solutions for chapter 9 end-of-chapter problemsquantum.bu.edu/.../text/answers/chapter09.pdfmarch...

March 2005 ACS Chemistry Cahpter 9 suggested solutions 1

Solutions for Chapter 9 End-of-Chapter Problems

Problem 9.1. (a) Balanced chemical equations for the formation of pale blue Cu(OH)2(s) and deep blue-violet [Cu(NH3)4]2+(aq) when NH3(aq) is slowly added to a light blue solution of Cu2+(aq) are: 2NH3(aq) + 2H2O(l) + Cu2+(aq) ⇔ Cu(OH)2(s) + 2NH4

+(aq) light blue pale blue Cu(OH)2(s) + 4NH3(aq) ⇔ [Cu(NH3)4]2+(aq) + 2OH–(aq) pale blue deep blue-violet

(b) To understand how the formation of Cu(OH)2(s) from Cu2+(aq) will be affected by basic conditions, we need to consider the reactions that make up the net reaction written in part (a): 2NH3(aq) + 2H2O(l) ⇔ 2NH4

+(aq) + 2OH–(aq)

Cu2+(aq) + 2OH–(aq) ⇔ Cu(OH)2(s) Additional OH–(aq) is a disturbance to these equilibria (Le Chatelier’s principle) and the system will respond by using up some of the additional OH–(aq). Thus, the first reaction will shift toward reactants, because OH–(aq) is a product, and the second reaction will shift toward products because OH–(aq) is a reactant. The net result will be to favor formation of Cu(OH)2(s) from Cu2+(aq). (c) The equation written in part (a) suggests that the formation of [Cu(NH3)4]2+(aq) from Cu(OH)2(s) will be favored under mildly acidic conditions, because H3O+(aq) will react with OH–(aq) ions to form water. This reaction is a disturbance to the equilibrium, because it removes a product of the reaction, The system should respond by trying to form more OH–(aq), which also would mean the formation of more [Cu(NH3)4]2+(aq). This analysis is flawed because it neglects the basicity of NH3(aq), which will also react with added H3O+(aq): NH3(aq) + H3O+(aq) ⇔ NH4

+(aq) This reaction removes a reactant, NH3(aq), necessary for formation of [Cu(NH3)4]2+(aq), so the presence of H3O+(aq) in the solution will not favor formation of [Cu(NH3)4]2+(aq). Note that the reaction of H3O+(aq) with OH–(aq) also removes a reactant in the second reaction in part (b), so Cu(OH)2(s) will tend to dissolve or not form under acidic conditions. (d) None of the reactions involves oxidation or reduction. Copper has an oxidation number of +2 in all cases.

Problem 9.2. If Fe3+(aq) and SCN–(aq) formed a one-to-two metal ion complex, the complex in Figure 9.1 would have to show the new ratio of two SCN–(aq) ions to every Fe3+(aq) ion. Twice as many SCN–(aq) ions would be used in forming the complex and there would be four more unreacted Fe3+(aq) ions in the right-hand part of the figure, as shown in this alternative figure:

Chemical Equilibria Chapter 9

2 ACS Chemistry Chapter 9 suggested solutions

Problem 9.3. (a) Addition of CN–(aq) to the equilibrium reaction, Cd2+(aq) + 6CN–(aq) ⇔ Cd(CN)6

2–(aq), will cause the reaction to shift toward formation of more of the product, Cd(CN)6

2–(aq). Addition of more of a reactant is a disturbance to the equilibrium system (Le Chatelier’s principle) and the system responds by minimizing the disturbance, using up some of the added CN–(aq) to form more of the product. (b) Identification of change in the CoCl2 aqueous equilibrium is possible because of color differences between the reactants and the products. Both species in this equilibrium are colorless, so there is no visual clue to the position of equilibrium.

Problem 9.4. The statements in this problem refer to the equilibrium reaction: H2(g) + Cl2(g) ⇔ 2HCl(g). (a) When this reaction has reached a state of equilibrium, no further reaction occurs. This is not a true statement. No further net reaction occurs, but there is a dynamic equilibrium in which molecules of HCl(g) continuously form, even as other HCl(g) molecules are decomposing into H2 and Cl2. While the macroscopic concentrations are not changing, there is constant change at the microscopic level. (b) When equilibrium is established, the number of moles of reactants equals the number of moles of products for this reaction. This is not a true statement. The stoichiometric coefficients that balance the equation do not give the relative number of moles present at equilibrium. It is necessary to know the position of the equilibrium and to use the equilibrium constant at a given temperature to find relative numbers of moles of reactants and products. (c) The concentration of each substance in the system will be constant at equilibrium. This is a true statement. The observable concentrations are constant if dynamic equilibrium has been established, but there still is constant change at the microscopic level.

Problem 9.5. The reaction of interest is: CO(g) + 2O2(g) ⇔ 2CO2(g). In this reaction, a C–O triple bond (BH = 1071 kJ·mol–1) and two O–O double bonds (BH = 498.7 kJ·mol–1) are broken, and four C–O double bonds in carbon dioxide (BH = 799 kJ·mol–1) are made. The overall change in enthalpy for the reaction is –1127 kJ·mol–1, that is, the reaction is exothermic. If the reaction is at equilibrium and is disturbed by adding energy to raise its temperature, the system will respond (Le Chatelier’s principle) by “using up” some of the added energy which it can do by proceeding in reverse (the endothermic direction) toward increase in the concentration of the reactants, CO(g) and O2(g), at the expense of some of the reactant, CO2(g).

Chapter 9 Chemical Equilibria

ACS Chemistry Chapter 9 suggested solutions 3

Problem 9.6. The reaction of interest is: CO(g) + 2H2(g) ⇔ CH3OH(g). (a) If some CH3OH(g) is removed while the volume of the system is held constant, the concentration of the product decreases. The system will respond to this disturbance (Le Chatelier’s principle) by producing more product at the expense of a decrease in the concentrations (amounts) of the reactants. (b) Adding CO(g) to the system while the volume is held constant increases the concentration of one of the reactants. The system will respond to this disturbance by reacting to reduce the concentration of this reactant with the consequence that the concentration of the other reactant will be reduced and the concentration of the product will increase. (c) If the system is compressed to a smaller volume, the concentrations of all the species will be increased. In response to this disturbance, the system will react to reduce the number of moles of gas, so as to try to bring the concentrations as near as possible to where they had been. The number of moles of gas will be reduced if reactants combine to form the product, so the result will be an increase in the concentration of the product and reductions in the concentrations of the reactants. (d) N2(g) has no role in the reaction and adding it at constant volume does not affect the concentrations of the reactants and products. There will be no change in the equilibrium system when this addition is made.

Problem 9.7. The reaction of interest is this Lewis acid-base equilibrium reaction between boric acid and water: B(OH)3(s) + 2H2O(aq) ⇔ [B(OH)4]–(aq) + H3O+(aq) (a) Addition of B(OH)3(s) will have no effect on the equilibrium. Addition of a solid reactant does not change its concentration. (b) Addition of H3O+(aq) will cause this equilibrium reaction to shift toward the reactants. Increasing the H3O+(aq) concentration is a disturbance to the equilibrium system (Le Chatelier’s principle) and the system will respond by using up some of the added H3O+(aq). It can do this by proceeding in reverse, using up some [B(OH)4]–(aq) and forming more B(OH)3(s). (c) Addition of OH–(aq) will cause this equilibrium reaction to shift toward the products. Added OH–(aq) will react with H3O+(aq), which will disturb the equilibrium system by decreasing the concentration of a product. The system will respond by producing more H3O+(aq) to replace part of what is lost, thus forming more [B(OH)4]–(aq) and using up some B(OH)3(s).

Problem 9.8. The reaction of interest is: HA(aq) + H2O(aq) ⇔ H3O+(aq) + A–(aq). (a) This reaction equation tells us that, in the forward direction, a molecule of HA, HA(aq), (solvated by the water) interacts with a molecule of water, H2O, to transfer a proton to the water to form a solvated hydronium ion, H3O+(aq), and a solvated anion, A–(aq), HA without its acidic proton. In the reverse direction a hydronium ion interacts with a solvated anion to transfer a proton to the anion to form a water molecule and a solvated molecule of the acid, HA(aq). The way the equation is written, it appears that the same hydronium and anion that form in the forward reaction react in the reverse reaction. This is an incorrect interpretation. The Web Companion animation illustrates the correct interpretation.



(b) The beginning (proton transfer from HA(aq) to H2O(aq)) and end (proton transfer from H3O+(aq) to A–(aq)) of the reaction shown in the Web Companion, Chapter 9, Section 9.4, page 1, animation are exactly consistent with the reaction written and described above. What we see in the “middle” of the animation is that protons can be transferred between hydronium ions and water molecules, that it is not the same proton transferred in each case, and that the anion to which a proton is transferred from a hydronium is not the same one from which a proton originally came. What is missing in the equation is a way to show: (1) that any one of the acid molecules in the mixture might transfer its proton to a water molecule, (2) that the hydronium ion formed is just one of many hydronium ions in the solution, (3) that the hydronium ions are continuously transferring protons to water molecules to form different hydronium ions, and (4) that any one of the acid anions and any one of the hydronium ions might be the ones that react to transfer a proton from hydronium ion to anion. Although reaction equations are a very compact way to represent the stoichiometry of reactions, they do not very well represent the details of what is occurring at the molecular level. As you use such equations, always try to use what you are learning about reactions to visualize what is going on at the molecular level as well.

Problem 9.9. The equilibrium constant expression for ammonia synthesis is:

Keq = NH3 (g )( )2

N 2(g )( ) H2 (g)( )3

Since all species in this equilibrium constant expression are gases, the dimensionless concentration ratios should be expressed in bar (standard state = 1 bar).

Problem 9.10. It requires 50 drops of a 1.0 M weak acid, HA(aq), solution added to 10.0 mL of distilled water to produce enough ions to light the bulb in a conductivity detector as brightly as 1 drop of 1 M HCl(aq) added to 10.0 mL of distilled water. We can conclude that there are only about 2%

= 1 drop50 drops( )⋅100%⎡

⎣⎢⎤⎦⎥

as many ions formed by the weak acid as by an equivalent number

of moles of HCl(aq), hydrochloric acid. We know that HCl(aq) reacts completely to form H3O+(aq) and Cl–(aq) ions, so we estimate that only about 2% of the weak acid reacts to transfer its proton to water. Thus, in a 1 M solution of HA(aq), the concentrations are [HA(aq)] = 1 M, [H3O+(aq)] = 0.02 M, and [A–(aq)] = 0.02 M. The proton transfer reaction and equilibrium constant (dissociation constant) for the weak acid are: HA(aq) + H2O(aq) ⇔ H3O+(aq) + A–(aq)

Keq = H3O

+(aq)( ) A– (aq)( )HA(aq)( ) H2O(aq)( )

⎛

⎝⎜

⎞

⎠⎟

eq

= (0.02)(0.02)

(1)(1)

⎛⎝⎜

⎞⎠⎟

= 4 × 10–4

Problem 9.11. (a) The reaction equation and equilibrium constant expression for the decomposition of calcium carbonate are: CaCO3(s) ⇔ CaO(s) + CO2(g)



Keq = CaO(s)( ) CO2 (g)( )

CaCO 3(s)( )

For pure solids the dimensionless concentration ratio is 1, unity. For the gas we need to use bar as the unit (standard state = 1 bar). (b) If the equilibrium pressure of CO2(g) is 0.37 bar, then:

Keq =

(0.37 bar)(1 bar)( )1( )

1( ) = 0.37

Here, we have substituted the values for the concentration ratios to get Keq.

Problem 9.12. (a) Addition of four more thiocyanate anions, SCN(aq)–, to the equilibrium solution depicted in the center of Figure 9.2 leads to the formation of a total of three iron-thiocyanate complexes at equilibrium in the solution. As in Worked Example 9.14, we count the number of each species represented in our solutions and assume the numbers are proportional to their molarities in the solution. After adding four more thiocyanate ions to the middle panel of Figure 9.2 and forming three iron-cyanate complexes in the resulting mixture, there will be nine thiocyanate ions unreacted (the original eight plus the four added minus the three in the complexes) and 17 iron ions (the original 20 minus the three in the complexes). The equilibrium constant for this system is then:

K = Fe(SCN)2 + (aq)[ ]

Fe3+(aq )[ ]SCN–

(aq)[ ] = 3c

(17c)(9c) =

1

51c

(b) The equilibrium constants in Worked Example 9.14 and Check This 9.15 are 1/54c and 1/55c, respectively. In part (a), we get 1/51c. Within the limitations of these simple pictures with countable numbers of “molecules,” the agreement among these three values is quite good.

Problem 9.13. (a) The pressure of each gas in the original mixture in the 1.00 L container at 500 K is given by P = nRT/V, where the number of moles of each gas is given in the problem statement:

P(HI) = (2.21 ×10–3 mol)(8.314 ×10–2 L ⋅ bar ⋅ K –1 ⋅ mol–1)(500K)

1.00 L = 9.19 × 10–2 bar

P(I2) = (1.46 ×10–3 mol)(8.314 ×10–2 L ⋅ bar ⋅ K–1 ⋅ mol–1)(500K)

1.00 L = 6.07 × 10–2 bar

P(H2) = (2.09 × 10–5 mol)(8.314 ×10–2 L ⋅ bar ⋅ K–1 ⋅ mol–1)(500K)

1.00 L = 8.69 × 10–4 bar

(b) The equilibrium constant expression for the reaction, H2(g) + I2(g) ⇔ 2HI(g), is:

Keq = HI(g)( )2

H 2(g)( ) I2 (g)( )

(c) Each of the dimensionless concentration terms in the equilibrium constant expression is a ratio of the pressure (in bar) of the reactant to the standard pressure, 1 bar. Therefore, we can



substitute the numeric values for the pressures in part (a) into the expression in part (b) to get the numeric value of the equilibrium constant:

Keq = 9.19 ×10–2( )2

8.69 ×10–4( ) 6.07 ×10–2( ) = 1.60 × 102

Problem 9.14. {NOTE: There is a typographical error in this problem. The concentrations should be [HIn(aq)] = 6.3 × 10–5 M and [In–(aq)] = 8.7 × 10–5 M. Both are labeled [HIn(aq)].} (a) For the reaction, HIn(aq) + H2O(aq) ⇔ H3O+(aq) + In–(aq), the equilibrium constant expression is:

Keq = H 3O

+(aq)( ) In– (aq)( )HIn(aq)( ) H 2O(aq)( )

(b) The molar concentrations of the acidic and basic forms of the indicator are given and we can use the pH to get the molar concentration of the hydronium ion, [H3O+(aq)] = 10–7.41 M = 3.9 × 10–8 M. The ratio of these molar concentrations to the standard concentration, 1 M, gives the dimensionless concentration terms for these species. Since the solution is dilute, the water concentration is essentially the same as in pure water, 55.5 M, so the dimensionless concentration term for water is unity. The numeric value for the equilibrium constant is:

Keq = 3.9 ×10–8( ) 8.7 ×10–5( )

6.3 ×10–5( )1( ) = 5.4 × 10–8

Problem 9.15. The information in this table is used in this problem and the next.

Acid name Formula Ka acetic acid CH3C(O)OH 1.8 × 10–5 chloroacetic acid ClCH2C(O)OH 1.4 × 10–3

(a) The acid equilibrium (proton transfer to water) equations for the acids are: CH3C(O)OH(aq) + H2O(l) ⇔ H3O+(aq) + CH3C(O)O–(aq) ClCH2C(O)OH(aq) + H2O(l) ⇔ H3O+(aq) + ClCH2C(O)O–(aq)

(b) The equilibrium constant expressions for Ka for the reactions in part (a) are:

Ka(acetic acid) = H3O

+(aq)( ) CH3C(O)O – (aq)( )(CH 3C(O)OH(aq)) H2O(aq)( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

eq

Ka(chloroacetic acid) = H3O

+(aq)( ) ClCH2C(O)O– (aq)( )(ClCH 2C(O)OH(aq)) H2O(aq)( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

eq



(c) The equilibrium constant for chloroacetic acid is nearly two orders of magnitude greater than that for acetic acid, that is, chloroacetic acid is a stronger acid than acetic acid (but still does not transfer all its protons to water). For solutions of equal concentration, chloroacetic acid will release more hydronium ions into solution, resulting in a lower pH than for acetic acid.

Problem 9.16. (a) Use the equilibrium constant expressions and equilibrium constants from the previous problem to find the pH of a 0.050 M solution of acetic acid:

Ka(acetic acid) = 1.8 ×10–5 = H3O

+(aq)( ) CH3C(O)O – (aq)( )(CH 3C(O)OH(aq)) H2O(aq)( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

eq

species H3O+(aq) CH3C(O)O–(aq) CH3C(O)OH(aq) initial mol 1.0 × 10–7 - 0 - 0.050 change in mol x formed x formed x reacts final mol x x 0.050 – x

Because Ka(acetic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.050:

Ka = 1.8 × 10–5 = x ⋅ x

0.050 – x⎛⎝⎜

⎞⎠⎟ ≈

x ⋅ x

0.050⎛⎝⎜

⎞⎠⎟ =

x2

0.050

x2 = 0.050·(1.8 × 10–5) = 9.0 × 10–7 x = 9.5 × 10–4 Check the assumption that the value of x is small and can be neglected relative to 0.050:

9.5 ×10–4

0.050

⎛⎝⎜

⎞⎠⎟

·100% = 1.9%

Our rule of thumb is that a 5% error is acceptable and this is well within the limit. pH = –log (H3O+(aq)) = –log (9.5 × 10–4) = 3.02 (b) Use the equilibrium constant expressions and equilibrium constants from the previous problem to find the pH of a 0.050 M solution of chloroacetic acid:

Ka(chloroacetic acid) = 1.4 ×10–3 = H3O

+(aq)( ) ClCH2C(O)O– (aq)( )(ClCH 2C(O)OH(aq)) H2O(aq)( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

eq

species H3O+(aq) ClCH2C(O)O–(aq) ClCH2C(O)OH(aq) initial mol 1.0 × 10–7 - 0 - 0.050 change in mol x formed x formed x reacts final mol x x 0.050 – x

Because Ka(chloroacetic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.050:



Ka = 1.4 × 10–3 = x ⋅ x

0.050 – x⎛⎝⎜

⎞⎠⎟ ≈

x ⋅ x

0.050⎛⎝⎜

⎞⎠⎟ =

x2

0.050


8.4 ×10–3

0.050

⎛⎝⎜

⎞⎠⎟

·100% = 17%

Our rule of thumb is that a 5% error is acceptable and this value is far outside the limit, so we will have to discard the assumption and solve the problem more exactly. One approach is to use successive approximations. We start with the result just obtained for x, and use it to give us an approximation for the concentration ratio of the unreacted acid: 0.050 – x = 0.050 – 0.0084 ≈ 0.042 Then we repeat the above calculation with this new value for the acid concentration:

Ka = 1.4 × 10–3 ≈ x ⋅ x

0.042⎛⎝⎜

⎞⎠⎟ =

x2

0.042

x2 = 0.042·(1.4 × 10–3) = 5.9 × 10–5 x = 7.7 × 10–3 If we consider a third approximation we would use this new value of x to give us another approximation for the concentration of the unreacted acid: 0.050 – x = 0.050 – 0.0077 ≈ 0.042 This value is identical (within the limits of uncertainty of the data given) to the one used in the previous calculation, so it will give the same result, x = 7.7 × 10–3. Thus, we take this result as the correct value for x = (H3O+(aq)). Another approach would have been to begin with the exact equation, rearrange it to quadratic form, ax2 +bx + c = 0, and then solve using the quadratic

formula: x = −b ± b2 − 4ac

2a. The result is x = 7.7 × 10–3, so both methods give the same value.

