce{AgOH (s) + HCl (aq) <=> AgCl (s) + H2O (l)}$
Q.21
Hydrogen selenide is classified as a binary acid that has similar properties as hydrogen sulfide. Write down the balanced equation of hydrogen selenide reacting with the base potassium hydroxide and name the salt product of the reaction.
Solution
When the acid $\ce{H2S}$ reacts with the base $\ce{KOH}$, the reaction goes as follows:
$\ce{H2S + 2 KOH -> K2S + 2 H2O} \nonumber$
Forming potassium sulfide.
Similarly, since $\ce{H2Se}$ has similar properties as $\ce{H2S}$, a similar reaction occurs:
$\ce{H2Se + 2 KOH -> K2Se + 2 H2O} \nonumber$
Potassium selenide is the salt product of the reaction.
Q.25
A student working with an unknown solution of hydrochloric acid is assigned the task of finding its concentration. The student places a 200.0 mL sample of the solution in a 350.0 mL Erlenmeyer flask and titrates the solution with a 0.1234 M solution of sodium hydroxide, where a phenolphthalein indicator is used to find the endpoint. The endpoint is found to be 31.416 mL, what is the concentration of the original hydrochloric acid solution sample?
Solution
Remember that in titration of a strong acid and base, when the endpoint is reached the number of moles used is equal, so the number of moles of NaOH at the endpoint are equal to that of HCl. Use this information to find moles of NaOH from the concentration and volume used to titrate.
$\text{Concentration of X} = \dfrac{\text{moles of X}}{\text{liters of X}}\nonumber$
Use dimensional analysis to convert concentration of NaOH to moles using the known amount of volume titrated, in liters:
$0.031416\; \cancel{\text{liter}}\; \ce{NaOH} =\dfrac{0.1234\: \text{mol}\, \ce{NaOH}}{1.00\; \cancel{\text{liter}}\; \ce{NaOH}}= 0.003877\; \text{mol} \nonumber$
Now because we know that the moles of NaOH are equal to the moles of HCl we simply plug in our calculated moles and known original sample volume, in liters, to find the concentration of HCl:
$\text{Concentration of HCl} = \dfrac{0.003877\: \text{mol}\ \ce{HCl}}{0.200\: \text{liters} \ \ce{HCl}}= 0.019385\; \text{M} \; \ce{HCl} \nonumber$
The concentration of the original hydrochloric acid solution sample was, therefore, 0.019385 M.
Q.41
The vapor pressure of salicylic acid $\ce{C7H6O3}$ at 200°C is 0.1598 atm. A 19.0 g sample of ethanol $\ce{C2H6O}$ is dissolved in 60.0 g of salicylic acid. Calculate the vapor pressure of salicylic acid above the resulting solution.
Solution
First find the molar masses of salicylic acid and ethanol. Then use them to compute the chemical amounts of the two in solution.
$n_{\ce{C2H6O}}=19.0\;g\; \ce{C2H6O} \times\dfrac{1\;mol\;\ce{C2H6O}}{46.068\;g\;\ce{C2H6O}}=0.4124\;mol\;\ce{C2H6O}\nonumber$
$n_{\ce{C7H6O3}}=60.0\;g\;\ce{C7H6O3}\times\dfrac{1\;mol\;\ce{C7H6O3}}{138.121\;g\;\ce{C7H6O3}}=0.4344\;mol\;\ce{C7H6O3}\nonumber$
The mole fraction of ethanol in the solution equals
$X_{\ce{C2H6O}}=\dfrac{0.4124}{0.4344+0.4124}=0.4870\nonumber$
The change in vapor pressure of the salicylic acid due to the presence of of the ethanol is
$\Delta{P_{\ce{C4H6O3}}} = -X_{\ce{C4H6O3}}\times{P_{\ce{C4H6O3}}^{\circ}} = -0.5130 \times 0.1518 \, atm= -0.08198$
The final vapor pressure of the salicylic acid equals its original vapor pressure P° plus the change. This is 0.1598 atm minus 0.08198 atm, which is 0.07782 atm.
Q.43
When 53 mol of an unknown compound was added to 100 g of water, the normal boiling point of water increased to 101.1oC. What is the boiling point elevation constant of water? The compound does not dissociate in solution.
Solution
The relevant relationship here is the colligative property associated with boiling point elevation
$\Delta T_{b}=iK_{b}m$
$\Delta T_{b}=101.1 ^{o}\,C-100 ^{o}\,C=1.1^{o}\,C$
Since the compound was told to not dissociation, we know that the van't Hoff factors is 1:
$i=1$
$m=\dfrac{\text{mol solute}}{\text{kg solvent}}=\dfrac{53\,mol}{0.100\,kg}=0.53\frac{mol}{g}$
$1.1^{o}C=(1)\times K_{b}\times (0.53\frac{mol}{g})$
$K_{b}=\dfrac{1.1^{o}C}{0.53\dfrac{mol}{kg}}=2.075\dfrac{^{o}C\ kg}{mol}$
This problem does not give the correct constant though 0.513:
https://chem.libretexts.org/Referenc...perties/B5%3A_Ebullioscopic_(Boiling_Point_Elevation)_Constants
Q.45
When 5.82 g of uniformly dissociating unknown salt, YX, is dissolved in 100.0 g of water, the boiling point of the water is raised by 0.20°C. When 7.09 g of uniformly dissociating unknown salt, ZX, is dissolved in 200.0 g of water, the boiling point of the water is raised by 0.30°C. Assuming both have a van't hoff factor of 2, identify the elements that X, Y, and Z represent.
Solution
For both salts, one can utilize the boiling point elevation expression (including the van't hoff factor):
$\Delta{T}= i\times K_b\times m$
and by substituting moles solute over mass solvent into the molality, m, one can derive that:
$n_{\text{salt}}= \dfrac{\Delta{T}\times m_{\text{water}}}{i\times K_b}$
From this we get the moles of each salt present, and can therefore get their molar mass by dividing their mass by moles present. Next, subtracting the heavier salt by the lighter salt will yield the molar mass of X, $\sim 35.5 \mathrm{\frac{g}{mol}}$. Subtracting this from the molar masses of each salt will then yield Y and Z respectively.
$\Delta T = i \times m\times K_b = 2\times m\times 0.512 = 0.2,\ \text{m of } \ce{YX} = 0.1953 \mathrm{\frac{mol}{kg}};\ m \times 0.1kg = 0.0195\ \mathrm{mol};\ \frac{5.82}{0.0195} = 298.46 \, \mathrm{\frac{g}{mol}}$
$\Delta T = i \times m\times K_b = 2\times m\times 0.512 = 0.3,\ m\ of\ \ce{ZX} = 0.2930 \mathrm{\frac{mol}{kg}};\ m\times 0.2 kg = 0.0586\ \text{mol}; \frac{7.09}{0.0586} = 120.99 \mathrm{\frac{g}{mol}}$
molar mass difference between Y and Z = $298.46 - 120.99 = 177.47 \mathrm{\frac{g}{mol}}$
Q.47
Camphor is a useful agent for determining solute’s molar mass. Camphor’s melting point and $K_f$ are given at $451.55 \, \mathrm{K}$ and $37.7 \mathrm{\frac{K\ kg}{mol}}$. Find the freezing point of the solution of 0.72 g of glucose is dissolved in the 30.0 g of camphor.
Solution
Since we are evaluating glucose as the solute, $i = 1$.
$\Delta T = -iK_{f}m$
First find number of moles of glucose in the solution using molar mass.
$\text{Moles}_{\text{glucose}} = \frac{(0.72g)}{(180.1559\frac{g}{mol})}= 0.0040mol$
Molality then can be determined.
$m= \frac{(0.0040mol)}{(0.030kg)}= 0.13\frac{mol}{kg}$
Freezing then can be easily found as such:
$\Delta T = -iK_{f}m=-(1)(37.7\frac{^oC\ kg}{mol})(0.13\frac{mol}{kg})=-4.9^oC$
The freezing is $451.55K - 4.9K = 446.65K = 173.5^oC$.
Q.48
It is observed that 2 grams of a non-electrolyte solute dissolved in 100 grams of benzene lowered the freezing point of the solution by 0.40°C compared to pure benzene. Given that the freezing point depression constant of benzene is 5.12 K kg/mol, find the molar mass of the solute. If it is given that the molecular formula of the solute is $\ce{X_8}$ where $\ce{X}$ is an element, what is the formula of the solute?
Solution
$\Delta T = -ik_fm \nonumber$
It is given that $\Delta T = 0.40°C$
Since the solution is a non-electrolyte, i=1
Hence,
$\dfrac{0.40}{K_f} = m\nonumber$
$m = \dfrac{\frac{2}{\text{Molecular Mass}}}{100\times10^{-3}} = \dfrac{\frac{2}{\text{Molecular Mass}}}{0.1}\nonumber$
Hence, we get
$\dfrac{0.40\times 0.1}{K_f} = \dfrac{2}{\text{Molecular Mass}}\nonumber$
$\text{Molecular Mass } = \dfrac{2K_f}{0.40\times0.1} \nonumber$
Hence, the Molecular Mass of the solute is $256 \,\frac{g}{mol}$
Since the Molecular Mass of the solute is $256 \,\frac{g}{mol}$, we can say that,
$\text{Molecular Mass of X} = \dfrac{256}{8} = 32\nonumber$
The solute is $\ce{S_8}$.
Q.49
At what temperature does the first ice crystals begin to form in a 27% salt (by mass) aqueous solution of NaCl? As the crystallization of water carries on, the remaining solution becomes more concentrated, so what happens to the freezing point of the solution? Given that the freezing point depression constant $(K_f)$ of water is $1.86\frac{^oC\ kg}{mol}$.
Solution
$27 \%\: \ce{NaCl (aq)} = \dfrac{27 \; g \; \ce{NaCl}}{73 \; g \; \ce{H2O}}$
$molality = \dfrac{27 \; g \; \ce{NaCl}}{73 \; g \; \ce{H2O}} \times \dfrac{mol \; \ce{NaCl}}{58.44 \; g} \times \dfrac{1000 \; g}{1 \; kg} = 6.33 \; m$
$\Delta T = -i\times m\times K_{p} = -2 \times 6.33 \dfrac{mol}{kg} \times 1.86 \dfrac {^{o}C \cdot kg}{mol} = -23.5^{o}C$
The first ice crystal begins to appear at $-23.5^oC$, as the solution becomes more concentrated, its freezing point decreases further because freezing point depression is a colligative property, so the freezing point of a solution is lower than that of the pure solvent and is directly proportional to the molality of the solute.
Q.50
When homemade ice cream is being made, the temperature ranging downward from -3°C are needed. Ice cubes from a freezer have a temperature of about -12°C, which is cold enough; however, when the ice cream mixture is mixed with the ice cubes, the liquid balances out to 0°C, which is too warm. To obtain a liquid that is cold enough, salt NaCl is dissolved in water and ice is added to the saltwater. The salt lowers the freezing point of the water enough so that it can freeze the liquid inside the ice cream maker. The instruction for the ice cream maker say to add one part salt to 11 parts water (by mass). What is the freezing point of this solution in °C? Assume the NaCl dissociates fully into ions, and that the solution is ideal.
Solution
$\Delta T_{f} = -i k_{f} m$
$\dfrac{1.00\ g\ \ce{NaCl}}{53.5\frac{g}{mol}}=0.01869\ mol\ \ce{NaCl}$
$m=\dfrac{0.01869\ mol}{11.00 g(\dfrac{1\ kg}{1000\ g})}=1.699 m$
$x - (-12°C) = - (2)\times (1.86\frac{°C}{m})\times (1.699 m)$
$x = -18.32 °C$
Q.53
A sample of a purified unknown compound is dissolved in toluene, diluting the solvent to a volume of 1.05 mL. The resulting solution has an osmotic pressure of 0.025 atm at 275 K. If the solute has a molar mass of 46.06 g/mol, how many grams of the unknown compound were added?
Solution
$\Pi$ = $\dfrac{nRT}{V}$
and we know $T= 275K$, $R=0.0821\dfrac{L\ atm}{mol\ K}$ and $\Pi$=0.025 atm
Thus V= 1.05 mL
Since we need V in liters because the units of R are $\dfrac{L\ atm}{mol\ K}$ , we use unit conversions:
$1.05\ mL\times \dfrac{1 L}{1000 mL} = 0.00105\ L$
Plug in the values we know to find n
$0.025= \dfrac{n\times 0.0821\times 275}{0.00105}$
$\dfrac{0.025\times 0.00105}{0.0821\times 275} = n$
$n=1.163\times 10^{-6}$
Now that we know number of moles we just have to multiply the number of moles by the molar mass to find grams of substance.
$(1.163\times 10^{-6}) \times 46.06844\frac{g}{mol}=5.356\times 10^{-5}g$
Q.57
An amount of hydrogen gas ($\ce{H_2}$) with a partial pressure of 4.6 atm is dissolved in water and this solution is sealed. The Henry's Law constant K for this solution at $25^oC$ is $7.8\times 10^{-4} \mathrm{\frac{M}{atm}}$.
1. How many moles of hydrogen gas will dissolve per liter of water? Assume the density of water is $1.0 \mathrm{\frac{grams}{cm^3}}$ and has not been affected by the temperature.
2. What will the dissolved gas molecules do if the sealed solution is unsealed?
Solution
a) According to Henry's Law,
$C = k \times P_{\text{gas}}\nonumber$
In this equation, C = concentration of gas (M), k = Henry's Law constant $(\frac{M}{atm})$, and $P_{\text{gas}}$ = partial pressure of gas.
So, given k and $P_{\text{gas}}$:
$C = 7.8 \times 10^{-4} \dfrac{M}{atm} \times 4.6 \; atm\nonumber$
$C = 0.003588 \dfrac{mol \; \ce{H_2}}{L_{\text{water}}}\nonumber$
b) The partial pressure of $\ce{H_2}$ in the Earth's atmosphere is less than 1 atm. When the solution is unsealed, the $\ce{H_2}$ molecules will no longer be under enough pressure to dissolve in water, so they will quickly bubble out of the solution and escape as gas into the surroundings.
Q.59
Propene is a colorless organic compound with a weak but unpleasant smell. However, a company can take advantage of its chemical properties to produce propanol:
$\ce{C3H6 +H2O -> CH3(CH2)2OH} \nonumber$
• The heat of formation of propene is $20.4 \mathrm{\frac{kJ}{mol}}$.
• The heat of formation of $\ce{H2O}$ is $-241.8 \mathrm{\frac{kJ}{mol}}$.
• The heat of formation of $\ce{CH3(CH2)2OH}$ is $-104.6 \mathrm{\frac{kJ}{mol}}$.
What suggestions can you give to the company regarding the conditions of pressure and temperature that will have a maximum yield of propanol at equilibrium without calculating the actual data?
Solution
$\Delta H^{o}_{f(\text{propene})}=20.4\dfrac{kJ}{mol}$
$\Delta H^{o}_{f(\text{water})}=-241.8\dfrac{kJ}{mol}$
$\Delta H^{o}_{f(\ce{CH3(CH2)2OH})}=-104.6\dfrac{kJ}{mol}$
$\Delta H^o = \Delta H^o_{f(\ce{CH3(CH2)2OH})} - \Delta H^o_{f(\text{propene})} - \Delta H^{o}_{f(\text{water})}$
$=-104.6\dfrac{kJ}{mol} - (-241.8\dfrac{kJ}{mol}) - 20.4\dfrac{kJ}{mol}\=116.8\dfrac{kJ}{mol}>0$
It is an endothermic reaction. As a result, a higher temperature is required to maximize the yield of product.
Also, according to the ideal gas law, because the moles of gas are proportional to the volume, and the volume is inverse proportional to the pressure; a higher pressure is required to maximize the yield of product.
Q.61
Suppose heptane and octane form an ideal solution. At $20.0\text{°C}$, pure heptane has a vapor pressure of $\ce{P_1^°}$= $0.0562\ \text{atm}$ while pure octane has a vapor pressure of $\ce{P_2^°}$= $0.0145\ \text{atm}$. What is the mole fraction of heptane vapor over a solution of heptane and octane if the solution contained $0.300\ \text{mol}$ heptane and $0.700 \ \text{mol}$ octane?
Solution
Since we have been told to assume the solution is ideal, we can use Raoult's law to obtain the partial pressures of heptane and octane, respectively.
$P_{\text{heptane}} = \left( \dfrac{0.300\ \text{mol}}{0.300\ \text{mol} + 0.700 \ \text{mol}} \right) \times \left( 0.0562\ \text{atm} \right) = 0.01686\ \text{atm}$
$P_{\text{octane}} = \left( \dfrac{0.700\ \text{mol}}{0.300\ \text{mol} + 0.700 \ \text{mol}} \right) \times \left( 0.0145\ \text{atm} \right) = 0.01015\ \text{atm}$
Using the relationship between partial pressure and mole fraction, we can obtain the mole fraction of heptane over the solution. Partial pressure and mole fraction can be related through the ideal gas law $P = n\dfrac{RT}{V}$. Which shows that under constant temperature and pressure, pressure and number of moles are directly related.
$\chi_{\text{heptane}} = \left( \dfrac{ 0.01686\ \text{atm}}{ 0.01686\ \text{atm} + 0.01015\ \text{atm}} \right) = 0.624$
Q.63
At 50oC, benzene ($\ce{C6H6}$) has a vapor pressure of 42.5 kPa and isobutene ($\ce{C3H8}$) has a vapor pressure of 0.83 torr. When 10.0 g of benzene and 30 g of isobutene are mixed in a solution, they form a nearly ideal solution.
1. Calculate the mole fraction of benzene in the solution.
2. What is the total vapor pressure above the solution in kPa at 50oC ?
3. Calculate the mole fraction of isobutene that exists above the solution in the vapor phase.
Solution
The mole fraction is equal to:
$\left (\dfrac{X_A\: mol}{X_A + X_B\: mol} \right )\nonumber$
Use dimensional analysis to convert grams to moles:
$(10.0\; \cancel{g\; \ce{C6H6}}) \left(\dfrac{1\: mol}{78.108\: \cancel{ g\; \ce{C6H6}}} \right) = 0.128\; mol \nonumber$
$(30.0\; \cancel{g \; \ce{C3H8}} ) \left (\dfrac{1\: mol}{44.094\: \cancel{g\; \ce{C3H8}}} \right ) = 0.680\; mol \nonumber$
Let:
$X_A = \text{ the moles of } \ce{C6H6} = 0.128\; mol$, $X_B = \text{ the moles of } \ce{C3H8} = 0.680\; mol$,
${\chi_B}=\dfrac{0.0929\;kPa}{6.81\;kPa} = 0.0136 \nonumber$
Plug into the equation and solve:
$\left (\dfrac{0.128\: mol}{0.128 + 0.680\: mol} \right )= 0.158 \nonumber$
b) Since the problem states that the result of mixing the two solutions is nearly ideal, we can assume an ideal solution is formed and use Raoult’s Law and we can choose to keep pressure in units of kPa OR torr.
lt’s Raoult's Law:
${P_A} = {\chi_A}\times{P^o_A}$
Here we have chosen to use units of kPa:
$P_{\ce{C6H6}} = \chi_{\ce{C6H6}}\times P^o_{\ce{C6H6}} \nonumber$
$P_{\ce{C6H6}} = 0.158\times 42.5\; kPa = 6.72\;kPa\nonumber$
To convert torr to kPa:
$(0.83\; \cancel{torr} ) \left (\dfrac{0.133\: kPa}{1\: \cancel{torr}} \right ) = 0.1104\; kPa\nonumber$
$P_{\ce{C3H8}} = \chi_{\ce{C3H8}} \times P^o_{\ce{C3H8}} \nonumber$
$P_{\ce{C3H8}} = 0.842\times 0.1104\; kPa= 0.0929\;kPa\nonumber$
Then we add these up to find the total pressure:
${P_{\text{Total}}} = {\chi_A}{P^o_A} + {\chi_B}{P^o_B} \nonumber$
${P_{\text{Total}}} = 6.72\;kPa + 0.0929\;kPa \ = 6.81\;kPa\nonumber$
c) To find the mole fraction of isobutene only in the vapor phase, we can use partial pressure to represent the amount of gas, becase in mixed gas $n \propto p$:
$\chi_{\text{isobutene}} = \frac{n_{\text{isobutene}}}{n_{\text{tot}}} = \frac{P_{\text{isobutene}}}{P_{\text{tot}}} = \frac{0.83 \, \text{torr}}{6.81\, \text{kPa}} = 0.0162 \nonumber$
Q.65
A solution is prepared by mixing 2g of $\ce{CaCl2 (s)}$, 2g of $\ce{MgCl2 (s)}$, and 3g of $\ce{MnO2 (s)}$ in enough water to make a total volume of 150 mL. Calculate the concentration of chlorine in the solution in grams per liter.
Solution
$m_{\ce{Cl}}=2g\ \ce{CaCl2} (\frac{2(35.5g\ \ce{Cl})}{110.98g\ \ce{CaCl2}}) + 2g\ \ce{MgCl2} (\frac{2(35.5g\ \ce{Cl})}{95.12g\ \ce{MgCl2}})$
$m_{\ce{Cl}}=2.77g\ \ce{Cl}$
$\text{mass of chlorine per liter: } \frac{2.77g\ \ce{Cl}}{150 \times 10^{-3} L}=18.5 \frac{g}{L}$
Q.69
Chromium ion, $\ce{Cr^{2+}}$, is a good reducing agent, often being itself turned into $\ce{Cr^{3+}}$. Suppose that 10 mL of 0.1 M $\ce{Cr(OH)2}$ was needed to reduce completely an 0.05 grams of unknown substance X.
1. If each molecule of X accept just one electron, what is the molecular weight of X?
2. If each molecule of X accept five electrons, what is the molecular weight of X?
Solution
(a) Number of moles of $\ce{Cr(OH)2}$ used = $M\times V$
$= 0.1\times 10\ \mathrm{mL} = 1\ \mathrm{mmol} = 10^{-3}\ \mathrm{moles}$
Since only one mole of X was used for every mole of Chromium Hydroxide, the number of moles reacted are equal.
$\frac{\text{Mass}}{\text{Molar mass}}$ = number of moles, hence for X,
$\frac{0.05\ \mathrm{g}}{(0.01 \mathrm{L} \times 0.1 \mathrm{M})}= 50 \mathrm{\frac{g}{mol}}$
(b) $\frac{0.05\ \mathrm{g}}{(0.01 \mathrm{L} \times 0.1 \mathrm{M} \times 5)}= 10 \mathrm{\frac{g}{mol}}$
Q.71
A 0.200 g sample of chloride with unknown purity was dissolved in 200 mL of water was then titrated with $\ce{AgNO3}$ solution with a molarity of 0.1 M. After the titration it was determined that 28 mL of $\ce{AgNO3}$ was required to fully titrate the sample of chloride. What's the mass percent of chloride in the original dry sample?
Solution
First write the precipitation reaction:
$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)} \nonumber$
Next we need to determine the amount of moles of $\ce{Ag}$ used during the titration:
$\textrm{mol Ag}^{+}=\textrm{28} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.1\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=2.80\times10^{-3}\textrm{ mol Ag}^{+}\nonumber$
Now that we know the amount of moles of Ag reacted because Ag and Cl combine at a one to one ration we can find the mass of Cl used:
$\textrm{mass Cl}^{-}=2.80\times10^{-3}\cancel{mol}\;\left(\dfrac{35.45\textrm{ g}}{\textrm{1} \cancel{mol}\;}\right )=9.92\times10^{-2}\textrm{ g Cl}^{-}\nonumber$
Now that we know the mass of Cl in the unknown we can calculate the mass percent:
$100\textrm{%}\;\left(\dfrac{0.0992\textrm{ g}}{\textrm{0.2} \textrm{g}\;}\right )=49.63\textrm{%}\nonumber$
Q.73
At 20°C, the vapor pressure of pure water is 17.5 mmHg. When 110 g of a substance $\ce{X}_y$ is dissolved in 500 g of water, the vapor pressure of the solution is 15.0 mmHg. Given that this substance has a molecular mass of 12.011 g/mol, what is the molecular formula of this substance $\ce{X}_y$. (i.e. what is the number value of subscript y).
Solution
For this question, we can use Raoult's Law
$P_{\text{solution}} = \chi_{\text{solvent}} \times P°_{\text{solvent}}\nonumber$
First, we plug in the known values of the pressure of the solution and the pressure of the pure solvent, water, to find the mole fraction.
$15.0 \; \mathrm{mmHg} = \chi_{\text{solvent}} \times 17.5 \; \mathrm{mmHg}$
$\chi_{\text{solvent}} = 0.8571$
Next, we can calculate the the number of moles of the solvent from its mass and molecular weight.
$\text{Moles of solvent} = 500 \; \mathrm{g} \times \dfrac {1}{18.015 \; \mathrm{g/mol}} = 27.75 \; \mathrm{moles}$
Next, we can calculate the moles of the solute from the mole fraction of the solvent
$0.8571 = \dfrac{27.75 \; \mathrm{moles}}{27.75 \; \mathrm{moles} \; + \text{ moles of } X_y}$
$\text{moles of } X_y = 4.63 \; \mathrm{moles}$
From here, we can calculate the molecular formula of the substance by dividing the amount of grams per mole used in the solution by the molecular mass of the substance.
$\dfrac {110 \; \mathrm{g}}{4.63 \; \mathrm{moles}} = 23.76 \; \mathrm{g/mol}$
$(23.76 \; \mathrm{g/mol}) \times (\dfrac {1}{12.011 \; \mathrm{g/mol}}) = 1.98$
$1.98 \approx 2$
Thus, the molecular formula of our substance is $\ce{X_2}$.
Q.81
The cell membrane of a plant cell is permeable to water. In a salt solution, the plant cells shrivel (lose water) when the concentration is high, and swell (take up water) when the concentration is low. Scientists perform an experiment at 30°C, where an aqueous solution of NaCl with a freezing point of -0.028°C is used to water the plant cells. This causes the plant cells to neither swell nor shrink indicating that the osmotic pressure of the plant cell contents is equal to that of the NaCl solution. Assuming the molarity and molality of the plant cell contents are equal, calculate the osmotic pressure of the solution inside the plant cells. The $K_f$ for the solution is $1.86 \mathrm{\frac{K\ kg}{mol}}$.
Solution
Plug in the given values into the equation for freezing point depression to solve for the molality of the solution
$\Delta T_{f}=-K_{f} m\nonumber$
$-0.0218=-1.86 K \cdot kg \cdot mol^{-1} m \nonumber$
$m=0.015\dfrac{kg}{mol}\nonumber$
Because the plant cell neither swells nor shrinks, the molality of the solution is equivalent to the molality of the plant cell contents. According to the question, assume the molarity and molality are equal so concentration
$c=0.15M\nonumber$
Use the following equation to calculate for the osmotic pressure, π.
\begin{align*} \pi&=cRT \[4pt] &=(0.15M)(0.08206\frac{L\ atm}{mol\ K})(303.15K) \[4pt] &=0.374atm \end{align*}
Q.83
Jim loves the smell of grass. He loves it so much that he frequently sniffs the fumes from 1-Hexanol, which is a chemical compound that is known to smell like grass trimmings. Because of his love, he decides to make his garage into a “gas den”, and aimed to have 60% of the air in it as 1-Hexanol gas. He achieves this by filling a pool with a mixture of 1-Hexanol and Water. Assume that the air in the “gas den” consists of only the fumes from this mixture. If the pool has 10000L of water in it, how many litres of 1-Hexanol does he need? Use Henry’s Law to determine this amount.
Relevant Information:
1-Hexanol:
• Density: $814\dfrac{kg}{m^3}$
• Molar mass: $107.1748\dfrac{g}{mol}$
• Henry’s Law Constant: $64\dfrac{molal}{bar}$
Water:
• Density: $1000\dfrac{kg}{m^3}$
Solution
This problem asks for the amount, in litres, of 1-Hexanol that is required to make Jim’s dreams come true. The information given is the density of 1-Hexanol and its molar mass, but since they don’t have very much to do with pressure, they can be set aside for now. The question explicitly mentions that Henry’s Law should be used, so start be identifying what Henry’s Law actually does:
$C = kP \nonumber$
Where C is concentration, k is Henry’s Law constant, and P is partial pressure. The question gives out the fact that Jim wants the air in his garage to be 60% 1-Hexanol vapor. Remember the total pressure of air is 1 atm ≈ 1 bar (ignore the error here). Therefore, the partial pressure of 1-hexanol is $1 \, \text{bar} \times 60\% = 0.6 \, \text{bar}$. The Henry’s Law constant is also provided, and by plugging those two in, the concentration of 1-Hexanol in the desired “solution” (more accurately, giant pool of chemicals) is found. This process is shown below:
$C= kP = (64\dfrac{molal}{bar})\times 0.6 \, \text{bar} = 38.4\nonumber$
This concentration, from the units of k, is in molality, which is $\dfrac{mol_{\text{solute}}}{kg_{\text{solvent}}}$. Now all that is left is to convert molality to litres. By utilizing the molar mass of 1-Hexanol:
$38.4\dfrac{mol_{\text{solute}}}{kg_{\text{solvent}}}\times{107.1748\dfrac{g}{mol}}\approx{4115.51\dfrac{g_{\text{solute}}}{kg_{\text{solvent}}}}\nonumber$
We know that the solvent is water, and that water has a density of $1000\dfrac{kg}{m^3}$. Also, the conversion rate of $m^{3}$ to $L$ is 1:1000. Therefore, the equation can be solved as follows.
$4115.51 \dfrac{g_{\text{solute}}}{kg_{\text{solvent}}} \times 1000 \dfrac{kg_{\ce{H2O}}}{m^3} \times \dfrac{m^3}{1000 \, L_{\ce{H2O}}} \approx 4115.51 \dfrac{g_{\text{solute}}}{L_{\text{solvent}}}\nonumber$
We now have the value for how many grams of solute (1-Hexanol) there should be per litre of solvent. simply divide this value by the density to obtain litres of 1-Hexanol per litre of water. Multiply this number by the amount of water in the pool to get the desired amount of solute that is needed for the solution! (note that the density was converted to $\dfrac{g}{L}$,
$4115.51 \dfrac{g_{\text{solute}}}{L_{\text{solvent}}} \div \left( 814000\dfrac{g_{\text{1-hexanol}}}{L} \right) \times 10000L_{\ce{H2O}}=50.56 \, \text{L} \nonumber$
Therefore, Jim needs 50,560L of 1-Hexanol to put in his pool. That's going to take a big pool!
