Chapter 15 471 15.1 The reaction proceeds, converting cis ...

Chapter 15

471

15.1 The reaction proceeds, converting cis-butene into trans-butene, but as the concentration of trans-butene builds up, the reverse reaction becomes increasingly important until, when the concentration ratio is 3, the rates are equal. Then there is no net change. A plot of concentration vs. time appears similar to Figure 15-2, with 0.75 atm for the final pressure of trans-butene and 0.25 atm for the final pressure of cis-butene:

15.2 The reaction proceeds, converting trans-butene into cis-butene, but as the concentration of cis-butene builds up, the reverse reaction becomes increasingly important until, when the concentration ratio is 1/3, the rates are equal. Then there is no net change. A plot of concentration vs. time appears similar to Figure 15-2, with 0.75 atm for the final pressure of trans-butene and 0.25 atm for the final pressure of cis-butene:

Chapter 15

472

15.3 (a) According to the principle of reversibility, every elementary reaction can run in either direction, so as the system approaches equilibrium, the reverse reactions become important:

Cl- (aq) + ClO2- (aq) k− 1 → ClO- (aq) + ClO- (aq)

Cl- (aq) + ClO3- (aq) k− 2 → ClO- (aq) + ClO2- (aq)

(b) The net reaction is 3 ClO- (aq) 2 Cl- (aq) + ClO3- (aq) . By inspection, the

equilibrium constant expression is

Keq =[ClO3

- ]eq[Cl - ]eq2

[ClO- ]eq3

(c) The net reaction is the sum of the two elementary steps, so the equilibrium constant is

the product of the rate ratios of the two steps: Keq = k1k2

k− 1k− 2.

15.4 (a) According to the principle of reversibility, every elementary reaction can run in either

direction, so the decomposition of H2CO is described by the reverse reactions:

H2CO k− 3 → HCO + H

HCO k− 2 → H + CO

2 H k− 1 → H2

(b) The net reaction is H2 + CO H2CO . By inspection, the equilibrium constant

expression is Keq =

[H2CO]eq

[H2 ]eq[CO]eq

(c) The net reaction is the sum of the three elementary steps, so the equilibrium constant

is the product of the rate ratios of the steps: Keq = k1k2k3

k− 1k− 2k− 3.

15.5 A molecular picture of an elementary reaction shows the reactants, the products, and (if

necessary) the intermediate collision complex.

Chapter 15

473

15.6 A molecular picture of an elementary reaction shows the reactants, the products, and (if necessary) the intermediate collision complex.

15.7 To test the reversibility of a reaction, set up a system containing the products and observe

whether or not reactants form. Here, a solution containing Cl- and ClO3- ions should react to form some ClO- ions. (If the reaction of ClO- to form Cl- and ClO3- goes virtually to completion, this experiment may not succeed.)

15.8 To test the reversibility of a reaction, set up a system containing the products and observe

whether or not reactants form. Here, if a reactor is charged with formaldehyde gas, H2CO, H2 and CO should form. (If the reaction between H2 and CO to form H2CO goes virtually to completion, this experiment may not succeed.)

15.9 Equilibrium constant expressions can be written by inspection of the overall

stoichiometry, remembering that pure liquids and solids do not appear in the expression. Equilibrium constant expressions contain product concentrations over reactant concentrations raised to their stoichiometric coefficient:

(a) 5eq2F

2eq5IF

eq )()(

pp

K = ; (b) ( )5

eq2Oeq

1p

K = ; (c) 2eqCOeq )( pK = ;

(d) 2eq2HeqCO

eq )()(1

ppK = ; (e)

eq43

eq-3

43eq

+3

eq ]POH[][PO]O[H

=K .


stoichiometry, remembering that pure liquids and solids do not appear in the expression. Equilibrium constant expressions contain product concentrations over reactant concentrations raised to their coefficient:

(a)

3eq2OeqS2H

eq2SO

eq )()()(pp

pK = ; (b)

3eqCO

3eq2CO

eq )()(

pp

K = ; (c) eq2Cl

3eq2F

2eq3ClF

eq )()()(pp

pK = ;

(d)

Keq =[NH4

+ ]eq

[H3O+ ]eq[NH3 ]eq; (e) 2

eq2Heq )(

1 p

K = .