The pH of the 0.050 M chloroacetic acid solution is: pH = –log (H3O+(aq)) = –log (7.7 × 10–3) = 2.11

(c) The results calculated in parts (a) and (b) confirm the prediction made in our solution to Problem 9.15(c). For acetic acid and chloroacetic acid at the same concentrations, the stronger acid, chloroacetic, transfers more protons to water and forms more hydronium ions. Thus, the pH is lower for chloroacetic acid.

Problem 9.17. (a) For the autoionization reaction, H2O(l) + H2O(l) ⇔ H3O+(aq) + OH–(aq), Kw = (H3O+(aq))(OH–(aq)) and pKa = –log{(H3O+(aq))(OH–(aq))}. If the value of pKa decreases as the temperature increases, this means that the value of Kw increases with increasing temperature: at 25 ° C: Kw = 10–pKw = 10–14.0 = 1 × 10–14 at 60 ° C: Kw = 10–pKw = 10–11.5 = 3 × 10–12



The product concentrations increase at the higher temperature. Le Chatelier’s principle predicts that equilibria will shift toward formation of more product if energy is added (the temperature is increased) in endothermic reaction. The autoionization of water is an endothermic reaction. (b) An aqueous solution with H3O+(aq) concentration equal to 1.0 × 10–7 M at 25 °C will have a pH of 7.0 [= –log(H3O+(aq)) = –log(1.0 × 10–7)]. To find out whether the solution is acidic, basic, or neutral, we need to know how the concentrations of hydronium ion, [H3O+(aq)], and hydroxide ion, [OH–(aq)], compare. Use Kw to find [OH–(aq)]: Kw = 1 × 10–14 = (H3O+(aq))(OH–(aq)) = (1.0 × 10–7)(OH–(aq)) (OH–(aq)) = 1 × 10–7 ∴[OH–(aq)] = 1 × 10–7 M Because [H3O+(aq)] = [OH–(aq)] (within the uncertainty of the data given), a solution in which [H3O+(aq)] = 1.0 × 10–7 M at 25 °C is neutral. (c) An aqueous solution with H3O+(aq) concentration equal to 1.0 × 10–7 M at 60 °C will have a pH of 7 [= –log(H3O+(aq)) = –log(1.0 × 10–7)]. To find out whether the solution is acidic, basic, or neutral, we need to know how the concentrations of hydronium ion, [H3O+(aq)], and hydroxide ion, [OH–(aq)], compare. Use Kw to find [OH–(aq)]: Kw = 3 × 10–12 = (H3O+(aq))(OH–(aq)) = (1.0 × 10–7)(OH–(aq)) (OH–(aq)) = 3 × 10–5 ∴[OH–(aq)] = 3 × 10–5 M Because [H3O+(aq)] < [OH–(aq)], a solution in which [H3O+(aq)] = 1.0 × 10–7 M at 60 °C is basic. Note that a pH of 7.0 at 60 °C means the solution is basic, but a pH of 7.0 at 25 °C indicates a neutral solution. The relationship between [H3O+(aq)] and [OH–(aq)] determines whether a solution is acidic, neutral, or basic. The “neutral pH” at a particular temperature depends upon the equilibrium constant at that temperature. Neutral pH at 60 °C is about 5.8.

Problem 9.18. We are asked to determine the order of acid strength for species acting as acids in these reactions, given that the position of equilibrium lies to the right (products are favored over reactants) in each reaction: (i) N2H5

+(aq) + NH3(aq) ⇔ NH4+(aq) + N2H4(aq)

(ii) NH3(aq) + HBr(aq) ⇔ NH4+(aq) + Br–(aq)

(iii) N2H4(aq) + HBr(aq) ⇔ N2H5+(aq) + Br–(aq)

It is convenient to use the Brønsted-Lowry approach to identify the acids, proton donors, in these reactions. The acids are: N2H5

+(aq), NH4+(aq), and HBr(aq). Recall that the position of

equilibrium in an acid-base reaction favors the weaker acid (and base). Since these equilibria all favor products, the weaker acid in each reaction is on the product side. Note that HBr(aq) is never on the product side in these reactions, so we can conclude that it is probably the strongest of the three acids. Conversely, NH4

+(aq) is always found on product side, so we can conclude that it is probably the weakest of the three acids. This leaves N2H5

+(aq) as the intermediate strength acid and you see in reaction (i) that it is stronger than NH4

+(aq). In reaction (iii), you see that N2H5

+(aq) is weaker than HBr(aq). Thus, the relative acid strengths (strongest to weakest) are: HBr(aq) > N2H5

+(aq) > NH4+(aq).



Problem 9.19. (a) The pH of a 0.1 M solution of lactic acid can be calculated from its equilibrium constant expression and equilibrium constant (let LacH = lactic acid and Lac– = lactate ion): LacH(aq) + H2O(aq) ⇔ Lac–(aq) + H3O+(aq)

Ka(lactic acid) = 1.4 × 10–4 = (H3O+(aq))(Lac– (aq))

(LacH(aq))(H2O(aq))

⎛

⎝⎜

⎞

⎠⎟

eq

species H3O+(aq) Lac–(aq) LacH(aq) initial mol 1.0 × 10–7 - 0 - 0.1 change in mol x formed x formed x reacts final mol x x 0.1 – x

Because Ka(lactic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1:

Ka = 1.4 × 10–4 = x ⋅ x

0.1 – x⎛⎝⎜

⎞⎠⎟

≈ x ⋅ x

0.1⎛⎝⎜

⎞⎠⎟

= x2

0.1

x2 = 0.1·(1.4 × 10–4) = 1.4 × 10–5 x = 3.7 × 10–3 (uncertainty is larger than the ±4% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1:

3.7 × 10–3

0.1

⎛⎝⎜

⎞⎠⎟

·100% = 3.7%

Our rule of thumb is that a 5% error is acceptable and this result is within the limit. pH = –log (H3O+(aq)) = –log (3.7 × 10–3) = 2.4

(b) The pH of a 0.1 M solution of benzoic acid can be calculated from its equilibrium constant expression and equilibrium constant (let BzOH = benzoic acid and BzO– = benzoate ion): BzOH(aq) + H2O(aq) ⇔ BzO–(aq) + H3O+(aq)

Ka(benzoic acid) = 6.5 × 10–5 = (H3O

+(aq))(BzO– (aq))

(BzOH(aq))(H2O(aq))

⎛⎝⎜

⎞⎠⎟

eq

species H3O+(aq) BzO–(aq) BzOH(aq) initial mol 1.0 × 10–7 - 0 - 0.1 change in mol x formed x formed x reacts final mol x x 0.1 – x

Because Ka(benzoic acid) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1:

Ka = 6.5 × 10–5 = x ⋅ x

0.1 – x⎛⎝⎜

⎞⎠⎟

≈ x ⋅ x

0.1⎛⎝⎜

⎞⎠⎟

= x2

0.1

x2 = 0.1·(6.5 × 10–5) = 6.5 × 10–6



x = 2.5 × 10–3 (uncertainty is larger than the ±4% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1:

2.5 × 10–3

0.1

⎛⎝⎜

⎞⎠⎟

·100% = 2.5%

Our rule of thumb is that a 5% error is acceptable and this result is within the limit. pH = –log (H3O+(aq)) = –log (2.5 × 10–3) = 2.6

(c) The pH of a 0.1 M solution of aniline can be calculated from its equilibrium constant expression and equilibrium constant (let An = aniline and AnH+ = protonated aniline): An(aq) + H2O(aq) ⇔ AnH+(aq) + OH–(aq)

Kb(aniline) = 4.3 × 10–10 = (OH– (aq))(AnH+(aq))

(An(aq))(H2O(aq))

⎛

⎝⎜

⎞

⎠⎟

eq

species OH–(aq) An(aq) AnH+(aq) initial mol 1.0 × 10–7 0.1 -0- change in mol x formed x reacts x formed final mol x 0.1 – x x

Because Kb(aniline) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1:

Kb = 4.3 × 10-10 = x ⋅ x

0.1 – x⎛⎝⎜

⎞⎠⎟

≈ x ⋅ x

0.1⎛⎝⎜

⎞⎠⎟

= x2

0.1

x2 = 0.1·(4.3 × 10-10) = 4.3 × 10-11 x = 6.6 × 10–6 (uncertainty is larger than the ±2% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1:

6.6 ×10–6

0.1

⎛

⎝⎜

⎞

⎠⎟ ·100% = 0.007%

Our rule of thumb is that a 5% error is acceptable and this result is well within the limit. pOH = –log (OH-(aq)) = –log (6.6 × 10–6) = 5.2 pH = 14.00 – pOH = 14.0 – 5.2 = 8.8

(d) The pH of a 0.1 M solution of hydrazine can be calculated from its equilibrium constant expression and equilibrium constant: N2H4(aq) + H2O(aq) ⇔ N2H5

+(aq) + OH–(aq)

Kb(hydrazine) = 8.9 × 10–7 = (OH– (aq))(N2H5

+(aq))

(N2H4 (aq))(H2O(aq))

⎛⎝⎜

⎞⎠⎟

eq



species OH–(aq) N2H4(aq) N2H5+(aq)

initial mol 1.0 × 10–7 0.1 -0- change in mol x formed x reacts x formed final mol x 0.1 – x x

Because Kb(hydrazine) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.1:

Kb = 8.9 × 10–7 = x ⋅ x

0.1 – x⎛⎝⎜

⎞⎠⎟

≈ x ⋅ x

0.1⎛⎝⎜

⎞⎠⎟

= x2

0.1

x2 = 0.1·(8.9 × 10–7) = 8.9 × 10–8 x = 3.0 × 10–4 (uncertainty is larger than the ±3% implied here) Check the assumption that the value of x is small and can be neglected relative to 0.1:

3.0 ×10–4

0.1

⎛

⎝⎜

⎞

⎠⎟ ·100% = 0.3%

Our rule of thumb is that a 5% error is acceptable and this result is well within the limit. pOH = –log (OH-(aq)) = –log (3.0 × 10–4) = 3.5 pH = 14.00 – pOH = 14.0 – 3.5 = 10.5

Problem 9.20. From Table 9.2, pKa = 9.24 for NH4

+(aq) + H2O ⇔ NH3(aq) + H3O+(aq). We are interested in the basic reaction of ammonia, NH3(aq) + H2O(aq) ⇔ NH4

+(aq) + OH–(aq), and its pKb, which is 4.76 (= 14.00 – pKa). Using this value, we can calculate the pH of a 0.250 M aqueous ammonia solution.

Kb(NH3) = 10–pKb = 10–4.76 = 1.7 × 10–5 = (OH– (aq))(NH4

+(aq))

(NH3(aq))(H2O(aq))

⎛⎝⎜

⎞⎠⎟

eq

species OH–(aq) NH3(aq) NH4+(aq)

initial mol 1.0 × 10–7 0.250 -0- change in mol x formed x reacts x formed final mol x 0.250 – x x

Because Kb(NH3) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.250:

Kb = 1.7 × 10–5 = x ⋅ x

0.250 – x⎛⎝⎜

⎞⎠⎟

≈ x ⋅ x

0.250⎛⎝⎜

⎞⎠⎟

= x2

0.250




2.1× 10–3

0.250

⎛⎝⎜

⎞⎠⎟

·100% = 0.84%

Our rule of thumb is that a 5% error is acceptable and this result is within the limit. pOH = –log (OH-(aq)) = –log (2.1 × 10–3) = 2.68 pH = 14.00 – pOH = 14.00 – 2.68 = 11.32

Problem 9.21. The [H3O+(aq)] in a 0.350 M solution of aqueous hydrogen fluoride, HF, solution can be calculated from its equilibrium constant expression and equilibrium constant: HF(aq) + H2O(aq) ⇔ F–(aq) + H3O+(aq)

Ka(HF) = 6.46 × 10–4 = (H3O+(aq))(F– (aq))

(HF(aq))(H2O(aq))

⎛

⎝⎜

⎞

⎠⎟

eq

species H3O+(aq) F–(aq) HF(aq) initial mol 1.0 × 10–7 - 0 - 0.350 change in mol x formed x formed x reacts final mol x x 0.350 – x

Because Ka(HF) is relatively small, start by assuming that the value of x is small and can be neglected relative to 0.350:

Ka = 6.46 × 10–4 = x ⋅ x

0.350 – x

⎛⎝⎜

⎞⎠⎟ ≈ x ⋅ x

0.350

⎛⎝⎜

⎞⎠⎟ = x2

0.350


1.50 ×10–2

0.350

⎛

⎝⎜

⎞

⎠⎟ ·100% = 4.30%

Our rule of thumb is that a 5% error is acceptable and this result is just within the limit. This percentage also represents the percentage of the HF(aq) that transfers its protons to water.

Problem 9.22. Assume that the only source of hydronium ions in pH 3.4 lemon juice is transfer of protons from citric acid to water, H3Cit(aq) + H2O(aq) ⇔ H3O+(aq) + H2Cit–(aq), with Ka = 7.4 × 10-4. From the reaction stoichiometry, (H3O+(aq)) = (H2Cit–(aq)) = 10–pH = 10–3.4 = 4 × 10–4. We can use the equilibrium constant expression and solve to find (H3Cit(aq)) in the solution:

K = 7.4 × 10–4 = (H3O+(aq))(H2Cit– (aq))

(H3Cit(aq)) = (4 ×10–4 )2

(H3Cit(aq))

(H3Cit(aq)) = (4 ×10–4 )2

7.4 ×10–4 = 2 × 10–4 ∴[H3Cit(aq)] = 2 × 10–4 M



This is the amount of citric acid left undissociated. The total concentration of citric acid present in the solution is the concentration left unreacted, [H3Cit(aq)], plus the amount present as the anion, [H2Cit–(aq)], that is, 2 × 10–4 M + 4 × 10–4 M = 6 × 10–4 M. Another way to do this problem is to let the original concentration (before proton transfer) of citric acid be c M. Then the equilibrium concentration is c – [H3O+(aq)] M, so we have:

7.4 × 10–4 = (4 ×10–4)2

(c – (4 × 10–4 ))

c = (7.4 ×10–4)(4 ×10–4) + (4 ×10–4)2

(7.4 ×10–4) = 6 × 10–4 M

We see that both methods give the same result for the total concentration of citric acid in the solution (unreacted and monoanionic base form).

Problem 9.23. (a) The aluminum cation, Al3+, is relatively small (it has three fewer electrons than an aluminum atom) and has a high charge. The electric field produced by the ion is quite strong; water molecules can get fairly close to the ion and interact strongly with the field. When a water molecule is part of the hydration layer of the cation, the positive charge on the cation attracts electron density from the oxygen atoms toward the cation and makes the bonds between the oxygen atom and hydrogen atoms in the water much more polar than in the water molecule alone. Thus, it is easier for the hydrogen atoms on the waters in the hydration layer to be transferred to adjacent water molecules, than for proton transfer between water molecules in pure water. The result is that there is more proton transfer and higher hydronium ion concentration in the cation solutions than in pure water. The sodium ion (with only one fewer electrons than its atom) is a larger ion with lower charge than the aluminum ion. Water molecules are not nearly so strongly held by this larger ion with lower charge, so there is negligible effect on the hydronium ion concentration of the solution. (b) For the reaction, Al(H2O)6

3+(aq) + H2O(aq) ⇔ H3O+(aq) + Al(OH)(H2O)52+(aq), the acid

equilibrium constant expression is:

Ka = H3O

+(aq)( ) Al(OH)(H2O)52+(aq)( )

Al(H2O)63+(aq)( ) H2O(aq)( )

To find the numeric value for the equilibrium constant, we need to find the molar concentrations of the reactants and products in this equation. The pH of the solution, 2.9, gives us: [H3O+(aq)] = 10-pH =10–2.9 M = 1 × 10–3 M The stoichiometry of the reaction tells us that: [H3O+(aq)] = [Al(OH)(H2O)5

2+(aq)] = 1 × 10–3 M The solution contains 0.1 M aluminum ion, so, assuming all of this ion gets hydrated, the stoichiometry gives: [Al(H2O)6

3+(aq)] = (0.1 M) – [Al(OH)(H2O)52+(aq)] = (0.1 M – (1 × 10–3 M) = 0.1 M

At these low solute concentrations, the water concentration is essentially the same as in pure water, 55.5 M, so the dimensionless concentration term for water is unity. The numeric value for the equilibrium constant is:



Ka = 1 ×10–3( )1 ×10–3( )

0.1( ) 1( ) = 1 × 10–5 ; pKa = 5.0

(c) The smaller the value of pKa, the larger the value of Ka. Since the pKa for the Fe(III) ion, 2.5, is much smaller than that for Al(III), 5.0 [from part (b)] the acid equilibrium reaction for Fe(III) goes further toward products and produces more hydronium ions. Fe(III) solutions are more acidic than Al(III) solutions.

Problem 9.24. (a) For the reaction, HA(aq) + H2O(aq) ⇔ H3O+(aq) + A–(aq), the acid equilibrium constant expression is:

Ka = H3O

+(aq)( ) A– (aq)( )HA(aq)( ) H2O(aq)( )

= H3O

+(aq)( ) A– (aq)( )HA(aq)( )

For a 0.215 M HA solution, with a pH of 4.66, we have: (H3O+(aq)]) = 10–pH = 10–4.66 = 2.2 × 10–5 and the stoichiometry of the reaction tells us that: (H3O+(aq)) = (A–(aq)) = 2.2 × 10–5 It is easy to observe that much less than 5% of the original acid, [HA(aq)] = 0.215 M, has reacted.

Ka = 2.2 ×10–5( )2.2 ×10–5( )

0.215( ) = 2.3 × 10–9

(b) For the reaction, A–(aq) + H2O(aq) ⇔ OH–(aq) + HA(aq), the base equilibrium constant expression is:

Kb = OH– (aq)( ) HA(aq)( )

A– (aq)( )

We know that Ka·Kb = Kw = 1.00 × 10–14, so:

Kb = KwKa

= (1.00 ×10–14 )(2.3×10–9 )

= 4.4 × 10–6

The stoichiometry of the base reaction shows us that [OH–(aq)] = [HA(aq)], so, assuming that a negligible amount of A–(aq) reacts, we can rearrange the equilibrium constant expression to find the hydroxide concentration and pH of the solution: (OH–(aq))(HA(aq)) = (OH–(aq))2 = Kb·(A–(aq)) = (4.4 × 10–6)(0.175) = 7.7 × 10–7 (OH–(aq)) = 8.8 × 10–4; pOH = 3.06 pH = 14.00 – pOH = 14.00 – 3.06 = 10.94 Check the assumption that the value of (OH–(aq))x is small and can be neglected relative to 0.175:

8.8 ×10–4

.175

⎛

⎝⎜

⎞

⎠⎟ ·100% = 0.50%



Our rule of thumb is that a 5% error is acceptable and this result is well within the limit.