Therefore, Jim needs 50.56L of 1-Hexanol to put in his pool.
Please note that, while various units do appear in the conversions, $\ce{H2O}$ can be considered as solute and 1-Hexanol as the solvent, so the units can cancel.
Abstract: Utilize henry's law to find concentration (molality). From that, convert molality to $g_{\text{solute}}$ to $L_{\text{solute}}$.
Q.87
A 3.5 gram sample is decomposed into a compound containing nitrogen, hydrogen, and oxygen, together creating 4.00 g of $\ce{NO2}$ and 7.00 g of $\ce{H2O}$. The molar mass of the sample is $78.5 \mathrm{\frac{g}{mol}}$. Find the molecular formula of this sample.
Solution
First, the moles of each Nitrogen, Hydrogen, and Oxygen in solution are calculated:
$mol_N = 4g\ \ce{NO2} \times (\frac{1\ mol\ \ce{NO2}}{46.005\frac{g}{mol}})\times (\frac{1\ mol\ N}{1\ mol\ \ce{NO2}}) = 0.0869\ mol$
$mass_N = (14.00\frac{g}{mol})(0.0869mol) = 1.2172g\ \ce{N}$
$mol_H = 7g\ \ce{H_2O} \times (\frac{1\ mol\ H_2O}{18.02\frac{g}{mol}})\times (\frac{2\ mol\ H}{1\ mol\ \ce{H_2O}}) = 0.7769\ mol$
$mass_H = (1.01\frac{g}{mol})(0.7769 mol) = 0.7847g\ \ce{H}$
From these values, moles of $O$ can be calculated to be:
$3.5\ gram\ sample = grams_N + grams_H + grams_O$
$3.5 = 1.2172 g + 0.7847 g + grams_O$
$grams_O = 1.4981g$
$mol_O = \frac{1.4981g}{15.9994\frac{g}{mol}} = 0.0936\ mol$
The mole values for each element create a ratio of $\mathrm{H}_{0.7769}\,\mathrm{N}_{0.0869}\,\mathrm{O}_{0.0936}$, creating an empirical formula $\ce{H9NO}$
From here, the molecular formula can be established using the molar mass:
$\text{Empirical formula mass} = 9(1.01\frac{g}{mol\ H}) + (14.00\frac{g}{mol\ N}) + (15.9994\frac{g}{mol\ O})$
$\text{Empirical formula mass} = 39.0894\frac{g}{mol}$
$\text{Sample Mass} = 78.5\frac{g}{mol}$
$\frac{78.5\frac{g}{mol}}{39.084\frac{g}{mol}} = 2.00$
$2(\ce{H9NO}) = \ce{H18N2O2} = \text{Molecular Formula}$
Q.87
5 grams of a substance containing hydrogen, carbon and oxygen were fully combusted and produced 3.00 grams of $\ce{H2O}$ and 7.33 grams of $\ce{CO2}$. When 10.64 grams of the substance was mixed with 0.5 kg of water, the freezing point was found to decrease by 0.0605°C. What is the substance?
Solution
The generic equation for this reaction is
$\ce{C_{x} H_{y} O_{z} + O_{2} -> H_{2}O + CO_{2}} \nonumber$
The molar masses of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol respectively. Therefore the molar masses of water and carbon dioxide are $18.016\frac{g}{mol}$ and $44.01\frac{g}{mol}$. These values can be used to determine the mass percent of hydrogen in water
$\dfrac {2 (1.008)\; \frac{g}{mol}} {18.016\; \frac{g}{mol}} \times 100 =11.2\% \nonumber$
And the mass percent of carbon in carbon dioxide
$\dfrac{12.01\; \frac{g}{mol}} {44.01\; \frac{g}{mol}} \times 100 =27.3\% \nonumber$
The mass percents can be used to determine the amount of hydrogen and carbon are present.
$3.00\; g \times 0.112 = 0.336\; g\; \text {of Hydrogen} \nonumber$
$7.33\; g \times 0.273 = 2.001\; g\; \text {of Carbon} \nonumber$
By subtracting the masses of Carbon and Hydrogen from the original mass of the substance, the mass of oxygen present in the substance can be found
$5\; g - (0.336\; g + 2.001\;g) = 2.663\; g\; \text {of Oxygen} \nonumber$
By dividing the masses present by the molar mass, the amount of moles of each element can be determined
$\dfrac{0.112\; g}{1.008\; \frac{g}{mol}} = 0.333\; mol\; \text {of Hydrogen} \nonumber$
$\dfrac{2.001\; g}{12.01\; \frac{g}{mol}} = 0.167\; mol\; \text {of Carbon} \nonumber$
$\dfrac{2.663\; g}{16.00\; \frac{g}{mol}} = 0.166\; mol\; \text {of Oxygen} \nonumber$
By looking at the amount of moles of each element, it can be noticed that there are equal amounts of carbon and oxygen and twice as much hydrogen as carbon or oxygen. Therefore the ratio of $\ce{C}:\ce{H}:\ce{O}$ in the substance is $1:2:1$. However, the molar mass of the substance remains unknown, but it can be found using the information about the freezing point depression.
From the formula for freezing point depression
$\Delta T=-i \times k_{f} \times m\nonumber$
$-0.0605°C = -1 \times 0.512°C\; m^{-1} \times m \nonumber$
$m=\dfrac{0.0605°C}{0.512\frac{°C}{m}}=0.118\; \frac{mol}{kg}\nonumber$
By multiply by the mass of solvent (water), we can determine the amount of moles that was dissolved in the water
$0.118\; \frac{mol}{kg} \times 0.5\; kg = 0.059\; mol\nonumber$
The last step in finding the molar mass of the substance is to divide the mass dissolved by the moles dissolved
$\dfrac{10.64\;g}{0.059\; mol} = 180.3\; \frac{g}{mol}\nonumber$
Now we can find the formula of the substance. The first step is to find the number of moles of the substance combusted
$\dfrac {5\;g} {180.3\; \frac{g}{mol}} = 0.0277\; mol\nonumber$
We can compare this to the number of carbon moles combusted to determine the number of carbon atoms per molecule of the substance
$\dfrac {0.167\; g \text{ of Carbon}\;} {0.0277\; mol} = 6 \text{ Carbon atoms} \nonumber$
When we apply the number of carbon atoms to the ratio of elements found earlier, the chemical formula comes to be $\ce{C_{6}H_{12}O_{6}}$ which is glucose.
Answer: Glucose | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.E%3A_Solutions_%28Exercises%29.txt |
This chapter introduces you to thermochemistry, a branch of chemistry that describes the energy changes that occur during chemical reactions. In some situations, the energy produced by chemical reactions is actually of greater interest to chemists than the material products of the reaction. For example, the controlled combustion of organic molecules, primarily sugars and fats, within our cells provides the energy for physical activity, thought, and other complex chemical transformations that occur in our bodies. Similarly, our energy-intensive society extracts energy from the combustion of fossil fuels, such as coal, petroleum, and natural gas, to manufacture clothing and furniture, heat your home in winter and cool it in summer, and power the car or bus that gets you to class and to the movies.
12: Thermodynamic Processes and Thermochemistry
Learning Objectives
• To understand the concept of energy and its various forms.
• To know the relationship between energy, work, and heat.
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
Forms of Energy
The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure $1$). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms.
Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.
Energy can be converted from one form to another (Figure $2$) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction.
Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of energy: Energy cannot be created or destroyed.
Kinetic and Potential Energy
The kinetic energy of an object is related to its mass $m$ and velocity $v$:
$KE=\dfrac{1}{2}mv^2 \label{12.1.4}$
For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is
\begin{align} KE &=\dfrac{1}{2}(1360\, kg)(26.8 \, ms)^2 \label{12.1.5} \[4pt] &= 4.88 \times 10^5 g \cdot m^2\nonumber \end{align}
Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is named after the British physicist James Joule (1818–1889), an early worker in the field of energy, is defined as 1 kilogram·meter2/second2 (kg·m2/s2). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 103 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 105 J or 4.88 × 102 kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.
To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is
$w = F\,d.$
The force ($F$) exerted by gravity on any object is equal to its mass ($m$, in this case, 1360 kg) times the acceleration ($a$) due to gravity (g, 9.81 m/s2 at Earth’s surface). The distance ($d$) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:
\begin{align} PE &= F\;d = m\,a\;d = m\,g\,h \label{12.1.6} \[4pt] &=(1360\, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m)\nonumber \[4pt] &= 4.88 \times 10^5\; \frac{Kg \cdot m^2}{s^2}\nonumber \[4pt] &=4.88 \times 10^5 J = 488\; kJ\nonumber \end{align}
The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h.
If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero.
Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.
Units of Energy
The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5 °C to 15.5 °C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1 °C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0 °C–100 °C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
\begin{align} 1 \;cal &\equiv 4.184 \;J \; \quad \text{(exactly)} \label{12.1.7a} \[4pt] 1 \;J &= 0.2390\; cal \label{12.1.7b} \end{align}
In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information.
Example $1$: Kinetic Energy of Baseballs
1. If the mass of a baseball is 149 g, what is the kinetic energy of a fastball clocked at 100 mi/h?
2. A batter hits a pop fly, and the baseball (with a mass of 149 g) reaches an altitude of 250 ft. If we assume that the ball was 3 ft above home plate when hit by the batter, what is the increase in its potential energy?
Given
• mass and velocity or height
Asked for
• kinetic and potential energy
Strategy
Use Equation $\ref{12.1.4}$ to calculate the kinetic energy and Equation $\ref{12.1.6}$ to calculate the potential energy, as appropriate.
Solution
1. The kinetic energy of an object is given by $\frac{1}{2} mv^2$ In this case, we know both the mass and the velocity, but we must convert the velocity to SI units: \begin{align*} v&= \left(\dfrac{100\; \cancel{mi}}{1\;\cancel{h}} \right) \left(\dfrac{1 \;\cancel{h}}{60 \;\cancel{min}} \right) \left(\dfrac{1 \; \cancel{min}}{60 \;s} \right)\left(\dfrac{1.61\; \cancel{km}}{1 \;\cancel{mi}} \right) \left(\dfrac{1000\; m}{1\; \cancel{km}}\right) \[4pt] &= 44.7 \;m/s \end{align*}
The kinetic energy of the baseball is therefore (via Equation $\ref{12.1.4}$) \begin{align*} KE &= \dfrac{1}{2} 149 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2 \[4pt] &= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2} \[4pt] &= 1.49 \times10^2\; J \end{align*}
2. The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus \begin{align*} PE &= 149\;\cancel{g} \left(\dfrac{1\; kg}{1000\; \cancel{g}} \right)\left(\dfrac{9.81\; m}{s^2} \right) \left(247\; \cancel{ft} \right) \left(\dfrac{0.3048\; m}{1 \;\cancel{ft}} \right) \[4pt] &= 1.10 \times 10^2 \dfrac{kg⋅m^2}{s^2} \[4pt] &= 1.10 \times 10^2\; J \end{align*}
Exercise $1$
1. In a bowling alley, the distance from the foul line to the head pin is $59 \,ft$ and $10 \,\frac{13}{16} in.$ (18.26 m). If a 16 lb (7.3 kg) bowling ball takes 2.0 s to reach the head pin, what is its kinetic energy at impact? (Assume its speed is constant.)
2. What is the potential energy of a 16 lb bowling ball held 3.0 ft above your foot?
Answer a
3.10 × 102 J
Answer b
65 J
Systems and Surroundings
To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure $3$). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa.
Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved.
The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure $4$). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose.
Summary
Thermochemistry is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. Energy is the capacity to do work. Mechanical work is the amount of energy required to move an object a given distance when opposed by a force. Thermal energy is due to the random motions of atoms, molecules, or ions in a substance. The temperature of an object is a measure of the amount of thermal energy it contains. Heat (q) is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of potential energy (PE), energy caused by the relative position or orientation of an object. Kinetic energy (KE) is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the law of conservation of energy states that energy can be neither created nor destroyed. The most common units of energy are the joule (J), defined as 1 (kg·m2)/s2, and the calorie, defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.1%3A_Systems_States_and_Processes.txt |
Learning Objectives
• To calculate changes in internal energy
• Distinguish the related properties of heat, thermal energy, and temperature
• Define and distinguish specific heat and heat capacity, and describe the physical implications of both
• Perform calculations involving heat, specific heat, and temperature change
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows:
$\Delta U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{12.4.1a}$
or
$\Delta{U_{sys}} =−ΔU_{surr} \label{12.4.1b}$
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the $\ce{CO_2}/\ce{H_2O}$ product mixture is less than that of the isooctane $\ce{O_2}$ reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred ($q$) and the work done ($w$):
$\underbrace{ΔU_{sys} = q + w}_{\text{First Law of Thermodynamics}} \label{12.4.2}$
Although $q$ and $w$ are not state functions on their own, their sum ($ΔU_{sys}$) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
Although $q$ and $w$ are not state functions, their sum ($ΔU_{sys}$) is independent of the path taken and therefore is a difference of a state function.
Example $1$
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy ($ΔU$) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
1. Determine the sign of $q$ to use in Equation $\ref{12.4.2}$.
2. From Equation $\ref{12.4.2}$ calculate $w$ from the values given. Substitute this value into Equation $\ref{12.4.2}$ to calculate $ΔU$.
Solution
A From Equation $\ref{12.4.2}$, we know that $ΔU = q + w$ (First Law of Thermodynamics). We are given the magnitude of $q$ (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, $q$ is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so $w$ must be positive. From Equation $\ref{12.4.2}$,
\begin{align} w&=-P_{\textrm{ext}}\Delta V\nonumber \[4pt] &=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)\nonumber \[4pt] &=284\textrm{ J}\nonumber \end{align}\nonumber
Thus
\begin{align} ΔU &= q + w\nonumber \[4pt] &= −140\; J + 284 \;J\nonumber \[4pt] &= 144\; J\nonumber \end{align}\nonumber
In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise $1$
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy ($ΔU$) of the gas in joules?
Answer
−216 J
By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Summary
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the $PV$ work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for $ΔU$. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.2%3A_The_First_Law_of_Thermodynamics_-_Internal_Energ.txt |
Learning Objectives
• Explain the technique of calorimetry
• Calculate and interpret heat and related properties using typical calorimetry data
• To use calorimetric data to calculate enthalpy changes.
Heat Capacity
We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity ($C$) of a body of matter is the quantity of heat ($q$) it absorbs or releases when it experiences a temperature change ($ΔT$) of 1 degree Celsius (or equivalently, 1 kelvin)
$C=\dfrac{q}{ΔT} \label{12.3.1}$
Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance.
For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast iron frying pan is found by observing that it takes 18,140 J of energy to raise the temperature of the pan by 50.0 °C
$C_{\text{small pan}}=\dfrac{18,140\, J}{50.0\, °C} =363\; J/°C \label{12.3.2} \nonumber$
The larger cast iron frying pan, while made of the same substance, requires 90,700 J of energy to raise its temperature by 50.0 °C. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:
$C_{\text{large pan}}=\dfrac{90,700\, J}{50.0\,°C}=1814\, J/°C \label{12.3.3} \nonumber$
The specific heat capacity ($c$) of a substance, commonly called its specific heat, is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):
$c = \dfrac{q}{m\Delta T} \label{12.3.4}$
Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property—the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g. The specific heat of iron (the material used to make the pan) is therefore:
$c_{iron}=\dfrac{18,140\; J}{(808\; g)(50.0\;°C)} = 0.449\; J/g\; °C \label{12.3.5} \nonumber$
The large frying pan has a mass of 4040 g. Using the data for this pan, we can also calculate the specific heat of iron:
$c_{iron}=\dfrac{90,700 J}{(4,040\; g)(50.0\;°C)}=0.449\; J/g\; °C \label{12.3.6} \nonumber$
Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of J/mol °C (Figure $1$).
The heat capacity of an object depends on both its mass and its composition. For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (Cp) is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of Cp are thus J/(mol•°C).The subscript p indicates that the value was measured at constant pressure. The specific heat ($c_s$) is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C).
We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change either via moles (Equation $\ref{12.3.7}$) or mass (Equation $\ref{12.3.8}$):
$q = nc_p ΔT \label{12.3.7}$
where
• $n$ is the number of moles of substance and
• $c_p$ is the molar heat capacity (i.e., heat capacity per mole of substance), and
• $ΔT= T_{final} − T_{initial}$ is the temperature change.
$q = mc_s ΔT \label{12.3.8}$
where
• $m$ is the mass of substance in grams,
• $c_s$ is the specific heat (i.e., heat capacity per gram of substance), and
• $ΔT= T_{final} − T_{initial}$ is the temperature change.
Both Equations \ref{12.3.7} and \ref{12.3.8} are under constant pressure (which matters) and both show that we know the amount of a substance and its specific heat (for mass) or molar heat capcity (for moles), we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost.
The specific heats of some common substances are given in Table $1$. Note that the specific heat values of most solids are less than 1 J/(g•°C), whereas those of most liquids are about 2 J/(g•°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g•°C), is one of the highest known. The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table $1$.
Table $1$: Specific Heats of Common Substances at 25 °C and 1 bar
Substance Symbol (state) Specific Heat (J/g °C)
helium He(g) 5.193
water H2O(l) 4.184
ethanol C2H6O(l) 2.376
ice H2O(s) 2.093 (at −10 °C)
water vapor H2O(g) 1.864
nitrogen N2(g) 1.040
air mixture 1.007
oxygen O2(g) 0.918
aluminum Al(s) 0.897
carbon dioxide CO2(g) 0.853
argon Ar(g) 0.522
iron Fe(s) 0.449
copper Cu(s) 0.385
lead Pb(s) 0.130
gold Au(s) 0.129
silicon Si(s) 0.712
quartz $\ce{SiO2 (s)}$ 0.730
The value of $C$ is intrinsically a positive number, but $ΔT$ and $q$ can be either positive or negative, and they both must have the same sign. If $ΔT$ and $q$ are positive, then heat flows from the surroundings into an object. If $ΔT$ and $q$ are negative, then heat flows from an object into its surroundings.
If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, then $ΔT>0$ and $q$ is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, so $ΔT<0$ and $q$ is negative.
Example $1$: Measuring Heat
A flask containing $8.0 \times 10^2\; g$ of water is heated, and the temperature of the water increases from $21\, °C$ to $85\, °C$. How much heat did the water absorb?
Solution
To answer this question, consider these factors:
• the specific heat of the substance being heated (in this case, water)
• the amount of substance being heated (in this case, 800 g)
• the magnitude of the temperature change (in this case, from 21 °C to 85 °C).
The specific heat of water is 4.184 J/g °C (Table $1$), so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C).
This can be summarized using Equation \ref{12.3.8}:
\begin{align*} q&=mc_sΔT \nonumber \[4pt] &= m c_s (T_\ce{final}−T_\ce{initial}) \[4pt] &=\mathrm{(4.184\:J/\cancel{g}°C)×(800\:\cancel{g})×(85−21)°C} \[4pt] &=\mathrm{(4.184\:J/\cancel{g}°\cancel{C})×(800\:\cancel{g})×(64)°\cancel{C}} \[4pt] &=\mathrm{210,000\: J(=210\: kJ)} \nonumber \end{align*}
Because the temperature increased, the water absorbed heat and $q$ is positive.
Exercise $1$
How much heat, in joules, must be added to a $5.00 \times 10^2 \;g$ iron skillet to increase its temperature from 25 °C to 250 °C? The specific heat of iron is 0.451 J/g °C.
Answer
$5.07 \times 10^4\; J$
Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced.
Example $2$: Determining Other Quantities
A piece of unknown metal weighs 348 g. When the metal piece absorbs 6.64 kJ of heat, its temperature increases from 22.4 °C to 43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity).
Solution
Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using Equation \ref{12.3.8}:
\begin{align*} q&=m c_s \Delta T &=m c_s (T_{final}−T_{initial}) \end{align*}
Substituting the known values:
$6,640\; J=(348\; g) c_s (43.6 − 22.4)\; °C \nonumber$
Solving:
$c=\dfrac{6,640\; J}{(348\; g)(21.2°C)} =0.900\; J/g\; °C \nonumber$
Comparing this value with the values in Table $1$, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.
Exercise $2$
A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5 °C to 39.1 °C. Determine the specific heat of this metal, and predict its identity.
Answer
$c = 0.45 \;J/g \;°C$; the metal is likely to be iron from checking Table $1$.
Example $3$: Solar Heating
A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.)
Given: volume and density of water and initial and final temperatures
Asked for: amount of energy stored
Strategy:
1. Use the density of water at 22.0°C to obtain the mass of water (m) that corresponds to 400 L of water. Then compute $ΔT$ for the water.
2. Determine the amount of heat absorbed by substituting values for $m$, $c_s$, and $ΔT$ into Equation \ref{12.3.1}.
Solution:
A The mass of water is
$mass \; of \; H_{2}O=400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) = 3.99\times 10^{5}g\; H_{2}O \nonumber$
The temperature change (ΔT) is 38.0°C − 22.0°C = +16.0°C.
B From Table $1$, the specific heat of water is 4.184 J/(g•°C). From Equation \ref{12.3.8}, the heat absorbed by the water is thus
$q=mc_s\Delta T=\left ( 3.99 \times 10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \bcancel{^{o}C}} \right ) \left ( 16.0 \; \bcancel{^{o}C} \right ) = 2.67 \times 10^{7}J = 2.67 \times 10^{4}kJ \nonumber$
Both q and ΔT are positive, consistent with the fact that the water has absorbed energy.
Exercise $3$: Solar Heating
Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO2) in Table $1$.
Answer
$2.7 × 10^4\, kJ$
Even though the mass of sandstone is more than six times the mass of the water in Example $1$, the amount of thermal energy stored is the same to two significant figures.
Heat "Flow" to Thermal Equilibrium
When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:
$q_{cold} + q_{hot} = 0 \label{12.3.9}$
The equation implies that the amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite:
$q_{cold} = −q_{hot} \label{12.3.10}$
Thus heat is conserved in any such process, consistent with the law of conservation of energy.
The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object.
Substituting for $q$ from Equation $\ref{12.3.8}$ gives
$\left [ mc_s \Delta T \right ] _{cold} + \left [ mc_s \Delta T \right ] _{hot}=0 \label{12.3.11} \nonumber$
which can be rearranged to give
$\left [ mc_s \Delta T \right ] _{cold} = - \left [ mc_s \Delta T \right ] _{hot} \label{12.3.12}$
When two objects initially at different temperatures are placed in contact, we can use Equation $\ref{12.3.12}$ to calculate the final temperature if we know the chemical composition and mass of the objects.
Example $4$: Thermal Equilibration of Copper and Water
If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings.
Given: mass and initial temperature of two objects
Asked for: final temperature
Strategy: Using Equation $\ref{12.3.12}$ and writing $ΔT= T_{final} − T_{initial}$ for both the copper and the water, substitute the appropriate values of $m$, $c_s$, and $T_{initial}$ into the equation and solve for $T_{final}$.
Solution
We can adapt Equation $\ref{12.3.12}$ to solve this problem, remembering that $ΔT= T_{final} − T_{initial}$:
$\left [ mc_s \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mc_s \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \nonumber$
Substituting the data provided in the problem and Table $1$ gives
\begin{align*} \left (30 \; g \right ) (0.385 \; J/ (g °C) ) (T_{final} - 80°C) + (100\;g) (4.184 \; J/ (g °C) ) (T_{final} - 27.0°C ) &= 0 \nonumber \[4pt] T_{final}\left ( 11.6 \; J/ ^{o}C \right ) -924 \; J + T_{final}\left ( 418.4 \; J/ ^{o}C \right ) -11,300 \; J &= 0 \[4pt] T_{final}\left ( 430 \; J/\left ( g\cdot ^{o}C \right ) \right ) &= 12,224 \; J \nonumber \[4pt] T_{final} &= 28.4 \; ^{o}C \end{align*}
Exercise $\PageIndex{4A}$: Thermal Equilibration of Gold and Water
If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings?
Answer
80.0°C
Exercise $\PageIndex{4B}$: Thermal Equilibration of Aluminum and Water
A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.)
Answer
90.6°C
Measuring Heat "Flow"
One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.
The thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (qrxn < 0), then heat is absorbed by the calorimeter (qcalorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (qrxn > 0), then heat is transferred from the calorimeter to the system (qcalorimeter < 0) and the temperature of the calorimeter decreases. In both cases, the amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction. The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants.
The amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction.
Constant-Pressure Calorimetry
Because $ΔH$ is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give $ΔH$ values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure $3$), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6 °C).
Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium. If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:
$q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{12.3.13}$
This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:
$q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{12.3.14}$
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{substance\; M}$ and $q_{substance\; W}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, $q_{substance\, M}$ is a negative value and qsubstance W is positive, since heat is transferred from M to W.
Example $5$: Heat between Substances at Different Temperatures
A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or:
$q_\ce{rebar}=−q_\ce{water} \nonumber$
Since we know how heat is related to other measurable quantities, we have:
$(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \nonumber$
Letting f = final and i = initial, in expanded form, this becomes:
$c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber$
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
$\mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=-(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \nonumber$
$\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \nonumber$
Solving this gives $T_{i,rebar}$= 248 °C, so the initial temperature of the rebar was 248 °C.
Exercise $\PageIndex{5A}$
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
Answer
The initial temperature of the copper was 335.6 °C.
Exercise $\PageIndex{5B}$
A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.
Answer
The final temperature (reached by both copper and water) is 38.8 °C.
This method can also be used to determine other quantities, such as the specific heat of an unknown metal.
Example $6$: Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:
$q_\ce{metal}=−q_\ce{water} \nonumber$
In expanded form, this is:
$c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \nonumber$
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
$\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \nonumber$
Solving this:
$\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C} \nonumber$
Comparing this with values in Table T4, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
Exercise $6$
A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).
Answer
$c_{metal}= 0.13 \;J/g\; °C$
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.
When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must add up to zero:
$q_\ce{reaction}+q_\ce{solution}=0\ \label{ 12.3.15}$
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:
$q_\ce{reaction}=−q_\ce{solution} \label{12.3.16}$
This concept lies at the heart of all calorimetry problems and calculations. Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and ΔHrxn is
$\Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mc_s \Delta T \label{12.3.17}$
The use of a constant-pressure calorimeter is illustrated in Example $7$.
Example $7$: Heat of Solution
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is $ΔH_{soln}$ (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:
1. Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
2. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation \ref{12.3.1}.
3. Use the molar mass of $\ce{KOH}$ to calculate ΔHsoln.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
$\left (100.0 \; \cancel{mL}\; \ce{H2O} \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \nonumber$
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
\begin{align*} q_{calorimater} &= mc_s \Delta T \nonumber \[4pt] &= \left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \bcancel{^{o}C}} \right )\left ( 11.7 \; \bcancel{^{o}C} \right ) \nonumber \[4pt] &= 5130 \; J \[4pt] &=5.13 \; kJ \end{align*}
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \ref{12.3.1}, we see that
$ΔH_{rxn} = −q_{calorimeter} = −5.13\, kJ \nonumber$
This experiment tells us that dissolving 5.03 g of $\ce{KOH}$ in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of $\ce{KOH}$ to calculate $ΔH_{soln}$ - the heat associated when dissolving 1 mol of $\ce{KOH}$:
\begin{align*} \Delta H_{soln} &= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right ) \nonumber \[4pt] &= -57.2 \; kJ/mol\end{align*}
Exercise $7$: Heat of Dissolving
A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example $7$, find $ΔH_{soln}$ for NH4Br (in kilojoules per mole).
Answer
16.6 kJ/mol
Constant-Volume Calorimetry
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter (A device used to measure energy changes in chemical processes. shown schematically in Figure $4$). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in internal energy ($ΔU$) rather than the enthalpy change (ΔH); ΔU is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that $ΔU < ΔH$, the relationship between the measured temperature change and ΔHcomb is given in Equation \ref{12.3.18}, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it
$\Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{12.3.18}$
To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example $8$.
Video $1$: Video of view how a bomb calorimeter is prepared for action.
Example $8$: Combustion of Glucose
The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample
Asked for: ΔHcomb of glucose
Strategy:
1. Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use Equation \ref{12.3.1} to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT.
2. Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.
Solution:
The first step is to use Equation \ref{12.3.1} and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate $q_{comb}$ from the mass of benzoic acid:
$q_{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ \nonumber$
From Equation \ref{12.3.1},
$-C_{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C \nonumber$
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
$q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ \nonumber$
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
$\Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol \nonumber$
This result is in good agreement (< 1% error) with the value of $ΔH_{comb} = −2803\, kJ/mol$ that calculated using enthalpies of formation.
Exercise $8$: Combustion of Benzoic Acid
When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer
−1.30 × 103 kJ/mol
Summary
Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The heat capacity (C) of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The specific heat ($c_s$) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the molar heat capacity ($c_p$) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a constant-pressure calorimeter, which gives $ΔH$ values directly, or a bomb calorimeter, which operates at constant volume and is particularly useful for measuring enthalpies of combustion.
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called calorimeters. To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.3%3A_Heat_Capacity_Enthalpy_and_Calorimetry.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q1A
A sample of $\ce{O_2}$ gas is under an external pressure of 17 atm and contained in a cylinder with a volume of 50 L. The gas is cooled and the resulting volume is 25 L. Calculate the amount of work done on the $\ce{O_2}$ gas.
Solution
\begin{align*} w &=-P\Delta V \nonumber \[5pt] &= -P_{\text{ext}}(V_2-V_1) \[5pt] &=-17atm\ (25\,L-50\,L)=425\, L\ atm \[5pt] &=425\ atm\ L\ (101.325\dfrac{J}{atm\ L}) \[5pt] &= 43, 063\,J \[5pt] & =43.06\, kJ \end{align*} \nonumber
Q1B
A system containing oxygen gas is heated at a constant pressure of 40.0 atm so that its volume increases 177 L to 458 L. Express the amount of work that the system did in kilo-joules.