Chapter 15

474


stoichiometry, remembering that pure liquids and solids do not appear in the expression: (a) 2IF5 (g) I2 (s) + 5 F2 (g)

2eq5IF

5eq2F

eq )()(

pp

K = ;

(b) P4O10(s) P4 (s) + 5 O2 (g) ( )5eq2Oeq pK =

;

(c) BaO (s) + 2 CO (g) BaCO3 (s) + C (s) 2eqCO

eq )(1

pK = ;

(d) CH3OH(l) CO (g) + 2 H2 (g) 2eq2HeqCOeq )()( ppK =

;

(e) PO43-(aq) + 3 H3O+(aq) H3PO4(aq) + 3 H2O(l) eq

-34

3eq

+3

eq43eq ][PO]O[H

]POH[=K .


stoichiometry, remembering that pure liquids and solids do not appear in the expression: (a) 2 H2O (l) + 2 SO2 (g) 2 H2S (g) + 3 O2 (g)

eq2SO

3eq2OeqS2H

eq )()()(

ppp

K = ;

(b) 2 Fe (s) + 3 CO2 (g) Fe2O3 (s) + 3 CO (g) 3eq2CO

3eqCO

eq )()(

pp

K = ;

(c) 2 ClF3 (g) Cl2 (g) + 3 F2 (g) 2eq3ClF

eq2Cl3eq2F

eq )()()(

ppp

K = ;

(d) NH4+ (aq) + H2O (l) NH3 (g) + H3O+ (aq)

Keq =[H3O+ ]eq[NH3 ]eq

[NH4+ ]eq

;

(e) Sn (s) + 2 H2O (l) SnO2 (s) + 2 H2 (g) 2eq2Heq )( pK = .

15.13 The standard states of gases are gases at 1 atm, the standard states of solutes are solutions

at 1 M, and the standard states of solvents and pure liquids and solids are unit mole fractions, X = 1. (a) p = 1 atm for F2 and IF5, X = 1 for I2; (b) p = 1 atm for O2, X = 1 for P4 and P4O10; (c) p = 1 atm for CO, X = 1 for others; (d) p = 1 atm for CO and H2, X = 1 for CH3OH;

(e) c = 1 M for H3PO4, H3O+, and PO43-, X = 1 for H2O. 15.14 The standard states of gases are gases at 1 atm, the standard states of solutes are solutions

at 1 M, and the standard states of solvents and pure liquids and solids are unit mole fractions, X = 1.

(a) p = 1 atm for SO2, O2, and H2S, X = 1 for H2O;

Chapter 15

475

(b) p = 1 atm for CO and CO2, X = 1 for Fe and Fe2O3; (c) p = 1 atm for each gas;

(d) c = 1 M for H3O+ and NH4+, P = 1 atm for NH3, X = 1 for H2O; (e) p = 1 atm for H2, X = 1 for others. 15.15 A molecular picture of an equilibrium between phases shows the concentrations of the

species and arrows indicating transfers between phases:

15.16 A molecular picture of an equilibrium between phases shows the concentrations of the

species and arrows indicating transfers between phases:

15.17 Table 15-1 indicates that the solubility of O2 in water increases as the temperature falls

from 25 °C to 0 °C. Use Henry's law to calculate the concentration at each temperature:

[O2(aq) ]eq

(pO2)eq

= KH, so [O2(aq) ]eq = (pO2)eq KH

The atmosphere is 22 % O2, so the partial pressure of O2 is 0.22 atm. At 25 °C, [O2(aq) ]eq = (0.22 atm)(1.3 x 10-3) = 2.9 x 10-4 M; At 0 °C, [O2(aq) ]eq = (0.22 atm)(2.5 x 10-3) = 5.5 x 10-4 M;

Chapter 15

476

The percentage change is: (100 %)M10 x 5.5

M10 x 2.9-M10 x 5.54-

-4-4 = 47%.