Problem 9.25. (a) For the reaction, HO(O)CC(O)O–(aq) + H2O(aq) ⇔ –O(O)CC(O)O–(aq) + H3O+(aq), pKa2 = 4.27 (“a2” because it is the second proton transferred from oxalic acid). Since pKa + pKb = 14.00 for a conjugate acid-base pair, the pKb for –O(O)CC(O)O–(aq) is 9.73. (b) The molar mass of disodium oxalate is 134 g·mol–1, so a solution containing 35 g of this salt in a liter of water (assumed to yield a liter of solution) gives [–O(O)CC(O)O–(aq)] = 0.26 M. The basic reaction in this solution is: –O(O)CC(O)O–(aq) + H2O(aq) ⇔ HO(O)CC(O)O–(aq) + OH–(aq) The stoichiometry gives [OH–(aq)] = [HO(O)CC(O)O–(aq)] and we can use the equilibrium constant value from part (a), with the assumption that a negligible amount of the oxalate ion reacts, to get the (OH–(aq)) and the pH of the solution:

Kb = 10–9.73 = 1.9 × 10–10 = OH– (aq)( ) HO(O)CC(O)O– (aq( )

– O(O)CC(O)O– (aq( ) = OH–

(aq)( )2

0.26( )

(OH–(aq)) = (1.9 ×10–10 )(0.26) = 7.0 × 10–6; pOH = 5.15

pH = 14.00 – pOH = 14.00 – 5.15 = 8.85 Checking the assumption, we see that 7.0 × 10–6 M (concentration of hydrogen oxalate ion) is much less than 5% of the original, 0.26 M, oxalate ion, so a negligible amount of the oxalate has reacted.

Problem 9.26. (a) For ease in writing, let HA = HC(O)OH (methanoic acid) and A– = HC(O)O– (methanoate anion). The proton transfer reaction, equilibrium reaction (for a dilute solution), and equilibrium constant are: HA(aq) + H2O(aq) ⇔ H3O+(aq) + A–(aq)

Ka = H3O

+(aq)( ) A– (aq)( )HA(aq)( ) H2O(aq)( )

= H3O

+(aq)( ) A– (aq)( )HA(aq)( )

= 1.8 × 10-4

From the reaction stoichiometry we see that (H3O+(aq)) = (A–(aq)), as in several of the previous problems. Substitute this equality, rearrange the equilibrium constant equation, and solve for (H3O+(aq)): (H3O+(aq))2 = Ka·( HA(aq))

(H3O+(aq)) = Ka ⋅ (HA(aq))

Assuming that a negligible amount of HA(aq) transfers its protons to water, we can begin with the approximation that (HA(aq)) = c = initial concentration of HA(aq) in the solution, so:

(H3O+(aq)) = Ka ⋅ c

In the present example, we have:

(H3O+(aq)) = Ka ⋅ c = (1.8 ×10–4 )(0.100) = 4.2 × 10–3



pH = –log(H3O+(aq)) = –log(4.2 × 10–3) = 2.38 We can check the assumption (approximation) and get the percent ionization of the acid in one operation:

% ionization = 4.2 ×10–3

0.100

⎛

⎝⎜

⎞

⎠⎟ ·100% = 4%

Although this result is within the limit we have set (5%) for accepting the approximation that a negligible amount of acid has transferred its protons, we should check to see whether this much ionization affects the result we get for (H3O+(aq)). One way to do this is to assume that 4% of the acid reacts and use the lower concentration in the equation to see what (H3O+(aq)) is:

(H3O+(aq)) = (1.8 ×10–4 )(0.096) = 4.1 × 10–3 Within the precision of the data, this result is the same as the previous one, so the 4% ionization result is acceptable. (b) Le Chatelier’s principle predicts that disturbing an equilibrium system by decreasing the concentration of a reactant results in the system responding to form more of the reactant, thereby decreasing the concentration of the product(s). Le Chatelier’s principle is, however, powerless to make a quantitative prediction of the result, so the best we can say is that the [H3O+(aq)] and [A–

(aq)] will be lower when [HA(aq)] is lower. (c) Begin by using the simplest approach and then checking to see whether the results make sense or whether further work is necessary. For a 0.050 M solution of HA(aq), we find:

(H3O+(aq)) = (1.8 ×10–4 )(0.050) = 3.0 × 10–3 Our prediction of a lower concentration of H3O+(aq) [and A-(aq)] is borne out. The percent ionization is:

% ionization = 3.0 ×10–3

0.050

⎛

⎝⎜

⎞

⎠⎟ ·100% = 6%

This result, 6% ionization, is larger than for the more concentrated acid, a result that Le Chatelier’s principle could not predict and is also larger than we usually are willing to accept without trying a less approximate calculation. We should again try what we did in part (b), substitute this tentative result back into the equation to see how it affects the outcome:

(H3O+(aq)) = (1.8 ×10–4 )(0.047) = 2.9 × 10–3 Again we find that our result, within the precision of the data, is not changed, so we can have reasonable confidence that our conclusion is valid: as the concentration of a weak acid is decreased, the percent that ionizes increases.

Problem 9.27. In essentially all the problems we have presented, the concentration of hydronium ion produced by the transfer of protons from an acid is a good deal larger than 10–7 M, the maximum concentration of hydronium ions that water autoionization can produce. In addition, the presence of the hydronium ion from the acid decreases the amount of hydronium ion produced by water autoionization because the added hydronium ion is a disturbance to the autoionization equilibrium and the response of the system is to react to try to use up the added product which reduces the amount of hydroxide ion and hydronium ion formed by autoionization. If the concentration of hydronium ion from the acid we add to the system is very low, comparable to



10–7 M, then we have to account for the autoionization, because it will be adding a significant amount of hydronium ion to the solution. The conditions under which we have to be account for the autoionization of water can be determined by looking at the equation developed in the solution for Problem 9.26(a): (H3O+(aq)) = Ka ⋅ c . When the added concentration of acid is very low and/or when the acid dissociation constant for the added acid is very small, the amount of hydronium ion from the added acid will be tiny and could be comparable to 10–7 M.

Problem 9.28. A solution contains a buffer system, if significant concentrations of the protonated and unprotonated forms of a weak acid-base conjugate pair are present. In case (i), a solution of HClO4 and NaClO4, no significant concentration of the protonated form, HClO4(aq) exists in the solution, because HClO4(aq) is a strong acid and transfers essentially all its protons to water. In case (ii), a solution of Na2CO3 and NaHCO3, the HCO3

–(aq) anion is a weak acid (also a weak base that is not relevant to this discussion) that is in equilibrium with its conjugate base: HOCO2

–(aq) + H2O(aq) ⇔ H3O+(aq) + CO32–(aq)

Since the solution contains both the conjugate acid and conjugate base, this equilibrium reaction provides buffering action.

Problem 9.29. The reaction that is relevant to this problem is: HOAc(aq) + H2O(aq) ⇔ H3O+(aq) + OAc–(aq) If NaOAc(s) is added to an aqueous solution of acetic acid, the concentration of OAc–(aq) will increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so the system will react by trying to decrease the effect of the disturbance, that is, by shifting toward reactants to use up some of the added product. The result will also be to decrease the concentration of H3O+(aq), because it is required to react with the added acetate ion. The result of addition of the conjugate base to a weak acid-base equilibrium system is to lower the (H3O+(aq)), that is, raise the pH of the solution.

Problem 9.30. The reaction that is relevant to this problem is: NH3(aq) + H2O(aq) ⇔ NH4

+(aq) + OH–(aq) If NH4Br(s) is added to an aqueous solution of ammonia, the concentration of NH4

+(aq) will increase as the solid dissolves. Adding a product is a disturbance to the equilibrium system, so the system will react by trying to decrease the effect of the disturbance, that is, by shifting toward reactants to use up some of the added product. The result will also be to decrease the concentration of OH–(aq), because it is required to react with the added ammonium ion. The result of addition of the conjugate acid to a weak acid-base equilibrium system is to lower the (OH–(aq)), that is, lower the pH of the solution.

Problem 9.31. The reaction that is relevant to this problem is: HOAc(aq) + H2O(aq) ⇔ H3O+(aq) + OAc–(aq)



Because HOAc(aq) is a weak acid, we know that only a tiny amount of it will transfer its protons to water in a 0.250 M solution, so (H3O+(aq)) = (OAc–(aq)) will be rather small in this solution. Using the equation in the solution for Problem 9.26 we have:

(H3O+(aq)) = Ka ⋅ c = (1.7 ×10–5 )(0.250) = 2.1 × 10–3

Addition of sufficient solid NaOAc to make the final sodium ion concentration, [Na+(aq)], 1.50 M will enormously increase (OAc–(aq)) and decrease the amount of HOAc(aq) that reacts by the above equation. We can start with the approximation that (HOAc(aq)) and (OAc–(aq)) have the values, 0.250 and 1.50, respectively, they would have, if no proton transfers took place. Thus, we have:

Ka = 1.7 × 10-5 = H3O

+(aq)( ) OAc– (aq)( )HOAc(aq)( )

≈ H2O(aq)( ) 1.50( )

0.250

⎛

⎝⎜

⎞

⎠⎟

(H3O+(aq)) = 2.8 × 10-6; pH = 5.55 As we said, addition of the OAc–(aq reduces the (H3O+(aq)) (by a factor of about 1000 in this case), which is also the direction of the change predicted in the solution for Problem 9.29. The hydronium ion that is no longer present reacted with the added OAc–(aq), the amount of OAc–

(aq) reacted is about 0.14% = (2.1×10–3)(1.50)

⎛⎝⎜

⎞⎠⎟ ⋅100%

⎡⎣⎢

⎤⎦⎥, so a negligible amount has been used

up and our assumptions about the concentrations of the conjugate acid and conjugate base are valid.

Problem 9.32. If x = (OAc–(aq)) in a 0.25 M acetic acid-acetate buffer solution, then (HOAc(aq)) = (0.25 – x) in this solution. If the pH = 5.36, then (H3O+(aq)) = 4.4 × 10-6. Substitute these values into the equilibrium constant expression and solve for x to find the concentrations of acetate and acetic acid that will produce this buffer solution:

Ka = 1.7 × 10-5 = x( ) 4.4 ×10-6( )

0.25 − x

⎛

⎝⎜⎜

⎞

⎠⎟⎟

x = (OAc–(aq)) = 0.20; [OAc–(aq)] = 0.20 M (HOAc(aq)) = (0.25 – x) = 0.05 [HOAc(aq)] = 0.05 M

Problem 9.33. (a) The equilibrium reaction, equilibrium constant expression, and equilibrium constant for lactic acid transferring a proton to water are: HC3H5O3(aq) + H2O(aq) ⇔ H3O+(aq) + C3H5O3

–(aq)

Ka = 1.4 × 10–4 = H3O

+(aq)( ) C3H5O3– (aq)( )

HOC3H5O2 (aq)( ) H2O(aq)( ) =

H3O+(aq)( ) C3H5O3

– (aq)( )HOC3H5O2 (aq)( )

For a buffer solution composed of 0.15 M lactic acid, HOC3H5O2, and 0.10 M sodium lactate, NaOC3H5O2, we begin by making the approximation that these concentrations are negligibly affected by the proton transfers, and solve the equilibrium constant expression to find (H3O+(aq)):



1.4 × 10–4 = H3O

+(aq)( ) 0.10( )0.15( )

(H3O+(aq)) = 2.1 × 10–4; pH = –log(2.1 × 10–4) = 3.68 Because (H3O+(aq)) is small compared to the initial concentrations of the conjugate acid-base pair, the approximation that their concentrations are not appreciably affected is valid. An alternative (equivalent) solution is to use the Henderson-Hasselbalch equation [with pKa = -logKa = –log(1.4 × 10–4) = 3.85]:

pH = pKa + logconjugate base( )conjugate acid( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 3.85 + log 0.10

0.15

⎛⎝⎜

⎞⎠⎟ = 3.85 + (–0.18) = 3.67

Within the round-off uncertainty of the calculations, the pH calculated by each route is the same. (b) Half way to the equivalence point in the titration between 0.15 M lactic acid and 0.15 M NaOH, the concentration of the lactic acid and its conjugate base will be equal, and the pH will equal the pKa, 3.85, as we see from the Henderson-Hasselbalch equation when (conjugate base) = (conjugate acid):

pH = pKa + logconjugate base( )conjugate acid( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 3.85 + log

conjugate base( )conjugate base( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 3.85 + log(1)

= 3.85 + 0 = 3.85

In part (a), where there was more conjugate acid than conjugate base, the pH of the buffer solution was lower, more acidic. The solution had a higher concentration of hydronium ion than in the solution here.

Problem 9.34. (a) This is a buffer solution containing a weak acid, HA = phenol, pKa = 9.98, and its conjugate base, A– = phenolate ion. The pH of the solution is:

pH = pKa + log (A– (aq))

(HA(aq))

⎡

⎣⎢

⎤

⎦⎥ = 9.98 + log

0.377

0.551⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 9.81

Note that this solution with a higher concentration of the conjugate acid than the conjugate base has a lower numeric value of pH (more acidic) than the numeric pKa value. (b) Assume that all the hydronium ion produced by the added HCl (0.100 M in the 1 L of solution) reacts with the phenolate anion to produce phenol. The new concentration of the phenolate anion will be 0.227 M (= 0.377 M – 0.100 M) and the concentration of phenol will be 0.651 M (= 0.551 M + 0.100 M). The new pH will be:

pH = 9.98 + log0.277

0.651⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 9.61

As we would expect, the addition of hydronium ion (from the HCl) lowers the pH, but by only 0.20 pH units in this buffered solution. (c) Assume that all the added hydroxide ion (0.125 M in 1 L of solution) from the NaOH reacts with phenol to produce phenolate anion. The new concentration of phenolate anion will be 0.502



M (= 0.377 M + 0.125 M) and the concentration of phenol will be 0.426 M (= 0.551 M – 0.125 M). The new pH will be:

pH = 9.98 + log0.502

0.426⎡ ⎣ ⎢

⎤ ⎦ ⎥ = 10.05

As we would expect, the addition of hydroxide ion (from the NaOH) raises the pH, but by only 0.24 pH units in this buffered solution.

Problem 9.35. (a) If the density of bleach and water are the same, then 1 L of bleach has a mass of 1.00 kg and 5% of this, about 50 g, is NaOCl. The molar mass of NaOCl is 74.5 g·mol–1, so the molarity of NaOCl in the solution and, hence, the concentration of the aqueous anion is:

[OCl–(aq)] = 50 g ⋅ L–1

74.5 g ⋅ mol–1 = 0.7 M

Only one significant figure is justified by the data (5% solution). (b) The reaction of the hypochlorite ion (a weak base) with water is: OCl–(aq) + H2O(aq) ⇔ HOCl(aq) + OH–(aq) The equilibrium constant expression for this reaction is:

Kb = OH– (aq)( ) HOCl(aq)( )OCl– (aq)( ) H2O(aq)( )

The symbol, Kb, we have used is the usual designation for the equilibrium constant that refers to the reaction of a base with water to form hydroxide ion. The dimensionless concentration ratio for water is included in the expression, but often (usually) it is left out, since, in dilute solutions, it is unity. The numeric value for Kb is easily obtained from the pKa = 7.75 value for hypochlorous acid, the conjugate acid of the hypochlorite anion: pKb = 14.00 – pKa = 6.25; Kb = 10–6.25 = 5.6 × 10–7

(c) Assuming that a negligible amount of the hypochlorite anion reacts and that (HOCl(aq)) = (OH–(aq)) in the resulting solution, we can write: (OH–(aq))2 = Kb·(OCl–(aq)) = (5.6 × 10–7)(0.7) = 4 × 10–7 (OH–(aq)) = 6 × 10–4

(H3O+(aq)) = 1.00 ×10–14

6 ×10–4 = 2 × 10–11; pH = –log(2 × 10–11) = 10.8

(d) To lower the pH of the bleach solution to 6.5, you would have to add hydrochloric acid to react with the hypochlorite ion to convert much of it to hypochlorous acid. This target pH is a lower numeric value than the pKa for hypochlorous acid, so the ratio of conjugate base to conjugate acid (assuming this is a buffer solution) is less than unity. (e) If we assume (as in part (d)) that the solution of hypochlorous acid and hypochlorite ion is a buffer, we can use the equation for the pH of a buffer solution (the Henderson-Hasselbalch equation) rearranged to find the ratio of the base to acid:



pH – pKa = log (OCl– (aq))

(HOCl(aq))

⎡

⎣⎢

⎤

⎦⎥ = 6.5 – 7.75 = –1.25

(OCl– (aq))

(HOCl(aq))

⎡

⎣⎢

⎤

⎦⎥ = 10–1.25 = 0.06

Just as we reasoned in part (d), the ratio of conjugate base to conjugate acid is quite low; almost all the hypochlorite ion has to be converted to its acid form to get the pH so low.

Problem 9.36. We will rearrange the Henderson-Hasselbalch equation to find the (HO)2CO(aq) concentration in human arterial blood, given that the pH of blood is 7.41, the HOCO2

–(aq) concentration is 26 × 10-3 M, and Ka1 for (HO)2CO(aq) is 8.0 × 10-7 at 37 ° C [pKa1 = 6.10].

pH – pKa1 = logHOCO2

– (aq)( )(HO)2CO(aq)( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

7.41– 6.10 = 1.31 = log 26 ×10–3

(HO)2CO(aq)( )⎛

⎝⎜⎜

⎞

⎠⎟⎟

26 ×10–3

(HO)2CO(aq)( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ = 101.31 = 2.0 × 101

((HO)2CO(aq)) = 1.3 × 10-3; [(HO)2CO(aq)] = 1.3 × 10-3 M

`Problem 9.37. (a) The pKa for acetic (ethanoic) acid, HOAc(aq), is 4.76. The desired pH is not too different, so we will be making a buffer solution by adding 6.0 M hydrochloric acid to 250. mL of 0.10 M aqueous sodium acetate solution to give a solution with a pH of 4.30. We are starting with a solution containing 0.10 M acetate anion, OAc–(aq), and adding hydronium ions (from the 6.0 M hydrochloric acid solution) to convert some of the acetate anion to acetic acid. We can use the buffer equation (Henderson-Hasselbalch equation) to determine the ratio of acetate anion to acetic acid necessary to obtain the desired pH and then calculate how much hydrochloric acid is needed to convert the requisite number of moles of acetate anion to acetic acid to give this ratio:

pH – pKa = log (OAc– (aq))

(HOAc(aq))

⎡

⎣⎢

⎤

⎦⎥ = 4.30 – 4.76 = –0.46

(OAc– (aq))

(HOAc(aq))

⎡

⎣⎢

⎤

⎦⎥ = 10–0.46 = 0.35

The values in this ratio can have any convenient units, so we will choose to use moles. The original solution contains 0.0250 mol acetate [= (0.250 L)·(0.100 M)]. If we form x mol of acetic acid, then we will have left (0.0250 – x) mol of acetate, and we use the ratio to find x:

(OAc– (aq))

(HOAc(aq))

⎡

⎣⎢

⎤

⎦⎥ = 0.35 =

(0.0250 – x) mol

x mol; x = [HOAc(aq)] = 0.0185

mol



In order to form 0.0185 mol of acetic acid, we have to add 0.0185 mol of hydronium ion (hydrochloric acid) to the original solution. The volume of 6.0 M hydrochloric acid required is:

volume 6.0 M HCl solution = 0.0185 mol

6.0 mol ⋅ L–1 = 0.0031 L

(b) The addition of the HCl solution brings the volume of the solution to 0.253 L, so we get:

[HOAc(aq)] = 0.0185 mol

0.253 L = 0.073 M

[OAc–(aq)] = (0.0250 – 0.0185) mol

0.253 L = 0.026 M

Note that we started with a solution that was 0.100 M in acetate anion and end with one that has been diluted by about one percent (250 to 253 mL) in which the sum of the molarities of acetate anion and acetic acid is 0.099 M.