Solution
The formula for work from the expansion of a gas at constant pressure
$w=-P_{ext} \Delta V\nonumber$
$w=-40\; atm \times (458\; L - 177\; L) = -11240\; L\; atm \nonumber$
Convert from L atm to joules
$-11240 \; L\; atm \times 101.325\; \dfrac{J}{L\ atm} = -1138893\; J \nonumber$
Convert from J to kJ and round to get the final answer
$-1138893\; J \times \dfrac{1\; kJ}{1000\; J}=-1140\; kJ \nonumber$
$w=-1140\; kJ \nonumber$
Q2
The gas mixture inside one of the cylinders of an airplane expands against a constant external pressure of 5.00 atm because of the growing altitude, from an initial volume of 500 mL (at the end of the compression stroke) to the final volume of 1200 mL. Calculate the work done on the gas mixture during this process and express it in Joules.
Solution
$w = -(5.00\ \mathrm{atm}) (1200\ \mathrm{mL} - 500\ \mathrm{mL}) (\dfrac{1\ \mathrm{L}}{1000 \, \mathrm{mL}}) (101.3 \, \mathrm{\dfrac{J}{L\times atm}})= -354.55 \, \mathrm{J} \nonumber$
Q3
The following events are based on a true story. A butcher for the local Chinese restaurant needs to defrost a large chunk of beef, which weights $50\,lb$ and is currently at $0^{\circ}C$. He wants to accomplish this by is going out onto the street and repeatedly drop it onto the ground. Suppose the potential energy of the meat completely transforms into heat each time it hits the ground, and that energy can be calculated from
$V = mg\Delta{h}\nonumber$
where $m$ is mass of the object, $g$ is the acceleration of gravity, and $Δh$ is change in height. If the man is $5\, \text{ft}$ tall and he wants to get the meat to room temperature ($25^{\circ}\text{C}$), how many times does he have to drop the piece of meat? Assume the environment has no effect on the meat, and that it does not lose any heat. (Specific Heat of the Meat: $0.25\mathrm{\dfrac{J}{g \, ^{\circ}C}}$).
Solution
There are two steps to solve this equation. First, to find the amount of heat absorbed by the meat from one drop. Second, to find the total amount of heat needed to warm the meat up from $0^{\circ}C$ to $25^{\circ}C$.
To find the total amount of heat required to bring the meat from $0^{\circ}C$ to $25^{\circ}C$, the equation for calculating heat has to be used:
$Q = mC_{p}\Delta{T}$
The mass of the meat can be calculated via conversion from $lb$ to $g$, and is as follows:
$50lb\times{453.592\dfrac{g}{lb}} = 22679.6g$
The change in temperature is just the final temperature minus the initial, so it is just $25^{\circ}C$.
Lastly, $C_{p}$ is given. Plugging in all of these values gives:
$Q = 22679.6g\times{0.25\dfrac{J}{g^{\circ}C}}\times{25^{\circ}C} = 141747.5 J\nonumber$
Thus, $141747.5 J$ is the total amount of heat needed to heat up the meat to $25^{\circ}C$.
Now to find the amount of heat transferred into the meat from one drop. This utilizes the potential energy equation. The question mentions how to calculate the potential energy, and all potential energy is translated into heat. Therefore.
$Q= V= mg\Delta{h}\nonumber$
The problem gives the mass, the acceleration of gravity is $9.8\dfrac{m}{s^2}$, and height. Therefore, since the potential energy for one drop is equal to the heat for one drop, the heat can be calculated as such:
$Q= 22.6796kg\times9.8\dfrac{m}{s^2}\times{1.524m}= 338.724\dfrac{kg\;m^2}{s^2} = 338.724 J\nonumber$
From there, since no heat is lost, it carries over for each drop. Therefore, the number of times he needs to drop the meat is found by simple division:
$141747.5 J\div{338.724 J}= 418.474 drops\nonumber$
Thus, the drops needed is around 419.
Abstract: Calculate heat from $Q = mC_{p}\Delta{T}$ and $Q= V= mg\Delta{h}$. Divide the two.
Q4
Suppose you have a ball $(C_p= 0.85\dfrac{J}{g^oC})$ at 25°C, what will its final temperature be if the amount of work equal to dropping it down from a height of 86.6m is done to it ? $(g=9.81\dfrac{m}{s^2})$
Solution
\begin{align} U &=mg\Delta h \[5pt] &=m_{ball}\cdot 86.6m\cdot 9.81\dfrac{m}{s^{2}}\[5pt] &=850m_{ball} \end{align} \nonumber
If we assuming all the kinetic energy at the time of collision converts to heat, then
\begin{align} \Delta T &=\dfrac{q}{Cp\cdot m_{ball}}\[5pt] &=\dfrac{U}{C_p\cdot m_{ball}} \[5pt] &= \dfrac{850 m_{ball}}{85\dfrac{J}{g^oC} \cdot m_{ball}} \dfrac{1\,kg}{1,000\,g} \[5pt] &= 1.0^{\circ}C \end{align} \nonumber
\begin{align} T_{final} &=25^{\circ}C+1^{\circ}C\[5pt] &=26^{\circ}C \end{align} \nonumber
Q7A
The rule of Dulong and Petit shows that the molar heat capacities of most metallic elements group around a certain value “X” $\dfrac{J}{K\ mol}$ at 25oC. By first calculating and showing the molar heat capacities of metals rhenium, silver, lead, tungsten, copper, molybdenum and hafnium (given that the respective specific heat capacities of these metals are 0.14$\dfrac{J}{K\ g}$, 0.23$\dfrac{J}{K\ g}$, 0.13$\dfrac{J}{K\ g}$, 0.13$\dfrac{J}{K\ g}$, 0.39$\dfrac{J}{K\ g}$, 0.25$\dfrac{J}{K\ g}$ and 0.14$\dfrac{J}{K\ g}$ ), find the value of “X”. (hint: take the average value of the calculated molar heat capacities and round off value to nearest whole number).
Solution
To first find the molar heat capacities of each metal, multiply the molar mass of each metal by their specific heat capacities:
Molar heat capacity of:
$\ce{Re} = 0.14\dfrac{J}{K\ g}\times186.207\dfrac{g}{mol}= 26.068\dfrac{J}{K\ mol}$ $\ce{Ag}= 0.23\dfrac{J}{K\ g}\times 107.868\dfrac{g}{mol}=24.809\dfrac{J}{K\ mol}$ $\ce{Pb}= 0.13\dfrac{J}{K\ g}\times 207.2\dfrac{g}{mol}= 26.936\dfrac{J}{K\ mol}$
calculate the average molar heat capacity:
Average molar heat capacity=$\dfrac{1}{7}(26.068\dfrac{J}{K\ mol} + 24.809\dfrac{J}{K\ mol} + 26.936\dfrac{J}{K\ mol} + 23.899\dfrac{J}{K\ mol} + 24.7829\dfrac{J}{K\ mol} + 23.99\dfrac{J}{K\ mol} + 24.988\dfrac{J}{K\ mol})=25.068\dfrac{J}{K\ mol}$
$\ce{W}= 0.13\dfrac{J}{K\ g}\times 183.84\dfrac{g}{mol}=23.899\dfrac{J}{K\ mol}$
$\ce{Cu}= 0.39\dfrac{J}{K\ g}\times 63.546\dfrac{g}{mol}=24.7829\dfrac{J}{K\ mol}$
$\ce{Mo}=0.25\dfrac{J}{K\ g}\times 95.96\dfrac{g}{mol}= 23.99\dfrac{J}{K\ mol}$
$\ce{Hf}= 0.14\dfrac{J}{K\ g}\times 178.49\dfrac{g}{mol}=24.988\dfrac{J}{K\ mol}$
Q7B
The specific heat capacities of metals aluminum, bismuth, copper, lead, and silver at 25°C are 0.900, 0.123, 0.386, 0.128, and 0.233$\dfrac{J}{g\ K}$. Calculate the molar heat capacities of these metals. According to the rule of Dulong and Petit, the molar heat capacities of metallic elements, like these, are approximately 25$\dfrac{J}{K\ mol}$.
Solution
Multiply each of the specific heat capacities by its corresponding molar mass to obtain their molar heat capacities.
Aluminum:
$0.900 \dfrac{J}{g\ K}\times 26.98 \dfrac{g}{mol} = 24.3 \dfrac{J}{mol\ K}\nonumber$
Bismuth:
$0.123 \dfrac{J}{g\ K}\times 208.98 \dfrac{g}{mol} = 25.7 \dfrac{J}{mol\ K}\nonumber$
Copper:
$0.386 \dfrac{J}{g\ K}\times 63.55 \dfrac{g}{mol} =24.5 \dfrac{J}{mol\ K}\nonumber$
Lead:
$0.128 \dfrac{J}{g\ K}\times 207.2 \dfrac{g}{mol} =26.5 \dfrac{J}{mol\ K}\nonumber$
Silver:
$0.223 \dfrac{J}{g\ K}\times 107.87 \dfrac{g}{mol} =25.1 \dfrac{J}{mol\ K}\nonumber$
Q9
An undisclosed volume of water is tightly sealed in a microwave-safe container at room temperature before it is placed in an ice bath where it is cooled by a student.
1. During the cooling process within the ice bath, state whether ($\Delta U$), Q, and W of the system are negative, zero, or positive. Explain your reasoning.
2. After being cooled, the student decides that the water seems too cold so it is placed in the microwave where the container of water is heated back to room temperature. What are the new signs of ($\Delta U$), Q, and W during the heating process? Explain your reasoning.
3. Now determine the signs of ($\Delta U_1 + \Delta U_2$), ($Q_1 + Q_2$) , and ($W_1 + W_2$), where possible, assuming that the cooling process was step 1 and the heating process was step 2.
Solution
1. W is zero, Q is negative, and ($\Delta U$) is negative. Work is zero for this step because neither volume nor pressure of the container is changing. Since the water is cooler than before heat has left the system so it is negative; and because ($\Delta U$) = Q + W and W = 0 then ($\Delta U$) = Q which is negative so, ($\Delta U$) is negative.
2. W is zero, Q is positive, and ($\Delta U$) is positive. Work is zero for this step because neither volume nor pressure of the container is changing. Since the water is hotter than before heat has entered the system so it is positive and because ($\Delta U$) = Q + W and W = 0 then ($\Delta U$) = Q which is positive so, ($\Delta U$) is positive.
3. Since W was equal to zero for both processes above (W1 + W2) is equal to zero. ( ($\Delta U$)1 + ($\Delta U$)2 ) = (Q1 + Q2) where they are both equal to zero.
If the water is cooled from room temperature to a lower temperature and then heated directly back to room temperature, wouldn't ( ($\Delta U$)1 + ($\Delta U$)2 ) be 0? Q is the only factor actually changing here and q(heating to cooling) = -q(cooling to heating) provided the change in t is the same. So both ( ($\Delta U$)1 + ($\Delta U$)2 ) and q should be 0.
Q11A
Your lab partner slipped a sample of unknown hot metal that is 40.0g, which is initially at 130.0°C into a 100.0 g water that is initially at 50.0°C. A temperature probe indicates that equilibrium is reached at 60.15°C. Using the specific heat capacity of water 4.18$\dfrac{J}{K}$, calculate the specific heat capacity of the unknown metal.
Solution
Imagine two sub-systems: the metal and the water. If mixing the hot metal and cool water inside a well-insulated container (that prevent leaks of heat), then the heat absorbed by the system will equal zero. Since the system is the sum of the two sub-systems:
$q_{\text{sys}}=0=q_{\text{metal}}+q_{\text{water}}\nonumber$
For both sub-systems, the amount of heat gained is equal to the specific heat capacity times the mass times the temperature change:
$q_{\text{metal}}+ q_{\text{water}}= m_{\text{water}} C_{s,\text{water}} \, \Delta T_{\text{metal}} = 0\nonumber$
Solving for the specific heat capacity of the metal:
$c_{s,\text{metal}}=\dfrac{-m_{\text{water}}c_{s,\text{water}} \, \Delta T_{\text{water}}}{m_{\text{metal}} \Delta T_{\text{metal}}}=\dfrac{-100.0\;\text{g} \times 4.18 \mathrm{\dfrac{J}{K\ g}} \times 10.15 \mathrm{^{\circ}C}} {40.0\; \text{g} \times -69.85 \mathrm{^{\circ}C}} = 1.52 \mathrm{\dfrac{J}{K\ g}} \nonumber$
Tip: Do not convert Celsius to Kelvin. The Kelvin and Celsius scales differ only in their location of their zero points. Temperature change in Celsius is the same temperature change of Kelvin.
Q11B
Let's say you 34.5 grams of some hot metal that is initially at 75°C and you put that metal into 64.0 grams of water that is initially at 25°C. If the the two objects reach thermal equilibrium at 39°C, what is the specific heat capacity of the metal when the specific heat capacity of water is $4.18 \dfrac {J}{K\ g}$.
Solution
The formula used to solve this question is $m_1c_1 \Delta T_1 = -m_2c_2 \Delta T_2\nonumber$
Plug in your known values and solve for $c_1$
$(34.15 \; g)(c_1)(36°C) = -(64 \; g)(4.18 \; \dfrac {J}{K\ g})(-14°C)$
$c_1 = 3.02 \; \dfrac {J}{K\ g}$
Q12
A 10.00 g sample of Aluminum at 60.0 °C and a 30.0 g sample of copper at a temperature of -20.0 °C were thrown simultaneously into a 50.0 g of water at a temperature of 25.0 °C. What will be the final temperature of the system consisting of the two metal samples and the water? Assuming that this system is completely isolated from the surroundings. Use the information below for your calculations.
• $c_{s(\ce{Al})}=0.900 J/(K \cdot g)$
• $c_{s(\ce{Cu})}=0.385 J/(K \cdot g)$
• $c_{s(\ce{H_2 O})}=4.184 J/(K \cdot g)$
Solution
Since our system is isolated, the thermal energy lost by one component is transferred to the other components.
$q_1=-q_2\nonumber$
which is equivalent to
$C_1 ∆T=-C_2 ∆T\nonumber$
Noting that
$C=m C_s\nonumber$
To avoid the complexity of handling three components, we can utilize the fact that temperature is a state function; we will simplify our calculations by choosing a different path to arrive to our final state.
We can do this in two steps:
Ignore the Aluminum sample and treat the copper and water as the only components of our system. After finding the equilibrium temperature, we add the Aluminum sample to the water and copper system, thereby reaching the same final state.
Which is the same answer we got before. This should make intuitive sense because regardless of the path we take, we end up with the exact same amount of thermal energy in our system.
Step 1
$m_{\ce{Cu}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})=-m_{\ce{H_2O}}C_{s(\ce{H_2 O})}(T_f-T_{i(\ce{H_2 O})})\nonumber$
$30.0\,g×0.385\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber$
Solving for $T_f$ for the copper and water system gives us
$T_f=22.6 °C\nonumber$
Step 2
$C_{(\ce{Cu}+\ce{H_2 O})}×(T_f-T_{i(\ce{Cu}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Al})}×(T_f-T_{i(\ce{Al})})\nonumber$
Noting that
$C_{(\ce{Cu}+\ce{H_2 O})}=m_{\ce{H_2O}}×C_{s(\ce{H_2 O})}+m_{\ce{Cu}}×C_{s(Cu)}\nonumber$
Combining the two equations and plugging the values gives us
$[(30.0\,g×0.385 \,J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g))]×(T_f-22.6 °C)= -10.0\,g×0.900\, J/(K \cdot g)×(T_f-60 °C)\nonumber$
Solving for $T_f$:
$T_f=24.1°C\nonumber$
Which is the final temperature of the whole system.
Alternative Approach
We can also choose a path where we add the Aluminum first, and then we add the copper. Again, temperature is a state function and choosing a different path will not affect the final answer. We also show the calculation for this path for the sake of completion.
Step 1
$m_{\ce{Cu}}×C_{s(\ce{Al})}×(T_f-T_{i(Al)})=-m_{(\ce{H_2 O})}×C_{s(\ce{H_2 O})}×(T_f-T_{i(\ce{H_2 O})})\nonumber$
$10.0\,g×0.900\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber$
Solving for $T_f$ for the aluminum and water system gives us
$T_f=26.4 °C\nonumber$
Step 2
$[(m_{(\ce{H_2 O})}×C_{s(\ce{H_2O})})+(m_{\ce{Al}}×C_{s(\ce{Al})}) )]×(T_f-T_{i(\ce{Al}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})\nonumber$
Plugging the numbers,
$[(10.0\,g×0.900\, J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g)]×(T_f-26.4 °C)=-30.0\,g×0.385 \,J/(K \cdot g)×(T_f+20 °C)\nonumber$
Solving for $T_f$
$T_f=24.1°C\nonumber$
Q15A
Calculated the heat required to melt 3.00 g of ice and the heat required to change the temperature of water from 0°C to 100°C. What is the proportionality of the heat necessary to melt ice compared to the heat required to change the temperature of water from 0°C to 100°C? Use values from Table S2 and assume $\Delta H_{f}$ =334 J g-1 for calculations? Why is the heat positive instead of negative?
Solution
First lets determine the amount of heat needed to melt ice:
$q = m\Delta H_{f}$
$q = 1,002\ \mathrm{J}$
Next lets determine the heat required to raise the temperature of water 100°C:
$q = mC_s \Delta{T}$
$q = 1,254\ \mathrm{J}$
So the proportionality was determined to be 4:5
The heat is positive because the heat is required. The heat is needed to make the states go from solid to liquid when melting and then liquid to gas when the temperature is raised from 0 to 100 degrees Celsius.
Q15B
An observation in the 18th century stated that the heat that raised a certain mass of water from its freezing point to boiling point is equal to four-thirds of the heat required to melt the same mass of ice. Using the theory behind the observation, estimate the heat required to melt 10 g of ice, know that the heat capacity of water is 4.18 J/oC.
Solution
Let the heat required to raise the temperature of water from its freezing point to its boiling point is $q_1$:
$q = m c \Delta T$
$m = \text{mass}$
$c = 4.18\dfrac{J}{g\ ^{o}C}$
$\Delta T = \text{freezing point} - \text{boiling point} = 100 - 0 = 100^{o}C$
$q_{1}=(10 g)(4.18\dfrac{J}{g\ ^{o}C})(100^{o}C)$
$q_{1} = 4180 \, \text{J} = 4.18 \, \text{kJ}$
Let the heat required to melt 10 g of ice is $q_2$:
$q_{1}=\dfrac{4}{3}q_{2}$
$q_{2}=\dfrac{3}{4}q_{1}$
$q_{2}=\dfrac{3}{4}(4.18 \, \text{kJ})=3.135 \, \text{kJ}$
Q17
For his birthday, his John's parents have given him a compressible oven filled with 0.250 mol argon. If he sets this compressible oven at 1.00 atm and 273 K and let it contract from a constant external pressure of 0.100 atm until the gas pressure reaches 10.00 atm and the temperature reaches 400 K, what is the work done on the gas, the internal energy change, and the heat absorbed by the gas?
Solution
Using the ideal gas law $P V = n R T$, we can see that:
$V_o = \dfrac{n R T_o}{P_o} \nonumber$
$V_f = \dfrac{n R T_f}{P_f} \nonumber$
n = 0.25 mol, R = 0.082$\dfrac{atm\ L}{mol\ K}$ Po = 1 atm, Pf = 10 atm To= 273K Tf = 400K
where $V_o$ and $V_f$ represent the initial and final volume of the chamber. Now, the pressure clearly changes inside the chamber, but outside the chamber, the pressure is held constant, so from the perspective of the surroundings:
$W= -P(V_f - V_o)\nonumber$
where $P$ represents the constant external pressure. Now substituting in Vf and Vo we can see that we've solved for work. Now, for any thermodynamic process:
$\Delta{U}= \dfrac{3}{2} n R (T_f - T_o) \nonumber$
So plugging in those values allows us to solve for ∆U. Now, for heat, we simply subtract W from ∆U, by the first law of thermodynamics.
First law of thermodynamics: the total energy of an isolated system is a constant, energy can be transformed from one form to another, but it can not be created or destroyed.
The solution is incomplete.
Q19
Take 4 moles of ideal, monatomic gas going through expansion processes. The gas was initially put at a pressure of 5.00 atm and a temperature of 30°C. The gas first goes through isothermal expansion until the volume doubled. An isochoric process follows as the pressure is halved. $C_v = \dfrac{3}{2}R$
1. Build $\Delta U$, W, and Q table for each process.
2. Find the final temperature.
Solution
Isothermal Process:
$\Delta U = 0$
$W = Q$
$W = nRT\ln \left( \dfrac{V_f}{V_i} \right) =(4\,\text{mol})(8.314\, \mathrm{J \cdot K^{-1} mol^{-1}})(303.15\, \text{K}) \ln(2) = 6988 \, \text{J} = Q$
Isochoric Process
$W = 0$
$\Delta U = Q$
First find the initial pressure of this process.
Use ideal law to relate pressure to volume. $PV=nRT$, in which P is inversely proportional to V.
Thus, when volume is doubled, pressure is halved so $P_f = 2.50 \,\text{atm}$. Now, find the final temperature of this process.
Once again, relate P to T using ideal gas law. T is found to be directly proportional to P. Thus, if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15K.
$\dfrac{1}{2}(303.15 \, \text{K}) = 151.58 \, \text{K}$
$Q= nC_v\Delta T=(4.00 \, \text{mol})(\dfrac{3}{2})(8.314 \mathrm{\dfrac{J}{K\, mol}})(-151.58K)=-7561 \, \text{J} = \Delta U$
Processes ΔU Q W
Isothermal 0 6988J 6988J
Isochoric -7561J -7561J 0
Q20
An apparatus is set up such that an ideal gas is released into vacuum by opening a stopcock, hence allowing it to freely expand (i.e., no external force is applied). Calculate $\Delta{U}$ of the system, and prove that the free expansion process is adiabatic i.e no heat transfer.
Solution
$\Delta U = \dfrac{3}{2}nR\Delta T$
As the expansion is isothermal, i.e. it occurs without a change in temperature:
$\Delta T = 0$
Hence,
$\Delta U = 0$
and
$\Delta U = q_p+w$
As the external force applied on the gas is zero, the work done by the gas is zero.
Hence,
$0 = q_p + 0$
$q_p = 0$
This proves that the process is adiabatic.
Q21
A 150 L vessel contains 8.00 moles of neon at 270 K is compressed adiabatically, so that there is no gain nor loss of any heat, and irreversibly until the final temperature is 470 K. Calculate the change in internal energy, the heat added to the gas, and the work done on the gas.
Solution
Since Ne is a monatomic ideal gas and the volume of the vessel remains constant, the heat capacity of Ne can be expressed as:
$C_{p} = \dfrac{3}{2} \times R = \dfrac{3}{2} \times 8.314 \mathrm{\dfrac{J}{K \cdot mol}} = 12.47 \mathrm{\dfrac{J}{K \cdot mol}}$
$\Delta U = nC_{p} \Delta T$
$\Delta U = 8.00 \times 12.47 \mathrm{\dfrac{J}{K \cdot mol}} \times (470-2700) \; \mathrm{K} = 19952 \; \mathrm{J} = 20.0 \; \mathrm{kJ}$
Since the vessel is adiabatically compressed, no heat is added, therefore $q = 0$
$\Delta U = q + w$
$20.0 \; \mathrm{kJ} = 0 + w$
$w = 20.0 \; \mathrm{kJ}$
Q22
A gas expands at constant external pressure of 3.00 atm until its volume has increased by 9.00 to 15.00 L. During this process, it absorbs 800J of heat from the surroundings.
1. Calculate the energy change of the gas, $\Delta U$
2. Calculate the work, w, done on the gas in an irreversible adiabatic (q = 0) process connecting the same initial and final state.
Solution
a) $\Delta U = q+w$
$\Delta U= 800J + -((3.00 \, \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}})) =-1023.4 \mathrm{J}$
b) $q=0$
$\Delta U = w$
$\Delta U= -((3.00 \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}}))= -1823.4 \mathrm{J}$
Q23
Using the theorem of equipartition of energy, calculate the specific heat capacity at constant pressure $C_p$ at 25ºC and 1 atm for $\ce{O_2}$ and $\ce{CO}$. Compare the calculated values to the experimental data $(\ce{O_2}=29.36, \ce{CO}=29.1)$ and thus calculate the percent of the experimental value that results from vibrational motions. Answer: The percent of $C_p$ due to vibrational motions is 0.897% for $\ce{O_2}$ and 0.141% for $\ce{CO}$.
Solution
The equipartition theorem states that each degree of freedom in a molecule contributes to ½ RT to the molar internal energy of a gas. To solve this problem, the number of degrees of freedom $DOF$ in each molecule must be identifies.
$\ce{O2}$ is a linear diatomic particle, thus it has 3 translational degrees of freedom and 2 rotational degrees of freedom. Since $\ce{CO}$ is a linear molecule it also has 3 translational and 2 rotational degrees of freedom. The molecules are at room temperature. Thus, it is assumed that there is no vibration in the bond (other than the zero point energy).
The low percentage of $C_p$ due to vibrational modes in both molecules indicates that the vibrational motion is extremely small and can be neglected.
$DOF({\ce{O2}}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber$
$DOF(\ce{CO}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber$
Using the equation
$U = \left( DOF\right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber$
find the internal energy of each gaseous molecule.
$U \left( \ce{O_2} \right) = \left( 5 \right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber$
Since $\ce{O_2}$ and $\ce{CO}$ have the same degrees of freedom,
$U\left( \ce{O_2} \right) = U\left( \ce{CO} \right) = \dfrac{5}{2}RT\nonumber$
The internal energy is related to $C_v$ by
$C_v= \left( \dfrac{\partial U}{\partial T} \right) = \left( \dfrac{\partial \dfrac{5}{2}RT}{\partial T} \right) = \dfrac{5}{2}R\nonumber$
$C_v$ is related to $C_p$ by
$C_p = C_v + R = \dfrac{5}{2}R + R = \dfrac{7}{2}R = \dfrac{7}{2} \times 8.314 \dfrac{J}{mol\ K} = 29.099\dfrac{J}{mol\ K}\nonumber$
The percent of the experimental value that results from vibration motion is the difference between calculated and experimental value as a percentage of the experimental value.
$\% \text{vibrational motion}(\ce{O_2}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.36 - 29.099 }{29.099} \times 100 \% = 0.897\% \nonumber$
$\% \text{vibrational motion} (\ce{CO}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.14 - 29.099}{ 29.099} \times 100 \% = 0.141\% \nonumber$
Q23B
Using the classical equipartition theorem, calculate the value of $C_p$ at 298 K and 1 atm for $\ce{HF (g)}$ and $\ce{F2 (g)}$, assuming that their pressure is constant. Then compare your calculation with the experiment values of $29.13\mathrm{\dfrac{J}{K\ mol}}$ and $31.30\mathrm{\dfrac{J}{K\ mol}}$respectively. What is the per cent of the measured value that arises from vibrational motions?
Solution
Diatomic molecules possess a total of 6 degrees of freedom:
• 3 degrees in translational motion
• 2 degrees in rotational motion
• 1 degree in their vibrational motions
$6.11 \text{ for } \ce{F2 (g)} = 0.735R$
The differences are due to the fact that some quantized energy levels are not available at certain temperatures, in this case room temperature. This means that $\ce{HF_{(g)}}$ contributed only 0.004R of its experimental value of $29.13\mathrm{\dfrac{J}{K\ mol}}$ to the vibrational degree of freedom, which is 0.014%. $\ce{F_ {2(g)}}$ contributed 0.265R of its experimental value $31.30\mathrm{\dfrac{J}{K\ mol}}$, which is 0.847%.
Each translational and rotational degree of freedom contributes $\dfrac{R}{2}$ to the heat capacity of a gas, while each vibrational degree of freedom contributes $\dfrac{2R}{2}$ (=R). Therefore the predicted heat capacity $(C_v)$ at a constant volume is:
$C_v = 3\times (\dfrac{R}{2}) + 2\times (\dfrac{R}{2}) + 1\times (\dfrac{2R}{2}) = 7(\dfrac{R}{2})$
Since the question is asking for the heat capacity of these gases that are under a constant pressure, the result is multiplied by an additional $\dfrac{2R}{2}$ because the volume can change in order to keep the pressure constant. Therefore:
$C_v= (\dfrac{2R}{2})+(\dfrac{7R}{2}) = \dfrac{9R}{2}$
$R = 8.3145\mathrm{\dfrac{J}{K\ mol}}$
$C_p=\dfrac{9(8.3145\mathrm{\dfrac{J}{K\ mol}})}{2}=37.41\mathrm{\dfrac{J}{K\ mol}}$
This is the calculated value for both of the two gases. Since the experiment values were $29.13\mathrm{\dfrac{J}{K\ mol}}$ for $\ce{HF (g)}$ and $31.30\mathrm{\dfrac{J}{K\ mol}}$ for $\ce{F2(g)}$, the differences between the calculations were:
$8.28 \text{ for } \ce{HF (g)} = 0.996R$
Q25
1. Calculate the change of enthalpy when a 69-grams sample of zinc is heated from 753 K to 927 K at a constant pressure of 1 atm.
2. Calculate the change of enthalpy when 3 moles of butane are heated from 204 K to 258 K at a constant pressure of 1 atm.
Solution
a)
$q=mC_s\Delta T$
$m=69\ \mathrm{g}$
$\Delta T=927-753=174\ \mathrm{K}$
$C_{\text{s(zinc)}}=0.39 \mathrm{\dfrac{J}{g\ ^oC}}$
$q=69\times 0.39\times 174=4682.34\ \mathrm{J}=4.56\ \mathrm{kJ}$
b)
$q=nC_p\Delta T$
$n=3$
$\Delta T=258-204=54\ \mathrm{K}$
$C_p=132.42$
$q=3\times 132.42\times 54=21452.04\ \mathrm{J}= 21.4504\ \mathrm{kJ}$
Q31
What is the change in enthalpy when 8.19 grams of ethane ($\ce{C_2H_6}$) vaporizes, assuming a normal boiling point and $\Delta H_{\text{vap}}= 14.72\ \mathrm{\dfrac{kJ}{mol}}$.
Solution
First, convert grams of ethane to moles:
$(8.19\; \cancel{g\; \ce{C_{2}H_{6}}}) \left(\frac{1\: mol}{30.07\: \cancel{g\; \ce{C_{2}H_{6}}}} \right) = 0.272\; mol\nonumber$
This step is to match quantity units of ethane to those in the given value of $\ce{\Delta H_{vap}}$.