15.18 To determine how much water a fish must process, first find the concentration of O2 at

the two temperatures. From Table 15-1, the solubility of O2 in water decreases as temperature increases. Use Henry's law to calculate the concentration at each temperature:

[O2(aq) ]eq

(pO2)eq

= KH, so [O2(aq) ]eq = (pO2)eq KH

The atmosphere is 22 % O2, so the partial pressure of O2 is 0.22 atm. At 25 °C, [O2(aq) ]eq = (0.22 atm)(1.3 x 10-3) = 2.86 x 10-4 M; At 30 °C, [O2(aq) ]eq = (0.22 atm)(8.9 x 10-4) = 1.96 x 10-4 M;

At 25 °C, 1.00 L of water contains 2.86 x 10-4 mol of O2. To obtain this amount of O2 at

30 °C, a fish requires

mol 10 x 1.96L 1

mol 10 x 2.86 4-4- = 1.5 L of water, a 50% increase.

15.19 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ∆Goreaction from tabulated values for ∆Gf

o and then use Equation 15-3:

∆Go = –RT ln Keq (a) ∆Goreaction = [1 mol(0 kJ/mol) + 1 mol(–394.4 kJ/mol)]

– [1 mol(–137.2 kJ/mol) + 1 mol(–237.1 kJ/mol)] = – 20.1 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J 10 x 2.01-1-1-

-14

= 8.11;

Keq = e8.11 = 3.3 x 103; (b) ∆Goreaction = 2 mol(–394.4 kJ/mol)

– [1 mol(0 kJ/mol) + 2 mol(–137.2 kJ/mol)] = – 514.4 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 5.144-1-1-

-15

= 207.6;

Keq = e207.6 =1 x 1090; (c) ∆Goreaction = [1 mol(–520.3 kJ/mol) + 2 mol(–137.2 kJ/mol)]

– [1 mol(0 kJ/mol) + 1 mol(–1134.4 kJ/mol)] = 339.7 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 3.3971-1-

-15

= –137.1;

Keq = e-137.1 = 3 x 10-60; (d) ∆Goreaction = [3 mol(–237.1 kJ/mol) + 1 mol(74.62 kJ/mol)]

– [6 mol(0 kJ/mol) + 3 mol(–137.2 kJ/mol)] = –225.1 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 2.251-1-1-

-15

= 90.86;

Chapter 15

477

Keq = e90.86 = 2.9 x 1039. 15.20 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ∆Goreaction from tabulated values for ∆Gf

o and then use Equation 15-3:

∆Go = –RT ln Keq (a) ∆Goreaction = [3 mol(0 kJ/mol) + 1 mol(–137.2 kJ/mol)]

– [1 mol(–50.5 kJ/mol) + 1 mol(–237.1 kJ/mol)] = 150.4 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 1.5041-1-

-15

=–60.7;

Keq = e-60.7 = 4 x 10-27; (b) ∆Goreaction = [4 mol(87.6 kJ/mol) + 6 mol(–237.1 kJ/mol)]

–[4 mol(–16.4 kJ/mol) + 5 mol(0 kJ/mol)] = – 1006.6 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J x101.0066-1-1-

-16

= 406.3;

Keq = e406.3 = 3 x 10176; (c) ∆Goreaction = [2 mol(–237.1 kJ/mol) + 1mol(0 kJ/mol)]

– [1 mol(–515.8 kJ/mol) + 2 mol(0 kJ/mol)] = 41.6 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 4.161-1-

-14

= –16.8;

Keq = e-16.8 = 5.1 x 10-8; (d) ∆Goreaction = [1 mol(–1015.4 kJ/mol) + 4 mol(0 kJ/mol)]

– [3 mol(0 kJ/mol) + 4 mol(–237.1 kJ/mol)] = –67.0 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J10 x 6.70-1-1-

-14

= 27.0;

Keq = e27.0 =5.3 x 1011. 15.21 To estimate the equilibrium constant at a temperature different from 298 K, calculate