Problem 9.38. Recall that the best buffers are those for which the pKa of the conjugate acid used is close to the desired buffer pH. In Table 9.2 we see that pKa1, pKa2, and pKa3 for phosphoric acid are, respectively, 2.15, 7.20, and 12.15. These values are close to the pHs of the buffers we are asked to make, so it looks like we can use conjugate acid-base pairs of phosphoric acid and/or its salts to prepare these buffers. The reactions to which each of these pKa values refers are these (with the (aq) omitted for simplicity): (1) pKa1 = 2.15 (HO)3PO + H2O ⇔ (HO)2PO2

– + H3O+ (2) pKa2 = 7.20 (HO)2PO2

– +H2O ⇔ (HO)PO32– + H3O+

(3) pKa3 = 12.15 (HO)PO32– + H2O ⇔ PO4

3– + H3O+ From the reagents available, we could use these conjugate acid-base combinations for the buffers: (1) pH 2.00 (HO)3PO(aq) and Na(HO)2PO2(s) (2) pH 7.00 Na(HO)2PO2(s) and K2(HO)PO3(s) (3) pH 12.00 K2(HO)PO3(s) and Na3PO4(s)

Problem 9.39. The isoelectric point (or isoelectric pH) is the pH of the solution in which a protein has a net charge of zero. Within the protein, the net positive and net negative electric charges are equal, so the overall net charge is zero. Generally, the protein will be least soluble at its isoelectric point and will not move in an electrophoretic analysis. The charges on a protein are caused by loss or gain on protons by acidic and basic side groups, respectively. The pH of the solution in which the protein is dissolved controls the gain and loss of protons by the side groups through its effect on the equilibria that characterize each of the acid-base side groups.

Problem 9.40. (a) The equilibrium reactions and equilibrium constant expressions for glycine, leaving off the (aq) designations and assuming that (H2O) = 1, are: +H3NCH2C(O)OH + H20 + ⇔ H3O+ + +H3NCH2C(O)O–



K1a = (+ H3NCH2C(O)O– )(H3O

+ )

(+ H3NCH2C(O)OH)

+H3NCH2C(O)O– + H20 + ⇔ H3O+ + H2NCH2C(O)O–

K2a = (H2NCH2C(O)O– )(H3O

+ )

(+ H3NCH2C(O)O– )

(b) At the isoelectric pH, (+H3NCH2C(O)OH) = (H2NCH2C(O)O–). Multiplying the two equilibrium constant expressions and canceling identical and equal terms gives:

K1a·K2a = (+ H3NCH2C(O)O– )(H3O

+ )

(+ H3NCH2C(O)OH)·(H2NCH2C(O)O– )(H3O

+ )

(+ H3NCH2C(O)O– ) = (H3O+)2

(H3O+) = K1a ⋅ K2a( ) = K1a ⋅ K2a( )12

isoelectric pH = –log(H3O+) = –log K1a ⋅ K2a( )12 = 1

2 –logK1a( )+ –logK2a( )⎡⎣ ⎤⎦

= 12 pK1a + pK2a( ) = 1

2 2.35 + 9.78( ) = 6.07

(c) The graphic in the Web Companion, Chapter 9, Section 9.5, page 3, is simplified somewhat to indicate that all of the dipeptide goes to the zero net charge form at the isoelectric pH. In fact, small, but equal, amounts of both the net positive and net negative forms also are present at this point, as we assumed in part (b) for glycine. The dipeptide is like glycine in that it has an amino and a carboxylic acid functional group. The mathematics to find the isoelectric pH is identical for the dipeptide and glycine and the result is the same: the isoelectric pH is “half the sum of the pKa values of the two acid-base groups.”

Problem 9.41. The large open circles in this sketch represent hemoglobin molecules with the positions of the two amino acids that are changed from glutamate to valine in HbS denoted by the dark dots. The red lozenge shapes represent water molecules.

1)

2)

3)



In normal hemoglobin, HbA, water molecules interact with the carboxylate anion of the glutamate side groups by hydrogen bonding, a relatively strong interaction that helps keep the protein in solution. When the glutamate is replaced by valine, which has a nonpolar hydrocarbon side group water molecules are oriented around the nonpolar side group as they are around any nonpolar hydrocarbon, but the intermolecular forces holding the water and side group together are relatively weak dispersion interactions. If the nonpolar side groups on two HbS molecules come together and “squeeze out” the water molecules between them, the entropy of the system is increased by the release of the water molecules that are now free, as shown in line 2) of the sketch, to move about anywhere in the solution. This is the same kind of interaction as is represented in Figures 8.20 and 8.23 and in the Web Companion, Chapter 8, Section 8.13, to show the formation of micelles and bilayer membranes. This process can continue, as shown in line 3) of the sketch, to form long strands of HbS molecules that are responsible for the crescent shape of the red blood cells in sickle-cell disease. Thus, the clumping of HbS molecules is largely driven by an increase in entropy of the water.

Problem 9.42. (a) Since protein B moves further toward the positive end of the gel, we can conclude that this protein has the higher net negative charge. Table 9.2 shows that the pKa for the histidine side group is 6.00, so, at pH 8.6 (the gel pH), it will be in its basic, uncharged form. The cysteine side group has a pKa = 8.36, so most of it will also be in its basic, negatively-charged form. The protein containing the cysteine will have a more negative charge at pH 8.6, so it is protein B that contains the cysteine and A that contains the histidine. (b) You can tell which protein is which from the electrophoresis data at pH 8.6. To help confirm your identification and your assumption about the amino acids, you could run an electrophoresis analysis at a more acidic pH such as 5.6, where the histidine side group would be positively charged and the cysteine side group would be uncharged. If the assumption and identification are correct, protein A would be more attracted to the negative end of the gel than would protein B, so protein A would be further toward the negative end of the gel. Note that this is the same as saying protein B would be further toward the positive end. The relative positions of the proteins are not changed by the change in pH, but the result is consistent with our assumption and identification, so this is further evidence that they are correct.

Problem 9.43. In sickle-cell disease, the substitution of an amino acid with a non-polar side for one with a polar side group changes the solubility of the hemoglobin, by making it easier for the protein molecules to stick to one another and precipitate, causing the sickling of the red blood cell. Similarly, in Investigate This 9.42, as the net charge on the casein molecules changes with pH, the protein molecules are better able to stick together and precipitate, as the hemoglobin molecules do in sickle-cell disease.

Problem 9.44. (a) A mixture of Fe(III) and Co(II) ions in an 8 M solution of hydrochloric acid was analyzed by paper electrophoresis and, after several minutes of electrophoresis, a yellow spot on the paper was closer to the positive end of the paper than a blue spot. Since the yellow spot, Fe(III) chloride complex, is closer to the positive end of the paper than the blue spot, Co(II) chloride complex, we can conclude that the net charge on the Fe(III) complex is more negative than the net charge on the Co(II) complex. This means that the Fe(III) complex has to contain at least two



more negative chloride ions than the Co(II) complex. One of these extra chlorides compensates for the difference in charge between the two cations (FeCl2+ and Co2+ have the same net charge) and the second makes the Fe(III) complex more negative. If our explanation for the colors in Investigate This 9.8 is correct, the Co(II) complex is CoCl4

2-(aq), so the Fe(III) complex must be FeCl6

3–(aq). (b) One of the most common complex-ion structures is the octahedral arrangement of the ligands about the central cation. Thus, the structure of FeCl6

3–(aq) is likely to be like the structure shown in Figure 6.5 with the water molecules replaced by chloride ions. For the CoCl4

2–(aq) complex, two structures are possible, either a square planar arrangement like that shown for the Pt(II) complexes in Figure 6.8 or a tetrahedral arrangement like that shown around Zn(II) in Figure 6.9. Other data indicate that the Co(II) complex is tetrahedral, which we might have predicted from its position near Zn in the periodic table.

Problem 9.45. (a) The equilibrium reaction, equilibrium constant expression, and equilibrium constant for the dissolution of barium sulfate are: BaSO4(s) ⇔ Ba2+(aq) + SO4

2-(aq)

Keq = Ba2+(aq)( ) SO4

2– (aq)( )BaSO4 (s)( )

⎛

⎝⎜

⎞

⎠⎟

eq

= [(Ba2+(aq))(SO42-(aq))]eq = Ksp = 1.1 × 10-10

From the stoichiometry of the reaction, we know that (Ba2+(aq)) = (SO42-(aq)) = s, where s is

numeric value of the molar solubility of the solid ionic compound. Substitution into the solubility product equation gives: Ksp = 1.1 × 10–10 = s2 s = 1.0 × 10–5 = (Ba2+(aq)) = (SO4

2-(aq)) For the rest of the problem, we will use the solubility products and the stoichiometries to find the concentrations of the ions in solution. (b) Ag2S(s) ⇔ 2Ag+(aq) + S2-(aq) (Ag+(aq)) = 2s; (S2-(aq)) = s Ksp = 8 × 10–51 = [(Ag+(aq))2(S2-(aq))]eq = (2s)2(s) = 4s3 s = 1.3 × 10–17 = (S2-(aq)); (Ag+(aq)) = 2.6 × 10–17 The uncertainty in the data (Ksp) justify only one significant figure in the results. Two are shown to make clear that the concentration of silver ion is twice that of sulfide ion in the solution. (c) Mg(OH)2(s) ⇔ Mg2+(aq) + 2OH-(aq) (Mg2+(aq)) = s; (OH-(aq)) = 2s Ksp = 7.1 × 10–12 = [(Mg2+(aq))(OH-(aq))2]eq = (s)(2s)2 = 4s3 s = 1.2 × 10–4 = (Mg2+(aq)); (OH-(aq)) = 2.4 × 10–4 (d) PbBr2(s) ⇔ Pb2+(aq) + 2Br-(aq) (Pb2+(aq)) = s; (Br-(aq)) = 2s Ksp = 2.1 × 10–6 = [(Pb2+(aq))(Br-(aq))2]eq = (s)(2s)2 = 4s3 s = 8.1 × 10–3 M = (Pb2+(aq)); (Br–(aq)) = 1.6 × 10–2 M



Problem 9.46. To find the moles and mass of solid calcium carbonate, CaCO3(s), that will be obtained if 100. mL of a solution saturated with calcium carbonate is evaporated to dryness, calculate the molar solubility of the solid, s, in water, convert to moles in 100. mL, and then to mass of CaCO3(s). CaCO3(s) ⇔ Ca2+(aq) + CO3

2-(aq) (Ca2+(aq)) = (CO3

2-(aq)) = s Ksp = 6.0 × 10–9 = [(Ca2+(aq))(CO3

2-(aq))]eq = (s)(s) = s2 s = 7.7 × 10–5 The molar solubility is 7.7 × 10–5 M, so the number of moles of CaCO3(s) dissolved in 100. mL is: mol in 100. mL = (7.7 × 10–5 mol·L-1)(0.100 L) = 7.7 × 10–6 mol The molar mass of CaCO3 = 100 g·mol-1 , so the mass of CaCO3(s) dissolved in 100. mL is: mass CaCO3(s) = (100 g·mol-1)(7.7 × 10–6 mol) = 7.7 × 10–4 g = 0.77 mg

Problem 9.47. To calculate the mass of calcium fluoride, CaF2(s), that will dissolve in 100. mL of water (assuming no volume change), calculate the molar solubility, s, and convert to mass (pKsp of CaF2 = 10.57): CaF2(s) ⇔ Ca2+(aq) + 2F–(aq) (Ca2+(aq)); = s (F–(aq)) = 2s Ksp = 10–10.57 = 2.7 × 10-11 = [(Ca2+(aq))(F–(aq))2]eq = (s)(2s)2 = 4s3 s = 1.9 × 10–4 The molar solubility is 1.9 × 10–4 M, so the number of moles of CaF2(s) dissolved in 100. mL is: mol in 100. mL = (1.9 × 10–4 mol·L-1)(0.100 L) = 1.9 × 10–5 mol The molar mass of CaF2 = 78 g·mol-1 , so the mass of CaF2(s) dissolved in 100. mL is: mass CaF2(s) = (78 g·mol-1)(1.9 × 10–5 mol) = 1.5 × 10–3 g = 1.5 mg

Problem 9.48. Assuming that the dissolution of 5.8 × 10–5 g of PbCrO4(s) in exactly 1 L of water does not change the volume of liquid, we can calculate the molar solubility for PbCrO4(s). Assuming complete dissociation of the ionic compound into ions, so that [Pb2+(aq)] = [CrO4

2–(s)] = molar solubility of the solid, we can calculate the Ksp for the dissolution. The molar mass of PbCrO4 is 323.2 g·mol–1.

molar solubility = 5.8 × 10–5 g ⋅ L–1

323.2 g ⋅ mol–1 = 1.8 × 10–7 mol·L–1

Ksp = (Pb2+(aq))(CrO42–(s)) = (1.8 × 10–7)2 = 3.2 × 10–14

Problem 9.49. To do this problem, assume that the dissolution of Fe(OH)2(s) is a simple solubility represented by this reaction equation:



Fe(OH)2(s) ⇔ Fe2+(aq) + 2OH–(aq) Solutions containing transition metal ions are usually more complicated than this with complexes (other than the aquo complex) often accounting for much of the metal ion species in the solution. In the absence of any information about complexes in this system, we will ignore them and assume that Fe(OH)2(s), Fe2+(aq) and OH–(aq) are the only species we need to account for in these solutions.

(a) For dissolution in water, we use the stoichiometry of the reaction and Ksp = 7.0 × 10–16, to find the solubility, (Fe2+(aq)), (OH–(aq)), and pH. (Fe2+(aq)); = s (OH–(aq)) = 2s Ksp = 7.0 × 10–16 = [(Fe2+(aq))(OH–(aq))2]eq = (s)(2s)2 = 4s3 s = 5.6 × 10-6 = (Fe2+(aq)); (OH-(aq)) = 1.12 × 10-5 ; pOH = 4.95 pH = 14.00 – pOH = 14.00 – 4.95 = 9.05

(b) For dissolution in 0.25 M NaOH solution, we use Ksp and the solubility product, with the (OH–(aq)) provided by the NaOH, to find the solubility, (Fe2+(aq)), and pH. (Fe2+(aq)); = s (OH–(aq)) = 0.25 Ksp = 7.0 × 10–16 = [(Fe2+(aq))(OH–(aq))2]eq = (s)(0.25)2 = 0.0625s s = 1.1 × 10-14 = (Fe2+(aq)); pOH = 0.60 pH = 14.00 – pOH = 14.00 – 0.60 = 13.40 As we would anticipate, the solubility of Fe(OH)2(s) in a solution that already contains hydroxide ion is much lower than in pure water. Hydroxide from this dissolution is about 13 orders of magnitude less than that from the NaOH.

(c) For dissolution in 0.25 M Fe(NO3)2 solution, we use Ksp and the solubility product, with the (Fe2+(aq)) provided by the Fe(NO3)2, to find the solubility, (OH–(aq)), and pH. (Fe2+(aq)); = 0.25 (OH–(aq)) = 2s Ksp = 7.0 × 10–16 = [(Fe2+(aq))(OH–(aq))2]eq = 0.25(2s)2 = 1.00s2 s = 2.6 × 10-8; (OH-(aq)) = 2s = 5.2 × 10-8 ; pOH = 7.28 pH = 14.00 – pOH = 14.00 – 7.28 = 6.72 Note that there is something peculiar about these results. The pH of the solution is less than the pH of pure water, so there is less hydroxide ion in the solution than in pure water. The analysis suggests that the Fe2+(aq) in a solution that is 0.25 M in Fe2+(aq) should react with hydroxide ion from the water self ionization to precipitate a little Fe(OH)2(s). This result shows that this system is more complex than our analysis indicates.

Problem 9.50. A precipitate will form when 0.0025 mol of Ba2+(aq) (from Ba(NO3)2) and 0.00030 mol of CO3

2–

(aq) (from Na2CO3) are mixed in 250. mL of distilled water, if the product of the ion concentrations in this solution exceeds the solubility product constant for dissolution of BaCO3(s). Calculate (Ba2+(aq)) and (CO3

2–(aq)), find the ion product, and compare it with the solubility product constant, Ksp = 6.0 × 10-9, for BaCO3(s) dissolution.

[Ba2+(aq)] = 0.0025 mol Ba2+(aq)

0.250 L = 1.0 × 10–2 M



[CO32– (aq)] = 0.0030 mol CO3

2– (aq)

0.250 L = 1.2 × 10–2 M

ion product = (Ba2+(aq))( CO32– (aq)) = (1.0 × 10–2)( 1.2 × 10–2) = 1.2 × 10–4

The ion product far exceeds Ksp (by about 5 orders of magnitude) so a precipitate will form in this case.

Problem 9.51. The minimum concentration of sulfate anion, [SO4

2–(aq)], required to begin the precipitation of either 2.5 × 10-3 M Pb2+(aq) or 0.05 M Ca2+(aq) is the concentration that will make the ion product just equal the solubility product constant for PbSO4(s) or CaSO4(s). Calculate the (SO4

2–

(aq)) required for each cation concentration and the one that requires the lower concentration will be the one to precipitate first. Ksp(PbSO4) = 6.3 × 10-7 = [(Pb2+(aq))(SO4

2-(aq))]eq = (2.5 × 10-3)( SO42-(aq))

( SO42-(aq)) = 2.5 × 10-4

Ksp(CaSO4) = 2.4 × 10-5 = [(Ca2+(aq))(SO42-(aq))]eq = (0.05)( SO4

2-(aq)) ( SO4

2-(aq)) = 4.8 × 10-4 Therefore, PbSO4(s) will form once [SO4

2-(aq)] reaches about 2.5 × 10-4 M.

Problem 9.52. (a) To determine how much Ba2+(aq) is in 200 mL of a saturated barium sulfate solution, we use the solubility product, Ksp = 1.1 × 10–10, to find the molar solubility (or its equivalent) and then convert to mass of Ba2+(aq) [from the stoichiometry of the solubility equilibrium, (Ba2+(aq)) = (SO4

2–(aq))]: Ksp = 1.1 × 10–10 = (Ba2+(aq))(SO4

2–(aq)) = (Ba2+(aq))2 (Ba2+(aq)) = 1.0 × 10–5; [Ba2+(aq)] = 1.0 × 10–5 M mol Ba2+(aq) = (1.0 × 10–5 M)(0.200 L) = 2.1 × 10–6 mol mass Ba2+(aq) = (2.1 × 10–6 mol)(137.3 g·mol–1) = 2.9 × 10–4 g = 0.29 mg (in 200 mL)

(b) The liquid the patient drinks for x-ray analysis contains a large amount of solid barium sulfate suspended in water, but the amount of barium cation in the solution is, as part (a) shows, quite small, so it is below the toxic limit for humans.

Problem 9.53. Possible ways to decrease the [Ba2+(aq)], in order to minimize allergic discomfort when ingesting the BaSO4 solution for x-ray analysis, are suggested and require a choice. (i) Heating the saturated solution is not a good approach for decreasing the [Ba2+(aq)] in the ingested solution, because the dissolution, BaSO4(s) ⇔ Ba2+(aq) + SO4

2-(aq), is endothermic. Since energy is required to cause the dissolution, adding energy will increase the solubility (Le Chatelier’s principle) and increase [Ba2+(aq)]. This is just the opposite of the effect we wish. (ii) Adding sodium sulfate, , to the solution is a reasonable approach for decreasing the [Ba2+(aq)] in the ingested solution. Because SO4

2-(aq) is a product of the dissolution, increasing its concentration is a disturbance to the system, which will react to try to use up some of the



added SO42-(aq) and this will require reaction with Ba2+(aq) , thus reducing its concentration, as

we desire. The same result can be obtained by analyzing the system in terms of the solubility product when the concentration of sulfate ion is increased. (iii) Adding additional solid BaSO4, will have no effect on the [Ba2+(aq)] in the ingested solution, because the solution is already saturated and there is a great deal of the solid ionic compound in the mixture.