Then, given the $\Delta H_{\text{vap}}$,
$(0.272\; \cancel{mol \;\ce{C_{2}H_{6}}} ) \left (\frac{14.72\: kJ}{1\: \cancel{mol\; \ce{C_{3}H_{8}}}} \right ) = 4.004\; kJ\nonumber$
$\ce{\Delta H}=4.004\; kJ$
This value is the change in enthalpy, or thermodynamic energy, when the specified amount of ethane undergoes the above reaction.
Q33
The heat capacity $C_p$ of ice is $38 \mathrm{\dfrac{J}{K\ mol}}$ and $C_p$ of water is $75 \mathrm{\dfrac{J}{K\ mol}}$. A 24.0 g ice cube at -15 oC is placed into 120 g of water at room temperature (25oC). What is the temperature of the water when it reaches equilibrium?
Solution
$q_{\mathrm{ice}} = -q_{\mathrm{water}}$
$m_{\mathrm{ice}}C_{\mathrm{ice}}(T_{f}-T_{\mathrm{ice}})=-m_{\mathrm{water}}C_{\mathrm{water}}(T_{f}-T_{\mathrm{water}})$
$\left(\dfrac{24g\ \mathrm{ice}}{18.02\dfrac{g}{mol}}\right)\times (38\dfrac{J}{K\ mol})\times (T_f - (-15^oC)) = -\left(\dfrac{120g\ \mathrm{water}}{18.02\dfrac{g}{mol}}\right)\times (75\dfrac{J}{K\ mol})\times (T_f - 25^oC)$
$T_{f}=21.316^{o}C$
Q35
Determine the change in enthalpy $\left(\Delta H\right)$ for the following chemical reaction
$\ce{4NH_{3(g)} + 5O_{2(g)} -> 4NO_{(g)} + 6H_2O_{(g)}}$
given that:
$\ce{O_{2(g)} + N_{2(g)} -> 2NO_{(g)}}$ $\Delta H = +180.5\ \mathrm{\dfrac{kJ}{mol}}$
$\ce{2H_2O_{(g)} -> 2H_{2\, (g)} + O_{2\, (g)}}$ $\Delta H = +483.64\ \mathrm{\dfrac{kJ}{mol}}$
$\ce{3H_{2(g)} + N_{2(g)} -> 2NH_{3(g)}}$ $\Delta H = -92.22\ \mathrm{\dfrac{kJ}{mol}}$.
Solution
Using Hess's Law, we can obtain the $\ce{\Delta}{H}$ for
$\ce{4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}}$
by adding and modifying the other equations provided until they resemble the equation above.
$\ce \left( {O_{2(g)} + N_{2(g)} \rightleftharpoons 2NO_{(g)}} \right) \times \left( 2 \right)$ $\ce{\Delta}{H}$ = $+180.5\ \dfrac{kJ}{mol} \times \left( 2 \right)$
$\ce \left( {2H_2O_{(g)} \rightleftharpoons 2H_{2\, (g)} + O_{2\, (g)}} \right) \times \left( -3 \right)$ $\ce{\Delta}{H}$ = $+483.64\ \dfrac{kJ}{mol} \left( -3 \right)$
$\ce \left( {3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)}} \right) \times \left(-2 \right)$ $\ce{\Delta}{H}$ = $-92.22\ \dfrac{kJ}{mol} \times \left(-2 \right)$
The above modifications simplify out to
$\ce {2O_{2(g)} + 2N_{2(g)} \rightleftharpoons 4NO_{(g)}}$ $\ce{\Delta}{H}$ = $+361.0\ \dfrac{kJ}{mol}$
$\ce {6H_{2\, (g)} + 3O_{2\, (g)} \rightleftharpoons 6H_2O_{(g)}}$ $\ce{\Delta}{H}$ = $-1450.92\ \dfrac{kJ}{mol}$
$\ce {4NH_{3\, (g)} \rightleftharpoons 6H_{2\, (g)} + 2N_{2\, (g)}}$ $\ce{\Delta}{H}$ = $184.4\ \dfrac{kJ}{mol}$
Species that are common to both sides can be cancelled out.
$\ce { 2O_{2(g)} + \require{cancel} \cancel{2N_{2(g)}} \rightleftharpoons 4NO_{(g)}}$ $\ce{\Delta}{H}$ = $+361.0\ \dfrac{kJ}{mol}$
$\ce { \require{cancel} \cancel{6H_{2\, (g)}} + 3O_{2\, (g)} \rightleftharpoons 6H_2O_{(g)}}$ $\ce{\Delta}{H}$ = $-1450.92\ \dfrac{kJ}{mol}$
$\ce { 4NH_{3\, (g)} \rightleftharpoons \require{cancel} \cancel{6H_{2\, (g)}} + \cancel{2N_{2\, (g)}}}$ $\ce{\Delta}{H}$ = $184.4\ \dfrac{kJ}{mol}$
Finally, after adding the remaining specimens and the enthalpy change that occurs with them we have:
$\ce{4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}}$ with $\ce{\Delta}{H}$ = $-905.5\ \dfrac{kJ}{mol}$
Q37
$\Delta H$ of $\ce{C6H6(g)}$ from $\ce{C6H6(l)}$ to its gaseous form is +33.9 kJ/mol. Which of these two substances absorbs off more heat when 10 lbs of the substance is burned? Which of these two substances require less energy when condensed to solid form?
Solution
An alternative way to view this is to inspect the reactions under comparison
$\ce{2C6H6(g) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber$
vs.
$\ce{2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber$
Recognizing that the difference between these equations is
$\ce{2C6H6(l) -> 2C6H6(g)} \nonumber$
The problem states that the $∆H$ of the conversion of $\ce{C6H6(g)}$ to $\ce{C6H6(l)}$ is positive, and therefore, is an endothermic process. Consequently, ten pounds of $\ce{C6H6(g)}$ contains more enthalpy than 10 pounds of $\ce{C6H6(l)}$. Since these will both produce carbon dioxide and water when burned, their products are of the same final energy level. As a result, $\ce{C6H6(g)}$ will give off more energy than $\ce{C6H6(l)}$ when burned and $\ce{C6H6(l)}$ will require less energy to form a solid.
Q39
Calculate the enthalpy change under standard conditions for the reaction:
$\ce{C3H6O(l) + 4 O2 (g) → 3 CO2 (g) + 3 H2O(l)} \nonumber$
given information below:
• ΔH°f, O2(g): 0 kJ/mol
• ΔH°f, CO2 (g): -393.5 kJ/mol
• ΔH°f, H2O(l): -285.83 kJ/mol
• ΔH°f, C3H6O(l): -248 kJ/mol
Solution
$ΔH^o_{rxn} = \sum ΔH^o_{f,\, products} – \sum ΔH^o_{f,\, reactants} \nonumber$
ΔH°rxn = [ 3 (-393.5 kJ/mol) + 3 (-285.83 kJ/mol) ] - [ (-248 kJ/mol) + 4 (0 kJ/mol) ] = -1790 kJ/mol
Q47
A 10 gram sample of pure hydrogen is burned completely with excess oxygen to generate liquid water in a constant volume calorimeter at 25 °C. The amount of heat evolved is 1,420 kJ.
1. Write and balance the chemical equation for the combustion reaction.
2. Calculate the standard change in internal energy for the combustion of 1.00 mole hydrogen to liquid water.
3. Calculate the standard enthalpy change per mole of hydrogen for the same reaction as in part (a).
4. Calculate the standard enthalpy of formation per mole of hydrogen, using data for the standard enthalpies of formation of liquid water (Table T1).
Solution
1. $\ce{2H2 (g) + O2 (g) -> 2 H2O (l)}$
2. $ΔU^o= q_v = \dfrac{-1420 \,kJ}{10/(1.008 \times 2 ) g/mol} = -286.3 \,kJ/(mol\, of\, H_2)$
3. $ΔH^o_{rxn}= ΔU^o + Δn_{gas} RT = -286.3 \,kJ/(mol\, of\, H_2) + (2 mol. \, of\, H_2)(-3) (8.314 J/{K/mol}) (298 \,K) = -580\, kJ/mol$
4. $ΔH^o_rxn= 2 \Delta H_f^o \{\ce{H2} \}$ so $\Delta H_f^o /2$.
A consequence of the constant-volume condition is that the heat released corresponds to $q_v$ and thus to the internal energy change $ΔU_{sys}$ rather than to $ΔH_{sys}$ under constant pressure conditions.
$\Delta U_{sys} = q_v\nonumber$
This comes from the definition of enthalpy
$H_{sys} = U_{sys} + PV\nonumber$
and associated change
$\Delta H_{sys} = \Delta U_{sys} + \Delta (PV)\nonumber$
or using the chain rule
$\Delta H_{sys} = \Delta U_{sys} + \cancelto{0}{P\Delta V} + V \Delta P \label{Change in Bomb}\nonumber$
which simplifies to
$\Delta H_{sys} = \Delta U_{sys} + V \Delta P\nonumber$
If we assuming ideal gas law for the gases
$PV=nRT\nonumber$
or
$\Delta P = \dfrac{\Delta n R T}{V}\nonumber$
substituting this into the equation for enthalpy change gives the enthalpy change under constant volume conditions in terms of changing number of moles
$ΔH_{sys} = \Delta U_{sys} + Δn_gRT\nonumber$
or
$ΔH_{sys} = q_v + Δn_gRT\nonumber$
where $Δn_g$ is the change in the number of moles of gases in the reaction.
Q49
Carbon dioxide ($\ce{CO_2}$) is a common byproduct of the combustion of fossil fuels. Estimate the standard enthalpy of formation ($\Delta H_f$) of carbon dioxide at 25°C, use Table T1.
Solution
First write the equation:
$\ce{C (s) + O2 (g) -> CO2 (g)}\nonumber$
The bond enthalpies in Table T3 allows us to calculate $\Delta H°$ because 2 C=O and 2 C-O bonds are formed:
$\Delta H°$ 2 mol (-192.0 kJ mol-1) + 2 mol (-85.5) = -555 kJ
Next we need to write the equations for "striping" the atoms from there standard state to single atoms. Each of the atomization has an enthalpy that was found in Table T1:
$\ce{C (s) -> C (g)}\nonumber$ $\Delta H°$ = 1 mol (716.7 kJ mol-1) = 716.7 kJ
$\ce{O2 (g) -> 2 O(g) }\nonumber$ $\Delta H°$ = 2 mol (249.2 kJ mol-1) = 498.4 kJ
Combine the results of all three equations to calculate $\Delta H_f$ of 1 mol of CO2:
-555 kJ + 716.7 kJ + 498.4 kJ = 660.1 kJ
This solution should be the same if finding the standard enthalpy of formation by using Table T1 which would be -393.5 kJ for one mole of CO2. Since this question is not clear, it is better if the students answering this question to just use Table T1 which will just use the standard heat of formation for and then have the sum of products minus the sum of reactants = the answer. So the answer would be -393.5 kJ.
Q51
Given the table of average bond enthalpies shown below, estimate the enthalpy change $\Delta H°$ for the following reaction:
$\ce{2H2 (g) + O2 (g) -> 2H2O (l)} \nonumber$
Bond Bond Enthalpy (kJ/mol)
$\ce{H-H}$ $436$
$\ce{H-O}$ $463$
$\ce{O=O}$ $498$
Solution
To calculate the total enthalpy change of the reaction, we use Hess's Law. Thus, the total enthalpy of formation is equal to the sum of the enthalpy of formation of the reactants and the enthalpy of formation for the products.
Step 1. $\Delta H_1$ = the bond enthalpies of $\ce{2 H2 (g) + O2 (g)}$
= $2(\ce{H-H}) + (\ce{O=O})$
= $2(436) \; \mathrm{kJ/mol} + 498 \; \mathrm{kJ/mol}$
= $+1370 \; \mathrm{kJ/mol}$
Step 2. $\Delta H_2$ = the bond enthalpies of $2 \ce{H2O} (l)$
= $2 \times 2 (\ce{H-O})$
= $4(463) \; \mathrm{kJ/mol}$
= $-1852 \; \mathrm{kJ/mol}$
Step 3. $\Delta H°$ = $\Delta H_1$ + $\Delta H_2$
= $1370 \; \mathrm{kJ/mol} - 1852 \; \mathrm{kJ/mol}$
= $-482 \; \mathrm{kJ/mol}$
Note in Step 1, $\Delta H_1$ is positive because the process of breaking the bonds into individual atoms is always an endothermic process while in Step 2, $\Delta H_2$ is negative because the process of creating bonds out of the individual atoms is always an exothermic process.
Q59
Two quantum states are separated by an energy of 0.6 × 10-21J. Calculate the relative populations of the two quantum states at a temperature of 33°C.
Solution
The equation for the relative populations of two quantum states, P2/P1 is given by the equation
$\dfrac{P_{2}}{P_{1}} = e^{-(E_{2} - E_{1}) / k_B T}\nonumber$
Where E2-E1 is the energy difference between the two states, $k_B$ is the Boltzmann constant, and T is the temperature.
Plug in the given values into the equation to reach the final answer.
$\dfrac{P_{2}}{P_{1}}=e^{-(0.6 \cdot 10^{-21} \, \mathrm{J})/1.38\cdot 10^{-23} \mathrm{K^{-1}} (298 \, \mathrm{K}))}\nonumber$
$\dfrac{P_{2}}{P_{1}}=0.86\nonumber$
Q63
By utilizing the Harmonic Oscillator Model, calculate the relative population of the first energy state and the ground state, both at $278.15 \, \text{K}$, for $\ce{H2}$. The force constant for $\ce{H2}$ is $510 \mathrm{\dfrac{N}{m}}$
Solution
A few pieces of information need to be understood before going about solving this equation. First, the question asks for the “relative population” of the first energy state at a given temperature ($278.15 \, \mathrm{K}$) to the energy state at $0 \, \mathrm{K}$. Recall that energy states are basically a way of saying that every chemical species (molecule, atom, etc.) can “have” a specific (discreet) amount of energy. Each of these values are a “state”, and the lowest- value state can be considered the “ground state”, and all states above are “excited states”. Also recall that the states can be named by quantum number; i.e if $n$ were the quantum number, than the ground state would be $n=0$ and the first state after that, an excited state, has a quantum number $n=1$. The next step is to understand what “relative population is”. Basically, this refers to the probability that the species, in this case $\ce{H2}$, will be found in one energy state versus another. The equation that describes the probability of a species being in energy level $n$ is as follows:
$P(n)= Ce^{-\varepsilon_{n}/k_{b}T}$
Where C is a constant, n is quantum number (energy level), $k_{b}$ is Boltzmann’s constant, T is temperature, and $\varepsilon$ is the energy of the molecule, which can be determined as such:
$\varepsilon_{n} = \left(n + \dfrac{1}{2}\right)hv$
$hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}$
n, as already seen, is the quantum number. $hv$ in the equation for $\varepsilon$ is defined as the product of planck’s constant ($h$) divided by $2\pi$ and the $\sqrt{\dfrac{k}{\mu}}$, where k is the force constant, which is given in the problem, and the $\mu$ is the reduced mass, which can be calculated as follows:
$\mu = \dfrac{m_{1}m_{2}}{m_{1} + m_{2}}$
Where $m$ is just the mass of each of the atoms. Since the mass of both hydrogens are 1 amu ($u$),
$\mu = \dfrac{1\times1}{1 + 1} = \dfrac{1}{2} u$
$\dfrac{1}{2} u = 8.3027\times10^{-28}kg$
Because the question is only interested in relative population, the actual number of molecules there are doesn’t really matter, since relative means to take the ratio between the probability of one energy state to another. Since the two energy states in comparison for this problem is $n=1, n=0$, (remember ground state is $n=0$), the equation is as follows.
$\dfrac{P(1)= Ce^{-\varepsilon_{1}/k_{b} 278.15 \, \mathrm{K}}}{P(0)= Ce^{-\varepsilon_{0}/k_{b} 278.15 \, \mathrm{K}}}$
By simplification,
$= e^{\dfrac{\varepsilon_{1} - \varepsilon_{0}}{K_{b}T}}$
And, substituting in the formula for energy($\varepsilon$),
$= e^{\dfrac{\left[\left(n + \dfrac{1}{2}\right)hv - \dfrac{1}{2}hv\right]}{K_{b}T}}$
$= e^{nhv/K_{b}T}$
Here, $n=1$ (because the 0 has already been taken into consideration) and $hv$ is as follows:
$hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}$
$hv = \dfrac{6.62607\times10^{-34}\dfrac{kgm^{2}}{s^2}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}$
$hv = \dfrac{6.62607\times10^{-34}Js^{-1}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}$
$hv= 8.2650\times 10^{-20} J$
All the variables have been found, so all there is left is to substitute and solve.
$= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{m^{2}kg}{s^{2}K}\left(278.15K\right)}}$
$= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{J}{K}\left(278.15K\right)}}$
$\approx e^{-21.5229}$
$\approx 4.495\times10^{-10}$
Thus, the probability of the $\ce{H2}$ molecules being in the 1st energy state when compared to the vibrational state of the lowest energy is $\approx 4.495\times10^{-10}$. This extremely low probability makes sense, because intuition says that the compound will most likely be found at its lowest energy.
Abstract: Find difference in energy, $\varepsilon$, divide by Boltzmann's constant ($K_{b}$) and Temperature. Plug as $e^{x}$ and solve.
Q69
A container holds 2 L of gas under 5.00 atm and a ball floating in 10 L of NaOH. As the volume of the gas expands to 20 L, the ball is turned upside down. This turn is caused by the temperature increase of the NaOH after the gas expands. Assuming that no heat is lost, the density of NaOH is 2.13 g/cm3 and the specific heat is 4.184 J/g K, calculate the increase in temperature of the NaOH.
Solution
The "work" completed by the ball to turn equals the negative value of the work the gas absorbs.
w = -(-P (V2-V1))
w = -(-5 atm) (20-2) L
w = 90 atm L (101.325 J/ 1 atm L)
w = 9119.25 J
The work carried out by the ball by the NaOH (in order to turn the ball) correlates to the increase in 10 L of NaOH temperature .
10 L of NaOH (1000 mL/1 L) = 10000 mL = 10000 cm3 (2.13 g/cm3) = 21300 g of NaOH
q = mCpΔT
ΔT = (q NaOH) / (Specific Heat NaOH * g NaOH)
ΔT = (9119.25 J) / (4.184 J/K g)(21300 g)
ΔT = 0.102 K gained by the NaOH to turn the ball
Q69
Some gas in a piston expands against a constant pressure of 1.2 atm from a volume of 3 L to 18 L. The piston turns an egg beater submerged in 150 g of water. If the water was originally at 25°C, what is its temperature once the gas stops expanding? Assume that all of heat goes into the water and the specific heat capacity of water is 4.184 J K-1 g-1.
Solution
From the first law of thermodynamics
$\Delta U = q + w\nonumber$
In this case,$\Delta U =0$, so
$q =w = P_{ext}\Delta V = P_{ext} (V_f - V_i)\nonumber$
$q = (1.2 \; \mathrm{atm} )(18\; \mathrm{L} - 3\; \mathrm{L}) = 14.4 \; \mathrm{L\; atm} \nonumber$
Convert From L atm to joules using the conversion factor
$14.4 \; \mathrm{L\; atm} \times 101.325\; \mathrm{J \; L^{-1} atm^{-1}} = 1459\; \mathrm{J} \nonumber$
Now connect the heat transferred to the temperature increased via the specific heat $c_{sp}$ via
$q = m c_{sp} \Delta T\nonumber$
$\Delta T = \dfrac{q}{m c_{sp}}\nonumber$
$\Delta T = \dfrac{1459\; \mathrm{J}}{(150\; \mathrm{g})(4.184\; \mathrm{J\; K^{-1} g^{-1}})} = 2.325 \; \mathrm{K} \nonumber$
Since initial temperatures is 25° = 298 K so
$T_{f} = 298 \; K + 2.325 \; \mathrm{K}\nonumber$
$T_{f}=300.325\; K = 27.325 \mathrm{°C} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.E%3A_Thermodynamic_Processes_%28Exercises%29.txt |
A spontaneous process is the time-evolution of a system in which it releases free energy and moves to a lower, more thermodynamically stable energy state.
• 13.1: The Nature of Spontaneous Processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur by force. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions and how relatively quickly or slowly that natural change proceeds. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
• 13.2: Entropy and Spontaneity - A Molecular Statistical Interpretation
These forms of motion are ways in which the molecule can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a perfect crystal lattice.
• 13.3: Entropy and Heat - Experimental Basis of the Second Law of Thermodynamics
A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The 2nd law states that in a reversible process, the entropy of the universe is constant and in an irreversible process, the entropy of the universe increases.
• 13.4: Entropy Changes in Reversible Processes
Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system. Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system.
• 13.5: Entropy Changes and Spontaneity
In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity.
• 13.6: The Third Law of Thermodynamics
This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal (Ω=1). According to the Boltzmann equation, the entropy of this system is zero.
• 13.7: The Gibbs Free Energy
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe. This is not particularly useful and a criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
• 13.8: Carnot Cycle, Efficiency, and Entropy
The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures.
• 13.E: Spontaneous Processes (Exercises)
13: Spontaneous Processes and Thermodynamic Equilibrium
Learning Objectives
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
Spontaneous and Nonspontaneous Processes
Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.
The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure $1$).
As another example, consider the conversion of diamond into graphite (Figure $2$).
$\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1}$
The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.
Dispersal of Matter and Energy
As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.
\begin{align} w&=−PΔV \[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align}
Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.
\begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \[4pt] &=0+0=0 \label{Eq3}\end{align}
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).
Now consider two objects at different temperatures: object X at temperature TX and object Y at temperature TY, with TX > TY (Figure $4$). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.
$q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4}$
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.
As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.
Example $1$: Redistribution of Matter during a Spontaneous Process
Describe how matter and energy are redistributed when the following spontaneous processes take place:
1. A solid sublimes.
2. A gas condenses.
3. A drop of food coloring added to a glass of water forms a solution with uniform color.
Solution
1. Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition. However, an input of energy from the surroundings ss required for the molecules to leave the solid phase and enter the gas phase.
2. Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the gas-to-liquid transition. As the gas molecules move together to form the droplets of liquid, they form intermolecular forces and thus release energy to the surroundings.
3. The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout. This process can occur with out a change in energy because the molecules have kinetic energy relative to the temperature of the water, and so will be constantly in motion.
Exercise $1$
Describe how matter and energy are redistributed when you empty a canister of compressed air into a room.
Answer
This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings.
Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging.
Glossary
nonspontaneous process
process that requires continual input of energy from an external source
spontaneous change
process that takes place without a continuous input of energy from an external source | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.1%3A_The_Nature_of_Spontaneous_Processes.txt |
Learning Objectives
• The Learning Objective of this Module is to understand the relationship between internal energy and entropy.
We know that the first law of thermodynamics governs changes in the state function we called internal energy ($U$) and that changes in the internal energy ($ΔU$) are closely related to changes in the enthalpy ($ΔH$), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Spontaneous Reactions
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., $ΔH_{soln} > 0$). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure $1$.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy ($S$), a thermodynamic property of all substances that is proportional to their degree of disorder.
The direction of a spontaneous process is not governed by the internal energy change nor enthalpy change and thus the First Law of Thermodynamics cannot predict the direction of a natural process
Probabilities and Microstates
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy ($ΔS > 0$) or a decrease in entropy ($ΔS < 0$), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states:
$ΔS = S_f − S_i$
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
How can we express disorder quantitatively? From the example of coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins.
Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its members are heads-up constitutes a macroscopic state of the system. Since we do not care which coins are heads-up, there are clearly numerous configurations of the individual coins which can result in this “macrostate”. Each of these configurations specifies a microscopic state of the system. The greater the number of microstates that correspond to a given macrostate (or configuration), the greater the probability of that macrostate. To see what this means, consider the possible outcomes of a toss of four coins (Table $1$):
Table $1$: Coin Toss Results
macrostate (Configuration) ways probability
microstates
0 heads 1 1/16 TTTT
1 head 4 4/16 = 1/4 HTTT THTT TTHT TTTH
2 heads 6 6/16 = 3/8 HHTT HTHT HTTH THHT TTHH THTH
3 heads 4 4/16 = 1/4 HHHT HTHH HHTH THHH
4 heads 1 1/16 HHHH
A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the table. The second column gives the number of “ways”— that is, the number of head/tail configurations of the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss resulting in a particular macrostate is proportional to the number of microstates corresponding to the macrostate, and is equal to this number, divided by the total number of possible microstates (in this example, 24 =16). An important assumption here is that all microstates are equally probable; that is, the toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an entirely random way.
The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate.
Entropy and Microstates
Following the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates possible for the system. A microstate ($\Omega$) is a specific configuration of the locations and energies of the atoms or molecules that comprise a system like the following:
$S=k \ln \Omega \label{Eq2}$
Here k is the Boltzmann constant and has a value of 1.38 × 10−23 J/K.
As for other state functions, the change in entropy for a process is the difference between its final ($S_f$) and initial ($S_i$) values:
\begin{align} ΔS &= S_\ce{f}−S_\ce{i} \[4pt] &=k \ln \Omega_\ce{f} − k \ln \Omega_\ce{i} \[4pt] &= k \ln\dfrac{\Omega_\ce{f}}{\Omega_\ce{i}} \label{Eq2a} \end{align}
For processes involving an increase in the number of microstates, $\Omega_f > \Omega_i$, the entropy of the system increases, $ΔS > 0$. Conversely, processes that reduce the number of microstates, $\Omega_f < \Omega_i$, yield a decrease in system entropy, $ΔS < 0$. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs.
Consider the general case of a system comprised of N particles distributed among n boxes. The number of microstates possible for such a system is nN. For example, distributing four particles among two boxes will result in 24 = 16 different microstates as illustrated in Figure $2$. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions. The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates (Equation \ref{Eq2}), the most probable distribution is therefore the one of greatest entropy.
For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is
$\dfrac{6}{16} = \dfrac{3}{8} \nonumber$
The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of
$\dfrac{1}{16} \nonumber$
The probability of finding all particles in only one box (either the left box or right box) is then
$\left(\dfrac{1}{16}+\dfrac{1}{16}\right)=\dfrac{2}{16} = \dfrac{1}{8} \nonumber$
As you add more particles to the system, the number of possible microstates increases exponentially (2N). A macroscopic (laboratory-sized) system would typically consist of moles of particles (N ~ 1023), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations.
The previous description of an ideal gas expanding into a vacuum is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system.
A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of energy (represented as “*”) in Figure $3$. The hot object is comprised of particles A and B and initially contains both energy units. The cold object is comprised of particles C and D, which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. And so, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is $\frac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\frac{7}{10}$. The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being $\frac{4}{10}$. As for the previous example of matter dispersal, extrapolating this treatment to macroscopic collections of particles dramatically increases the probability of the uniform distribution relative to the other distributions. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy.
Example $1$: Determination of ΔS
Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)?
Solution
We are interested in the following change:
The initial number of microstates is one, the final six:
\begin{align*} ΔS &=k \ln\dfrac{\Omega_\ce{c}}{\Omega_\ce{a}}\nonumber \[4pt] &= 1.38×10^{−23}\:J/K × \ln\dfrac{6}{1} \[4pt] &= 2.47×10^{−23}\:J/K \end{align*}
The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive.
Exercise $1$
Consider the system shown in Figure $3$. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)?
Answer
0 J/K
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Definition: Entropy
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Disorder is more probable than order because there are so many more ways of achieving it. Thus coins and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to become more scattered about a teenager’s room during the course of daily living. But there are some important differences between these large-scale mechanical, or macro systems, and the collections of sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically as molecules. Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move around. Thus, in chemical systems,
1. We are dealing with huge numbers of particles. This is important because statistical predictions are always more accurate for larger samples. Thus although for the four tosses there is a good chance (62%) that the H/T ratio will fall outside the range of 0.45 - 0.55, this probability becomes almost zero for 1000 tosses. To express this in a different way, the chances that 1000 gas molecules moving about randomly in a container would at any instant be distributed in a sufficiently non-uniform manner to produce a detectable pressure difference between the two halves of a container will be extremely small. If we increase the number of molecules to a chemically significant number (around 1020, say), then the same probability becomes indistinguishable from zero.
2. Once the change begins, it proceeds spontaneously. That is, no external agent (a tosser, shuffler, or teenager) is needed to keep the process going. Gases will spontaneously expand if they are allowed to, and reactions, once started, will proceed toward equilibrium.
3. Thermal energy is continually being exchanged between the particles of the system, and between the system and the surroundings. Collisions between molecules result in exchanges of momentum (and thus of kinetic energy) amongst the particles of the system, and (through collisions with the walls of a container, for example) with the surroundings.
4. Thermal energy spreads rapidly and randomly throughout the various energetically accessible microstates of the system. The direction of spontaneous change is that which results in the maximum possible spreading and sharing of thermal energy.
The importance of these last two points is far greater than you might at first think, but to fully appreciate this, you must recall the various ways in which thermal energy is stored in molecules.
Sign of Entropy Change
The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure $4$. In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, $S_{liquid} > S_{solid}$ and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, $ΔS > 0$. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, $ΔS < 0$.
Now consider the vapor or gas phase. The atoms or molecules occupy a much greater volume than in the liquid phase; therefore each atom or molecule can be found in many more locations than in the liquid (or solid) phase. Consequently, for any substance,
$S_{gas} \gg S_{liquid} > S_{solid}$
and the processes of vaporization and sublimation likewise involve increases in entropy, $ΔS > 0$. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, $ΔS < 0$.
According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure $5$ ).
The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (which is a topic beyond the scope of our treatment). For molecules, greater numbers of atoms (regardless of their masses) increase the ways in which the molecules can vibrate and thus the number of possible microstates and the system entropy.
Finally, variations in the types of particles affect the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, $ΔS > 0$.
Considering the various factors that affect entropy allows us to make informed predictions of the sign of $ΔS$ for various chemical and physical processes as illustrated in Example $2$.
Example $2$: Predicting the Sign of ∆S
Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions.
1. One mole liquid water at room temperature $⟶$ one mole liquid water at 50 °C
2. $\ce{Ag+}(aq)+\ce{Cl-}(aq)⟶\ce{AgCl}(s)$
3. $\ce{C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l)}$
4. $\ce{NH3}(s)⟶\ce{NH3}(l)$
Solution
1. positive, temperature increases
2. negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter
3. negative, net decrease in the amount of gaseous species
4. positive, phase transition from solid to liquid, net increase in dispersal of matter
Exercise $2$
Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.