∆Horeaction and ∆Soreaction at 298 K and then use Equations 13-9 and 15-3: ∆Go = ∆Ho– T∆So ∆Go = –RT ln Keq

15.19 (b) ∆Horeaction = 2 mol(–393.5 kJ/mol)

– [1 mol(0 kJ/mol) + 2 mol(–110.5 kJ/mol)] = – 566.0 kJ/mol; ∆Soreaction = 2 mol(213.8 J/mol K)

– [1 mol(205.15 J/mol K) + 2 mol(197.7 J/mol K)] = – 173.0 J/mol K; ∆Goreaction, 250K = (–566.0 kJ/mol) – (250 K)(–0.1730 kJ/mol K) = –522.8 kJ/mol

ln Keq = – K) )(250K mol J (8.314

mol J x105.228-1-1-

-15

= 251.5;

Keq = e251.5 =2 x 10109;

Chapter 15

478

15.19 (c) ∆Horeaction = 1 mol(–548.0 kJ/mol) + 2 mol(–110.5 kJ/mol)

– [1 mol(0 kJ/mol) + 1mol(–1213.0 kJ/mol)] = 444.0 kJ/mol; ∆Soreaction = [1 mol(72.1 J/mol K) + 2 mol(197.7 J/mol K)]

–[ 1 mol(5.7 J/mol K) + 1 mol(112.1 J/mol K)] = 349.7 J/mol K; ∆Goreaction, 250K = (444.0 kJ/mol) – (250 K)(0.3497 J/mol K) = 356.6 kJ/mol

ln Keq = – K) )(250K mol J (8.314

mol J10 x 356.61-1-

-13

= –171.6;

Keq = e-171.6 =3 x 10-75 15.22 To estimate the equilibrium constant at a temperature different from 298 K, calculate


15.20 (c) ∆Horeaction = [2 mol(–285.83 kJ/mol) + 1 mol(0 kJ/mol)] – [1 mol(–577.6 kJ/mol) + 2 mol(0 kJ/mol)] = 5.9 kJ/mol;

∆Soreaction = [2 mol(69.95 kJ/mol) + 1 mol(51.2 kJ/mol)] – [1 mol(49.0 J/mol K) + 2 mol(130.680 J/mol K)] = -119.3 J/mol K;

∆Goreaction, 350K = (5.9 kJ/mol) – (350 K)(–0.1193 J/mol K) = 47.7 kJ/mol

ln Keq = – K) K)(350 J/mol (8.314

J/kJ)kJ/mol)(10 (47.7 3

= – 16.4;

Keq = e-16.4 = 7.5 x 10-8;

15.20 (d) ∆Horeaction = [1 mol(–1118.4 kJ/mol) + 4 mol(0 kJ/mol)] – [3 mol(0 kJ/mol) + 4 mol(–285.83 kJ/mol)] = 24.9 kJ/mol;

∆Soreaction = [1 mol(146.4 J/mol K) + 4 mol(130.680 J/mol K)] – [3 mol(27.3 J/mol K) + 4 mol(69.95 J/mol K)] = 307.4 J/mol K;

∆Goreaction, 350K = (24.9 kJ/mol) – (350 K)(0.3074 J/mol K) = –82.7 kJ/mol

ln Keq = – K) K)(350 J/mol (8.314

J/kJ)kJ/mol)(10 (-82.7 3

= 28.4;

Keq = e28.4 = 2.2 x 1012 15.23 To estimate the equilibrium constant at a temperature different from 298 K, calculate

∆Horeaction and ∆S rxn

o at 298 K and then use Equations 13-9 and 15-3:

∆Go = ∆Ho– T∆So ∆Go = –RT ln Keq 15.19 (a)

∆Horeaction = [1 mol(–393.5 kJ/mol) + 1 mol(0 kJ/mol)] – [1 mol(–110.5 kJ/mol) + 1 mol(–241.83 kJ/mol)] = – 41.2 kJ/mol;

∆Soreaction = [1 mol(213.8 J/mol K) + 1 mol(130.680 J/mol K)] – [1 mol(197.7 J/mol K) + 1 mol(188.835 J/mol K)] = –42.1 J/mol K;