Problem 9.54. Assume that the metal salt, M(OH)2 dissociates completely to it cations and hydroxide ions when it dissolves in water. From the stoichiometry of the dissolution, two moles of hydroxide ion for each mole of metal cation dissolved, we can calculate the concentration of the metal ion in terms of the concentration of the hydroxide ion:

[M2+(aq)] = 1 mol M2+

2 mol OH–

⎛ ⎝ ⎜

⎞ ⎠ ⎟ [OH–(aq)] = 0.5[OH–(aq)]

The pH of the solution is determined by the hydroxide from the dissolved salt. We can use the pH to calculate the concentration of hydroxide in the solution and then the Ksp: pOH = 14.00 – pH = 14.00 – 8.11 = 5.89; (OH–(aq)) = 10–5.89 = 1.3 × 10–6 Ksp = (M2+(aq))(OH–(aq))2 = [0.5(OH–(aq))](OH–(aq))2 = 0.5(OH–(aq))3 = 0.5(1.3 × 10–6)3 = 1.1 × 10–18

Problem 9.55. This problem is like all others for which you have a g·L–1 solubility (0.000562 g Ag3AsO4(s) dissolves in 1 liter of water) and wish to find the molar solubility and/or Ksp (pKsp). The only difference in this case is the slightly greater complexity of the stoichiometry: Ag3AsO4(s) ⇔ 3Ag+(aq) + AsO4

3–(aq) In the solution, therefore:

[Ag+(aq)] = 3 mol Ag+

1 mol AsO43–

⎛ ⎝ ⎜

⎞ ⎠ ⎟ [AsO4

3–(aq)] = 3[AsO43–(aq)]

The molar mass of Ag3AsO4 is 462.5 g·mol–1. Thus, the molar solubility of the salt and the molar concentration of the arsenate ion are:

molar solubility = 0.000562 g ⋅ L–1

462.5 g ⋅ mol–1 [AsO43–(aq)] = 1.22 × 10–6 M

Therefore, Ksp = (Ag+(aq))3(AsO4

3–(aq)) = [3(AsO43–(aq))]3(AsO4

3–(aq)) = 27(AsO43–(aq))4

Ksp = 27(1.22 × 10–6)4 = 5.98 × 10–23; pKsp = –log(5.98 × 10–23) = 22.223

Problem 9.56. The solubility of silver chromate is somewhat greater than the solubility of silver chloride, so essentially all the chloride in a solution will be precipitated by added silver ion before any red Ag2CrO4(s) is formed to indicate that the titration of chloride ion is complete. Let us calculate the ratio of (CrO4

2–(aq)) to (Cl–(aq)) at the point where the Ag2CrO4(s) just begins to form. At this point, both these equilibria need to be satisfied:



AgCl(s) ⇔ Ag+(aq) + Cl–(aq) Ksp(AgCl) = 1.8 × 10–10 Ag2CrO4(s) ⇔ 2Ag+(aq) + CrO4

2–(aq) Ksp(Ag2CrO4) = 1.2 × 10–12 Since all species are together in the same solution, we can solve both equilibrium constant equations for (Ag+(aq)) and set them equal to find the desired ratio of anions in the solution:

(Ag+(aq)) = Ksp (AgCl)

(Cl– (aq)) =

Ksp (Ag2CrO4 )

(CrO42– (aq))

(CrO4

2– (aq))

(Cl– (aq)) =

Ksp (Ag2CrO4 )

Ksp (AgCl) =

1.2 × 10–12

1.8 × 10–10 = 6.1 × 104

Assume that the few drops of the potassium chromate solution that is added to the chloride solution to be titrated gives (CrO4

2–(aq)) ≈ 0.01. The red color of Ag2CrO4(s) will then start to appear when:

(Cl–(aq)) = (CrO4

2– (aq))

6.1× 104 =

0.01

6.1× 104 = 1.6 × 10–6

This is quite a low value for the concentration of Cl–(aq) remaining in the solution, so we can be pretty confident that essentially all the original Cl–(aq) has precipitated as AgCl(s). Thus, the titration is complete, that is, the equivalence point between the added silver ion and chloride ion in solution has been reached. [Note that the assumption about the concentration of CrO4

2–(aq) in the solution is not critical. Assuming a higher—but reasonable—concentration, say 0.1, still gives a low chloride ion concentration when the red precipitate begins to form. Lower concentrations of CrO4

2–(aq) give lower chloride concentrations. From a practical point of view, the concentration of CrO4

2–(aq) has to be low enough to give the color change when essentially all the chloride has been titrated, but high enough to form a visibly detectable reddish-pink color with only a tiny excess of the silver ion titrant.]

Problem 9.57. Solve the Ni(OH)2(s) (pKsp = 17.2) and Fe(OH)2(s) (pKsp = 13.8) solubility product expressions for the hydroxide anion concentration that would be in equilibrium with the cations at their specified concentrations: [Ni2+(aq)] = 0.001 M and [Fe2+(aq)] = 0.06 M. The cation that requires the lower concentration of hydroxide anion to just be in equilibrium with the solid salt will be the first to precipitate. We convert this lower hydroxide ion concentration to pH to answer the question asked. Ksp(Ni(OH)2) = 10–17.2 = 6.3 × 10–18 = (Ni2+(aq))(OH–(aq))2

(OH–(aq)) = 6.3 × 10−18

Ni2+(aq)( ) = 6.3 × 10−18

0.001 = 7.9 × 10–8; pOH = 7.10

Ksp(Fe(OH)2) = 10–13.8 = 1.6 × 10–14 = (Fe2+(aq))(OH–(aq))2

(OH–(aq)) = 1.6 × 10−14

Fe2+(aq)( ) = 1.6 × 10−14

0.06 = 5.1 × 10–7; pOH = 6.29



Thus, the nickel cation will begin to precipitate first at a pOH = 7.10, a pH of 6.90. Note that both of these hydroxide salts will begin to precipitate at about pH 7. In order to keep Ni(II) and Fe(II) cations in solution, the solutions have to be somewhat acidic.

Problem 9.58. The focus of this problem is this reaction (at equilibrium in a constant volume reaction vessel): H2(g) + I2(g) ⇔ 2HI(g)

(a) If the pressure of I2(g) is increased (by adding more I2 to the vessel), the equilibrium system will act to decrease the disturbance by reacting to use up some of the added I2(g), which will also use up some of the H2(g) and form more HI(g). Thus, the concentration of HI(g) would increase while the concentration of H2(g) would decrease. (b) If the pressure of HI(g) is reduced (by removing some HI from the vessel), the equilibrium system will act to decrease the disturbance by reacting to produce more HI(g) to compensate for what has been removed. In the system, both the forward and reverse reactions are always going on. At equilibrium, the forward and reverse rates are the same, so there is no net change in the concentrations. When the equilibrium is disturbed, however, the rates change and are no longer the same. In order to produce more HI(g), the rate of the forward reaction has to be greater than the rate of the reverse reaction. As the concentrations approach their new equilibrium values, the rates of the forward and reverse reactions again approach one another, and, at equilibrium are again the same. At equilibrium, the H2(g) and HI(g) concentrations will be lower than they were before the disturbance. The equilibrium will have shifted toward the products. The concentration of HI(g) will have increased from what it was just after some was removed, but not back to its original value before the disturbance.

Problem 9.59. To determine which of these two reactions has the larger equilibrium constant, we need to compare the ΔG° values for the two reactions: (i) H2(g) + 1/2O2(g) ⇔ H2O(l)

(ii) H2(g) + 1/2O2(g) ⇔ H2O(g) These reactions are the reactions forming (i) liquid and (ii) gaseous water from the elements. The standard free energies for these reactions are (i) ΔG° = ΔG° f for liquid water and (ii) ΔG° = ΔG° f for gaseous water. The relationship between ΔG° for a reaction and the equilibrium constant is: ΔG° = –RTln(Keq) R is the gas law constant in units of J·K-1mol-1 and T is the Kelvin temperature. The interpretation of this relationship is that the more negative the value of ΔG° , the larger are ln(Keq) and Keq. Since ΔG° f for liquid water is more negative than ΔG° f for gaseous water, the equilibrium constant for the formation of liquid water from its elements in their standard states, reaction (i), will have a larger equilibrium constant.

Problem 9.60. Use the data in Appendix B to determine ΔG°rxn at 298 K for the formation of urea from carbon dioxide and ammonia, CO2(g) + 2NH3(g) ⇔ (NH2)2CO(s) + H2O(l:



ΔG° rxn = (1 mol)ΔG° f(urea(s)) + (1 mol)ΔG° f(H2O(l)) – (1 mol)ΔG° f(CO2(g)) – (2 mol)ΔG° f(NH3(g)) ΔG° rxn = (1)(–197.15 kJ) + (1)(–237.13 kJ) – (1)(–394.36 kJ) – (2)(–16.45 kJ) ΔG° rxn = –7.02 kJ Under standard conditions at 298 K, ΔGrxn = ΔG° rxn. Thus, ΔGrxn = –7.02 kJ < 0 and the reaction is spontaneous. ΔGrxn < 0 is the criterion for spontaneity.

Problem 9.61. We can use the equilibrium constant for H2(g) + I2(g) ⇔ 2HI(g) at 500 K, from Problem 9.13, Keq = 1.60 × 102, to calculate ΔG° rxn for the reaction forming hydrogen iodide gas from its gaseous elements at 500 K: ΔG° rxn = –RTlnKeq = –(8.314 J·K–1·mol–1)(500 K)ln(1.60 × 102) ΔG° rxn = –21.1 kJ·mol–1 Note that the units for the standard free energy change are kJ·mol–1, and you should ask: “Per mole of what?” This means “per mole of reaction,” that is, per Avogadro’s number of the reaction events represented by the reaction equation.

Problem 9.62. (a) For the oxidation of ammonia, 4NH3(g) + 5O2(g) ⇔ 4NO(g) + 6H2O(l), we have, from the standard enthalpies of formation: ΔH° rxn = (4 mol)ΔH° f(NO(g)) + (6 mol)ΔH° f(H2O(l)) – (4 mol)ΔH° f(NH3(g)) – (5 mol)ΔH° f(O2(g)) ΔH° rxn = (4)(90.25 kJ) + (6)(–285.83 kJ) – (4)(–46.11 kJ) – (5)(0 kJ) ΔH° rxn = –1169.54 kJ To calculate ΔH° rxn from bond enthalpies we have to break 12 N–H bonds and 5 O=O bonds and we get back the enthalpy of formation of 4 of the bonds in NO and 12 H–O bonds. Table 7.3 has all these bond enthalpies except for NO, which is given as 630 kJ·mol–1 in the problem. [NOTE: The bond enthalpy for NO is the enthalpy of reaction for NO(g) → N(g) + O(g), which from Appendix B is: ΔH° = (472.70 kJ·mol–1) + (249.4 kJ·mol–1) – (90.25 kJ·mol–1) = 632 kJ·mol–1. This is the value we will use in this problem.] Therefore, the enthalpy required to break the requisite bonds is: ΔH° (breaking) = (12 mol)(BHN–H) + (5 mol)(BHO=O) ΔH° (breaking) = (12)(393 kJ) + (5)(498.7 kJ) ΔH° (breaking) = 7210 kJ ΔH° (making) = (4 mol)(BHNO) + (12 mol)(BHH-O) ΔH° (making) = (4)(632 kJ) + (12)(460 kJ) ΔH° (making) = 8048 kJ For the overall reaction: ΔH° rxn = ΔH° (breaking) – ΔH° (making) = –838 kJ This result from bond enthalpies seems to predict that quite a bit less energy is released in the ammonia oxidation reaction than we calculated from enthalpies of formation. However, recall



that bond enthalpies refer to molecules in the gas phase and the product water in this reaction is a liquid. You can see in Appendix B that 44 kJ·mol–1 is released when water condenses from gas to liquid and this energy is included in the calculation based on enthalpies of formation. If (6 mol)(–44 kJ·mol–1) = –264 kJ is added to the result from the bond enthalpy calculation, we get –1102 kJ for the overall enthalpy of the reaction and this value is within 10% of the value from enthalpies of formation. The two calculations give comparable results when the states of the reactants and products are accounted for. (b) For the standard entropy change of the reaction, we get: ΔS° rxn = (4 mol)S° (NO(g)) + (6 mol)S° (H2O(l)) – (4 mol)S° (NH3(g)) – (5 mol)S° (O2(g)) ΔS° rxn = (4)(210.76 J·K–1) + (6)(69.91 J·K–1) – (4)(192.45 J·K–1) – (5)(205.14 J·K–1) ΔS° rxn = –533.64 J·K–1 This negative value of the standard entropy change for the reaction suggests that the reaction is unfavorable as far as the entropy of the system is concerned. We see from the reaction equation that 9 moles of gaseous reactants react to give 4 moles of a gaseous product and 6 moles of a liquid product. We have seen that gases have higher entropies than liquids (larger volume per molecule), so when the moles of gas decrease in the reaction, we expect a decrease in the entropy, as we have calculated. (c) The standard free energy change for the reaction, based on the results in parts (a) and (b), is: ΔG° rxn = ΔH° rxn – TΔS° rxn = –1169.54 kJ – (298 K)(–533.64 J·K–1) ΔG° rxn = –1169.54 kJ + 159.02 kJ = – 1010.52 kJ Using the data in Appendix B, we get: ΔG° rxn = (4 mol)ΔG° f(NO(g)) + (6 mol)ΔG° f(H2O(l)) – (4 mol)ΔG° f(NH3(g)) – (5 mol)ΔG° f(O2(g)) ΔG° rxn = (4)(86.55 kJ) + (6)(–237.13 kJ) – (4)(–16.45 kJ) – (5)(0 kJ) ΔG° rxn = –1010.78 kJ Within the round-off errors of the data and calculations, the two results are identical. (d) We can use the relationship between the equilibrium constant and the standard free energy of reaction to calculate the equilibrium constant. In order to do this, we need to specify how many “moles of reaction” we are considering for the standard free energy change. It makes sense to think about one mole of ammonia being oxidized, so the thermodynamic quantities in parts (a) – (c) have to be divided by four to get the values for one mole of ammonia in the balanced reaction equation. With this understanding, the equilibrium constant for the ammonia oxidation reaction is:

lnKeq = –ΔGrxn

o

RT = –

–252.70 × 103 J ⋅ (mol NH3)–1

8.314 J ⋅ K–1 ⋅ mol–1( ) 298 K( )

lnKeq = 102

Keq = 1.97 × 1044 = NO(g)( ) H2O(l)( )3/2

NH3(g)( ) O2 (g)( )5 /4



The equilibrium constant is very large; products are enormously favored over reactants (as the standard free energy change also suggests).

Problem 9.63. (a) ΔHreaction for the ethyne-forming reaction, CaC2(s) + 2H2O(l) ⇔ Ca(OH)2(s) + C2H2(g), must be negative (reaction is exothermic), since the problem statement says “the solution gets quite hot.” (b) ΔSreaction for the ethyne-forming reaction is almost certainly positive, since the reactants are both condensed phases (liquid and solid) and one of the products is a gas (ethyne). The positional entropy of the system increases. (c) ΔGreaction = ΔHreaction –TΔSreaction for the ethyne-forming reaction must be negative, since ΔHreaction is negative and the positive ΔSreaction means that –TΔSreaction will be negative as well. The reaction is spontaneous under standard conditions. Experimentally we note that the reaction is spontaneous and the conditions are close to standard. The two solids and the liquid are pretty much in their standard states and, if the gas is close to atmospheric pressure, it will also be approximately in its standard state. (d) For this reaction, the data from Appendix B give: ΔH° rxn = (1 mol)ΔH° f(Ca(OH)2(s)) + (1 mol)ΔH° f(C2H2(g)) – (1 mol)ΔH° f(CaC2(s)) – (2 mol)ΔH° f(H2O(l)) ΔH° rxn = (1)(–986.09 kJ) + (1)(226.73 kJ) – (1)(–59.8 kJ) – (2)(–285.83 kJ) ΔH° rxn = –127.9 kJ

ΔS° rxn = (1 mol)S° (Ca(OH)2(s)) + (1 mol)S° (C2H2(g)) – (1 mol)S° (CaC2(s)) – (2 mol)S° (H2O(l)) ΔS° rxn = (1)(83.39 J·K–1) + (1)(200.94 J·K–1) – (1)(69.96 J·K–1) – (2)(69.91 J·K–1) ΔS° rxn = 74.55 J·K–1

ΔG° rxn = (1 mol)ΔG° f(Ca(OH)2(s)) + (1 mol)ΔG° f(C2H2(g)) – (1 mol)ΔG° f(CaC2(s)) – (2 mol)ΔG° f(H2O(l)) ΔG° rxn = (1l)(–898.49 kJ) + (1)(209.20 kJ) – (1)(–64.9 kJ) – (2)(–237.13 kJ) ΔG° rxn = –150.1 kJ Our reasoning in parts (a) – (c) is reinforced by these results: the signs we predicted were correct. (e) We see that the standard enthalpy change accounts for more than 80% of the standard free energy change, so it is the exothermicity of the reaction, which increases the thermal entropy of the surroundings, that is the major factor contributing to the large negative free energy change for this reaction.