1. $\ce{NaNO3}(s)⟶\ce{Na+}(aq)+\ce{NO3-}(aq)$
2. the freezing of liquid water
3. $\ce{CO2}(s)⟶\ce{CO2}(g)$
4. $\ce{CaCO}(s)⟶\ce{CaO}(s)+\ce{CO2}(g)$
Answer a
Positive; The solid dissolves to give an increase of mobile ions in solution.
Answer b
Negative; The liquid becomes a more ordered solid.
Answer c
Positive; The relatively ordered solid becomes a gas.
Answer d
Positive; There is a net production of one mole of gas.
Summary
Entropy ($S$) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system. For a given substance, $S_{solid} < S_{liquid} \ll S_{gas}$ in a given physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions may be reliably predicted. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.2%3A_Entropy_and_Spontaneity_-_A_Molecular_St.txt |
Learning Objectives
• The Learning Objective of this Module is to understand the relationship between internal energy and entropy.
Thermodynamic Definition of Entropy
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure \(1\), the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of \(\ce{NaCl}\) and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an \(\ce{NaCl}\) solution, disruption of the crystalline \(\ce{NaCl}\) structure and the hydrogen-bonded interactions in water is quantitatively more important, so \(ΔS_{soln} > 0\).
Dissolving \(\ce{NaCl}\) in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Example \(1\)
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
2. 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
1. Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
2. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Exercise \(1\)
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm or a sample of 2 mol of NH3(g) at 25°C and 1 atm
Answer a
1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
Answer b
a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm (more molecules of gas are present)
Video Solution
Summary
A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.3%3A_Entropy_and_Heat_-_Experimental_Basis_of.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in bold.
• A reversible process is one carried out in infinitesimal steps after which, when undone, both the system and surroundings (that is, the world) remain unchanged (see the example of gas expansion-compression below). Although true reversible change cannot be realized in practice, it can always be approximated.
• As a process is carried out in a more reversible manner, the value of w approaches its maximum possible value, and q approaches its minimum possible value.
• Although q is not a state function, the quotient qrev/T is, and is known as the entropy.
A change is said to occur reversibly when it can be carried out in a series of infinitesimal steps, each one of which can be undone by making a similarly minute change to the conditions that bring the change about. For example, the reversible expansion of a gas can be achieved by reducing the external pressure in a series of infinitesimal steps; reversing any step will restore the system and the surroundings to their previous state. Similarly, heat can be transferred reversibly between two bodies by changing the temperature difference between them in infinitesimal steps each of which can be undone by reversing the temperature difference.
The most widely cited example of an irreversible change is the free expansion of a gas into a vacuum. Although the system can always be restored to its original state by recompressing the gas, this would require that the surroundings perform work on the gas. Since the gas does no work on the surrounding in a free expansion (the external pressure is zero, so $PΔV = 0$,) there will be a permanent change in the surroundings. Another example of irreversible change is the conversion of mechanical work into frictional heat; there is no way, by reversing the motion of a weight along a surface, that the heat released due to friction can be restored to the system.
These diagrams show the same expansion and compression ±ΔV carried out in different numbers of steps ranging from a single step at the top to an "infinite" number of steps at the bottom. As the number of steps increases, the processes become less irreversible; that is, the difference between the work done in expansion and that required to re-compress the gas diminishes. In the limit of an ”infinite” number of steps (bottom), these work terms are identical, and both the system and surroundings (the “world”) are unchanged by the expansion-compression cycle. In all other cases the system (the gas) is restored to its initial state, but the surroundings are forever changed.
Definition: Reversible Changes
A reversible change is one carried out in such as way that, when undone, both the system and surroundings (that is, the world) remain unchanged.
It should go without saying, of course, that any process that proceeds in infinitesimal steps would take infinitely long to occur, so thermodynamic reversibility is an idealization that is never achieved in real processes, except when the system is already at equilibrium, in which case no change will occur anyway! So why is the concept of a reversible process so important?
The answer can be seen by recalling that the change in the internal energy that characterizes any process can be distributed in an infinity of ways between heat flow across the boundaries of the system and work done on or by the system, as expressed by the First Law of thermodynamics
$ΔU = q + w. \label{first}$
Each combination of q and w represents a different pathway between the initial and final states. It can be shown that as a process such as the expansion of a gas is carried out in successively longer series of smaller steps, the absolute value of q approaches a minimum, and that of w approaches a maximum that is characteristic of the particular process. Thus when a process is carried out reversibly, the w-term in Equation \ref{first} has its greatest possible value, and the q-term is at its smallest. These special quantities $w_{max}$ and $q_{min}$ (which we denote as $q_{rev}$ and pronounce “q-reversible”) have unique values for any given process and are therefore state functions.
Work and Reversibility
Changes in entropy ($ΔS$), together with changes in enthalpy ($ΔH$), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum ($P_{ext} = 0$) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure ($w = P_{ext}ΔV$), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: $w_{rev} ≥ w_{irrev}$. Whether a process is reversible or irreversible, the first law of thermodynamics holds (Equation \ref{first}). Because $U$ is a state function, the magnitude of $ΔU$ does not depend on reversibility and is independent of the path taken. So
\begin{align} ΔU &= q_{rev} + w_{rev} \[4pt] &= q_{irrev} + w_{irrev} \label{Eq1} \end{align}
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process:
$w_{rev} \ge w_{irrev}$
In other words, $ΔU$ for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process ($q_{rev}$) to define entropy quantitatively.
For a process that reversibly exchanges a quantity of heat $q_{rev}$ with the surroundings, the entropy change is defined as
$\color{red} \Delta S = \dfrac{q_{rev}}{T} \label{23.2.1}$
This is the basic way of evaluating $ΔS$ for constant-temperature processes such as phase changes, or the isothermal expansion of a gas. For processes in which the temperature is not constant such as heating or cooling of a substance, the equation must be integrated over the required temperature range, as discussed below.
Q&A: If no real process can take place reversibly, what use is an expression involving $q_{rev}$?
This is a rather fine point that you should understand: although transfer of heat between the system and surroundings is impossible to achieve in a truly reversible manner, this idealized pathway is only crucial for the definition of $ΔS$; by virtue of its being a state function, the same value of $ΔS$ will apply when the system undergoes the same net change via any pathway. For example, the entropy change a gas undergoes when its volume is doubled at constant temperature will be the same regardless of whether the expansion is carried out in 1000 tiny steps (as reversible as patience is likely to allow) or by a single-step (as irreversible a pathway as you can get!) expansion into a vacuum (Figure $1$).
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Entropy and "disorder"
Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical sense this is correct, since the spreading and sharing of thermal energy does have the effect of randomizing the disposition of thermal energy within a system. But to simply equate entropy with “disorder” without further qualification is extremely misleading because it is far too easy to forget that entropy (and thermodynamics in general) applies only to molecular-level systems capable of exchanging thermal energy with the surroundings. Carrying these concepts over to macro systems may yield compelling analogies, but it is no longer science. It is far better to avoid the term “disorder” altogether in discussing entropy.
$\Delta S_{sys}$ for an Isothermal Expansion (or Compression)
As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy per mole of solute (the molar entropy) will of course always increase as the solution becomes more dilute.
For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by
$\Delta S = R \ln \left( \dfrac{V_2}{V_1} \right) \label{23.2.4}$
Note: If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.
Because the pressure of an ideal gas is inversely proportional to its volume, i.e.,
$P = \dfrac{nRT}{V}$
we can easily alter Equation \ref{23.2.4} to express the entropy change associated with a change in the pressure of an ideal gas:
$\Delta S = R \ln \left( \dfrac{P_1}{P_2} \right) \label{23.2.5}$
Also the concentration $c=n/V$ for an ideal gas is proportional to pressure
$P = c RT$
we can expressing the entropy change directly in concentrations, we have the similar relation
$\Delta S = R \ln \left( \dfrac{c_1}{c_2} \right) \label{23.2.6}$
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
The Relationship between Internal Energy and Entropy
Because the quantity of heat transferred ($q_{rev}$) is directly proportional to the absolute temperature of an object ($T$) ($q_{rev} \propto T$), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder ($ΔS ∝ q_{rev}$). Combining these relationships for any reversible process,
$q_{\textrm{rev}}=T\Delta S$
and
$\Delta S=\dfrac{q_{\textrm{rev}}}{T} \label{Eq2}$
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is
$w_{rev} = −PΔV,$
we can express Equation $\ref{Eq1}$ as follows:
\begin{align} ΔU &= q_{rev} + w_{rev} \[4pt] &= TΔS − PΔV \label{Eq3} \end{align}
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the $PV$ work done.
To illustrate the use of Equation $\ref{Eq2}$ and Equation $\ref{Eq3}$, we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its T constant (Figure $5$). The internal energy of the gas does not change because the temperature of the gas does not change; that is, ΔU = 0 and qrev = −wrev. During expansion, ΔV > 0, so the gas performs work on its surroundings:
$w_{rev} = −PΔV < 0.$
According to Equation $\ref{Eq3}$, this means that $q_{rev}$ must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore
$ΔS_{sys} = +\dfrac{q_{rev}}{T}$
and the entropy change of the surroundings is
$ΔS_{surr} = −\dfrac{q_{rev}}{T}.$
The corresponding change in entropy of the universe is then as follows:
\begin{align} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \[4pt] &=\dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \[4pt] &= 0 \label{Eq4} \end{align}
Thus no change in $ΔS_{univ}$ has occurred.
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure $6$. The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation $\ref{Eq2}$, we see that the entropy of fusion of ice can be written as follows:
$\Delta S_{\textrm{fus}}=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{\Delta H_{\textrm{fus}}}{T} \label{Eq5}$
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case,
\begin{align} ΔS_{fus} &= \dfrac{6.01\, kJ/mol}{273\, K} \[4pt] &= 22.0\, J/(mol•K) \[4pt] &= ΔS_{sys}.\end{align}
The amount of heat lost by the surroundings is the same as the amount gained by the ice, so
\begin{align} ΔS_{surr} &= \dfrac{q_{rev}}{T} \[4pt] &= \dfrac{ −6.01\, kJ/mol}{273\, K} \[4pt] &= −22.0\, J/(mol•K).\end{align}
Once again, we see that the entropy of the universe does not change:
\begin{align} ΔS_{univ} &= ΔS_{sys} + ΔS_{surr} \[4pt] &= 22.0 \,J/(mol•K) − 22.0\, J/(mol•K) \[4pt] &= 0 \end{align}
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The Second Law of Thermodynamics
The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔScold = q/Tcold. Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔShot = −q/Thot, where Tcold and Thot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
\begin{align} \Delta S_{\textrm{univ}} &= \Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}} \[4pt] &= \dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6}\end{align}
The numerators on the right side of Equation $\ref{Eq6}$ are the same in magnitude but opposite in sign. Whether ΔSuniv is positive or negative depends on the relative magnitudes of the denominators. By definition, Thot > Tcold, so −q/Thot must be less than q/Tcold, and ΔSuniv must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔSuniv is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔSuniv approaches zero will not occur spontaneously as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure $7$).
When molten lava flows into cold ocean water, so much heat is spontaneously transferred to the water that steam is produced.
Example $1$: Tin Pest
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
1. What is ΔS for this process?
2. Which is the more highly ordered form of tin—white or gray?
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy:
Use Equation $\ref{Eq2}$ to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution
1. We know from Equation $\ref{Eq2}$ that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{(-2.1\;\mathrm{kJ/mol})(1000\;\mathrm{J/kJ})}{\textrm{286.4 K}}=-7.3\;\mathrm{J/(mol\cdot K)}$
1. The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C) to facilitate tin pest.
Exercise $1$: Sulfur
Elemental sulfur exists in two forms: an orthorhombic form (Sα), which is stable below 95.3°C, and a monoclinic form (Sβ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with $ΔH = 0.401\, kJ/mol$ at 1 atm.
1. What is ΔS for this process?
2. Which is the more highly ordered form of sulfur—Sα or Sβ?
Answer a
1.09 J/(mol•K)
Answer b
Sα
Summary
During a spontaneous process, the entropy of the universe increases. $\Delta S=\dfrac{q_{\textrm{rev}}}{T}\nonumber$ A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.4%3A_Entropy_Changes_in_Reversible_Processes.txt |
Learning Objectives
• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
Connecting Entropy and Heat to Spontaneity
In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the system ($ΔS_{sys} > 0$) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:
$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{$1$}$
To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}$ and $ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{$2$}$The arithmetic signs of qrev denote the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of ΔSsys and ΔSsurr will yield a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}$ and $ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{$3$}$ The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The temperature difference between the objects is infinitesimally small, $T_{sys} ≈ T_{surr}$, and so the heat flow is thermodynamically reversible. See the previous section’s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.
These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $1$.
Table $1$: The Second Law of Thermodynamics
$ΔS_{univ} > 0$ spontaneous
$ΔS_{univ} < 0$ nonspontaneous (spontaneous in opposite direction)
$ΔS_{univ} = 0$ reversible (system is at equilibrium)
Definition: The Second Law of Thermodynamics
All spontaneous changes cause an increase in the entropy of the universe, i.e., $ΔS_\ce{univ} > 0.$
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, $q_{surr}$ is a good approximation of qrev, and the second law may be stated as the following:
\begin{align} ΔS_\ce{univ} &=ΔS_\ce{sys}+ΔS_\ce{surr} \[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4} \end{align}
We may use this equation to predict the spontaneity of a process as illustrated in Example $1$.
Example $1$: Will Ice Spontaneously Melt?
The entropy change for the process
$\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber$
is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?
Solution
We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.
At −10.00 °C (263.15 K), the following is true:
\begin{align*} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr} \[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \[4pt] &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}} \[4pt] &=−0.7\:J/K \end{align*}
Suniv < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
\begin{align*} ΔS_\ce{univ} &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \[4pt] &= 22.1 \:J/K + \dfrac{−6.00×10^3\:J}{283.15\: K} \[4pt] &=+0.9\: J/K \end{align*}
$\Delta S_{univ} > 0$, so melting is spontaneous at 10.00 °C.
Exercise $1$
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of $\Delta S_{univ}$?
Answer
Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
Summary
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, $S_{univ} > 0$. If $ΔS_{univ} < 0$, the process is nonspontaneous, and if $ΔS_{univ} = 0$, the system is at equilibrium.
Key Equations
• $ΔS^\circ=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants})$
• $ΔS=\dfrac{q_\ce{rev}}{T}$
• $ΔS_{univ} = ΔS_{sys} + ΔS_{surr}$
• $ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.5%3A_Entropy_Changes_and_Spontaneity.txt |
Learning Objectives
• The absolute entropy of a pure substance at a given temperature is the sum of all the entropy it would acquire on warming from absolute zero (where $S=0$) to the particular temperature.
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $1$). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum zero point motion).
This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal ($\Omega = 1$). According to the Boltzmann equation, the entropy of this system is zero.
\begin{align*} S&=k\ln \Omega \[4pt] &= k\ln(1) \[4pt] &=0 \label{$5$} \end{align*}
In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Definition: Third Law of Thermodynamics
The entropy of a pure, perfect crystalline substance at 0 K is zero.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Standard-State Entropies
One way of calculating $ΔS$ for a reaction is to use tabulated values of the standard molar entropy ($S^o$), which is the entropy of 1 mol of a substance under standard pressure (1 bar). Often the standard molar entropy is given at 298 K and is often demarked as $ΔS^o_{298}$. The units of $S^o$ are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K (corresponding to $S = 0$) and 298 K (Tables T1 and T2).
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, $S^o$ values fall in the order
$S^o(\text{gas}) \gg S^o(\text{liquid}) > S^o(\text{solid}).$
For instance, $S^o$ for liquid water is 70.0 J/(mol•K), whereas $S^o$ for water vapor is 188.8 J/(mol•K). Likewise, $S^o$ is 260.7 J/(mol•K) for gaseous $\ce{I2}$ and 116.1 J/(mol•K) for solid $\ce{I2}$. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases (Figure $1$). The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
The Third Law Lets us Calculate Absolute Entropies
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat $q$ required to bring the substance from 0 K to the temperature of interest, and then summing the ratios $q/T$. Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with transition at temperature $T$ is $\dfrac{ΔH_{fusion}}{T}.$
2. The heat capacity $C$ of a phase expresses the quantity of heat required to change the temperature by a small amount $ΔT$, or more precisely, by an infinitesimal amount $dT$. Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities $C \frac{dT}{T}$ for each increment of temperature $dT$. This is of course just the integral
$S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p}{T} dt \label{eq20}$
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of $C$ on $T$ be used in the integral in Equation \ref{eq20}, i.e.,:
$S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p(T)}{T} dt. \label{eq21}$
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments $ΔT$ and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval $ΔT$. To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes.
Values of $C_p$ for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (Figure $3$) are then plotted as a function of $T$, and any phase-change entropies such as
$S_{vap} = \dfrac{H_{vap}}{T_b}$
are added to obtain the absolute entropy at temperature $T$. As shown in Figure $2$ above, the entropy of a substance increases with temperature, and it does so for two reasons:
• As the temperature rises, more microstates become accessible, allowing thermal energy to be more widely dispersed. This is reflected in the gradual increase of entropy with temperature.
• The molecules of solids, liquids, and gases have increasingly greater freedom to move around, facilitating the spreading and sharing of thermal energy. Phase changes are therefore accompanied by massive and discontinuous increase in the entropy.
Calculating $\Delta S_{sys}$
We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label $S^o_{298}$ for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The standard entropy change ($ΔS^o$) for any process may be computed from the standard entropies of its reactant and product species like the following:
$ΔS^o=\sum νS^o_{298}(\ce{products})−\sum νS^o_{298}(\ce{reactants}) \label{$6$}$
Here, $ν$ represents stoichiometric coefficients in the balanced equation representing the process. For example, $ΔS^o$ for the following reaction at room temperature
$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D} \label{$7$}$
is computed as the following:
$ΔS^o=[xS^o_{298}(\ce{C})+yS^o_{298}(\ce{D})]−[mS^o_{298}(\ce{A})+nS^o_{298}(\ce{B})] \label{$8$}$
Table $1$ lists some standard entropies at 298.15 K. You can find additional standard entropies in Tables T1 and T2
Gases Liquids Solids
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Substance $S^o$ [J/(mol•K)] Substance $S^o$ [J/(mol•K)] Substance $S^o$ [J/(mol•K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar $S^o$ values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond ($S^o = 2.4 \,J/(mol•K)$). In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher $S^o$ (5.7 J/(mol•K)) due to more disorder (microstates) in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the $S^o$ values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of $S^o$, which reflects a more ordered structure.
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
To calculate $ΔS^o$ for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $1$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane ($\ce{C8H18}$; 2,2,4-trimethylpentane).
Example $1$
Use the data in Table $1$ to calculate $ΔS^o$ for the reaction of liquid isooctane with $\ce{O2(g)}$ to give $\ce{CO2(g)}$ and $\ce{H2O(g)}$ at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table $1$. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain $ΔS^o$ for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane ($\ce{C8H18}$) is as follows:
$\ce{C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(g)} \nonumber$
We calculate $ΔS^o$ for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align*}\Delta S^o_{\textrm{rxn}}&=\sum mS^o(\textrm{products})-\sum nS^o(\textrm{reactants}) \[4pt] &=[8S^o(\mathrm{CO_2})+9S^o(\mathrm{H_2O})]-[S^o(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^o(\mathrm{O_2})] \[4pt] &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \} \[4pt] & \,\,\, -\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} \[4pt] &=515.3\;\mathrm{J/K}\end{align*}
$ΔS^o$ is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $1$
Use the data in Table $1$ to calculate $ΔS^o$ for the reaction of $\ce{H2(g)}$ with liquid benzene ($\ce{C6H6}$) to give cyclohexane ($\ce{C6H12}$) at 298 K.
Answer
-361.1 J/K
Example $2$: Determination of ΔS°
Calculate the standard entropy change for the following process at 298 K:
$\ce{H2O}(g)⟶\ce{H2O}(l)\nonumber$
Solution
The value of the standard entropy change at room temperature, $ΔS^o_{298}$, is the difference between the standard entropy of the product, H2O(l), and the standard entropy of the reactant, H2O(g).
\begin{align*} ΔS^o_{298} &=S^o_{298}(\ce{H2O (l)})−S^o_{298}(\ce{H2O(g)})\nonumber \[4pt] &= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})\nonumber \[4pt] &=−118.8\:J\:mol^{−1}K^{−1} \end{align*}
The value for $ΔS^o_{298}$ is negative, as expected for this phase transition (condensation), which the previous section discussed.
Exercise $2$
Calculate the standard entropy change for the following process at 298 K:
$\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g)\nonumber$
Answer
−120.6 J mol−1 K−1
Example $3$: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH at 298 K:
$\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l)\nonumber$
Solution
The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients. The standard entropy of formations are found in Table $1$.
\begin{align*} ΔS^o &=ΔS^o_{298} \[4pt] &= ∑νS^o_{298}(\ce{products})−∑νS^o_{298} (\ce{reactants}) \[4pt] & = 2S^o_{298}(\ce{CO2}(g))+4S^o_{298}(\ce{H2O}(l))]−[2S^o_{298}(\ce{CH3OH}(l))+3S^o_{298}(\ce{O2}(g))]\nonumber \[4pt] &= [(2 \times 213.8) + (4×70.0)]−[ (2 \times 126.8) + (3 \times 205.03) ]\nonumber \[4pt] &= −161.6 \:J/mol⋅K\nonumber \end{align*}
Exercise $3$
Calculate the standard entropy change for the following reaction at 298 K:
$\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l)\nonumber$
Answer
24.7 J/mol•K
Summary
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298 K and 1 atm pressure as zero. The same is not true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the Third law of thermodynamics, which states that the entropy of a perfectly-ordered solid at 0 K is zero.
In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity ($C_p$) as a function of temperature and then plotting the quantity $C_p/T$ versus $T$. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at $T$. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms.
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.6%3A_The_Third_Law_of_Thermodynamics.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if $ΔS_{univ} > 0$ for a process or a reaction, then the process will occur spontaneously as written. Conversely, if $ΔS_{univ} < 0$, a process cannot occur spontaneously; if $ΔS_{univ} = 0$, the system is at equilibrium. The sign of $ΔS_{univ}$ is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using $ΔS_{univ}$ requires that we calculate $ΔS$ for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Energy and Spontaneity
The Gibbs energy (also known as the Gibbs function) is defined as
$G_{sys} = H_{sys} – T S_{sys} \label{23.4.1}$
in which $S$ refers to the entropy of the system. Since $H$, $T$ and $S$ are all state functions, so is $G$. Thus for any change in state, we can expand Equation \ref{23.4.1} to
$ΔG_{sys} = ΔH_{sys} – T ΔS_{sys} \label{23.4.2}$
How does this simple equation relate to the entropy change of the universe $ΔS_{univ}$ that we know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition
$ΔS_{univ} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}$
we would first like to get rid of $ΔS_{surr}$. How can a chemical reaction (a change in the system) affect the entropy of the surroundings? Because most reactions are either exothermic or endothermic, they are accompanied by heat $q_p$ across the system boundary (we are considering constant pressure processes). The enthalpy change of the reaction $ΔH_{sys}$ is the "flow" of heat into the system from the surroundings under constant pressure, so the heat "withdrawn" from the surroundings will be $–q_p$.
From the thermodynamic definition of entropy, the change of the entropy of the surroundings will be
$ΔS_{surr} = –\dfrac{q_p}{T} = –\dfrac{ΔH_{sys}}{T}.$
We can therefore rewrite Equation $\ref{23.4.3}$ as
$ΔS_{univ} = \dfrac{– ΔH_{sys}}{T} + ΔS_{sys} \label{23.4.4}$
Multiplying through by $–T$ , we obtain
$–T ΔS_{univ} = ΔH_{sys} – T ΔS_{sys} \label{23.4.5}$
which expresses the entropy change of the universe in terms of thermodynamic properties of the system exclusively.
If $–TΔS_{univ}$ is denoted by $ΔG$, then we have Equation \ref{23.4.2} which defines the Gibbs energy change for the process.
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and
$TΔS = q_{rev}.$
Using these expressions, we can reduce Equation $\ref{Eq2}$ to
$ΔG = q − q_{rev}.$
Thus $ΔG$ is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, $q = q_{rev}$ and $ΔG = 0$. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which $ΔG ≠ 0$? To understand how the sign of $ΔG$ for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T} \nonumber$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH \nonumber$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align*} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align*}
Multiplying both sides of the inequality by $−T$ reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG_{sys} < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the entropy of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of $ΔG$ dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor.
Example $1$: Vaporizing Water
At Room Temperature (100 °C)
If the conversion of liquid water to water vapor is carried out at 1 atm and the normal boiling point of 100.00 °C (373.15 K), we can calculate $ΔG$ from the experimentally measured value of $ΔH_{vap}$ (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of $ΔS$ (Equation $\ref{Eq3}$), we know that for 1 mol of water,
\begin{align*} \Delta S_{\textrm{vap}} &= \dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \[4pt] &=\textrm{108.96 J/K} \end{align*}
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\begin{align*} \Delta G_{100^\circ\textrm C} &=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{0 J} \end{align*}
The energy required for vaporization offsets the increase in entropy of the system. Thus $ΔG = 0$, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Above Room Temperature (110 °C)
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of $ΔG$ for the vaporization of 1 mol of water at 110°C, assuming that $ΔH$ and $ΔS$ do not change significantly with temperature, becomes
\begin{align*} \Delta G_{110^\circ\textrm C} &=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \[4pt] &=-\textrm{1,091 J} \end{align*}
Since $ΔG < 0$, the vaporization of water is predicted to occur spontaneously and irreversibly at 110 °C.
Below Room Temperature (90 °C)
We can also calculate $ΔG$ for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\begin{align*}\Delta G_{90^\circ\textrm C} &= \Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]\[4pt] &=\textrm{1,088 J} \end{align*}
Since $ΔG > 0$, water does not spontaneously convert to water vapor at 90 °C. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Equilibrium Temperature
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of $ΔH$ and $ΔS$ into the definition of $ΔG$ (Equation $\ref{Eq2}$), setting $ΔG = 0$, and solving for $T$,
\begin{align*} 0 &=40,657\, J−T(108.96\, J/K) \[4pt] T&=373.15 \,K \end{align*}
Thus $ΔG = 0$ at $T = 373.15\, K$ and $1\, atm$, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water.
At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, $ΔG$ is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the $ΔH$ and $TΔS$ terms vary with temperature for the vaporization of water. When the two lines cross, $ΔG = 0$, and $ΔH = TΔS.$
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy ($ΔH > 0$), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid ($ΔS > 0$). At temperatures greater than 373 K, the $TΔS$ term dominates, and $ΔG < 0$, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
Figure $1$
$\Delta{H}$ $\Delta{S}$ $-T\Delta S$ $\Delta{G}$
-
(exothermic)
+
(products more disordered)
-
(favors spontaneity)
-
(spontaneous at all T)
-
(exothermic)
-
(products less disordered)
+
(opposes spontaneity)
-
(spontaneous) at low T
+
(non-spontaneous) at high T "Enthalpically-driven process"
+
(endothermic)
+
(products more disordered)
-
(favors spontaneity)
+
(non-spontaneous) at low T
-
(spontaneous) at high T "Entropically-driven process"
+
(endothermic)
-
(products less disordered)
+
(opposes spontaneity)
+
(non-spontaneous at all T)
Enthalpically vs. Entropically Driving Reactions
Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether $ΔH$ or the $TΔS$ term dominates. This is technically correct, but misleading because it disguises the important fact that $ΔS_{total}$, which this equation expresses in an indirect way, is the only criterion of spontaneous change.
Consider the following possible states for two different types of molecules with some attractive force:
There would appear to be greater entropy on the left (state 1) than on the right (state 2). Thus the entropic change for the reaction as written (i.e. going to the right) would be (-) in magnitude, and the energetic contribution to the free energy change would be (+) (i.e. unfavorable) for the reaction as written.
In going to the right, there is an attractive force and the molecules adjacent to each other is a lower energy state (heat energy, $q$, is liberated). To go to the left, we have to overcome this attractive force (input heat energy) and the left direction is unfavorable with regard to heat energy q. The change in enthalpy is (-) in going to the right (q released), and this enthalpy change is negative (-) in going to the right (and (+) in going to the left). This reaction as written, is therefore, enthalpically favorable and entropically unfavorable. Hence, It is enthalpically driven.
From Table $1$, it would appear that we might be able to get the reaction to go to the right at low temperatures (lower temperature would minimize the energetic contribution of the entropic change). Looking at the same process from an opposite direction:
This reaction as written, is entropically favorable, and enthalpically unfavorable; it is entropically driven. From Table $1$, it would appear that we might be able to get the reaction to go to the right at high temperatures (high temperature would increase the energetic contribution of the entropic change).
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of $ΔG$ allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of $ΔG$ for a process provides other important information. The change in Gibbs energy is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure):
$ΔG = w_{max}.$
To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process ($ΔH < 0$), such as during the combustion of a fuel, some of that energy can be used to do work ($ΔG < 0$), while some is used to increase the entropy of the universe ($TΔS > 0$). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, $ΔH$ will never equal $ΔG$ because $ΔS$ has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $2$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $2$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of $ΔG^o$ depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of $ΔG^o$ for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive $ΔG^o$ means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$\ce{ H2(g) + O2(g) <=> H2O2(l)} \nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate $ΔG^o$ for the reaction.
3. Use the value of $ΔG^o$ to determine whether the reaction is spontaneous as written.
Solution
A To calculate $ΔG^o$ for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times 205.2\;\mathrm{J/(mol\cdot K)}] \right \} \[4pt] &=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*}
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}] \[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \[4pt] &=-120.31\textrm{ kJ/mol} \nonumber \end{align*}
The negative value of $ΔG^o$ indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of $ΔG^o$ depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$\ce{ 2H2(g) + N2(g) <=> N2H4(l)}. \nonumber$
At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N2H4) = 121.2 J/(mol•K), S°(N2) = 191.6 J/(mol•K), and S°(H2) = 130.7 J/(mol•K). Is the reaction spontaneous as written?
Answer
149.5 kJ/mol; no
Tabulated values of standard free energies of formation allow chemists to calculate the values of $ΔG^o$ for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $ΔG^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$ΔG^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum n ΔG^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative $ΔG^o$ indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
"Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated $ΔG^o_f$ values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate $ΔG^o$ for the reaction of isooctane with oxygen gas to give carbon dioxide and water. Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol.
Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\ce{ C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(l)} \nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \[4pt] &-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}}) \end{align*}
Because $ΔG^o$ is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of $ΔG^o$ is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate $ΔG^o$ for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Calculated values of $ΔG^o$ are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of $ΔG^o$ may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate $ΔG^o$ at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) $ΔG^o$ and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: $ΔG^o$ and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain $ΔG^o$ for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate $ΔG^o$ for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \[4pt] &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \[4pt] &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \} \[4pt] &=-198.1\textrm{ J/K (per mole of N}_2)\end{align*}
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})] \[4pt] &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})] \[4pt] &=-91.8\textrm{ kJ(per mole of N}_2) \end{align*}
B Inserting the appropriate values into Equation $\ref{Eq5}$
\begin{align*} \Delta G^\circ_{\textrm{rxn}} &=\Delta H^\circ-T\Delta S^\circ \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt] &=-\textrm{32.7 kJ (per mole of N}_2) \end{align*}
C To calculate $ΔG$ for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt] &=21.7\textrm{ kJ (per mole of N}_2) \end{align*}
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$\ce{2NO(g) + O2 (g) <=> 2NO2 (g)} \nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting $ΔG^o$ = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example $3$ to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set $ΔG^o$ equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to $ΔG^o$ = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting $ΔG^o$ equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \[4pt] \Delta H^\circ &=T\Delta S^\circ \[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*}
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example $3$, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change
$ΔG = ΔH − TΔS \nonumber$
• Standard free-energy change
$ΔG° = ΔH° − TΔS° \nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate $ΔG^o$ for a reaction. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.7%3A_The_Gibbs_Free_Energy.txt |
In the early 19th century, steam engines came to play an increasingly important role in industry and transportation. However, a systematic set of theories of the conversion of thermal energy to motive power by steam engines had not yet been developed. Nicolas Léonard Sadi Carnot (1796-1832), a French military engineer, published Reflections on the Motive Power of Fire in 1824. The book proposed a generalized theory of heat engines, as well as an idealized model of a thermodynamic system for a heat engine that is now known as the Carnot cycle. Carnot developed the foundation of the second law of thermodynamics, and is often described as the "Father of thermodynamics."
The Carnot Cycle
The Carnot cycle consists of the following four processes:
1. A reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs $q_{in}$ amount heat from a heat source at a high temperature $T_{high}$, expands and does work on surroundings.
2. A reversible adiabatic gas expansion process. In this process, the system is thermally insulated. The gas continues to expand and do work on surroundings, which causes the system to cool to a lower temperature, $T_{low}$.
3. A reversible isothermal gas compression process. In this process, surroundings do work to the gas at $T_{low}$, and causes a loss of heat, $q_{out}$.
4. A reversible adiabatic gas compression process. In this process, the system is thermally insulated. Surroundings continue to do work to the gas, which causes the temperature to rise back to $T_{high}$.
P-V Diagram
The P-V diagram of the Carnot cycle is shown in Figure $2$. In isothermal processes I and III, ∆U=0 because ∆T=0. In adiabatic processes II and IV, q=0. Work, heat, ∆U, and ∆H of each process in the Carnot cycle are summarized in Table $1$.
Table $1$: Work, heat, ∆U, and ∆H in the P-V diagram of the Carnot Cycle.
Process w q ΔU ΔH
I $-nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)$ $nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)$ 0 0
II $n\bar{C_{v}}(T_{low}-T_{high})$ 0 $n\bar{C_{v}}(T_{low}-T_{high})$ $n\bar{C_{p}}(T_{low}-T_{high})$
III $-nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ $nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ 0 0
IV $n\bar{C_{v}}(T_{high}-T_{low})$ 0 $n\bar{C_{v}}(T_{hight}-T_{low})$ $n\bar{C_{p}}(T_{high}-T_{low})$
Full Cycle $-nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ $nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ 0 0
T-S Diagram
The T-S diagram of the Carnot cycle is shown in Figure $3$. In isothermal processes I and III, ∆T=0. In adiabatic processes II and IV, ∆S=0 because dq=0. ∆T and ∆S of each process in the Carnot cycle are shown in Table $2$.
Table $1$: Work, heat, and ∆U in the T-S diagram of the Carnot Cycle.
Process ΔT ΔS
I 0 $-nR\ln\left(\dfrac{V_{2}}{V_{1}}\right)$
II $T_{low}-T_{high}$ 0
III 0 $-nR\ln\left(\dfrac{V_{4}}{V_{3}}\right)$
IV $T_{high}-T_{low}$ 0
Full Cycle 0 0
Efficiency
The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the carnot engine is defined as the ratio of the energy output to the energy input.
\begin{align*} \text{efficiency} &=\dfrac{\text{net work done by heat engine}}{\text{heat absorbed by heat engine}} =\dfrac{-w_{sys}}{q_{high}} \[4pt] &=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln \left(\dfrac{V_{4}}{V_{3}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)} \end{align*}
Since processes II (2-3) and IV (4-1) are adiabatic,
$\left(\dfrac{T_{2}}{T_{3}}\right)^{C_{V}/R}=\dfrac{V_{3}}{V_{2}}$
and
$\left(\dfrac{T_{1}}{T_{4}}\right)^{C_{V}/R}=\dfrac{V_{4}}{V_{1}}$
And since T1 = T2 and T3 = T4,
$\dfrac{V_{3}}{V_{4}}=\dfrac{V_{2}}{V_{1}}$
Therefore,
$\text{efficiency}=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}$
$\boxed{\text{efficiency}=\dfrac{T_{high}-T_{low}}{T_{high}}}$
Summary
The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures.
Problems
1. You are now operating a Carnot engine at 40% efficiency, which exhausts heat into a heat sink at 298 K. If you want to increase the efficiency of the engine to 65%, to what temperature would you have to raise the heat reservoir?
2. A Carnot engine absorbed 1.0 kJ of heat at 300 K, and exhausted 400 J of heat at the end of the cycle. What is the temperature at the end of the cycle?
3. An indoor heater operating on the Carnot cycle is warming the house up at a rate of 30 kJ/s to maintain the indoor temperature at 72 ºF. What is the power operating the heater if the outdoor temperature is 30 ºF? | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.8%3A_Carnot_Cycle_Efficiency_and_Entropy.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q3
Write detailed calculation down for each question involved in gambling,
1. How many “microstates” are there for the cards that come up from five random cards each from a separate deck?
2. What is the probability of getting 5 queens of hearts?
3. What is the probability of getting any specific hand of five cards?
Solution
Microstates are specific configurations in which each particle is distinguishable. The number of ways that they can be arranged to describe the macrostate is the weight of that macrostate and used in the definition of entropy.
1. The number of possible outcomes (microstates) of any repeated independent situation is equal to the number of possibilities in one iteration to the power of the repetitions. So in this case, the answer is 52 to the power of 5.
2. One of these $\mathrm{52^5}$ possibilities.
3. One of these $\mathrm{52^5}$ possibilities.
Q5
Is the above reaction spontaneous under 1 atm and room temperature? If so, what are the driving factors?
$\ce{3NO(g) <=> N2O(g) + NO2(g)} \nonumber$
Solution
From a superficial overview of the reaction, we would conclude that the enthalpy change would be negative (i.e., $\Delta H < 0$) (why? more bonding in the products than the reactants). We would also expect a negative value of entropy ($\Delta n_{gas} < 0$. It is difficult to argue which would win from just looking at the reaction and we need to do a Hess's law like approach to solve numerically using the Gibbs energies of formation from Table T1:
$\Delta^o G_{rxn} = \Delta^o G_f \{ N_2O \} + \Delta G^o_f \{ NO_2 \} - 3 \Delta G^o_f \{ NO \}\nonumber$
Q9
For each of the following scenarios predict the system's entropy change whether it is $\Delta S < 0$, $\Delta S = 0$, or $\Delta S > 0$.
1. Increasing the amount of quarters being flipped from 2 to 4.
2. Moving a child from a 30 ft. × 30 ft. room to a 5 ft. × 5 ft. room to play in .
3. Increasing the temperature of a non-ideal polyatomic gas.
Solution
1. Increasing the amount of flipped quarters leads to ($\Delta S > 0$) because it increases the number of microstates available. For example: $2^2 = 4$ so 4 microstates possible with 2 quarters but $2^4 = 16$ so 16 microstates are available with four quarters. 16 > 4 so entropy would increase.
2. If you divided the room in 1 sq. ft cubes as the places where the child may stand/play, there are only 25 squares in which the child could stand so 25 microstates versus 900 squares/microstates in the other room. Therefore $\Delta S < 0$ because there are less microstates available.
3. Increasing the temperature of a non-ideal gas would lead to ($\Delta S > 0$) because the increased temperature would "unfreeze" "frozen" modes in the gas, which would allow it more degrees of freedom and therefore microstates.
Q10
Determine whether the change in entropy for the processes listed down below are positive or negative:
1. condensing of iodine gas into a solid
2. diffusion of an ideal gas
Solution
1. Change in entropy is negative, as gaseous iodine turns solid, the decreased freedom of movement restricts the possible particle locations and hence there is a decrease in the number of microstates.
2. Change of entropy is positive. As diffusion occurs, there is increased freedom of movement and increase in the possible particle locations and an increase in the number of microstates.
Q11
Osmium melts at 3,027 °C and has an enthalpy change of fusion of 31.0 kJ/mol. Calculate the entropy of fusion of osmium.
Solution
Define the system at 1 mole of solid osmium at its melting point of 3027°C (3300 K). Imagine adding 31.0 kJ of heat infinitely slowly in such a way that the temperature remains constant, as the osmium melts. The heat is then equal to the qrev for the melting. Substitute the value of heat supplied and T into the equation that defines the entropy of the change of the entropy of a system. Because the change occurs at a constant temperature, T may be outside the integral sign
$\Delta S=\int\frac{dq_{rev}}{T}=\frac{1}{T}\int dq_{rev}=\frac{1}{T}q_{rev}=\frac{31.0\times10^{3}\;J\;mol^{-1}}{3300\;K}=9.39\;J\;K^{-1}\;mol^{-1}\nonumber$
Tip. The temperature must be an absolute temperature (in Kelvins, for example).
Q13
Trouton's Rule is used to estimate the molar enthalpy of vaporization. What is iodine's enthalpy of vaporization if its normal boiling point is 184.4oC?
Solution
Trouton's Rule:
$\Delta H_{vap}=\Delta S_{vap}T_{b}$
$\Delta S_{vap}\approx 88\frac{J}{Kmol}$
$T_{b}=184.4^{o}C=457.55K$
$\Delta H_{vap}=(88\frac{J}{Kmol})(457.55K)=40264.4\frac{J}{mol}=40.26\frac{kJ}{mol}$
Enthalpy of vaporization of iodine is 40.26 kJ/mol
Q15
If 0.250 mol Argon is expanded reversibly and isothermally at 400 K in its compressible oven from an initial volume of 12.0 L to a final volume of 30.0 L, what will the ∆U, q, w, ∆H, and ∆S for the gas?
Solution
For an isothermal process, ∆U = 0, and consequently, ∆H = 0. However, for the work:
$w = \int_{v_o}^{v_f} P \, dV \nonumber$
Where:
$P = \dfrac{nRT}{V} \nonumber$
meaning that if we put P into terms of V using PV=nRT, then we can get the work, which comes out look like:
$w= nRT \ln{\dfrac{V_f}{V_o}} \nonumber$
$w = (0.250 \; mol) \times (8.314 \; \mathrm{\dfrac{J}{K mol}} ) \times (400 \; \mathrm{K}) \times \ln{ \dfrac{30.0 \; L}{12.0 \; L}} = 761.8 \; \mathrm{J} \nonumber$
$PV = nRT$ is the ideal gas law. it implies that the molecules or atoms of the gas are point masses, they have no volume and undergo only elastic collision.
Therefore, by the first law, the heat, q, must be be the negative of the work. Now, as for ∆S, we know:
$\Delta{S}= \dfrac{-q}{T}\nonumber$
which implies that:
$\Delta{S}= nR\ln{\dfrac{V_f}{V_o}}\nonumber$
$\Delta{S}= (0.250 \; \mathrm{mol}) \times (8.314 \; \mathrm{\dfrac{J}{K mol}} ) \times \ln{ \dfrac{30.0 \; \mathrm{L}}{12.0 \; \mathrm{L}}} =1.90 \; \mathrm{\dfrac{J}{K}}\nonumber$
Q17
Consider this. Exactly 2 moles of ice undergoes three different processes.
1. heated reversibly at atmospheric pressure from temperature X to 0 °C, then
2. melted reversibly at 0 °C. Finally
3. it was heated reversibly at atmospheric pressure up to an unknown temperature Y.
Find temperature X & Y and identify $ΔS_{\text{surr}}$.
Given:
• $ΔH_{\text{fus}}$ = 6007 J.mol-1
• $C_p$ (ice) = 38 J.K-1.mol-1
• $C_p$ (water) = 75 J.K-1.mol-1
• $ΔS_{\text{I}}$ = 12.034 J.K-1
• $ΔS_{\text{III}}$ = 136 J.K-1
Solution
Since entropy changes are additive (thanks for them being state functions) and we have the entropy changes for Step I and Step III, all we need to do is calculate $\Delta S_{II}$. However, we need to calculate temperatures first.
The equation below can be used to calculate $ΔS_{sys}$ for a temperature change.
$\Delta S = nC_{p} \ln \left(\dfrac{T_{2}}{T_{1}}\right)$
To solve for unknown temperature, the equation can be arranged in this way.
$\exp(\dfrac{\Delta S}{nC_{p}}) = \dfrac{T_{2}}{T_{1}}$
Assuming the $C_p$ stays constant throughout all the temperature change. Initial temperature X and final temperature Y can be found using the formula below.
$T_{x}=T_{2}(\exp(\dfrac{\Delta S_{\text{I}}}{nC_{p}}))^{-1}=(273.15K)(\exp(\dfrac{12.034 J.K^{-1}}{(2.00 mol)(38 J.K^{-1}.mol^{-1})}))^{-1}=233.15K$
$T_{y}=T_{1}(\exp(\dfrac{\Delta S_{\text{III}}}{nC_{p}}))=(273.15K)(\exp(\dfrac{136 J.K^{-1}}{(2.00 mol)(75 J.K^{-1}.mol^{-1})}))=298.15K$
ΔSII must also be determined to obtain the ΔSsystem.
Since the temperature remains constant in the second step, $ΔS = q_{rev}$.
$\Delta S_{\text{II}}= 2\,mol \times \dfrac{6007\,J/mol}{273.15\,K}=43.984\, J.K^{-1}$
Since the entire process is reversible,
$\mathrm{\Delta {S_{\text{surr}}} + \Delta {S_{\text{system}}} = 0}\nonumber$
so
$\mathrm{\Delta {S_{\text{surr}}} = - \Delta {S_{\text{system}}} }.\nonumber$
Now we add the three entropy changes:
\begin{align} -\Delta S_{\text{system}} &= -(\Delta S_{\text{I}} + \Delta S_{\text{II}} +\Delta S_{\text{III}}) \[5pt] &= -(12.034 + 54.98 + 136)J.K^{-1} \[5pt] &= -80.15\, J.K^{-1} \[5pt] &= \Delta S_{\text{surr}} \end{align}
Q18
Suppose 2 moles of water at standard temperature (25°C) and pressure is spontaneously evaporated by allowing it to fall onto a nickel plate maintained at 125°C. Calculate $\Delta S$ for the water, $\Delta S$ for the nickel plate, and $\Delta S_{total}$ if $C_{p\ce{(H2O)(l)}}$= 75.4 J/(K.mol) and $C_{p\ce{(H2O) (g)}}$ = 36.0 J/(K.mol). Take $\Delta H_{vap}$ = 40.68 KJ/mol for water and its boiling point of 100°C.
Solution
$q_{1} = \Delta H = mc\Delta T = 2 \times 75.4 \dfrac{J}{K.mol} \times (100-25) = 11310 \; J$
$q_{2} = 40680 \dfrac{J}{mol} \times 2 = 81.36 \times 10^{3} \; J$
$q_{3} = \Delta H = mc\Delta T = 2 \times 36 \dfrac{J}{K.mol} \times (125-100) = 1800 \; J$
$q_{total} = 11310 + 81.36 \times 10^{3} + 1800 = 94470 \; J$
$\Delta S_{\ce{H2O}} = nC_{p(\ce{H2O(l)}) }) \ln{\dfrac{T_{2}}{T_{1}}} + nC_{p}(H_{2}O(g)) \ln {\dfrac{T_{2}}{T_{1}}} + n\dfrac{\Delta H}{T}$
$= (2\times 75.4 \times \ln{\dfrac{373}{298}}) + (2 \times 36 \times \ln {\dfrac{398}{373}}) + (2\times \dfrac{40680}{373}) = 256.6 \dfrac{J}{K}$
$\Delta S_{iron} = \dfrac{-94470 J}{398 K} = -237.4 \dfrac{J}{K}$
$\Delta S_{tot} = \Delta S_{\ce{H2O}} + \Delta S_{\text{iron}} = 19.2 \dfrac{J}{K}$
Q19
A 181.49 g sample of lead at 97.0oC initially is added to a coffee cup calorimeter that contains 150.0 g water which is at 24.7oC. The equilibrium temperature is 29.4oC, assuming that there is no heat lost to the calorimeter or the environment. The molar heat capacity of lead (Cp(Pb)) is 26.4 J K-1 mol-1 and that of water (Cp(H2O)) is 75.2 J K-1 mol-1. What is $\Delta S$ for the lead sample, $\Delta S$ for the water sample, and $\Delta S_{total}$ for this process?
Solution
Since this progress is carried out at a constant pressure, and the temperature change for this progress doesn't include a phase transition, the relationship between the $\Delta S$ and the $\Delta T$ of a system can be described as:
$\Delta S = nc_{p}ln\left(\frac{T_{2}}{T_{1}}\right)$
Convert the temperatures to Kelvin:
$T_{equilibrium} = (29.4 + 273.15)K = 302.55 \; K$
$T_{initial, \; Pb} = (97.0 + 273.15) K = 370.15 \; K$
$T_{initial, \; H_{2}O} = (24.7 + 273.15) K = 297.85 \; K$
Calculate the amount of moles of the substance present:
$n_{Pb} = 181.49 \; g \; Pb \times \frac{1 \; mol \; Pb}{207.2 \; g \; Pb} = 0.8759 \; mol \; Pb$
$n_{H_{2}O} = 150.0 \; g \; H_{2}O \times \frac{1 \; mol \; H_{2}O}{18.02 \; g \; H_{2}O} = 8.324 \; mol \; H_{2}O$
The given molar heat capacities:
$C_{p, \; Pb} = 26.4 \frac {J}{K \cdot mol}$
$C_{p, \; H_{2}O} = 75.2 \frac {J}{K \cdot mol}$
Substitute the known variables into the equation: $\Delta S = nc_{p}ln\left(\frac{T_{2}}{T_{1}}\right)$
$\Delta S_{Pb} = nc_{p,Pb}ln\left(\frac{T_{equilibrium}}{T_{inital, \; Pb}}\right)$
$\Delta S_{Pb} = 0.8759 \; mol \; Pb \times 26.4 \frac {J}{K \cdot mol} \times ln\left(\frac{302.55 \; K}{370.15 \; K}\right) = -4.66 \; \frac{J}{K}$
$\Delta S_{H_{2}O} = nc_{p,H_{2}O}ln\left(\frac{T_{equilibrium}}{T_{inital, \; H_{2}O}}\right)$
$\Delta S_{H_{2}O} = 8.324 \; mol \; H_{2}O \times 75.2 \frac {J}{K \cdot mol} \times ln\left(\frac{302.55 \; K}{297.85 \; K }\right) = 9.80 \; \frac{J}{K}$
$\Delta S_{total} = \Delta S_{Pb} + \Delta S_{H_{2}O} = (-4.66 + 9.80) \; \frac{J}{K} = 5.14 \; \frac{J}{K}$
Q20
Copper has a heat capacity of 38.5 J K-1 mol-1, approximately independent of temperature between 0°C to 100°C. Calculate the enthalpy and entropy change of 5.00 moles of copper as it is cooled at atmospheric pressure from 100°C to 0°C.
Solution
$\Delta H = q=nC_{p}\Delta T$
$= (5.00 \; mol) (38.5 \; \dfrac{J}{K.mol}) (273 \; K – 373 \; K)\nonumber$
$\Delta H = -19250 \; J\nonumber$
$\Delta S = nC_{p}ln(\frac{T_{2}}{T_{1}})$
$= (38.5 \dfrac{J}{K.mol}) \ln {(\dfrac{273 \; K}{373 \; K})} (5.00 \; mol)\nonumber$
$= -60.1 \dfrac{J}{K}\nonumber$
Q23
The alkaline earth metals react with oxygen to give the following compounds:
$\ce{2 Be (g) + O2 (g) -> 2 BeO (s) } \nonumber$
$\ce{2 Mg (g) + O2 (g) -> 2 MgO (s) } \nonumber$
$\ce{2 Ca (g) + O2 (g) -> 2 CaO (s) }\nonumber$
Compute ΔS for each reaction, and identify a periodic trend about the entropy.
Solution
All the values you need to solve this problem are in the back of the Oxtoby textbook, so just find the numbers needed and use the formula:
$\sum{S^o_{products}} -\sum{S^o_{reactants}} = ΔS^o_{rxn}\nonumber$
$2 Be_{(g)} + O_{2(g)} \rightarrow 2 BeO_{(s)}\nonumber$
$\mathrm{2(14.14) - (2(136.16)+205.03)= -449.07 \dfrac{J}{K.mol}}\nonumber$
$2 Mg_{(g)} + O_{2(g)} \rightarrow 2 MgO_{(s)} \nonumber$
$\mathrm{2(26.92) - (2(148.54)+205.03)= -448.27 \dfrac{J}{K.mol}}\nonumber$
$2 Ca_{(g)} + O_{2(g)} \rightarrow 2 CaO_{(s)} \nonumber$
$\mathrm{2(39.75) - (2(154.77)+205.03)= -435.07 \dfrac{J}{K.mol}}\nonumber$
The entropy of the reaction decreases as you go up the periodic table.
Q27
Is the entropy change in the reaction positive, negative or zero and why?
$\ce{CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)} \nonumber$
Hint: You don't need to actually calculate the change in entropy to determine whether the change in entropy is positive or negative. Think about it conceptually.
Solution
The answer is negative because the number of gas molecules decreases. Gas molecules have more entropy than liquid molecules since they have more energy in the form of degrees of freedom of motion (translational, rotational, and vibrational). Therefore, fewer gas molecules mean less entropy.
Q33
At $25.0\text{°C}$ the reaction below is not spontaneous.
$\ce{2 H2O (g) -> 2H2 (g) + O2 (g)} \nonumber$
with $\ce{\Delta}{G}$ = $+228.59\ \frac{kJ}{mol}$
If the above reaction were coupled with the following nonspontaneous reaction, could it be made to proceed? Why or Why not?
$\ce{3 H2 (g) + N2 (g) <=> 2 NH3 (g)} \nonumber$
with $\ce{\Delta}{G}$ = $-16.48\ \frac{kJ}{mol}$
Solution
Coupling the nonspontaneous reaction
$\ce{2H_2O_{(g)} \rightleftharpoons 2H_{2\, (g)} + O_{2\, (g)}}\nonumber$
with $\ce{\Delta}{G} = +228.59\ \frac{kJ}{mol}$
for a spontaneous reaction $\left( \ce{\Delta}{G} \right)$ is negative.
$\ce{3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)}}\nonumber$
with $\ce{\Delta}{G} = -16.48\ \frac{kJ}{mol}$
yields
$\ce{2H_2O_{(g)} + \require{cancel} \cancel{3}H_{2(g)} + N_{2(g)} \rightleftharpoons \cancel{2H_{2\, (g)}} + O_{2\, (g)} + 2NH_{3(g)}}\nonumber$
with $\ce{\Delta}{G}$ = $+212.11\ \frac{kJ}{mol}$
Since $\ce{\Delta}{G}$ is positive, we know the reaction is still not spontaneous after being coupled with a spontaneous reaction
Q35A
A reaction at constant temperature and pressure is spontaneous if $\Delta G<0$ and nonspontaneous if $\Delta G>0$. Over what range of temperatures is each of the following processes spontaneous? Assume that all gases are at a pressure of 1 atm. (Hint: Use Tables T1 to calculate $\Delta{H}$ and $\Delta{S}$ (assumed independent of temperature and equal to $\Delta{H}^{\circ}$ and $\Delta{S}^{\circ}$, respectively, and then use the definition of $\Delta{G}$).
1. Photosynthesis, a reaction of how plants produce food for themselves and animals, and convert carbon dioxide into water: $\ce{6CO2(g) + 6H2O(l) + light -> C6H12O6(s) + 6O2(g) }\nonumber$
2. The combustion reaction of propane, found in gas grills and some fireplaces: $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) }\nonumber$
3. Methane burning in $\ce{O2}$ gas. $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \nonumber$
Solution
a. First calculate $\Delta H^{\circ}$ and $\Delta S^{\circ}$of the reaction $\ce{6 CO2 (g) + 6 H2O(l) + light -> C6H12O6 (s) + 6 O2 (g)}$ from the data in Tables T1.
$\Delta H^{\circ}=-1273.3+(6\times0)-(6\times-393.5)-(6\times-285.8)=2802.5\;\dfrac{kJ}{mol}\nonumber$
$\Delta S^{\circ}=212.1+(6\times205.2)-(6 \times 213.8)-(6\times70)=-259.5\; \dfrac{J}{mol.K}\nonumber$
Since the problem asks for the temperature range in which the reaction is spontaneous. The changeover from spontaneity to non-spontaneity occurs at $\Delta{G^{\circ}}=0$. To find the temperature the makes $\Delta{G^{\circ}}=0 \;$, the relationship of $\Delta{G^{\circ}}=\Delta{H^{\circ}}-T\;\Delta{S^{\circ}} \;$ will be used. Remember to convert $\Delta{S^{\circ}} \;$ to $kJ\;mol^{-1} \;$ (or $\Delta{H^{\circ}} \;$ to $J\;mol^{-1} \;$) so that the units cancel out properly.
$T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{2802.5 \;\dfrac{kJ}{mol}}{0.2595\; \dfrac{kJ}{mol.K}}=10799.6\;K\nonumber$
Because $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are both positive, the reaction is spontaneous at temperatures above 10799.6 K.
Reviewer Note: The solution is incorrect here. Because $\Delta H^{\circ} > 0$ and $\Delta S^{\circ} < 0$, the reaction is never spontaneous.
b. Perform similar calculations for the reaction $\ce{C3H8 (g) + 5 O2 (g) -> 3CO2 (g) + 4 H2O (l)}$
$\Delta H^{\circ}=(3\times-393.5)+(4\times-285.83)+103.8-(5\times0)=-2220.02\;\dfrac{kJ}{mol}\nonumber$
$\Delta S^{\circ}=(3\times28)+(4\times70)-270.3-(5\times 205.2)=-374.9\;\dfrac{J}{mol.K}\nonumber$
$T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{2220.02\;\dfrac{kJ}{mol}}{0.3749\;\dfrac{kJ}{mol.K}}=5922\;K\nonumber$
Since $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are both negative, the reaction is spontaneous below 5922 K.
c. The reaction $\ce{CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)}$
$\Delta H^{\circ}=(-393.5)+(2\times-285.8)+74.6-(2\times0)=-890.5\;\dfrac{kJ}{mol}\nonumber$
$\Delta S^{\circ}=(28)+(2\times70)-186.3-(2\times205.2)=-242.9\; \dfrac{J}{mol.K}\nonumber$
$T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\dfrac{-890.5\;\dfrac{kJ}{mol}}{-0.2429\;\dfrac{kJ}{mol.K}}=3666\;K\nonumber$
Since $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are both negative, the reaction is spontaneous below 3666 K.
Q35B
Over what temperatures are these reactions spontaneous (under constant pressure and temperature)? You may need to use Table 1.
1. $\ce{S (s, rhombic) + Zn (s) \rightarrow ZnS (s, sphalerite)}$
2. $\ce{2Cu(s) + Cl2 \rightarrow 2CuCl (s)}$
3. $\ce{4Al(s) + 3O2(g) \rightarrow 2Al2O3(s)}$
Answer
By calculating $∆H_{rxn}$ and $∆S_{rxn}$ un der standard conditions and at 25oC, we can calculate the range of temperature in which ∆G is positive and therefore spontaneous. To calculate $∆H_{rxn}$ and $∆S_{rxn}$, we use Hess’s Law , in which:
$\Delta{H_{rxn}} = \sum{nH_{products}} - \sum{nH_{reactants}} \nonumber$
$\Delta{S_{rxn}} = \sum{nS_{products}} - \sum{nS_{reactants}} \nonumber$
a)
$\mathrm{\Delta{H_{rxn}} = -206.0 – (0 + 0) = -206.0 \dfrac{kJ}{mol}}\nonumber$
$\mathrm{\Delta{S_{rxn}} = 57.7 – (41.6 + 32.1) = -16 \dfrac{J}{K.mol} }\nonumber$
$\mathrm{\Delta{G_{rxn}} = -206.0 - T(-0.016)}\nonumber$
From here, we can set T = 0 because we know that ∆G must be < 0 to be a spontaneous reaction.
T<12875K
b) $\mathrm{\Delta{H_{rxn}} = (2 \times -137.2) – (0+0) = -274.4 \dfrac{kJ}{mol}}\nonumber$
$\mathrm{\Delta{S_{rxn}} = (2 \times 86.2) – ((2 \times 33.2) + 223.1) = -117.1 \dfrac{J}{K.mol}}\nonumber$
$\mathrm{\Delta{G_{rxn}} = -274.4- T(-0.1171)}\nonumber$
T < 2343.3K
c) $\mathrm{\Delta{H_{rxn}} = (2(-1675.7)) - (0+0) = -3351.4 \dfrac{kJ}{mol} }\nonumber$
$\mathrm{\Delta{S_{rxn}} = (2(50.92) - (4(28.3) + (3(205.2)) = -626.96 \dfrac{J}{K.mol}}\nonumber$
$\mathrm{\Delta{G_{rxn}} = -3351.4 - T(-0.62696)}\nonumber$
T < 5345.5K
Q37
Determine if the following reaction is spontaneous at 25°C by evaluating ∆H°rxn, ∆S°rxn and ∆G°rxn.