Chapter 15

479

∆Goreaction, 395K = (–41.2 kJ/mol) – (395 K)(–0.0421 kJ/mol K) = –24.6 kJ/mol

ln Keq = – K) )(395K mol J (8.314

mol J x102.46-1-1-

-14

= 7.49;

Keq = e7.49 =1.8 x 103;

15.19 (d) ∆Horeaction = 1 mol(20.0 kJ/mol) + 3 mol(–241.83 kJ/mol)

– [3 mol(–110.5 kJ/mol) + 6mol(0 kJ/mol)] = -374.0 kJ/mol; ∆Soreaction = [1 mol(226.9 J/mol K) + 3 mol(188.835 J/mol K)]

–[3 mol(197.7 J/mol K) + 6 mol(130.680 J/mol K)] = -583.8 J/mol K; ∆Goreaction, 395K = (-374.0 kJ/mol) – (395 K)(-0.5838 J/mol K) = -143.4 kJ/mol

ln Keq = – K) )(395K mol J (8.314

mol J10 x 1.434-1-1-

-15

= 43.67;

Keq = e43.67 = 9.2 x 1018 15.24 To estimate the equilibrium constant at a temperature different from 298 K, calculate


15.20 (c)

∆Horeaction = [1 mol(0 kJ/mol) + 2 mol(–241.83 kJ/mol)] – [1 mol(–577.6 kJ/mol) + 2 mol(0 kJ/mol)] = 93.9 kJ/mol;

∆Soreaction = [1 mol(51.2 J/mol K) + 1 mol(188.835 J/mol K)] – [1 mol(49.0 J/mol K) + 2 mol(130.680 J/mol K)] = –70.3 J/mol K;

∆Goreaction, 425K = (93.9 kJ/mol) – (425 K)(–0.0703 kJ/mol K) = 123.8 kJ/mol

ln Keq = – K) )(425K mol J (8.314

mol J x101.2381-1-

-15

= –35.0;

Keq = e–35.0 =6.3 x 10-16;

15.20 (d) ∆Horeaction = 1 mol(-1118.4 kJ/mol) + 4 mol(0 kJ/mol)

– [3 mol(0 kJ/mol) + 4 mol(-241.83 kJ/mol)] = -151.1 kJ/mol; ∆Soreaction = [1 mol(146.4 J/mol K) + 4 mol(130.680 J/mol K)]

–[3 mol(27.3 J/mol K) + 4 mol(188.835 J/mol K)] = -168.1 J/mol K; ∆Goreaction, 425K = (-151.1 kJ/mol) – (425 K)(-0.1681 J/mol K)

= -79.7 kJ/mol

ln Keq = – K) )(425K mol J (8.314

mol J10 x 7.97-1-1-

-14

= 22.6;

Keq = e22.6 = 6.5 x 109

Chapter 15

480

15.25 Use Le Châtelier's principle to determine the effect on an equilibrium position caused by adding one reagent:

In reactions (a), (b), and (d), CO is a reactant, so adding CO causes the reaction to go to the right, forming products; in reaction (c), CO is a product, so adding CO causes the reaction to proceed to the left, forming reactants.

15.26 Use Le Châtelier's principle to determine the effect on an equilibrium position caused by

adding one reagent: In reactions (a) and (d), H2O is a reactant, so adding H2O causes the reaction to go to the

right, forming products; in reactions (b) and (c), H2O is a product, so adding H2O causes the reaction to proceed to the left, forming reactants.