Problem 9.64. This is a subtle problem. The solution is not obvious and needs careful consideration of the meaning of ΔG. Starting from pure reactants or pure products, ΔG always decreases going from either pure reactants or pure products toward equilibrium. That is, reactions are spontaneous proceeding toward equilibrium from either side. Since ΔG decreases beginning from either reactants or products, at some intermediate composition, ΔG must reach a minimum. This is the



equilibrium composition for the given system. Since the reactants and products are in equilibrium at this composition, very tiny changes one way or the other (toward the reactant side or toward the product side) will result in no change in ΔG. This is the criterion for equilibrium: the change in ΔG is zero for the change from reactants to products at this composition. In the language of the calculus we say that the derivative of ΔG with respect to the composition of the reaction mixture (usually called the extent of reaction) is zero, which indicates that ΔG has reached a minimum. The dependence of ΔG on reaction mixture composition looks like this (where the position of equilibrium, the minimum in ΔG, varies from one reaction to another):

Problem 9.65. (a) The reaction equation and equilibrium constant expression for the calcite decomposition

reaction are: CaCO3(s) ⇔ CaO(s) + CO2(g)

Keq = 1.6 × 10–23 = CaO(s)( ) CO2 (g)( )

CaCO3 (s )( ) = (CO2(g)) = P(CO2 (g))

1 bar

The dimensionless concentration ratios for the pure solids are unity, so the expression reduces to the pressure of carbon dioxide, 1.6 × 10–23 bar, in equilibrium with these solids. The amount of CO2(g) in the atmosphere varies somewhat but is in the neighborhood of about 10–2 bar. Obviously, the CO2(g) contribution from chalk decomposition is negligible [and the decomposition is inhibited by the presence of the high atmospheric content of CO2(g)]. (b) From Keq we can get ΔG° reaction for the calcite decomposition. Combine this value with the ΔG° f values for calcite and carbon dioxide and solve to get the ΔG° f(CaO(s)). ΔG° reaction = –RTlnKeq = –(8.314 J·K–1·mol–1)( 298 K) ln(1.6 × 10–23) ΔG° reaction = 1.3 × 102 kJ·mol–1 The reaction, as written, involves one mole of the reactant, so we have: ΔG° reaction = (1 mol)ΔG° f(CaO(s)) + (1 mol)ΔG° f(CO2(g)) – (1 mol)ΔG° f(CaCO3(s)) 1.3 × 102 kJ = (1 mol)ΔG° f(CaO(s)) + (1)(–394.36 kJ) – (1)(–1128.8 kJ) ΔG° f(CaO(s)) = –604 kJ·mol–1



Problem 9.66. Compounds A, B, and C were dissolved in water in a reaction vessel and allowed to come to equilibrium. The graph shows the variation of concentration with time for this reaction: 3A(aq) ⇔ B(aq) + 2C(aq) (a) The reaction quotient, Q, and its value at time t1, the time the compounds were mixed, are:

Q = B(aq)( ) C(aq)( )2

A(aq)( )3 =

0.01( ) 0.02( )2

0.08( )3 = 0.008

(b) The graph shows that the concentration of A is decreasing, while the concentrations of B and C are increasing to attain equilibrium. This means that the reaction is shifting to the right, favoring the formation of additional product molecules. (c) Macroscopic concentrations of A, B, and C do not appear to be changing after t2, so it is reasonable to suppose that equilibrium has been reached in this system at t2. (d) At t2, we assume that equilibrium has been reached, so the reaction quotient equals K for this reaction:

Q = K = B(aq)( ) C(aq)( )2

A(aq)( )3

⎛

⎝⎜⎜

⎞

⎠⎟⎟

eq

= 0.02( ) 0.04( )2

0.05( )3 = 0.3

Note that the stoichiometry of the reaction is included in the plot. As the concentration of A decreases by 0.03 M, the sum of the concentrations of B and C increases by 0.03 M, so mass is conserved. (e) Use the relationship, ΔG° rxn = –RTlnK, and the K value from part (d) to find the standard free energy change for this reaction at 298 K: ΔG° rxn = –RTlnKeq = –(8.314 J·K–1·mol–1)( 298 K) ln(0.3) = 3 kJ·mol–1

Problem 9.67. (a) For the reaction, H2(g) + I2(g) ⇔ 2HI(g), we use the equilibrium constant at 700 K to find ΔG° reaction at 700 K: ΔG° reaction = –RTlnK = –(8.314 J·K–1·mol–1)( 700 K) ln(54) ΔG° reaction = –23 kJ·mol–1 (for reaction of one mole of each reactant)

(b) To get the reaction quotient, Q, in a mixture in which the pressures of H2(g), I2(g), and HI(g) are, respectively, 0.040 bar, 0.35 bar, and 1.05 bar, we write the reaction quotient expression and substitute these pressure values (as ratios to the standard pressure, 1 bar):

Q = HI(g)( )2

H2 (g)( ) I2 (g)( ) =

1.05( )2

0.040( ) 0.35( ) = 79

(c) The free energy of a reaction is related to the standard free energy and the reaction quotient under the conditions in the reaction system: ΔGreaction = ΔG° reaction + RTlnQ = –23 kJ·mol–1 + (8.314 J·K–1·mol–1)( 700 K) ln(79) ΔGreaction = 2.4 kJ·mol–1

t1 t2

A

B

C

time, min

M

0.080.070.060.050.040.030.020.010.00



(This result should really be reported only to the nearest ±1 kJ·mol–1, because the value for ΔG° reaction is only known to the nearest kJ·mol–1,) The positive sign of ΔGreaction is consistent with Q > K. When Q > K, the concentration of products compared to reactants is larger than the equilibrium value. Going from reactants to products under this circumstance is not favored, which is what the positive ΔGreaction tells us. (d) To attain equilibrium in the system described in part (b), some of the HI(g) has to be converted to its elements, H2(g) and I2(g),so that the value of Q will be reduced to K. The reaction proceeds to produce more reactants at the expense of the product.

Problem 9.68. We calculate ΔG° reaction for methanol decomposition, CH3OH(l) ⇔ CO(g) + 2H2(g), and use its sign to determine whether the reaction will occur spontaneously under standard conditions at 298 K: ΔG° reaction = (1 mol)ΔG° f(CO(g)) + (2 mol)ΔG° f(H2(g)) – (1 mol)ΔG° f(CH3OH(l)) ΔG° reaction = (1)(–137.17 kJ) + (2)(0 kJ) – (1)(–166.27 kJ) ΔG° reaction = 29.10 kJ Since ΔG° reaction > 0, methanol decomposition to carbon monoxide and hydrogen under standard conditions is not spontaneous.

Problem 9.69. Calculate ΔG° reaction for the reaction, CH2CH2(g) + H2O(l) ⇔ CH3CH2OH(l), and use its sign to determine whether the reaction will occur spontaneously under standard conditions at 298 K: ΔG° reaction = (1 mol)ΔGo

f(CH3CH2OH(l)) – (1 mol)ΔG° f(C2H4(g)) – (1 mol)ΔG° f(H2O(l)) ΔG° reaction = (1)(–174.78 kJ) – (1)(68.15 kJ) – (1)(–237.13 kJ) ΔG° reaction = –5.80 kJ The reaction of water with ethene (ethylene) to form ethanol under standard conditions is favored with an equilibrium constant of 10.4. Although the reaction is slow, its rate can be increased by catalysis and a good deal of ethanol is manufactured by this reaction.

Problem 9.70. To explain how temperature, pressure, and concentration affect a system at equilibrium, we can use le Chatelier’s principle: a system at equilibrium responds to a disturbance by adjusting to decrease the effect of the disturbance. The application to changes in temperature is through the energetics of the equilibrium reaction. If the temperature of an equilibrium system is increased, it gains energy, so the reaction shifts in a direction to use energy. An endothermic reaction requires energy, so raising its temperature will shift the equilibrium toward more products, in order to use energy. An exothermic reaction releases energy, so raising its temperature will shift the equilibrium toward more reactants, in order to use energy (since the reverse reaction must be endothermic). Pressure and concentration effects are related. If more reactant is added to a constant volume system, the concentration of reactant is increased and the equilibrium shifts toward formation of products to reduce the added reactant concentration. If a reactant is removed from a constant volume system, the equilibrium shifts toward formation of more reactants to



compensate for the loss of the reactant that was removed. Similar arguments can be made about adding or removing a product of the reaction. The effect of pressure changes is a special case of concentration change for gaseous species. If the pressure of a reactant is increased by adding more (or decreased by removing some) in a constant volume system, the effects are the same as just noted for any concentration change. The more subtle effect is a change in the volume of a system in which some or all of the reactants and products are gases. In this case, the concentrations of all gases in the system are increased. To try to lower the concentration of gases, the reaction will shift toward the side with the lower number of moles of gas. A third way of changing the pressure in a system is by adding a gas that is not involved in the reaction. This addition will have no affect on the equilibrium, because it does not change the concentration of any reactant or product.

Problem 9.71. [NOTE: The reaction equation in the problem statement is not balanced. The balanced equation is used here.]

For this exothermic reaction, 2SO2(g) + O2(g) ⇔ 2SO3(g), initially at equilibrium: (a) Removing some SO3(g) shifts the position of equilibrium to the right, forming more product SO3(g) to compensate for that which has been removed. The concentrations (pressures) of SO2(g) and O2(g) will decrease (assuming that the volume remains constant) as they react to form more SO3(g). (b) Decreasing the pressure (presumably by increasing the volume of the system) shifts the position of equilibrium to the left, forming more reactant, in order to form the largest number of moles of gas possible to compensate for the decrease in pressure (concentration) of all the gases. Thus more SO2(g) and O2(g) will form and some SO3(g) will be used up by the back reaction. (c) Decreasing the temperature, by removing energy from the system, shifts the position of equilibrium to the right, forming more product SO3(g), because the reaction, as written is exothermic and releases energy. If energy has been removed, the system reacts to release more energy to compensate for the loss. The final concentrations (pressures) in the system cannot be predicted without knowing the actual temperature change that occurs (as well as the volume and initial pressures of the reactants and product), because the pressures of gases are temperature dependent. (d) Adding some O2(g) is added shifts the position of equilibrium to the right, forming more product SO3(g), as the system compensates for the increase in concentration of a reactant by using up some of it, which will also decrease the concentration of SO2(g). The new equilibrium mixture will have more O2(g) and SO3(g) and less SO2(g), than the original equilibrium mixture.

Problem 9.72. [NOTE: The problem statement gives the bond energy for NO as 630 kJ·mol–1, but a better value is 632 kJ·mol–1 (See Problem 9.62), which we will use in this problem.]

(a) For the equilibrium reaction N2(g) + O2(g) ⇔ 2NO(g), higher temperatures favor the formation of NO(g). Since the reaction favors the products as temperature increases (energy is added), we predict, using Le Chatelier’s principle, that the reaction is endothermic, that is requires energy, ΔH° rxn > 0. An endothermic reaction responds to added energy by forming more product to use up some of the added energy and this is the direction of the observed effect.



(b) The bonds in N2 and O2 have to be broken for the reaction to proceed and this requires 941.4 kJ·mol–1 + 498.7 kJ·mol–1 = 1440.1 kJ·mol–1. Formation of the bonds in two moles of NO releases 1264 kJ·mol–1, so the reaction overall is endothermic by 176 kJ·mol–1. This result is consistent with our prediction in part (a), based on the Le Chatelier’s principle and the effect of temperature on the equilibrium reaction. (c) Since the formation of products is favored as temperature increases, the numerator of the equilibrium constant expression increases at the expense of the denominator and, hence, the equilibrium constant increases with temperature. (d) If the equilibrium constant for this reaction increases as temperature increases, as we reasoned in part (c), the standard free energy of reaction, ΔG° rxn = –RTlnK, must become more negative (or less positive) as the temperature increases. We know ΔG° rxn = ΔH° rxn – TΔS° rxn, so, since the standard enthalpy of reaction is positive, the only way that the standard free energy can get more negative (or less positive) is for the standard entropy change for the reaction to be positive. From Appendix B we have: ΔS° rxn = (2 mol)S° (NO) – (1 mol)S° (N2) – (1 mol)S° (O2) ΔS° rxn = 2(210.76 J·K–1) – (191.61 J·K–1) – (205.14 J·K–1) = 24.77 J·K–1 Thus, the thermodynamic data confirm the deductions we make based solely on the experimental information that the NO yield increases with increasing temperature.

Problem 9.73. (a) To find the solubility product for PbI2(s) at 20. ° C (293 K), use the temperature dependence of the solubility product constant, plus its value at one of the temperatures (273 K) and ΔH° rxn from Worked Example 9.67:

lnK293

K273

⎛⎝⎜

⎞⎠⎟

= lnK293

3.5 × 10–9

⎛⎝⎜

⎞⎠⎟

= ΔH rxn

o

R

⎛⎝⎜

⎞⎠⎟

T2 – T1

T2 ⋅T1

⎛⎝⎜

⎞⎠⎟

lnK293

3.5 × 10–9

⎛⎝⎜

⎞⎠⎟

= 56 kJ ⋅ mol–1

8.314 J ⋅ K–1 ⋅ mol –1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

293 K – 273 K

(293 K)(273 K)1

⎛

⎝ ⎜

⎞

⎠ ⎟ = 1.68

K293

3.5 × 10–9

⎛⎝⎜

⎞⎠⎟

= 5.4; ∴ K293 = 5.4(3.5 × 10–9) = 1.9 × 10–8

In Worked Example 9.67, the molar solubility is labeled s and the equilibrium constant is 4s3. Therefore, at 20 ° C, the molar solubility (moles of PbI2 dissolved in 1 L of water at 20 ° C) is:

s = K293

4⎛⎝⎜

⎞⎠⎟

13

= 1.9 × 10–8

4

⎛⎝⎜

⎞⎠⎟

13

= 1.7 × 10–3

Use the molar solubility (mol·L–1) and the molar mass of PbI2 to find the solubility in g·L–1: solubility in g·L–1 = (1.7 × 10–3 mol·L–1)(461 g·mol–1) = 0.77 g L–1

(b) The handbook value of 0.63 g L–1 for the solubility of PbI2 at 20 ° C is in the right ballpark with our calculated value, although 20% lower. This is another example of the non-ideality of aqueous ionic solutions. Our calculations are usually (but not always) of the correct order of magnitude (compared to experiment), but rarely right on the money. This is why we have



stressed that these calculations are only approximations to the true state of affairs in the solutions where many interactions of charged species are involved.

Problem 9.74. (a) The reaction of interest is: CaSO4(s) ⇔ Ca2+(aq) + SO4

2–(aq). If the molar solubility of CaSO4 is s, then Ksp = s2. The molar mass of CaSO4 is 135 g mol–1 so the molar solubilities and solubility products are:

at 303 K: s = 2.09 g ⋅ L–1

135 g ⋅ mol–1 = 1.55 × 10–2 mol L–1; Ksp = 2.40 × 10–4

at 373 K: s = 1.62 g ⋅ L–1

135 g ⋅ mol–1 = 1.20 × 10–2 mol L–1; Ksp = 1.44 × 10–4

(b) Use the temperature dependence of equilibrium to get ΔH° rxn (assuming that the enthalpy change is not temperature dependent over this range of temperature):

lnK

sp(373)

Ksp

(303)

⎛

⎝ ⎜ ⎞

⎠ ⎟ = ln1.44 ×10–4

2.40 ×10–4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ΔH rxno

8.314 J ⋅ K–1 ⋅ mol–1

⎛⎝⎜

⎞⎠⎟

373 K – 303 K

(373 K)(303 K)

⎛⎝⎜

⎞⎠⎟

ΔH° rxn = –6.86 kJ·mol–1 Note that the standard enthalpy change is negative, which we could have predicted from Le Chatelier’s principle and the decrease in solubility as temperature increases. To get ΔS° rxn (assuming that it also is not temperature dependent), we use ΔH° rxn and the equilibrium constant at one of the temperatures: ΔG° rxn = –RTlnKsp = ΔH° rxn – TΔS° rxn –(8.314 J·K–1·mol–1)( 303 K) ln(2.40 × 10–4) = –6.86 × 103 J·mol–1 – (303 K)ΔS° rxn ΔS° rxn = –91.9 J·K–1·mol–1

(c) Use the relationship, ΔG° rxn = ΔH° rxn – TΔS° rxn, to find ΔG° rxn at 298 K and thence Ksp at 298 K: ΔG° rxn = ΔH° rxn – TΔS° rxn = –6.86 × 103 J·mol–1 – (298 K)(–91.9 J·K–1·mol–1) ΔG° rxn = 20.5 × 103 J·mol–1 (for one mole of CaSO4(s) dissolving) –RTlnKsp = ΔG° rxn = –(8.314 J·K–1·mol–1)(298 K) ln(Ksp) = 20.5 × 103 J·mol–1 Ksp = 2.55 × 10–4 Note that the solubility product at 298 K is slightly greater than at 303 K, which is consistent with the exothermicity of the reaction [part (b)]. We can also calculate ΔG° rxn from the data in Appendix B: ΔG° rxn = (1 mol)ΔG° f(Ca2+(aq)) + (1 mol)ΔG° f(SO4

2–(aq)) – (1 mol)ΔG° f(CaSO4(s)) ΔG° rxn = (1)(–553.58 kJ) + (1)(–744.53 kJ) – (1)(–1321.79 kJ) ΔG° rxn = 23.68 kJ This value compares favorably (within about 10%) with that calculated from the solubility data. The pKsp from our calculations is 3.60 compared with the value of 4.62 in Table 9.4. This is about an order of magnitude difference in the solubility products with our value calculated from



the experimental solubilities being higher. The values given in tables like Table 9.4 are usually obtained from data that are extrapolated from real solutions to high dilutions where interionic interactions are lower and often the solubility products are lower than those calculated from experimental solubilities.

Problem 9.75. (a) For the coal gas reaction, C(s) + H2O(g) ⇔ CO(g) + H2(g), use the data in Appendix B to calculate ΔH° rxn, ΔS° rxn, and ΔG° rxn at 298 K (assuming the carbon is graphite): ΔH° rxn = (1 mol)ΔH° f(CO(g)) + (1 mol)ΔH° f(H2(g)) – (1 mol)ΔH° f(C(s)) – (1 mol)ΔH° f(H2O(g)) ΔH° rxn = (1)(–110.53 kJ) + (1)(0 kJ) – (1)(0 kJ) – (1)(–241.82 kJ) = 131.29 kJ

ΔS° rxn = (1 mol)S° (CO(g)) + (1 mol)S° (H2(g)) – (1 mol)S° (C(s)) – (1 mol)S° (H2O(g)) ΔS° rxn = (1)(197.67 J·K–1) + (1)(130.68 J·K–1) – (1)(5.74 J·K–1) – (1)(188.83 J·K–1) ΔS° rxn = 133.78 J·K–1

ΔG° rxn = (1 mol)ΔG° f(CO(s)) + (1 mol)ΔG° f(H2(g)) – (1 mol)ΔG° f(C(s)) – (1 mol)ΔG° f(H2O(g)) ΔG° rxn = (1)(–137.17 kJ) + (1)(0 kJ) – (1)(0 kJ) – (1)(–228.57 kJ) = 91.40 kJ

(b) Since both ΔH° rxn and ΔS° rxn are positive, there should be a temperature at which TΔS° rxn = ΔH° rxn, and ΔG° rxn = 0. Assuming that ΔH° rxn and ΔS° rxn are not temperature dependent over the temperature range of interest:

T = ΔH rxn

o

ΔSrxno

⎛⎝⎜

⎞⎠⎟

= 131.29 kJ

0.13378 kJ ⋅ K–1 = 980 K

We really ought to report this as ∼1000 K, since the assumptions of constant ΔH° rxn and ΔS° rxn are surely incorrect over such a large range.

(c) Since, from part (b), ΔG°rxn = 0 at 980 K, the equilibrium constant at 980 K must be unity, ln(K) = ln(1) = 0, in order to satisfy the equation, ΔG° rxn = –RTlnK.

Problem 9.76. (a) For a phase change, such as, liquid ⇔ gas, at equilibrium at one bar pressure, the dimensionless concentration ratios for the reactant and product are both unity, since they are pure substances. Thus, the reaction quotient (the ratio of the two concentrations) is also unity and lnQ = 0. Therefore: ΔGrxn = ΔG° rxn + RTlnQ = ΔG° rxn Since the reaction is in equilibrium, ΔGrxn = 0 = ΔG° rxn. (b) To estimate the temperature at which rhombic and monoclinic sulfur are in equilibrium, S(s, rhombic) ⇔ S(s, monoclinic), at one bar, assume that ΔH°rxn and ΔS°rxn are not temperature dependent over the temperature range of interest. Then, at the equilibrium temperature, where ΔG° rxn = 0, we have TΔS° rxn = ΔH° rxn. We can use the values of ΔH° rxn and ΔS° rxn to estimate



the equilibrium temperature. For the phase change of one mole of rhombic sulfur to monoclinic sulfur, Appendix B gives ΔH° rxn = 0.33 kJ [= (0.33kJ) – (0 kJ)) and ΔS° rxn = 0.8 J·K–1 [= (32.6 J·K–1) – (31.80 J·K–1)]. Thus:

T = ΔH rxn

o

ΔSrxno

⎛⎝⎜

⎞⎠⎟

= 0.33 kJ

0.0008 kJ ⋅ K–1 ≈ 410 K (~140 ° C)

This value is a bit high. Rhombic sulfur melts at 112 ° C and the triclinic form crystallizes out from the melt. Our calculation does not account for the solid-liquid phase changes, but the answer is not unreasonably out of line.