$\ce{N2H4 (l) + O2 (g) -> N2 (g) + 2 H2O (l)} \nonumber$
Solution
• $\mathrm{\Delta{H^{\circ}_{rxn}} = 0 +(2 \times -285.8) – 0 - 50.6 = 622.2 \frac{kJ}{mol}}$
• $\mathrm{\Delta{S^{\circ}_{rxn}}=(2 \times 70.0)+ 191.6 – 205.2 – 121.2 = 5.2 \frac{J}{mol.K}}$
• $\mathrm{\Delta{G^{\circ}_{rxn}} = -622.2 – (25 + 273.15)(0.0052) =-623.8 \frac{kJ}{mol}}$
=> Spontaneous ($\Delta{G^{\circ}_{rxn}} < 0$ )
Q39
A thermodynamic engine operates cyclically and reversibly between two temperatures reservoirs, absorbing heat from the high-temperature bath at 600 K and discarding heat to low-temperature bath at 300 K.
1. What is the thermodynamic efficiency of the engine?
2. How much heat is absorbed from the high-temperature bath if -1800 J of heat is discarded to the low-temperature bath during each cycle?
3. How much work does the engine perform in one cycle of operation?
Solution
(a) $\mathrm{Thermodynamic \; efficiency = \dfrac{(T_1 – T_2)}{T_1}}\nonumber$
$\mathrm{\dfrac{(600 \; K- 300 \; K)}{ (600 \; K)} \times 100 \%= 50 \% }\nonumber$
(b)$\mathrm{\dfrac{1800 \; J}{ 50 \%}= 3600 \; J }\nonumber$
(c) $\mathrm{-3600 \; J \times 50 \%= -1800 \; J}\nonumber$
Q41
Acetone ($\ce{C_3H_6O}$) is an volatile liquid with a normal boiling point of 56°C and a molar enthalpy of vaporization of 29.1 kJ mol-1. What is the molar entropy of vaporization of acetone under 1 atm of pressure?
Solution
First we need the write the equation relating entropy and enthalpy:
$\Delta G_{vap}= \Delta H_{vap} - T\Delta S_{vap}\nonumber$
As a result of the process of normal boiling being at quilibrium $\Delta G = 0$ so all that needs to be done is T in Kelvin and $\Delta H_{vap}$ needs to be plugged in and then we solve for $\Delta S_{vap}$:
$\mathrm{0 = 29.1 \; \dfrac{kJ}{mol} – (329 \; K)( \Delta{S_{vap}} )}\nonumber$
$\mathrm{\Delta{S_{vap}} = 88.45 \dfrac{J}{K.mol}}\nonumber$
Q53
Under standard conditions and 25°C, you have this reversible process $\ce{Sn(s, white) -> Sn(s,gray)} \nonumber$
1. Is the change in entropy ($\Delta S$) for this reaction positive or negative? Calculate the change in entropy using Table T1.
2. Calculate ($\Delta S$) for the reaction if $\Delta H$ for this reaction is -2.1 kJ/mol.
Solution
For part a), we use the formula $\Delta S° = \sum nS^\circ (\ce{products})-\sum nS^\circ (\ce{reactants})\nonumber$
Plugging in the values from Table T1 gives us $\Delta S = 44.1 - 55.2 = -7.1 \frac {J}{mol \cdot K}\nonumber$
For part b), we use the formula for a reversible process $\Delta S = \frac {\Delta H}{T}\nonumber$
$\Delta S = \frac {-2100 \; J/mol}{298.15 \; K}$
$\Delta S = -7.04 \frac {J}{mol \cdot K}$
Q59
Compute the $\Delta{G_{f}^{\circ}}$ for the following reaction the reaction. The $\Delta{G_{f}^{\circ}}$ of H2SO(aq) is -537.81 kJ mol-1, the $\Delta{G_{f}^{\circ}}$ of SO2 (g) is -300.19 kJ mol-1, and the $\Delta{G_{f}^{\circ}}$ of H2O (g) is -120.42 kJ mol-1.
$\ce{H2SO3 (aq) -> H2O (g) + SO2 (g)} \nonumber$
Solution
∆G° for the reaction is equal to the sum of the Gf° for the products minus the sum of the Gf° for the reactants.
$\Delta G^{\circ}= \left(-300.19 \dfrac{kJ}{mol}+-120.42\dfrac{kJ}{mol}\right)- \left(-537.81\dfrac{kJ}{mol} \right)\nonumber$
$\Delta G^{\circ}=117.2\dfrac{kJ}{mol}\nonumber$
Q61
Professor Nesral wants to create some water through the combustion of hydrogen. This reaction is depicted as follows:
$\ce{2 H2 (g) + O2 (g) -> 2 H2O (g)}\nonumber$
Is this reaction spontaneous? Prove it by calculating $\Delta{G^{\circ}}$ at 298.15K. Suppose that Nesral asks his friend in El Azizia, Libya, where the temperature is around $57^{\circ}C$ (330K), to perform the same reaction. Calculate the Gibbs energy for this value and compare it to $\Delta{G^{\circ}}$.
Relevant Information:
Entropy Values ($S^{\circ}$; $\dfrac{J}{mol\,K}$)
• Hydrogen: $130.6$
• Water: $188.7$
• Oxygen: $205.6$
Enthalpy Values($H^{\circ}$; $\dfrac{kJ}{mol}$)
• $\Delta{H_{f}^{\circ}}$ Water: $-241.826$
Solution
From intuition, the equation shows that 3 moles of gas, 2 moles of $\ce{H2}$ and 1 mole of $\ce{O2}$, are reacting to form 2 moles of $\ce{H2O}$ gas. This is a decrease in entropy, which may hint at the reaction being non spontaneous. $\Delta{G}^{\circ}$ should still be calculated to make sure.
The information provided is the entropic values of the product and reactants, and the enthalpy of water. If the $\Delta{S}^{\circ}$ is calculated, then the $\Delta{G}^{\circ}$ can also be found using the following equation:
$\Delta{G} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}$.
Where $\Delta{H_{rxn}^{\circ}}$ is the Heat and $\Delta{S_{rxn}^{\circ}}$ is the entropy of the reaction under standard conditions (constant pressure), and T is temperature. Knowing that $\Delta{S_{rxn}^{\circ}}$ is a state variable, it can be calculated as the difference between the sum of the entropy values of the products and reactants multiplied by their coefficients (This is known as Hess’s Law). In other words,
$\Delta{S^{\circ}}=\sum{n_{products}\times{S_{f_{products}}^{\circ}}} - \sum{n_{reactants}\times{S_{f_{reactants}}^{\circ}}}$
From this, $\Delta{S_{rxn}^{\circ}}$ can be calculated as such:
$\Delta{S^{\circ}} = \left(2\times188.7\dfrac{J}{molK}\right)- \left(\left[2\times130.6\dfrac{J}{molK}\right]+\left[205.6\dfrac{J}{molK}\right]\right)$
$= -89.4\dfrac{J}{mol.K}$
Since both $H_{2}$ and $O_{2}$ are in their natural states, the $H_{f}^{\circ}$ associated with them equals to zero. Since enthalpy ($\Delta{H_{rxn}^{\circ}}$) is also a state variable, it can be calculated with Hess's Law as well.
$\Delta{H^{\circ}} = \left(2\times-241.826\dfrac{kJ}{mol}\right)- \left(\left[0\dfrac{kJ}{mol}\right]+\left[0\dfrac{kJ}{mol}\right]\right)$
$\Delta{H^{\circ}}= -483.652\dfrac{kJ}{mol}$
Plugging in these values, as well as the temperature, $298.15K$, will yield $\Delta{G}^{\circ}$.
$\Delta{G_{rxn}^{\circ}} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}$.
$\Delta{G_{rxn}^{\circ}} = -483.652\dfrac{kJ}{mol} – 298.15K\left(-0.0894\dfrac{kJ}{mol.K}\right)$.
$\Delta{G_{rxn}^{\circ}}= -456.99739\dfrac{kJ}{mol}$
The $\Delta{G_{rxn}}$ is therefore $-456.99739\dfrac{kJ}{mol}$. This very negative value means that the reaction is DEFINITELY spontaneous. It also makes sense on a conceptual level because combustion reaction like these release a lot of heat, which means that the process if enthalpically driven! Calculating the $\Delta{G_{rxn}}$ at a different temperature, as mentioned in the problem, has the same method:
$\Delta{G_{rxn}} = \Delta{H_{rxn}^{\circ}} – T\Delta{S_{rxn}^{\circ}}$.
$\Delta{G_{rxn}} = -483.652\dfrac{kJ}{mol} – 330K\left(-0.0894\dfrac{kJ}{mol.K}\right)$.
$\Delta{G_{rxn}}= -454.15\dfrac{kJ}{mol}$
In both situations the $\Delta{G_{rxn}}$ is very negative, but in the hotter climate it is slightly less so. What this shows is that, at a higher temperatures the entropic factors contribute more to the gibbs energy than at standard conditions. This showcases the temperature dependence of entropy. In another reaction that released less energy, this difference may have made the reaction non-spontaneous!
Abstract : Hess's Law to find $\Delta{G_{rxn}}$.
Q63A
The concentration of sodium in the plasma is approximately 0.14 M. While in the fluid outside of the plasma, sodium concentration is about 0.5 M.
1. In what direction will the sodium ions spontaneously move past the cell wall?
2. The spontaneous process of movement in part a is called "passive transport". Movement in the opposite direction is called "active transport" and requires work to happen. Calculate the amount of free energy required to move 5 moles of Na+ by active transport (non-spontaneous direction) at 273 K.
Solution
a) The Na+ ions will spontaneously flow into the plasma because it flows from high to low concentration.
b) $\mathrm{\Delta{G} = nRT \ln{\frac{c_2}{c_1}} }$ where c2 is the destination of Na+ after active transport occurs while c1 is the original concentration of the Na+ ions.
$\mathrm{\Delta{G} = (5 \; mole)(8.3145 \frac{J}{K.mol})(273 \; K)\ln{\frac{0.5}{0.14}}}$
$\mathrm{\Delta{G} = 14447.26 \; J}$
Q63B
For a hypothetical nerve cell, the sodium ion concentration is 0.015 M outside the cell and 0.00045 M inside the cell. Active transport involves using proteins and chemical energy stored in ATP to move the ions in a thermodynamically unfavorable direction. Assuming that conditions in the cell allow the hydrolysis of a single ATP molecule to give 2.05×10-18 J of usable enegy. How many molecules of ATP would be needed to move 0.005 moles of sodium ions using active transport at standard conditions? You can assume that the sodium ion concentrations remain constant.
Solution
To answer this question, the change of free energy is needed which means the following equation is needed
$\Delta G = - R T lnQ \nonumber$
$R = 8.314\; \dfrac{J}{mol.K} \; \; \; \; \; \; \; T=298K \; \; \; \; \; \; \; Q = \dfrac{[Na^{+}]_{in}}{[Na^{+}]_{out}} \nonumber$
$\Delta G = - 8.314\; \dfrac{J}{mol.K} \times 298K \times ln(\dfrac{0.00045\; M}{0.015\; M}) \nonumber$
$\Delta G = -2424\; \dfrac{J}{mol} \times ln(0.03) = 8688\; \dfrac{J}{mol} \nonumber$
Now that we have the the free energy, we can determine the number of ATP molecules needed
$8688\; \dfrac{J}{mol} \times 0.005\; mol = 43.4\; J\nonumber$
$\dfrac {43.4\; J} {2.05 \times 10^{-18}\; J\; molecule^{-1}} = 2.119 \times 10^{19}\; \text{ ATP molecules} \nonumber$
Answer: $2.119 \times 10^{19}\; \text{ ATP molecules}$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/13%3A_Spontaneous_Processes_and_Thermodynamic_Equilibrium/13.E%3A_Spontaneous_Processes_%28Exercises%29.txt |
Chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. This results when the forward reaction proceeds at the same rate as the reverse reaction. Thus, no net changes in the concentrations of the reactant(s) and product(s) are observed. This is known as dynamic equilibrium.
• 14.1: The Nature of Chemical Equilibrium
At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.
• 14.2: The Empirical Law of Mass Action
The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants. For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
• 14.3: Thermodynamic Description of the Equilibrium State
In this unit we introduce a new thermodynamic function, the free energy, which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. As we will explain near the bottom of this page, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as the "Gibbs function" or "Gibbs energy."
• 14.4: The Law of Mass Action for Related and Simultaneous Equilibria
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
• 14.5: Equilibrium Calculations for Gas-Phase and Heterogenous Reactions
An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
• 14.6: Reaction Directions (Empirical Explanation)
The reaction Quotient (Q) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. The reaction Quotient (Qc or Qp ) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, Q=.
• 14.7: Reaction Directions (Thermodynamic Explanation)
• 14.8: Distribution of a Single Species between Immiscible Phases - Extraction and Separation
A partitioning of a compound exist between a mixture of two immiscible phases at equilibrium, which is a measure of the difference in solubility of the compound in these two phases. If one of the solvents is a gas and the other a liquid, the "gas/liquid partition coefficient" is the same as the dimensionless form of the Henry's law constant. A solute can partition when one or both solvents is a solid (e.g., solid solution).
• 14.E: Chemical Equilibria (Exercises)
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al.
14: Chemical Equilibrium
Learning Objectives
• To understand what is meant by chemical equilibrium.
In the last chapter, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time.
Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide ($\ce{N_2O_4}$) to nitrogen dioxide ($\ce{NO_2}$). You may recall that $\ce{NO_2}$ is responsible for the brown color we associate with smog. When a sealed tube containing solid $\ce{N_2O_4}$ (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of $\ce{NO_2}$ appears (Figure $1$). The reaction can be followed visually because the product ($\ce{NO_2}$) is colored, whereas the reactant ($\ce{N_2O_4}$) is colorless:
$\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f][k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1}$
The double arrow indicates that both the forward reaction
$\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B}$
and reverse reaction
$\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C}$
occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates.
Figure $2$ shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of $\ce{NO_2}$ were zero, then it increases as the concentration of $\ce{N_2O_4}$ decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no $\ce{N_2O_4}$ but an initial $\ce{NO_2}$ concentration twice the initial concentration of $\ce{N_2O_4}$ (Figure $\PageIndex{2a}$), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure $\PageIndex{2b}$). Thus equilibrium can be approached from either direction in a chemical reaction.
Figure $3$ shows the forward and reverse reaction rates for a sample that initially contains pure $\ce{NO_2}$. Because the initial concentration of $\ce{N_2O_4}$ is zero, the forward reaction rate (dissociation of $\ce{N_2O_4}$) is initially zero as well. In contrast, the reverse reaction rate (dimerization of $\ce{NO_2}$) is initially very high ($2.0 \times 10^6\, M/s$), but it decreases rapidly as the concentration of $\ce{NO_2}$ decreases. As the concentration of $\ce{N_2O_4}$ increases, the rate of dissociation of $\ce{N_2O_4}$ increases—but more slowly than the dimerization of $\ce{NO_2}$—because the reaction is only first order in $\ce{N_2O_4}$ (rate = $k_f[N_2O_4]$, where $k_f$ is the rate constant for the forward reaction in Equations $\ref{Eq1}$ and $\ref{eq1B}$). Eventually, the forward and reverse reaction rates become identical, $k_f = k_r$, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.
The rate of dimerization of $\ce{NO_2}$ (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of $\ce{N_2O_4}$ is zero, the rate of the dissociation reaction (forward reaction) at $t = 0$ is also zero. As the dimerization reaction proceeds, the $\ce{N_2O_4}$ concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of $\ce{N_2O_4}$ and $\ce{NO_2}$ no longer change.
At equilibrium, the forward reaction rate is equal to the reverse reaction rate.
Example $1$
The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation:
$2A \rightleftharpoons B \nonumber$
where the blue circles are $A$ and the purple ovals are $B$. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?
Given: three reaction systems
Asked for: relative time to reach chemical equilibrium
Strategy:
Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.
Solution:
In systems 1 and 3, the concentration of A decreases from $t_0$ through $t_2$ but is the same at both $t_2$ and $t_3$. Thus systems 1 and 3 are at equilibrium by $t_3$. In system 2, the concentrations of A and B are still changing between $t_2$ and $t_3$, so system 2 may not yet have reached equilibrium by $t_3$. Thus system 2 took the longest to reach chemical equilibrium.
Exercise $1$
In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?
Answer
system 2
A Video Introduction to Dynamic Equilibrium: Introduction to Dynamic Equilibrium(opens in new window) [youtu.be]
Summary
At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.1%3A_The_Nature_of_Chemical_Equilibrium.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To write an equilibrium constant expression for any reaction.
• To understand how different phases affect equilibria.
Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the decomposition reaction of $\ce{N_2O_4}$ to $\ce{NO_2}$. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
$\text{forward rate} = k_f[\ce{N2O4}] \label{Eq1}$
and
$\text{reverse rate} = k_r[\ce{NO2}]^2 \label{Eq2}$
At equilibrium, the forward rate equals the reverse rate (definition of equilibrium):
$k_f[\ce{N2O4}] = k_r[\ce{NO2}]^2 \label{Eq3}$
so
$\dfrac{k_f}{k_r}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq4}$
The ratio of the rate constants gives us a new constant, the equilibrium constant ($K$), which is defined as follows:
$K=\dfrac{k_f}{k_r} \label{Eq5}$
Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.
The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
Table $1$ lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation $\ref{Eq1}$. At equilibrium the magnitude of the quantity $[\ce{NO2}]^2/[\ce{N2O4}]$ is essentially the same for all five experiments. In fact, no matter what the initial concentrations of $\ce{NO2}$ and $\ce{N2O4}$ are, at equilibrium the quantity $[\ce{NO2}]^2/[\ce{N2O4}]$ will always be $6.53 \pm 0.03 \times 10^{-3}$ at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions (Equation \ref{Eq5}). That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.
Table $1$: Initial and Equilibrium Concentrations for $NO_2:N_2O_4$ Mixtures at 25°C
Initial Concentrations Concentrations at Equilibrium
Experiment [$N_2O_4$] (M) [$NO_2$] (M) [$N_2O_4$] (M) [$NO_2$] (M) $K = [NO_2]^2/[N_2O_4]$
1 0.0500 0.0000 0.0417 0.0165 $6.54 \times 10^{−3}$
2 0.0000 0.1000 0.0417 0.0165 $6.54 \times 10^{−3}$
3 0.0750 0.0000 0.0647 0.0206 $6.56 \times 10^{−3}$
4 0.0000 0.0750 0.0304 0.0141 $6.54 \times 10^{−3}$
5 0.0250 0.0750 0.0532 0.0186 $6.50 \times 10^{−3}$
Developing an Equilibrium Constant Expression
In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form
$aA+bB \rightleftharpoons cC+dD \label{Eq6}$
where $A$ and $B$ are reactants, $C$ and $D$ are products, and $a$, $b$, $c$, and $d$ are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action and can be stated as follows:
$K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq7}$
where $K$ is the equilibrium constant for the reaction. Equation $\ref{Eq6}$ is called the equilibrium equation, and the right side of Equation $\ref{Eq7}$ is called the equilibrium constant expression. The relationship shown in Equation $\ref{Eq7}$ is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.
The equilibrium constant can vary over a wide range of values. The values of $K$ shown in Table $2$, for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than $10^3$ indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between $\ce{H_2}$ and $\ce{Cl_2}$ to produce $\ce{HCl}$, which has an equilibrium constant of $1.6 \times 10^{33}$ at 300 K. Because $\ce{H_2}$ is a good reductant and $\ce{Cl_2}$ is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of $K$ less than $10^{-3}$ indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.
Table $2 \label{Table14.2.2}$: Equilibrium Constants for Selected Reactions*
Reaction Temperature (K) Equilibrium Constant (K)
*Equilibrium constants vary with temperature. The $K$ values shown are for systems at the indicated temperatures.
$\ce{S(s) + O2(g) \rightleftharpoons SO2(g)}$ 300 $4.4 \times 10^{53}$
$\ce{2H2(g) +O2(g) \rightleftharpoons 2H2O(g)}$ 500 $2.4 \times 10^{47}$
$\ce{H2(g) + Cl2(g) \rightleftharpoons 2 HCl(g)}$ 300 $1.6 \times 10^{33}$
$\ce{H2(g) + Br2(g) \rightleftharpoons 2HBr(g)}$ 300 $4.1 \times 10^{18}$
$\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)}$ 300 $4.2 \times 10^{13}$
$\ce{3H2(g) + N2(g) \rightleftharpoons 2NH3(g)}$ 300 $2.7 \times 10^{8}$
$\ce{H2(g) + D2(g) \rightleftharpoons 2HD(g)}$ 100 $1.92$
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}$ 300 $2.9 \times 10^{−1}$
$\ce{I2(g) \rightleftharpoons 2I(g)}$ 800 $4.6 \times 10^{−7}$
$\ce{Br2(g) \rightleftharpoons 2Br(g)}$ 1000 $4.0 \times 10^{−7}$
$\ce{Cl2(g) \rightleftharpoons 2Cl(g)}$ 1000 $1.8 \times 10^{−9}$
$\ce{F2(g) \rightleftharpoons 2F(g)}$ 500 $7.4 \times 10^{−13}$
You will also notice in Table $2$ that equilibrium constants have no units, even though Equation $\ref{Eq7}$ suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation $\ref{Eq8}$, the units of concentration cancel, which makes $K$ unitless as well:
$\dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{Eq8}$
Many reactions have equilibrium constants between 1000 and 0.001 ($10^3 \ge K \ge 10^{−3}$), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form $\ce{HD}$:
$\ce{H2(g) + D2(g) \rightleftharpoons 2HD(g)} \label{Eq9}$
The equilibrium constant expression for this reaction is
$K= \dfrac{[HD]^2}{[H_2][D_2]}$
with $K$ varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of $\ce{H_2}$, $\ce{D_2}$, and $\ce{HD}$ contains significant concentrations of both product and reactants.
Figure $1$ summarizes the relationship between the magnitude of $K$ and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants \rightleftharpoons products. Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations $\ref{Eq8}$ and $\ref{Eq7}$), when $k_f \gg k_r$, $K$ is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when $k_f \ll k_r$, $K$ is a very small number, and the reaction produces almost no products as written. Systems for which $k_f ≈ k_r$ have significant concentrations of both reactants and products at equilibrium.
A large value of the equilibrium constant $K$ means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.
Example $1$
Write the equilibrium constant expression for each reaction.
• $\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}$
• $\ce{CO(g) + 1/2 O2(g) \rightleftharpoons CO2(g)}$
• $\ce{2CO2(g) \rightleftharpoons 2CO(g)+O2(g)}$
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
Refer to Equation $\ref{Eq7}$. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.
Solution:
The only product is ammonia, which has a coefficient of 2. For the reactants, $N_2$ has a coefficient of 1 and $\ce{H2}$ has a coefficient of 3. The equilibrium constant expression is as follows:
$\dfrac{[NH_3]^2}{[N_2][H_2]^3}\nonumber$
The only product is carbon dioxide, which has a coefficient of 1. The reactants are $CO$, with a coefficient of 1, and $\ce{O_2}$, with a coefficient of $\ce{1/2}$. Thus the equilibrium constant expression is as follows:
$\dfrac{[CO_2]}{[CO][O_2]^{1/2}}\nonumber$
This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for $\ce{O_2}$. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2:
$\dfrac{[CO]^2[O_2]}{[CO_2]^2}\nonumber$
Exercise $1$
Write the equilibrium constant expression for each reaction.
1. $\ce{N2O(g) \rightleftharpoons N2(g) + 1/2O2(g)}$
2. $\ce{2C8H18(g) + 25O2(g) \rightleftharpoons 16CO2(g) + 18H2O(g)}$
3. $\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}$
Answer a
$K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}\nonumber$
Answer b
$K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}\nonumber$
Answer c
$K=\dfrac{[HI]^2}{[H_2][I_2]}\nonumber$
Example $2$
Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54$
2. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{−18}$
3. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97$
4. $2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}$
Given: systems and values of $K$
Asked for: composition of systems at equilibrium
Strategy:
Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.
Solution:
1. Only system 4 has $K >> 10^3$, so at equilibrium it will consist of essentially only products.
2. System 2 has $K << 10^{−3}$, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
3. Both systems 1 and 3 have equilibrium constants in the range $10^3 \ge K \ge 10^{−3}$, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise $2$
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\nonumber$
Values of the equilibrium constant at various temperatures were reported as
• $K_{25°C} = 3.3 \times 10^8$,
• $K_{177°C} = 2.6 \times 10^3$, and
• $K_{327°C} = 4.1$.
1. At which temperature would you expect to find the highest proportion of $H_2$ and $N_2$ in the equilibrium mixture?
2. Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?
Answer a
327°C, where $K$ is smallest
Answer b
25°C
Variations in the Form of the Equilibrium Constant Expression
Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation $\ref{Eq6}$ in reverse, we obtain the following:
$cC+dD \rightleftharpoons aA+bB \label{Eq10}$
The corresponding equilibrium constant $K′$ is as follows:
$K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{Eq11}$
This expression is the inverse of the expression for the original equilibrium constant, so $K′ = 1/K$. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is as follows:
$K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{Eq12}$
but for the opposite reaction, $2 NO_2 \rightleftharpoons N_2O_4$, the equilibrium constant K′ is given by the inverse expression:
$K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{Eq13}$
Consider another example, the formation of water: $2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}$. Because $H_2$ is a good reductant and $O_2$ is a good oxidant, this reaction has a very large equilibrium constant ($K = 2.4 \times 10^{47}$ at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form $O_2$ and $H_2$, is very small: $K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}$. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into $H_2$ and $O_2$.
The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.
Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction
$2NO_2 \rightleftharpoons N_2O_4$
as
$NO_2 \rightleftharpoons \frac{1}{2}N_2O_4$
with the equilibrium constant K″ is as follows:
$K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{Eq14}$
The values for K′ (Equation $\ref{Eq13}$) and K″ are related as follows:
$K′′=(K')^{1/2}=\sqrt{K'} \label{Eq15}$
In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by $n$, then the new equilibrium constant is the original equilibrium constant raised to the $n^{th}$ power.
Example $3$: The Haber Process
At 745 K, K is 0.118 for the following reaction:
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}\nonumber$
What is the equilibrium constant for each related reaction at 745 K?
1. $\ce{2NH3(g) \rightleftharpoons N2(g) + 3H2(g)}$
2. $\ce{ 1/2 N2(g) + 3/2 H2(g) \rightleftharpoons NH3(g)}$
Given: balanced equilibrium equation, $K$ at a given temperature, and equations of related reactions
Asked for: values of $K$ for related reactions
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate $K$ for each reaction.
Solution:
The equilibrium constant expression for the given reaction of $\ce{N2(g)}$ with $\ce{H2(g)}$ to produce $\ce{NH3(g)}$ at 745 K is as follows:
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118\nonumber$
This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:
$K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47\nonumber$
In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:
$K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344\nonumber$
Exercise
At 527°C, the equilibrium constant for the reaction
$\ce{2SO2(g) + O2(g) \rightleftharpoons 2SO3(g) }\nonumber$
is $7.9 \times 10^4$. Calculate the equilibrium constant for the following reaction at the same temperature:
$\ce{SO3(g) \rightleftharpoons SO2(g) + 1/2O2(g)}\nonumber$
Answer
$3.6 \times 10^{−3}$
Law of Mass Action for Gas-Phase Reactions
For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol $K_p$ is used to denote equilibrium constants calculated from partial pressures. For the general reaction
$\ce{aA + bB <=> cC + dD} \nonumber$
in which all the components are gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
$K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{Eq16}$
Thus $K_p$ for the decomposition of $N_2O_4$ (Equation $\ref{Eq1}$) is as follows:
$K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \label{Eq17}$
Like $K$, $K_p$ is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity. The “effective pressure” is called the fugacity, just as activity is the effective concentration.
Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of $K$ and $K_p$ are usually different. They are, however, related by the ideal gas constant ($R$) and the absolute temperature ($T$):
$K_p = K(RT)^{Δn} \label{Eq18}$
where $K$ is the equilibrium constant expressed in units of concentration and $Δn$ is the difference between the numbers of moles of gaseous products and gaseous reactants ($n_p − n_r$). The temperature is expressed as the absolute temperature in Kelvin. According to Equation $\ref{Eq18}$, $K_p = K$ only if the moles of gaseous products and gaseous reactants are the same (i.e., $Δn = 0$). For the decomposition of $N_2O_4$, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so $Δn = 1$. Thus, for this reaction,
$K_p = K(RT)^1 = KRT$
Example $4$: The Haber Process (again)
The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:
$N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\nonumber$
What is $K_p$ for this reaction at the same temperature?
Given: equilibrium equation, equilibrium constant, and temperature
Asked for: $K_p$
Strategy:
Use the coefficients in the balanced chemical equation to calculate $Δn$. Then use Equation $\ref{Eq18}$ to calculate $K$ from $K_p$.
Solution:
This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so $\Delta{n} = (2 − 4) = −2$. We know $K$, and $T = 745\; K$. Thus, from Equation $\ref{Eq15}$, we have the following:
$K_p=K(RT)^{−2}=\dfrac{K}{(RT)^2}=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K)][745\; K]\}^2}=3.16 \times 10^{−5}\nonumber$
Because $K_p$ is a unitless quantity, the answer is $K_p = 3.16 \times 10^{−5}$.
Exercise $4$:
Calculate $K_p$ for the reaction
$2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}\nonumber$
at 527°C, if $K = 7.9 \times 10^4$ at this temperature.
Answer
$K_p = 1.2 \times 10^3$
Law of Mass Action for Reactions Involving Pure Substances and Multiple Phases
When the products and reactants of a reaction at equilibrium in a single phase (e.g., liquid, gas or solids of different lattices), the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system where the reactants and products are in two or more phase is called a heterogeneous equilibrium (e.g, the reaction of a gas with a solid or liquid or two different solid lattices in co-existing).
Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are assigned an activity of 1. (Recall that the density of water, and thus its volume, changes by only a few percentage points between 0 °C and 100 °C.)