Problem 9.77. (a) Given that the equilibrium constant for the reaction, H2(g) + I2(g) ⇔ 2HI(g), is 794 at 298 K and using other results from Problems 9.13 and 9.61, find ΔH° rxn and ΔS° rxn. There are at least two ways to solve this problem and the assumptions in each are the same, that ΔH° rxn and ΔS° rxn are not temperature dependent over the temperature range of interest. One way to do the problem is to use the equilibrium constant values at 500 K (from Problem 9.13) and 298 K (from this problem statement) with equation (9.59) to solve for ΔH° rxn. Another way is to subtract the values for ΔG° rxn at 500 K (from Problem 9.61) and 298 K (calculated from the equilibrium constant in this problem), written in terms of ΔH° rxn and ΔS° rxn, so that the ΔH° rxn term cancels out and then solve for ΔS° rxn. In either case, we then substitute our known values into the expression for ΔG° rxn to get the unknown. Let’s use the second method. ΔG° rxn(298) = –RTlnK = –(8.314 J·K–1·mol–1)(298 K) ln(794) = –16.5 kJ·mol–1 ΔG° rxn(500) = – 21.1 kJ·mol–1 Now we write the equations for ΔG°rxn in terms of ΔH° rxn and ΔS° rxn and subtract: ΔG° rxn(500) = – 21.1 kJ·mol–1 = ΔH° rxn – (500 K)ΔS° rxn – {ΔG° rxn(298) = – 16.5 kJ·mol–1 = ΔH° rxn – (298 K)ΔS° rxn} (– 21.1 kJ·mol–1) – (– 16.5 kJ·mol–1) = – (500 K – 298 K)ΔS° rxn ΔS° rxn = 22.8 J·K·mol–1 Now substitute this value for ΔS° rxn into either of the ΔG° rxn equations to get ΔH° rxn: ΔG° rxn(298) = – 16.5 kJ·mol–1 = ΔH° rxn – (298 K)(0.0228 kJ·K·mol–1) ΔH° rxn = –9.71 kJ·mol–1

(b) If we had done part (a) by the first method (using equilibrium constants) we would not know ΔG° rxn(298) at this point and would have to calculate it, either as we did in part (a) or by combining the ΔH° rxn and ΔS° rxn we got without knowing ΔG° rxn(298). The answers would all be the same (within some calculational round-off uncertainty), no matter which method you choose. For comparison, using the data from Appendix B, we get: ΔG° rxn(298) = (2 mol)ΔG° f(HI(g)) – (1 mol)ΔG° f(H2(g)) – (1 mol)ΔG° f(I2(g)) ΔG° rxn(298) = (2)(1.70 kJ) – (1)(0 k) – (1)(19.3 kJ) = –15.9 kJ·mol–1 ΔG° rxn(298) = –15.9 kJ·mol–1



This value for ΔG° rxn(298) is almost the same as the one we found above. It appears that our assumptions are good (or that there are compensating effects that cancel out). Slight discrepancies between values determined in different ways, free energies from thermodynamic measurements and from equilibrium constants, for example, are usually due to small uncertainties in each experimental value that combine to yield slightly different numeric results for the same variable.

[For comparison, here are the values calculated by the other method.

lnK(500)

K(298)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln

160

794⎛ ⎝

⎞ ⎠ =

ΔHrxno

8.314 J ⋅ K–1 ⋅ mol –1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

500 K – 298 K

(500 K)(298 K)

⎛ ⎝ ⎜

⎞ ⎠ ⎟

ΔH° rxn = –9.82 kJ·mol–1 ΔG° rxn(500) = – 21.1 kJ·mol–1 = –9.82 kJ·mol–1 – (500 K)ΔS° rxn ΔS° rxn = 22.6 J·K·mol–1 ΔG° rxn(298) = –9.82 kJ·mol–1 – (298 K)(0.0226 kJ·K·mol–1) = –16.6 kJ·mol–1 The small discrepancies are due to rounding off and retaining only three significant figures in the intermediate values.]

Problem 9.78. (a) Estimate ΔH° rxn for ethanol vaporization. At its normal boiling point, a liquid is in equilibrium, ΔGrxn = 0, with its vapor at one bar pressure. Under these conditions, ΔHrxn = TΔSrxn, where T is the boiling point. Since the total pressure is one bar, ΔHrxn and ΔSrxn are equal to ΔH° rxn and ΔS° rxn, respectively. For ethanol at 351.6 K (its normal boiling point) we get: ΔH° rxn = TΔS° rxn = (351.6 K)(122 J·K–1·mol–1) = 42.9 kJ·mol–1

(b) The equilibrium constant expression for ethanol vaporization is:

Keq = (C2H5OH(g))

(C2H5OH(l)) =

PC2 H5OH( )1 bar( )

1

The denominator of the equilibrium constant expression is unity because the component is a pure liquid. The numerator is a ratio of the vapor pressure of ethanol (in bar) to the standard pressure, 1 bar. At the normal boiling temperature, 78.5 ° C, the vapor pressure of the ethanol is 1 bar, so both top and bottom of the equilibrium constant expression are unity and Keq = 1. Thus, ΔG° rxn = –RTlnKeq = –RTln(1) = 0. Also, since ΔG° rxn refers to the reaction at one bar total pressure and the system is in equilibrium at the normal boiling point and one bar total pressure, we can make the connection that ΔG° rxn = ΔGrxn = 0. (c) At 298 K (25 ° C), we can use ΔH° rxn and ΔS° rxn from part (a), assuming that they are independent of temperature, to find ΔG° rxn and Keq: ΔG° rxn = ΔH° rxn – TΔS° rxn = 42.9 kJ·mol–1 – (298 K)(0.122 kJ·K–1·mol–1) ΔG° rxn = 6.54 kJ·mol–1 ΔG° rxn = –RTlnKeq = 6.54 kJ·mol–1 = – (8.314 J·K–1·mol–1)(298 K)lnKeq Keq = 0.0714



(d) From part (b), you see that the numeric value of Keq is the vapor pressure of ethanol (in bar). Thus, the vapor pressure of ethanol at 298 K (25 ° C) is 0.0714 bar (= 7.14 × 103 Pa = 53.6 torr = 0.0705 atm).

Problem 9.79. (a) To determine ΔH° rxn and ΔS° rxn for the vaporization of 1-propanol (1-PrOH) and 2-propanol (2-PrOH), use the vapor pressure vs. temperature data given, the equivalence between vapor pressure and the equilibrium constant for vaporization, equation (9.61), and the relationship among K, ΔH° rxn, and ΔS° rxn, equation 9.55, as was done for similar data in Worked Example 9.71.

v. p., torr 1.0 10.0 40.0 100. 400. 760. v. p., bar 0.0013 0.0133 0.0533 0.133 0.533 1.013

ln K –6.65 –4.320 –2.932 –2.017 –0.629 0.0129 1-PrOH, T, K 258.1 287.8 309.5 325.9 355.1 370.9

1/T, K–1 0.00387 0.00347 0.00323 0.00307 0.00282 0.00270 2-PrOH, T, K 247.0 275.5 296.9 312.6 340.9 355.6

1/T, K–1 0.00405 0.00363 0.00337 0.00320 0.00293 0.00281

The plots of these data (and the computer-generated equations of the lines through the data points) are shown on this graph:

-7

-6

-5

-4

-3

-2

-1

0

1

0.0025 0.0030 0.0035 0.0040 0.0045

1-PrOH

2-PrOH

lnK = -5.70x103(1/T) + 15.46

lnK = -5.36x103(1/T) + 15.10

1/T, K–1

lnK

Values determined from the plot for 1-PrOH are: ΔH° rxn = –R·(slope) = – (8.314 J·K–1·mol–1)·(–5.70 × 103 K) = 47.4 kJ·mol–1 ΔS° rxn = R·(intercept) = (8.314 J·K–1·mol–1)·(15.46) = 129 J·K–1·mol–1

Values determined from the plot for 2-PrOH are: ΔH° rxn = – (8.314 J·K–1·mol–1)·(–5.36 × 103 K) = 44.6 kJ·mol–1 ΔS° rxn = (8.314 J·K–1·mol–1)·(15.10) = 126 J·K–1·mol–1

(b) The slightly higher standard enthalpy change for vaporization of 1-propanol compared to 2-propanol probably reflects a bit greater bonding interactions (both hydrogen bonding and



dispersion forces) among the linear-chain 1-propanol molecules compared to the more compact structure of 2-propanol. (c) If we assume that these isomeric molecules in the gas phase have about the same entropy, then the smaller change in entropy for vaporization of 2-propanol suggests that its liquid phase entropy is a bit closer to the gas phase entropy, that is the liquid phase entropy of 2-propanol is a bit higher than that of 1-propanol. If the molecules of the more compact isomer are freer to move about in the liquid [as our reasoning in part (b) suggests], higher liquid phase entropy for 2-propanol would be expected (higher positional entropy). (d) The 2-propanol is more volatile. The temperature required to produce any given vapor pressure is lower for 2-propanol. This is largely a result of the lower enthalpy of vaporization, which gives a less positive free energy for the vaporization (at temperatures below the boiling point). The positive entropies of vaporization also contribute to lowering the free energy, but contribute less for 2-propanol, because the entropy change is a bit smaller.

Problem 9.80. Assuming that the changes in standard enthalpy and entropy are not temperature dependent, we can use the temperature dependence of the equilibrium constant for water autoionization, pKw 14.00 at 25 ° C and 12.97 at 60 ° C, to find ΔH° rxn and then combine this with one of the equilibrium constants to get ΔS° rxn, and thence ΔG° rxn: Kw(298) = 1.00 × 10–14; Kw(333) = 1.07 × 10–13

lnK(333)

K(298)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln

1.07 ×10–13

1.00 ×10–14

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ΔHreactiono

8.314 J ⋅ K–1 ⋅mol –1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

333 K – 298 K

(333 K)(298 K)

⎛ ⎝ ⎜

⎞ ⎠ ⎟

ΔH° rxn = 55.9 kJ·mol–1

–RTlnKw(T) = ΔH° rxn – TΔS° rxn –(8.314 J·K–1·mol–1)(298 K) ln(1.00 × 10–14) = (55.9 × 103 J·mol–1) – (298 K)ΔS° rxn ΔS° rxn = –80.4 J·K–1·mol–1 ΔG° rxn can be determined either by combining ΔH° rxn and ΔS° rxn or from one of the equilibrium constants. There is no difference mathematically, since an equilibrium constant was used to get ΔS° rxn in the first place. ΔG° rxn = –RTlnKw(T) = 79.9 kJ·mol–1 The reaction of interest here is H2O(l) ⇔ H+(aq) + OH–(aq), for which we can calculate ΔH° rxn, ΔS° rxn, and ΔG° rxn from Appendix B to compare with the results above: ΔH° rxn = (1 mol)ΔH° f(H+(aq)) + (1 mol)ΔH° f(OH–(aq)) – (1 mol)ΔH° f(H2O(l)) ΔH° rxn = (1)(0 kJ) + (1)(–229.29 kJ) – (1)(–285.83 kJ) = 55.84 kJ

ΔS° rxn = (1 mol)S° (H+(aq)) + (1 mol)S° (OH–(aq)) – (1 mol)S° (H2O(l)) ΔS° rxn = (1)(0 J·K–1) + (1)(–10.75 J·K–1) – (1)(69.91 J·K–1) = 80.66 J·K–1

ΔG° rxn = (1 mol)ΔG° f(H+(aq)) + (1 mol)ΔG° f(OH–(aq)) – (1 mol)ΔG° f(H2O(l)) ΔG° rxn = (1)(0 kJ) + (1)(–157.24 kJ) – (1)(–237.13 kJ) = 79.89 kJ



The thermodynamic values derived from the data in Appendix B are the same (for one mole of reaction) as those determined from the temperature dependence of the equilibrium constant. This makes sense, because it is likely to be equilibrium constant data like these that are used to determine the thermodynamic values for the hydroxide ion. The values for the hydronium ion are assigned and all other ions are given relative to the hydronium ion to make them consistent with equilibrium data.

Problem 9.81. [NOTE: The relationship, K = ° C + 273.1, should be added to this problem.] (a) The numerical value of the equilibrium constant for a phase change from a condensed phase to the gas phase, such as, PCl5(s) ⇔ PCl5(g), is equal to the equilibrium pressure of the gas (in bar), equation (9.61). Assuming that the changes in standard enthalpy and entropy are not temperature dependent, we can use the temperature dependence of the equilibrium constant to find ΔH° rxn and then combine this with one of the equilibrium constants to get ΔS° rxn. For phosphorus pentachloride sublimation, the equilibrium constants are:

K(375.6) =

40 torr( )750 torr ⋅ bar–1( )1 bar

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 0.053

K(420.3) =

400 torr( )750 torr ⋅ bar–1( )1 bar

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 0.533

lnK(420.3)

K(375.6)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln

0.533

0.053⎛ ⎝

⎞ ⎠ =

ΔHrxno

8.314 J ⋅ K –1 ⋅ mol–1

⎛⎝⎜

⎞⎠⎟

420.3 K – 375.6 K

(420.3 K)(375.6 K)

⎛⎝⎜

⎞⎠⎟

ΔH° rxn = 67.6 kJ·mol–1

–RTlnK(T) = ΔH° rxn – TΔS° rxn – (8.314 J·K–1·mol–1)(420.3 K)ln(0.533) = 67.6 × 103 J·mol–1 – (420.3 K)ΔS° rxn ΔS° rxn = 156 J·K–1·mol–1

(b) When the equilibrium sublimation pressure is 1 bar, K = 1 and lnK = 0, so

T = ΔHrxn

o

ΔSrxno =

67.6 ×103 J ⋅ mol–1

156 ×103 J ⋅ K –1 ⋅ mol–1 = 433 K = 160 ° C

Problem 9.82. (a) A spontaneous endothermic reaction during which the entropy of the system increases must have a negative free energy change, because all spontaneous reactions have ΔG < 0. (b) A spontaneous exothermic reaction during which the entropy of the system decreases must have a negative free energy change, because all spontaneous reactions have ΔG < 0.



(c) In part (a), ΔH > 0 and ΔS > 0. If the temperature is decreased, the –TΔS term in the free energy expression contributes less to ΔG. At low enough T, it is possible that ΔH > |TΔS|, so that ΔG > 0 and the reaction would become non-spontaneous. In part (b), ΔH < 0 and ΔS < 0. If the temperature is increased, the –TΔS term in the free energy expression contributes more to ΔG. At high enough T, it is possible that |ΔH| < –TΔS, so that ΔG > 0 and the reaction would become non-spontaneous.

Problem 9.83. To determine whether the reaction, CH3OH(g) ⇔ CO(g) + 2H2(g), might occur spontaneously at 1000 K and 1 bar, we can determine ΔH° rxn and ΔS° rxn at 298 K and, assuming they are not dependent on temperature, find ΔH° rxn –TΔS° rxn at 1000 K to see if ΔG° rxn < 0. ΔH° rxn = (1 mol)ΔH°f(CO(g)) + (2 mol)ΔH°f(H2(g)) – (1 mol)ΔH°f(CH3OH(g)) ΔH° rxn = (1)(–110.53 kJ) + (2)(0 kJ) – (1)(–200.66 kJ) = 90.13 kJ

ΔS° rxn = (1 mol)S° (CO(g)) + (2 mol)S° (H2(g)) – (1 mol)S° (CH3OH(g)) ΔS° rxn = (1)(197.67 J·K–1) + (2)(130.68 J·K–1) – (1)(239.81 J·K–1) = 219.22 J·K–1

ΔG° rxn = ΔH° rxn – TΔS° rxn = (90.13 × 103 J) – (1000 K)(219.22 J·K–1) ΔG° rxn = – 129 kJ At 1 bar total pressure, the reaction is likely be spontaneous at 1000 K, (if the molecules do not dissociate or undergo reactions in other ways) since, under these conditions, ΔGrxn = ΔG° rxn has a large negative value.

Problem 9.84. (a) From Problem 9.79, we have ΔH° rxn = 44.6 kJ·mol–1 and ΔS° rxn = 126 J·K–1·mol–1 for the vaporization of 2-propanol. We can use these values to calculate K at 35 ° C and hence the vapor pressure (in bar) for 2-propanol: –RTlnK(T) = ΔH° rxn – TΔS° rxn – (8.314 J·K–1·mol–1)·(308 K)lnK(308) = 44.6 × 103 J·mol–1 – (308 K)(126 J·K–1·mol–1) lnK(308) = –2.262; K(308) = 0.104 The vapor pressure of 2-propanol at skin temperature, 35 ° C, is 0.104 bar [= 78 torr), which makes it evaporate (vaporize) relatively rapidly from the skin. The process requires the enthalpy of vaporization, which is furnished by the skin and, hence, we feel a cooling sensation. See Chapter 1, Section 1.12.

(b) 12 g of 2-propanol, CH3CHOHCH3, is 12 g( )60 g ⋅ mol–1( ) = 0.20 mol. The amount of

energy that is required to vaporize this sample is (0.20 mol)·(44.6 kJ·mol–1) = 8.9 kJ. Thus, the thermal energy of your body is decreased by this amount when the alcohol evaporates.

Problem 9.85.



This data table for Problems 9.85 through 9.88 is repeated here for easy reference. The free energies are for the hydrolysis at pH 7 of the phosphorylated compounds to give the non-phosphorylated compound plus phosphate: compound-P + H2O ⇔ compound + Pi

(a) The hydrolysis reactions for glu-6-P and glu-1-P and the combination to give the isomerization reaction are: glu-6-P + H2O ⇔ glu + Pi – [glu-1-P + H2O ⇔ glu + Pi] ––––––––––––––––––––––––– glu-6-P ⇔ glu-1-P (b) We can combine the standard free energy changes for the hydrolysis reactions, just as we combined the reactions, to give ΔG° ′ rxn for the isomerization reaction: ΔG° ′ rxn = (–14 kJ·mol–1) – (–21 kJ·mol–1) = 7 kJ·mol–1 (c) The equilibrium constant and equilibrium constant expression for the isomerization can be obtained as:

lnK = ln(glu-1-P)

(glu-6-P)

⎛⎝⎜

⎞⎠⎟

= –ΔGrxn

′o

RT =

–7 ×103 J ⋅ mol–1

(8.314 J ⋅ K–1 ⋅mol –1)(298 K)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –2.8

K = (glu-1-P)

(glu-6-P) = 0.06

Thus the equilibrium ratio of glu-6-P to glu-1-P is 1/0.06 or about 17 to 1. Since the free energy change is positive for the isomerization, it makes sense that the reactant, glu-6-P should predominate at equilibrium.

Problem 9.86. (a) The hydrolysis reactions, their combination, ATP + Pyr ⇔ ADP + PEP (Pyr = pyruvate), the ΔG° ′ rxn for the combination, and the equilibrium constant are: ATP + H2O ⇔ ADP + Pi – [PEP + H2O ⇔ Pyr + Pi] ––––––––––––––––––––––––– ATP + Pyr ⇔ ADP + PEP

ΔG° ′ rxn = (–30 kJ·mol–1) – (–62 kJ·mol–1) = 32 kJ·mol–1

lnK = ln(ADP)(PEP)

(ATP)(Pyr)

⎛⎝⎜

⎞⎠⎟

= –ΔGrxn

′o

RT =

–32 ×103 J ⋅ mol –1

(8.314 J ⋅ K–1 ⋅mol –1)(298 K)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –12.9

K = (ADP)(PEP)

(ATP)(Pyr) = 2.5 × 10–6

Free energy of hydrolysis at pH 7 and 25 ° C

Compound ΔG° ′ , kJ·mol–

1

phosphoenolpyruvate, PEP –62

creatine phosphate –43

ATP (gives ADP) –30

glucose-1-phosphate, Glu-1-P –21

glucose-6-phosphate, Glu-6-P –14

glycerol 3 phosphate G3P 9



(b) If the (ATP)/(ADP) ratio in the cell is 10/1, and the reaction in part (a) is at equilibrium, we

can substitute this ratio in the equilibrium constant expression and solve for the (Pyr)(PEP) ratio.

The result is:

(Pyr)(PEP) = 4 × 104

The reaction in part (a) heavily favors the reactants, as the substantial positive free energy suggests. Indeed, in the glycolysis pathway in cells, ATP is produced by reaction with PEP (the favored reverse reaction).

Problem 9.87. For each of these reactions, imagine starting with a mixture of all the reactants and products at the same concentration in a pH 7 solution at 25 ° C. For each case, tell which direction the reaction will go in order to approach equilibrium. (i) ATP + creatine ⇔ ADP + creatine phosphate (ii) ATP + Glu ⇔ ADP + Glu-6-P For each case, we combine the hydrolysis reactions to give the desired reaction and then the free energies in the same way to find out whether ΔG° ′ rxn is positive or negative. If ΔG° ′ rxn is positive, the reaction favors reactants (is not spontaneous), so the mixture will go from equal concentrations of reactants and products toward a state with higher concentrations of reactants (that is, it will proceed in reverse of the direction it is written). If ΔG° ′ rxn is negative, the reaction favors products (is spontaneous), so the mixture will go from equal concentrations of reactants and products toward a state with higher concentrations of products (that is, it will proceed in the direction it is written). (i) ATP + H2O ⇔ ADP + Pi – [creatine phosphate + H2O ⇔ creatine + Pi] ––––––––––––––––––––––––––––––––––––––– ATP + creatine ⇔ ADP + creatine phosphate

ΔG° ′ rxn = (–30 kJ·mol–1) – (–43 kJ·mol–1) = 13 kJ·mol–1 This mixture will react to form reactants from products under the conditions stated.

(ii) ATP + H2O ⇔ ADP + Pi – [Glu-6-P + H2O ⇔ Glu + Pi] –––––––––––––––––––––––––––– ATP + Glu ⇔ ADP + Glu-6-P

ΔG° ′ rxn = (–30 kJ·mol–1) – (–14 kJ·mol–1) = –16 kJ·mol–1 This mixture will react to form products from reactants under the conditions stated.

Problem 9.88. [NOTE: The reactions written in this problem should be: Glu +2ADP + 2Pi + 2NAD+ ⇔ 2Pyr +2ATP + 2NADH + 2H2O 2Pyr + 6ATP + 2NADH + 6H2O ⇔ Glu + 6ADP + 6Pi + 2NAD+ + 4H+



This stoichiometry is consistent with reaction (9.62) for the hydrolysis of ATP and with half reaction (6.77) for reduction of NAD+. Using the charges on the species in equations (9.62) and (6.77), you can show that the equations here are balanced in both atoms and charge.]

(a) The stoichiometry of the glycolysis pathway in organisms is: Glu +2ADP + 2Pi + 2NAD+ ⇔ 2Pyr +2ATP + 2NADH + 2H2O The ΔG° ′ rxn, for glycolysis is about –84 kJ at 25 ° C. When a reaction is reversed, the signs of its thermodynamic changes are reversed. Thus, if the glycolysis reaction is considered in reverse, ΔG° ′ rxn would be about 84 kJ. This is a very unfavorable change. Glycolysis is not reversible, so gluconeogenesis must use a different pathway, for which ΔG° ′ rxn = –38 kJ at 25 ° C: 2Pyr + 6ATP + 2NADH + 6H2O ⇔ Glu + 6ADP + 6Pi + 2NAD+ + 4H+

(b) If we sum the glycolysis and gluconeogenesis reactions, we get: Glu +2ADP + 2Pi + 2NAD+ ⇔ 2Pyr +2ATP + 2NADH + 2H2O 2Pyr + 6ATP + 2NADH + 6H2O ⇔ Glu + 6ADP + 6Pi + 2NAD+ + 4H+ ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 4ATP + 4H2O ⇔ 4ADP + 4Pi + 4H+ The standard free energy change for the net reaction (hydrolysis of 4 moles of ATP per mole of glucose broken down and remade) is the sum of the standard free energy changes for the individual reactions: ΔG° ′ net rxn = (–84 kJ) + (–38 kJ) = –122 kJ The standard free energy change for hydrolysis of one mole of ATP is about –29 kJ, so the free energy change for hydrolysis of four moles is –116 kJ, which is just about what we get for the sum of the glycolysis and gluconeogenesis pathways. The difference between the two overall reactions is consistent with their free energy changes. (c) In Figure 7.17, the downward facing arrow on the far left and the upward facing arrow next to it could represent glycolysis, a reaction pathway that has an overall negative free energy and produces ATP. The upward facing arrow on the far right and the downward facing arrow next to it could represent gluconeogenesis, a reaction that has an overall negative free energy and uses (hydrolyzes) a good deal of ATP (note that the downward arrow is longer than the upward arrow) to produce this overall negative free energy.

Problem 9.89. (a) The reduction of pyruvate to lactate, Pyr + NADH + H+ ⇔ Lac + NAD+, is a fermentation reaction and is required to enable glycolysis to proceed in the absence of oxygen (which is required form the complete glucose oxidation pathway). As you find in Chapter 6, Section 6.11, and in Problem 9.88, glycolysis requires the presence of NAD+ for the oxidation of glucose to pyruvate. There is only a small amount NAD+ in a cell, so it has to be recycled from NADH in order to keep glycolysis going. Usually, NADH is oxidized back to NAD+ in reactions that involve oxygen as the ultimate oxidizing agent. If there is an insufficient supply of oxygen, as considered in this problem, then some other pathway is required to oxidize the NADH and fermentation is such a pathway. (b) The combination of glycolysis and this pyruvate fermentation (taken twice, because each glucose produces two pyruvate molecules to be reduced) is:



Glu +2ADP + 2Pi + 2NAD+ ⇔ 2Pyr +2ATP + 2NADH + 2H2O 2Pyr + 2NADH + 2H+ ⇔ 2Lac + 2NAD+ –––––––––––––––––––––––––––––––––––––––––––––––––––– Glu + 2ADP + 2Pi + 2H+ ⇔ 2Lac + 2ATP + 2H2O

(c) We had to combine two moles of the fermentation reaction to give the net reaction of one mole of glucose to lactate. Thus, with ΔG° ′ rxn = –84 kJ for glycolysis (from Problem 9.88) and ΔG° ′ rxn = –25 kJ for this fermentation, we find: ΔG° ′ net rxn = (–84 kJ) + 2(–25 kJ) = –134 kJ (for one mole of glucose reacting)

(d) The equilibrium constant expression for the net reaction in part (b), with the standard state for H+(aq) taken as pH 7, and also assuming (H2O) = 1, is:

K′ = (Lac)2(ATP)2

(Glu)(ADP)2(Pi)2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

eq

(e) We get the free energy change for the reaction from the relationship: ΔG′ net rxn = ΔG° ′ net rxn + RTlnQ′ Q′ is the reaction quotient (at pH 7), which has the same form as the equilibrium constant expression in part (d), but with the actual system dimensionless concentration ratios, not necessarily equilibrium concentrations. ΔG′ net rxn = (–134 × 103 J·mol–1

+ (8.314 J·K–1·mol–1)·(298 K)ln(5 ×10–5)2 (2 ×10–3)2

(5 ×10–3)(2 ×10–4)2(1 ×10–3)2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

ΔG′ net rxn = –124 kJ·mol–1 Since ΔG′ net rxn < 0, the reaction converting reactants at the specified concentrations to products at the specified concentrations is spontaneous. Glycolysis followed by fermentation is a favored process under cellular conditions.

Problem 9.90. (a) We can write the entropy for a gas at pressures P1 and P2, equation (9.83), and take the difference to get ΔS for the change of pressure. S2 = So – RlnP2 – [S1 = So – RlnP1] –––––––––––––––––––––––––––––––––– S2 –S1 = ΔS = R(lnP1 – lnP2) = Rln(P1/P2)

(b) When a gas is compressed, the final pressure, P2, is higher than the initial pressure, P1. Thus, P1/P2 < 1 and ln(P1/P2) < 0, that is, the entropy change [part (a)] is negative. When a gas is compressed at constant temperature, the final volume of the gas is less than the initial volume. We found in Chapter 8 that positional entropy decreases as the volume available to molecules decreases, so the entropy will decrease upon compression, just as the equation here also shows.

Problem 9.91. (a) The equilibrium constant expression for the reaction, 2NO2(g) ⇔ N2O4(g), and the equilibrium constant when the system is at equilibrium with NO2(g) and N2O4(g) pressures of 0.225 bar and 0.438 bar, respectively, are:



K = PN2O 4

1 bar( )PNO2

1 bar( )2 = 0.438

0.225( )2 = 8.65

(b) If the volume is doubled, the pressure is halved (at constant temperature). Thus, P(NO2) = 0.113 bar and P(N2O4) = 0.219 bar, and the reaction quotient is:

Q = 0.219

0.113( )2 = 17.2

(c) In order to move the reaction quotient toward the equilibrium value, the quotient has to decrease. This means that the numerator, P(N2O4), will decrease and the denominator, P(NO2), will increase. In other words, the reaction will go in reverse to produce more reactant. The stress on this system is lowering the pressure of the gases. The response of the system, according to Le Chatelier’s principle, will be to try to produce more moles of gas to compensate for the lowering of the concentrations (pressures). That means going in reverse to form two molecules of NO2(g) from one molecule of N2O4(g). The predictions from thermodynamics (reaction quotients) and Le Chatelier are the same. (d) If N2O4(g) reacts to give NO2(g), P(N2O4) will be reduced and P(NO2) will increase. After this occurs, we can write the pressures of the gases as P(N2O4) = 0.219 – x and P(NO2) = 0.113 + 2x, since the stoichiometry produces two NO2(g) for each N2O4(g) that reacts. Substitute these values into the equilibrium constant expression from part (a) and solve for x and hence the new equilibrium pressures.

8.65 = 0.219 – x( )

0.113 + 2x( )2 = 0.219 – x

0.0127 + 0.452x + 4x2

This equation can be rearranged to a quadratic form and solved: 34.6x2 + 4.91x – 0.109 = 0

x = −4.91 ± (4.91)2 – 4 ⋅ 34.6 ⋅ (–0.109)

2 ⋅ 34.6 = 0.0195 (the positive root)

Thus, P(N2O4) = 0.219 – 0.0195 = 0.1995 and P(NO2) = 0.113 + 2(0.0195) = 0.152. This calculation carries too many significant figures, but the result is quite sensitive to round-off uncertainties, so they are retained. Check to see if these are correct by substituting into the equilibrium expression:

K = 0.1995

0.152( )2 = 8.63

Within the round-off uncertainties of the calculation, this value for K is the same as in part (a), so we can have some confidence that the pressures we calculated are correct.

Problem 9.92. From Problem 9.13, K = 1.60 × 102 for the reaction, H2(g) + I2(g) ⇔ 2HI(g), at 500 K. If 4.00 × 10–3 mol of HI are placed in a 450. mL reaction vessel heated to 500 K and no reaction occurred, the pressure of HI(g) in the reaction vessel would be:

P(HI) = nRT

V =

(4.00 ×10−3 mol)(8.314 ×10-2 L ⋅ bar ⋅ K-1mol -1)(500 K)

0.450 L

P(HI) = 0.3695 bar



After reaction occurs and equilibrium is established, let the pressures of H2(g) and I2(g) be x bar. From the stoichiometry of the reaction, we know that P(HI) = (0.3695 – 2x) bar at equilibrium. Therefore, we have:

1.60 × 102 = 0.3695 − 2x( )2

x ⋅ x =

0.1365 –1.478x + 4x2

x2

156x2 + 1.478x – 0.1365 = 0

x = −1.478 ± (1.478)2 – 4 ⋅156 ⋅(–0.1365)

2 ⋅156 = 0.0252

You can check this result by substituting the pressures back into the equilibrium constant expression and finding that the numeric value is 160. The numeric value of x is the pressure of I2(g) (and H2(g)) in bar. Use the ideal gas equation once again to find what we are asked for, the number of moles of I2(g):

n = PV

RT =

(0.0252 bar)(0.450 L)

(8.314 ×10-2 L ⋅bar ⋅ K-1mol-1)(500 K) = 2.73 × 10–4 mol

Problem 9.93. The equilibrium proton transfer reaction, HOAc(aq) + H2O(aq) ⇔ OAc–(aq) + H3O+(aq), for acetic acid will be upset by the addition of hydrochloric acid, HCl(aq), to the system. The hydrochloric acid solution contains H3O+(aq) [and Cl–(aq)] that will increase the concentration of a product of the acetic acid system. The system will respond by reacting to use up some of the added H3O+(aq) by going in reverse. This will, of necessity, use up some of the OAc–(aq), so the equilibrium concentration of OAc–(aq) will decrease. In the new equilibrium state, the concentration of H3O+(aq) will be higher than in the initial equilibrium. This change can be characterized as common ion effect, where the common ion is H3O+(aq), which is present in both the acetic acid and hydrochloric acid solutions that are mixed.

Problem 9.94. [NOTE: The data in this problem are from Science, 2002, 297, 1665 and also reported in C&EN, September 9, 2002, page 35.]

(a) The reaction and its equilibrium constant expression (with BzH and BzF used to symbolize benzene and fluorobenzene, respectively) are: BzH(g) + CuF2(s) ⇔ BzF(g) + Cu(s) + HF(g)

K = (BzF(g))(Cu(s))(HF(g))

(BzH(g))(CuF2 (s)) = (BzF(g))(HF(g))

(BzH(g)) =

PBzF ⋅ PHF

PBzH

The dimensionless concentration ratios for the solids are unity (Table 9.1) and the numeric values of the concentration ratios for the gases are equal to the pressures of each gas in bar. From the stoichiometry of the reaction we know that PBzF = PHF in the reaction mixture. We also know, from the problem statement, that the total pressure of gases in the reactor is about 0.1 bar, that is, PBzH + PBzF + PHF = PBzH + 2PBzF = 0.1 bar. At 350 ° C (623 K) there is a 5% conversion of BzH to BzF, so PBzF/PBzH = 5%/95% = 1/19; PBzF = (1/19)PBzH 0.1 bar = PBzH + 2(1/19)PBzH = (21/19)PBzH



PBzH = (19/21)(0.1 bar) = .09 bar PBzF = PHF = (1/19)PBzH = 0.005 bar

K(623) = PBzF ⋅ PHF

PBzH

= (0.005)(0.005)

(0.09) = 0.0003

At 450 ° C (723 K) there is a 30% conversion of BzH to BzF, so PBzF/PBzH = 30%/70% = 3/7; PBzF = (3/7)PBzH 0.1 bar = PBzH + 2(3/7)PBzH = (13/7)PBzH PBzH = (7/13)(0.1 bar) = .054 bar PBzF = PHF = (3/7)PBzH = 0.023 bar

K(723) = PBzF ⋅ PHF

PBzH

= (0.023)(0.023)

(0.054) = 0.01

(b) We get the standard enthalpy change, ΔH° rxn, for this process from the temperature dependence of the equilibrium constant [see equation (9.59)], assuming that ΔH° rxn is independent of temperature over this temperature range:

lnK(723)

K(623)

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln

0.01

0.0003⎛ ⎝

⎞ ⎠ = ΔH rxn

o

8.314 J ⋅ K–1 ⋅ mol–1

⎛

⎝⎜

⎞

⎠⎟

723 K – 623 K

(723 K)(623 K)

⎛

⎝⎜

⎞

⎠⎟

ΔH° rxn = 1.3 × 102 kJ·mol–1

(c) We get the standard entropy change, ΔS° rxn, for this process from ΔH° rxn and one of the equilibrium constant values [see equation (9.54)], assuming that ΔS° rxn is independent of temperature over this temperature range: –RTlnK(T) = ΔH° rxn – TΔS° rxn – (8.314 J·K–1·mol–1)·(723 K)ln(0.01) = 1.3 × 105 J·mol–1 – (723 K)ΔS° rxn ΔS° rxn = 1.4 × 102 J·K–1·mol–1

(d) For 50% conversion, we have: PBzF/PBzH = 50%/50% = 1/1; PBzF = PBzH 0.1 bar = PBzH + 2PBzH = 3PBzH PBzH = (1/3)(0.1 bar) = .033 bar PBzF = PHF = PBzH = 0.033 bar

K(T) = PBzF ⋅ PHF

PBzH

= (0.033)(0.033)

(0.033) = 0.03

We need to find the temperature, T, at which the equilibrium constant is 0.03. We can do this by rearranging equation (9.54): –RTlnK = ΔH° rxn – TΔS° rxn TΔS° rxn – RTlnK = ΔH° rxn

T = ΔH rxno

ΔSrxno – R ln K

= 1.3 ×105 J ⋅ mol–1

(1.4 ×102 J ⋅ K –1 ⋅ mol–1) – (8.314 J ⋅ K–1 ⋅ mol–1)ln(0.03)



T = 770 K (≈ 500 ° C)

Problem 9.95. \The equilibrium constant expression for our generalized reaction, A + B ⇔ C + D, is

K = (C) c(D)d

(A)a(B)b

⎧ ⎨ ⎩

⎫ ⎬ ⎭ eq

[equation (9.47)]

Le Chatelier’s principle says that if we disturb the system at equilibrium, that is, when the concentrations give the correct K, by increasing the concentration of a reactant, the system will respond by reacting to use up some of this reactant (the disturbance), in order to restore equilibrium. In the equation, you see that increasing the concentration of a reactant, say A, without changing the concentrations of any other species, results in a reaction quotient, Q, with a numeric value less than K. In order to restore the system to a state where Q equals K, the numerator must increase and the denominator decrease. That is, some of the added A (as well as some B) will react to form more C and D. You can make similar arguments for any change in the concentrations, so Le Chatelier’s principle and the equilibrium constant expression give the same directionality for the change that occurs to restore equilibrium. The power of the equilibrium constant is that it allows you also to quantify the change.

When a system at equilibrium is disturbed by changing its temperature, Le Chatelier’s principle says that the reaction will proceed to form more products if the reaction is endothermic, ΔH° rxn > 0, and will proceed in reverse to form more reactants if the reaction is exothermic, ΔH° rxn < 0. The thermodynamic relationship between the equilibrium constant and ΔH° rxn is:

lnK = – ΔH rxno

R

⎛

⎝⎜

⎞

⎠⎟

1

T + ΔSrxn

o

R [equation (9.55)]

Here we see that lnK is a linear function of 1/T, if ΔH° rxn and ΔS° rxn are constants, independent of temperature. As T increases, 1/T decreases and we can ask what happens to lnK (or K). The direction of change of lnK, depends on the sign of the coefficient of 1/T, that is, –ΔH° rxn/R. It’s easiest to see the direction of the effect on plots of lnK vs. 1/T, as shown here.

0

1

0 11/T, K–1

lnK

ΔHo < 0

ΔHo > 0

If ΔH° rxn > 0, the coefficient is negative and the slope of the line is negative; lnK increases as T increases (as 1/T decreases). Conversely, if ΔH° rxn < 0, the coefficient is positive and the slope



of the line is positive; lnK increases as T decreases (as 1/T increases). Here we see that increasing the temperature of an endothermic reaction system at equilibrium, increases its equilibrium constant, that is, increases the concentrations of products relative to the concentrations of reactants. This is the direction of change predicted by Le Chatelier’s principle. Increasing the temperature of an exothermic reaction system at equilibrium, decreases its equilibrium constant, that is, increases the concentrations of reactants relative to the concentrations of products, just as Le Chatelier’s principle also predicts.

solutions for chapter 9 end-of-chapter problemsquantum.bu.edu/.../text/answers/chapter09.pdfmarch...

Documents