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$\ce{CO2(g) + C(s) <=> 2CO(g)} \label{Eq14.4.1}$
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{[CO]^2}{[CO_2][C]} \label{Eq14.4.2}$
Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:
$[C] =\dfrac{2.26 \cancel{g}/{\cancel{cm^3}}}{12.01\; \cancel{g}/mol} \times 1000 \; \cancel{cm^3}/L = 188 \; mol/L = 188\;M \label{Eq14.4.3}$
We can rearrange Equation $\ref{Eq14.4.3}$ so that the constant terms are on one side:
$K[C]=K(188)=\dfrac{[CO]^2}{[CO_2]} \label{Eq14.4.4}$
Incorporating the constant value of $[C]$ into the equilibrium equation for the reaction in Equation $\ref{Eq14.4.4}$,
$K'=\dfrac{[CO]^2}{[CO_2]} \label{Eq14.4.5}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.6}$
Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction.
Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $CO$ and $CO_2$, the system described in Equation $\ref{Eq14.4.1}$ will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $2$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.
Example $1$
Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions.
1. $PCl_{3(l)}+Cl_{2(g)} \rightleftharpoons PCl_{5(s)}$
2. $Fe_3O_{4(s)}+4H_{2(g)} \rightleftharpoons 3Fe_{(s)}+4H_2O_{(g)}$
Given: balanced equilibrium equations
Asked for: expressions for $K$ and $K_p$
Strategy:
Find $K$ by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express $K_p$ as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation.
Solution
This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their concentrations do not appear in the equilibrium constant expression because they do not change significantly. So
$K=\dfrac{1}{[Cl_2]}\nonumber$
and
$K_p=\dfrac{1}{P_{Cl_2}}\nonumber$
This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which do not appear in the equilibrium constant expressions. The two gases do, however, appear in the expressions:
$K=\dfrac{[H_2O]^4}{[H_2]^4}\nonumber$
and
$K_p=\dfrac{(P_{H_2O})^4}{(P_{H_2})^4}\nonumber$
Exercise $1$
Write the expressions for $K$ and $K_p$ for the following reactions.
1. $CaCO_{3(s)} \rightleftharpoons CaO_{(s)}+CO_{2(g)}$
2. $\underset{glucose}{C_6H_{12}O_{6(s)}} + 6O_{2(g)} \rightleftharpoons 6CO_{2(g)}+6H_2O_{(g)}$
Answer a
$K = [CO_2]$ and $K_p = P_{CO_2}$
Answer b
$K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction.
The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium.
Summary
• The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
• For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
• Definition of equilibrium constant in terms of forward and reverse rate constants: $K=\dfrac{k_f}{k_r}\nonumber$
• Equilibrium constant expression (law of mass action): $K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}\nonumber$
• Equilibrium constant expression for reactions involving gases using partial pressures: $K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\nonumber$
• Relationship between $K_p$ and $K$: $K_p = K(RT)^{Δn}\nonumber$
The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same $K$. For a system at equilibrium, the law of mass action relates $K$ to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, $K$ and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures ($K_p$) is related to $K$ by the ideal gas constant ($R$), the temperature ($T$), and the change in the number of moles of gas during the reaction. An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.2%3A_The_Empirical_Law_of_Mass_Action.txt |
Learning Objectives
• Gibbs Energy is a state function defined as $G = H – TS$.
• The practical utility of the Gibbs function is that $ΔG$ for any process is negative if it leads to an increase in the entropy of the world. Thus spontaneous change at a given temperature and pressure can only occur when it would lead to a decrease in $G$.
• The sign of the standard free energy change $ΔG^o$ of a chemical reaction determines whether the reaction will tend to proceed in the forward or reverse direction.
• Similarly, the relative signs of $ΔH^o$ and $ΔH^o$ determine whether the spontaneity of a chemical reaction will be affected by the temperature, and if so, in what way.
• The existence of sharp melting and boiling points reflects the differing temperature dependancies of the free energies of the solid, liquid, and vapor phases of a pure substance, which are in turn reflect their differing entropies.
Previously, we saw that it is the sum of the entropy changes of the system and surroundings that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to focus our attention on the system rather than the surroundings, and would like to avoid having to calculate the entropy change of the surroundings explicitly.
In this unit we introduce a new thermodynamic function, the free energy, which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. However, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as "Gibbs energy." The free energy enables us to do this for changes that occur at a constant temperature and pressure (the Gibbs energy) or constant temperature and volume (the Helmholtz energy.)
Free energy: the Gibbs function
The Gibbs energy (also known as the Gibbs function or Gibbs Potential) is defined as
$G = H – T S \label{23.4.1}$
in which $S$ refers to the entropy of the system. Since $H$, $T$ and $S$ are all state functions, so is $G$. Thus for any change in state (under constant temperature), we can write the extremely important relation
$ΔG = ΔH – T ΔS \label{23.4.2}$
How does this simple equation encompass the entropy change of the world $ΔS_{total}$, which we already know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition
$ΔS_{total} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}$
we would first like to get rid of $ΔS_{surr}$. How can a chemical reaction (a change in the system) affect the entropy of the surroundings? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat qp across the system boundary. The enthalpy change of the reaction $ΔH$ is defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be $–q_p$ which will cause the entropy of the surroundings to change by $–q_p / T = –ΔH/T$. We can therefore rewrite Equation $\ref{23.4.3}$ as
$ΔS_{total} = \dfrac{- ΔH}{T} + ΔS_{sys} \label{23.4.4}$
Multiplying each side by $-T$, we obtain
$-TΔS_{total} = ΔH - TΔS_{sys} \label{23.4.5}$
which expresses the entropy change of the world in terms of thermodynamic properties of the system exclusively. If $-TΔS_{total}$ is denoted by $ΔG$, then we have Equation $\ref{23.4.2}$ which defines the Gibbs energy change for the process.
From the foregoing, you should convince yourself that $G$ will decrease in any process occurring at constant temperature and pressure which is accompanied by an overall increase in the entropy. The constant temperature is a consequence of the temperature and the enthalpy appearing in the preceding Equation $\ref{23.4.5}$. Since most chemical and phase changes of interest to chemists take place under such conditions, the Gibbs energy is the most useful of all the thermodynamic properties of a substance, and (as we shall see in the lesson that follows this one) it is closely linked to the equilibrium constant.
Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether $ΔH$ or the $TΔS$ term dominates. This is technically correct, but misleading because it disguises the important fact that $ΔS_{total}$, which this equation expresses in an indirect way, is the only criterion of spontaneous change.
Helmholtz Energy is also a "Free Energy"
We will deal only with the Gibbs energy in this course. The Helmholtz free energy is of interest mainly to chemical engineers (whose industrial-scale processes are often confined to tanks and reactors of fixed volume) and some geochemists whose interest is centered on the chemistry that occurs deep within the earth's surface.
Gibbs Energy and Chemical Change
Remember that $ΔG$ is meaningful only for changes in which the temperature and pressure remain constant. These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction.
$ΔG$ serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Moreover, it determines the direction and extent of chemical change.
In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the $TΔS$ term).
$\ce{H_2O(l) \rightarrow H2O(s)} \label{23.5.6}$
water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H2O diminishes, and the process proceeds spontaneously.
In a spontaneous change, Gibbs energy always decreases and never increases.
An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows:
• $ΔG < 0$: reaction can spontaneously proceed to the right: $A \rightarrow B \nonumber$
• $ΔG > 0$: reaction can spontaneously proceed to the left: $A \leftarrow B \nonumber$
• $ΔG = 0$: the reaction is at equilibrium and both $[A]$ and $[B]$ will not change: $A \rightleftharpoons B. \nonumber$
No need to find the value of ΔG for a Specific Reaction!
This might seem strange, given the key importance $ΔG$ in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate $ΔG$. As we will show in the lesson that follows this one, it is far more convenient to work with the equilibrium constant of a reaction, within which $ΔG$ is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of $ΔG$ depends on the proportions of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with $ΔH$.
Recalling the condition for spontaneous change
$ΔG = ΔH – TΔS < 0$
it is apparent that the temperature dependence of ΔG depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of $ΔH$ and $ΔS$ are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation the sign of the entropy change determines whether the reaction becomes more or less spontaneous as the temperature is raised. For any given reaction, the sign of $ΔH$ can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process:
Case 1: ΔH < 0 and ΔS > 0
Both enthalpic $\Delta H$ and entropic $-T\Delta S$ terms will be negative, so $ΔG$ will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures.
Case 2: ΔH < 0 and ΔS < 0
If the reaction is sufficiently exothermic it can force $ΔG$ negative only at temperatures below which $|TΔS| < |ΔH|$. This means that there is a temperature $T = ΔH / ΔS$ at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition.
Case 3: ΔH > 0 and ΔS > 0
This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that $TΔS > ΔH$. Since the effect of the temperature is to "magnify" the influence of a positive $ΔS$, the process will be spontaneous at temperatures above $T = ΔH / ΔS$. (Think of melting and boiling.)
Case 4: ΔH > 0 and ΔS < 0
With both $ΔH$ and $ΔS$ working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form.
The plots above are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of ΔH and ΔS.
• Their most important differentiating features are the position of the ΔH line (above or below the is TΔS line), and the slope of the latter, which of course depends on the sign of ΔS.
• The reaction A → B will occur spontaneously only when ΔG is negative (blue arrows pointing down.)
• TΔS plots are not quite straight lines as shown here. Similarly, the lines representing ΔH are even more curved.
The other two plots on each diagram are only for the chemistry-committed.
• Each pair of energy-level diagrams depicts the relative spacing of the microscopic energy levels in the reactants and products as reflected by the value of ΔS°. (The greater the entropy, the more closely-spaced are the quantized microstates.)
• The red shading indicates the range of energy levels that are accessible to the system at each temperature. The spontaneous direction of the reaction will always be in the direction in which the red shading overlaps the greater number of energy levels, resulting in the maximum dispersal of thermal energy.
• Note that the vertical offsets correspond to ΔH° for the reaction.
• Never forget that it is the ability of thermal energy to spread into as many of these states as possible that determines the tendency of the process to take place. None of this is to scale, of course!
The Standard Gibbs Energy
You have already been introduced to the terms such as $ΔU^o$ and $ΔH^o$ in which the $^o$ sign indicates that all components (reactants and products) are in their standard states. This concept of standard states is especially important in the case of the free energy, so let's take a few moments to review it. More exact definitions of the conventional standard states can be found in most physical chemistry textbooks. In specialized fields such as biochemistry and oceanography, alternative definitions may apply. For example, the "standard pH" of zero (corresponding to $[H^{+}] = 1\,M$) is impractical in biochemistry, so pH = 7 is commonly employed. For most practical purposes, the following definitions are good enough:
• gases: 1 atmosphere partial pressure
• pure liquids: the liquid under a total (hydrostatic) pressure of 1 atm.
• solutes: an effective concentration of 1 mol L–1 (1 mol dm–3). ("Effective" concentrations approach real concentrations as the latter approach zero; for practical purposes, these can be considered identical at real concentrations smaller than about 10–4 molar.)
• solids: the pure solid under 1 atm pressure
Reminder on Standard States
• There is actually no "standard temperature", but because most thermodynamics tables list values for 298.15 K (25° C), this temperature is usually implied.
• These same definitions apply to standard enthalpies and internal energies.
• Do not confuse these thermodynamic standard states with the "standard temperature and pressure" (STP) widely employed in gas law calculations.
To make use of Gibbs energies to predict chemical changes, we need to know the free energies of the individual components of the reaction. For this purpose we can combine the standard enthalpy of formation and the standard entropy of a substance to get its standard free energy of formation
$ΔG_f^o = ΔH_f^o – TΔS_f^o \label{23.4.7}$
Recall that the symbol ° refers to the standard state of a substance measured under the conditions of 1 atm pressure or an effective concentration of 1 mol L–1 and a temperature of 298 K. Then determine the standard Gibbs energy of the reaction according to
$ΔG^o = \sum ΔG_f^o \;(\text{products})– \sum ΔG_f^o \;(\text{reactants}) \label{24.4.8}$
As with standard heats of formation, the standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 K. Standard Gibbs free energies of formation are normally found directly from tables. Once the values for all the reactants and products are known, the standard Gibbs energy change for the reaction is found by Equation $\ref{23.4.7}$. Most tables of thermodynamic values list $ΔG_f^o$ values for common substances (e.g., Table T2), which can, of course, always be found from values of ΔHf° and ΔSf°.
Example $1$
Find the standard Gibbs energy change for the reaction
$\ce{CaCO3(s) \rightarrow CaO (s) + CO2(g)} \nonumber$
The $ΔG_f^o°$ values for the three components of this reaction system are $\ce{CaCO3(s)}$: –1128 kJ mol–1, CaO(s): –603.5 kJ mol–1, CO2(g): –137.2 kJ mol–1.
Solution
Substituting into Equation $\ref{23.4.7}$, we have
$ΔG^o = (–603.5 –137.2) – (–1128) kJ\, mol^{–1} = +130.9\, kJ\, mol^{–1} \nonumber$
This indicates that the process is not spontaneous under standard conditions (i.e., solid calcium carbone will not form solid calcium oxide and CO2 at 1 atm partial pressure at 25° C).
Comment: This reaction is carried out on a huge scale to manufacture cement, so it is obvious that the process can be spontaneous under different conditions.
The practical importance of the Gibbs energy is that it allows us to make predictions based on the properties (Δ values) of the reactants and products themselves, eliminating the need to experiment. But bear in mind that while thermodynamics always correctly predicts whether a given process can take place (is spontaneous in the thermodynamic sense), it is unable to tell us if it will take place at an observable rate.
When thermodynamics says "no", it means exactly that. When it says "yes", it means "maybe".
Example $2$
The reaction
$\ce{ 1/2 O2(g) + H2(g) → H2O(l)} \nonumber$
is used in fuel cells to produce an electrical current. The reaction can also be carried out by direct combustion.
Thermodynamic data: molar entropies in J mol–1 K–1: O2(g) 205.0; H2(g)130.6; H2O(l) 70.0; H2O(l) ΔH°f = –285.9 kJ mol–1.
Use this information to find
1. The amount of heat released when the reaction takes place by direct combustion;
2. The amount of electrical work the same reaction can perform when carried out in a fuel cell at 298 K under reversible conditions;
3. The amount of heat released under the same conditions.
Solution
First, we need to find $ΔH^o$ and $ΔS^o$ for the process. Recalling that the standard enthalpy of formation of the elements is zero,
\begin{align*} ΔH^o &= ΔH^p_f(\text{products}) – ΔH^°_f(\text{reactants}) \[4pt] &= –285.9\, kJ\, mol^{–1} – 0 \[4pt] &= –285.9 \,kJ \,mol^{–1} \end{align*}.
Similarly,
\begin{align*} ΔS^o &= S^o_f(\text{products}) – S^o_f(\text{reactants}) \[4pt] &= (70.0) – (½ \times 205.0 + 130.6) \[4pt] &= –163\, J\, K^{–1}mol^{–1} \end{align*}
1. When the hydrogen and oxygen are combined directly, the heat released will be $ΔH^o = –285.9\, kJ\, mol^{–1}$.
2. The maximum electrical work the fuel cell can perform is given by \begin{align*}ΔG^o &= ΔH^o – TΔS^o \[4pt] &= –285.9 \,kJ\, mol^{–1} – (298\, K)(–163\, JK^{–1}mol^{–1}) \[4pt] &= –237.2 \,kJ\, mol^{–1}.\end{align*}.
3. The heat released in the fuel cell reaction is the difference between the enthalpy change (the total energy available) and the reversible work that was expended: \begin{align*} ΔH^o – ΔG^o &= TΔS^o \[4pt] &= (298\, K)(–163\, JK^{–1}mol^{–1}) \[4pt] &= –48,800\, J\, mol^{–1} \[4pt] &=–48.8 \,kJ\, mol^{–1}.\end{align*}.
The foregoing example illustrates an important advantage of fuel cells. Although direct combustion of a mole of hydrogen gas yields more energy than is produced by the same net reaction within the fuel cell, the latter, in the form of electrical energy, can be utilized at nearly 100-percent energy efficiency by a motor or some other electrical device. If the thermal energy released by direct combustion were supplied to a heat engine, second-law considerations would require that at least half of this energy be "wasted" to the surroundings.
ΔG vs. ΔG°: what's the difference?
Δrefer to single, specific chemical changes in which all components (reactants and products) are in their standard states.
The $ΔG_f^o$ of a substance, like $ΔH_f^o$, refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. Both of these terms are by definition zero for the elements in their standard states. There are only a few common cases in which this might create some ambiguity:
Table $1$: Standard Gibbs Energies of Select Substances
Stable Form $ΔG_f^o$ (kJ mol–1) Unstable Form $ΔG_f^o$ (kJ mol–1)
$\ce{O2(g)}$ 0 $\ce{O3(g)}$ 163.2
$\ce{C(graphite)}$ 0 $\ce{C(diamond)}$ 2.9
$\ce{S(rhombic)}$ 0 $\ce{S(monoclinic)}$ 0.1
$\ce{P(white)}$ 0 $\ce{P4(g)}$ 24.4
Ions in aqueous solution are a special case; their standard free energies are relative to the hydrated hydrogen ion $\ce{H^{+}(aq)}$ which is assigned $ΔG_f^o = 0$.
$ΔG$ is very different from ΔG°. The distinction is nicely illustrated in Figure $5$ in which ΔG is plotted on a vertical axis for two hypothetical reactions having opposite signs of Δ. The horizontal axis schematically expresses the relative concentrations of reactants and products at any point of the process. Note that the origin corresponds to the composition at which half of the reactants have been converted into products.
Take careful note of the following:
• for the Δ > 0 reaction. Notice that there are an infinite number of these values, depending on the progress of the reaction. In contrast there is only a single value of Δ, corresponding to the composition at which ΔG = 0 ().
• At this point, some products have been formed, but the composition is still dominated by reactants.
• If we begin at a composition to the left of , ΔG will be negative and the composition will move to the right. Similarly if we begin with a composition to the right of , ΔG will be positive and the composition will move to the left.
• The plot on the right is for the ΔG° < 0 reaction, for which ΔG° is shown at . At its equilibrium point , there are more products than reactants. If we start at a composition to the right of , the composition will tend to move to the left. If the initial composition is to the left of , the reaction will tend to proceed to the right.
• What would happen if Δ were 0? The equilibrium point of such a reaction would be at the origin, corresponding to half the reactants being converted to products.
The important principle you should understand from this is that a negative Δ does not mean that the reactants will be completely transformed into products. By the same token, a positive Δ does not mean that no products are formed at all.
It should now be clear from the discussion above that a given reaction carried out under standard conditions is characterized by a single value of Δ.
The reason for the Gibbs energy minimum at equilibrium relates to the increase in entropy when products and reactants coexist in the same phase. As seen in the plot, even a minute amount of "contamination" of products by reactants reduces the free energy below that of the pure products. In contrast, composition of a chemical reaction system undergoes continual change until the equilibrium state is reached. So the a single reaction can have an infinite number of ΔG values, reflecting the infinite possible compositions between the extremes of pure reactants (zero extent of reaction) and pure products (unity extent of reaction).
In the example of a reaction A → B, depicted in the above diagram, the standard free energy of the products is smaller than that of the reactants , so the reaction will take place spontaneously. This does not mean that each mole of pure A will be converted into one mole of pure B. For reactions in which products and reactants occupy a single phase (gas or solution), the meaning of "spontaneous" is that the equilibrium composition will correspond to an extent of reaction greater than 0.5 but smaller than unity.Note, however, that for Δ values in excess of about ±50 kJ mol–1, the equilibrium composition will be negligibly different from zero or unity extent-of-reaction. The physical meaning of ΔG is that it tells us how far the free energy of the system has changed from G° of the pure reactants . As the reaction proceeds to the right, the composition changes, and ΔG begins to fall. When the composition reaches , ΔG reaches its minimum value and further reaction would cause it to rise. But because free energy can only decrease but never increase, this does not happen. The composition of the system remains permanently at its equilibrium value.
A G vs. extent-of-reaction diagram for a non-spontaneous reaction can be interpreted in a similar way; the equilibrium composition will correspond to an extent of reaction greater than zero but less than 0.5. In this case, the minimum at reflects the increase in entropy when the reactants are "contaminated" by a small quantity of products.
If all this detail about ΔG seems a bit overwhelming, do not worry: it all gets hidden in the equilibrium constant and reaction quotient that we discuss in the next lesson!
Interpretation of Standard Gibbs energy changes
Although it is $ΔG$ rather than $ΔG^o$ that serves as a criterion for spontaneous change at constant temperature and pressure, $ΔG^o$ values are so readily available that they are often used to get a rough idea of whether a given chemical change is possible. This is practical to do in some cases, but not in others:
Example
It generally works for reactions such as
$\ce{4 NH_3(g) + 5 O_2(g) → 4 NO(g) + 6 H_2O(g)} \nonumber$
with $ΔG^o = –1,010\, kJ$.
(industrially important for the manufacture of nitric acid) because $ΔG^o$ is so negative that the reaction will be spontaneous and virtually complete under just about any reasonable set of conditions.
Example
The following reaction expresses the fact that the water molecule is thermodynamically stable:
$\ce{2 H_2(g) + 1/2 O_2(g)→ H_2O(l)} \nonumber$
with $ΔG^o = –237.2 \,kJ$.
Note that this refers to liquid water (the standard state of H2O at 25°). If you think about it, a negative standard Gibbs energy of formation (of which this is an example) can in fact be considered a definition of molecular stability.
Example
Similarly, dissociation of dihydrogen into its atoms is highly unlikely under standard conditions:
$\ce{H_2O(g) → 2 H(g) + O(g)} \nonumber$
with $ΔG^o = +406.6\, kJ$.
Again, an analogous situation would apply to any stable molecule.
Example
Now consider the dissociation of dinitrogen tetroxide
$\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber$
with $ΔG^o = +2.8 kJ$.
in which the positive value of Δ tells us that N2O4 at 1 atm pressure will not change into two moles of NO2 at the same pressure, but owing to the small absolute value of Δ, we can expect the spontaneity of the process to be quite sensitive to both the temperature (as shown in the table below) and to the pressure in exactly the way the Le Chatelier principle predicts.
Example
For reactions involving dissolved ions, one has to be quite careful. Thus for the dissociation of the weak hydrofluoric acid
$\ce{HF(aq) → H^+(aq) + F^–(aq)} \nonumber$
with $ΔG^o = –317 \,kJ$.
it is clear that a 1 mol/L solution of HF will not dissociate into 1M ions, but this fact is not very useful because if the HF is added to water, the initial concentration of the fluoride ion will be zero (and that of H+ very close to zero), and the Le Chatelier principle again predicts that some dissociation will be spontaneous.
Example
It is common knowledge that dissociation of water into hydrogen- and hydroxyl ions occurs only very sparingly:
$\ce{H_2O(l) → H^+(aq) + OH^–(aq) } \nonumber$
with $ΔG^o = 79.9 \,kJ$.
which correctly predicts that the water will not form 1M (effective concentration) of the ions, but this is hardly news if you already know that the product of these ion concentrations can never exceed 10–14 at 298K.
Example
Finally, consider this most familiar of all phase change processes, the vaporization of liquid water:
$\ce{H_2O(l) → H_2O(g)} \nonumber$
with $ΔG^o = 8.58 \,kJ$.
Conversion of liquid water to its vapor at 1 atm partial pressure does not take place at 25° C, at which temperature the equilibrium partial pressure of the vapor (the "vapor pressure") is only 0.031 atm (23.8 torr.) Gaseous H2O at a pressure of 1 atm can only exist at 100° C. Of course, water left in an open container at room temperature will spontaneously evaporate if the partial pressure of water vapor in the air is less than 0.031 atm, corresponding to a relative humidity of under 100%
Finding the Equilibrium Temperature
A reaction is in its equilibrium state when
$ΔG = ΔH – TΔS = 0 \label{23.4.1a}$
The temperature at which this occurs is given by
$T = \dfrac{ΔH}{TΔS} \label{23.4.1b}$
If we approximate $ΔH$ by $ΔH^o$ and $ΔS$ by $ΔS^o$, so Equation \ref{23.4.1a} would be
$ΔG \approx ΔH^o – TΔS^o = 0 \label{23.4.1aa}$
We can then estimate the normal boiling point of a liquid. From the following thermodynamic data for water:
Caution!
Because ΔH° values are normally expressed in kilojoules while ΔS° is given in joules, a very common student error is to overlook the need to express both in the same units.
We find that liquid water is in equilibrium with water vapor at a partial pressure of 1 atm when the temperature is
$T = \dfrac{44,100\, J}{118.7\, J\, K^{–1}} = 371.5\, K$
But "the normal boiling point of water is 373 K", you say? Very true. The reason we are off here is that both ΔH ° and ΔS ° have their own temperature dependencies; we are using the "standard" 25° values without correcting them to 100° C. Nevertheless, if you think about it, the fact that we can estimate the boiling point of a liquid from a table of thermodynamic data should be rather impressive! Of course, the farther one gets from 298 K, the more unreliable will be the result. Thus for the dissociation of dihydrogen into its atoms,
All one can say here is that H2 will break down at something over 3000 K or so. (You may already know that all molecules will dissociate into their atoms at high temperatures.) We tend to think of high temperatures as somehow "forcing" molecules to dissociate into their atoms, but this is wrong. In order to get the H–H bond to vibrate so violently through purely thermal excitation that the atoms would fly apart, a temperature more like 30,000 K would be required. The proper interpretation is at the temperature corresponding to ΔH/TΔS, the molecule spontaneously absorbs energy from the surroundings sufficient to overcome the H-H bond strength.
Predicting the Effects of Temperature
The $T\Delta S$ term interacts with the $ΔH$ term in $\Delta G$ to determine whether the reaction can take place at a given temperature. This can be more clearly understood by examining plots of $TΔS^o$ and $ΔH^o$ as functions of the temperature for some actual reactions. Of course these parameters refer to standard states that generally do not correspond to the temperatures, pressures, or concentrations that might be of interest in an actual case. Nevertheless, these quantities are easily found and they can usefully predict the way that temperature affects these systems.
Case 1: Exothermic reaction, ΔS° > 0
$\ce{C(graphite) + O_2(g) → CO_2(g)} \nonumber$
• $ΔH^o = –393\, kJ$
• $ΔG^o = –394 \,kJ$ at $298\, K$
This combustion reaction, like most such reactions, is spontaneous at all temperatures. The positive entropy change is due mainly to the greater mass of $\ce{CO2}$ molecules compared to those of $\ce{O2}$.
Case 2: Exothermic reaction, ΔS° < 0
$\ce{3 H_2 + N_2 → 2 NH_3(g) } \nonumber$
• $ΔH^o = –46.2\, kJ$
• $ΔG^o = –16.4\, kJ$ at $298\, K$
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
Case 3: Endothermic reaction, ΔS° > 0
$\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber$
• $ΔH^o = 55.3\, kJ$
• $ΔG^o = +2.8\, kJ$ at $298\, K$
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures. Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
Case 4: Endothermic Reaction, ΔS° < 0
$\ce{ 1/2 N_2 (g) + O_2 (g)→ NO_2(g)} \nonumber$
• $ΔH^° = 33.2\, kJ$
• $ΔS^o = –249\, J\, K^{–1}$
• $ΔG^o = +51.3\, kJ$ at $298\, K$
This reaction is not spontaneous at any temperature, meaning that its reverse is always spontaneous. But because the reverse reaction is kinetically inhibited, NO2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Concluding remarks on Gibbs Energy
The appellation “free energy” for $G$ has led to so much confusion that many scientists now refer to it simply as the Gibbs energy. The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: $ΔG$ is the maximum amount of energy, which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible.
A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although $G$ has the units of energy (joules, or in its intensive form, J mol–1), it lacks one of the most important attributes of energy in that it is not conserved. Thus, although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy anywhere else. Referring to $G$ as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy. The quantity $–ΔG$ associated with a process represents the quantity of energy that is “shared and spread”, which as we have already explained is the meaning of the increase in the entropy. The quotient $–ΔG/T$ is in fact identical with $ΔS_{total}$, the entropy change of the world, whose increase is the primary criterion for any kind of change.
$G$ differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas $H$ and $S$ can be related to the quantity and distribution of energy in a collection of molecules. The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.
What Gibbs Energy is not...
• Gibbs Energy is not free energy
• Gibbs Energy is not energy
• Gibbs Energy is not even "real" | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.3%3A_Thermodynamic_Description_of_the_Equilibrium_State.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To write an equilibrium constant expression for any reaction.
Relationship among Equilibrium Expressions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of $N_2$ with $O_2$ to give $NO_2$. This reaction is an important source of the $NO_2$ that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), $N_2$ reacts with $O_2$ at the high temperatures inside an internal combustion engine to give $NO$. The released $NO$ then reacts with additional $O_2$ to give $NO_2$ (2). The equilibrium constant for each reaction at 100°C is also given.
1. $N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\;\; K_1=2.0 \times 10^{−25}$
2. $2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9$
Summing reactions (1) and (2) gives the overall reaction of $N_2$ with $O_2$:
1. $N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=?$
The equilibrium constant expressions for the reactions are as follows:
$K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2}$
What is the relationship between $K_1$, $K_2$, and $K_3$, all at 100°C? The expression for $K_1$ has $[NO]^2$ in the numerator, the expression for $K_2$ has $[NO]^2$ in the denominator, and $[NO]^2$ does not appear in the expression for $K_3$. Multiplying $K_1$ by $K_2$ and canceling the $[NO]^2$ terms,
$K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3$
Thus the product of the equilibrium constant expressions for $K_1$ and $K_2$ is the same as the equilibrium constant expression for $K_3$:
$K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15}$
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, $ΔH$ for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
Note
To determine $K$ for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example $6$
The following reactions occur at 1200°C:
1. $CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{−2}$
2. $CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4$
Calculate the equilibrium constant for the following reaction at the same temperature.
1. $CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?$
Given: two balanced equilibrium equations, values of $K$, and an equilibrium equation for the overall reaction
Asked for: equilibrium constant for the overall reaction
Strategy:
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of $K$ for that equation. Calculate $K$ for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
$CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}$
$\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}$
$CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}$
The values for $K_1$ and $K_2$ are given, so it is straightforward to calculate $K_3$:
$K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3$
Exercise $6$
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
1. $\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}$
2. $SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}$
3. $\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?$
Answer
$K_3 = 1.1 \times 10^{66}$
Summary
An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.4%3A_The_Law_of_Mass_Action_for_Related_and_Simultaneous_Equilibria.txt |
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