15.27 Use Le Châtelier's principle to determine the effect on an equilibrium position caused by

a change in conditions: (a) Solid reactants do not appear in the equilibrium constant expression, so adding PbCl2

has no effect; (b) Addition of water dilutes the solution, reducing the concentrations of the product ions,

so more of the solid will dissolve; (c) Addition of NaCl increases the concentration of one of the product ions, Cl-, so some

solid will precipitate; (d) Addition of KNO3 does not change the concentrations in the equilibrium constant

expression, so adding this solid has no effect. 15.28 Use Le Châtelier's principle to determine the effect on an equilibrium position caused by

a change in conditions: (a) Increasing the temperature causes the equilibrium position to shift in the endothermic

direction. The reaction is exothermic in the forward direction, so increasing the temperature shifts the equilibrium position in the direction of reactants, and the amount of SO3 drops;

(b) Addition of O2 increases the concentration of one of the reactants, so more SO3 will form;

(c) Addition of Ar gas leaves the concentrations of all reagents unchanged, so this addition has no effect on the amount of SO3 present.

15.29 Use Le Châtelier's principle to determine what changes in conditions will drive an

equilibrium in any given direction. This reaction can be driven to the left by removing SO2, removing Cl2, adding SO2Cl2, or increasing the temperature.

15.30 Use Le Châtelier's principle to determine what changes in conditions will drive an

equilibrium in any given direction. This reaction can be driven to the left by removing PCl5, adding Cl2, adding PCl3, or decreasing the temperature.

15.31 To calculate an equilibrium constant when amounts are available, convert amounts data

into concentrations or pressures using stoichiometric reasoning, then complete an amounts table and substitute into the equilibrium constant expression. Generally,

Chapter 15

481

pressures in atmospheres are required, but in this problem, the equilibrium constant expression contains p2 in the numerator and denominator, so units cancel and amounts can be used directly. Set up an amounts table, using stoichiometric reasoning and the fact that the change is 0.49 mol for CO:

Reaction: H2 + CO2 H2O + CO

Initial (mol) 1.00 1.00 0 0

Change (mol) –0.49 –0.49 + 0.49 + 0.49

Equilibrium (mol) 0.51 0.51 0.49 0.49

Now substitute into the equilibrium constant expression and evaluate K:

Keq =

( pH2O )eq ( pCO )eq

( pH2)eq ( pCO2

)eq=

(nH2O)eq (nCO )eq

(nH2)eq (nCO2

)eq= (0.49)2

(0.51)2 = 0.92

15.32 To calculate an equilibrium constant when experimental data concerning amounts are

available, convert amounts data into concentration or pressures using stoichiometric reasoning, then complete a concentration table and substitute into the equilibrium constant expression.

For gases, pressures in atmospheres are required. Set up a concentration table, using stoichiometric reasoning and the fact that the change is 0.93 atm for Cl2:

Reaction: 4 HCl + O2 2 Cl2 + 2 H2O

Initial (atm) 2.30 1.00 0 0

Change (atm) –1.86 –0.465 + 0.93 + 0.93

Equilibrium (atm) 0.44 0.535 0.93 0.93


Keq = )535.0((0.44)

)93.0()93.0()()()()(

4

22

eqO4eqHCl

2eqCl

2eqOH

2

22 =pppp = 37

15.33 To calculate an equilibrium constant when experimental data concerning amounts are

available, identify the reaction, convert amounts data into concentration using stoichiometric reasoning, then complete a concentration table and substitute into the equilibrium constant expression. For solutes, concentrations must be in mol/L.

[C2H5CO2H]initial = L 0.500mol 0.0500 = 0.100 M

To complete the concentration table, use stoichiometric reasoning and the fact that the change is 1.15 x 10-3 M for H3O+:

Chapter 15

482

Reaction: H2O + C2H5CO2H C2H5CO2- + H3O+

Initial (M) ---- 0.100 0 0

Change (M) ---- –1.15 x 10-3 + 1.15 x 10-3 +1.15 x 10-3

Equilibrium (M) ---- 0.099 1.15 x 10-3 1.15 x 10-3


Ka =

[C2H5CO2- ]eq[H3O

+ ]eq

[C2H5CO2H]eq= (1.15 x 10-3)2

(0.099) = 1.3 x 10-5

15.34 To calculate an equilibrium constant when experimental data concerning concentrations

are available, identify the reaction, then complete a concentration table and substitute into the equilibrium constant expression. To complete the concentration table, use stoichiometric reasoning and the fact that the change is 6.5 x 10-3 M for H3O+:

Reaction: H2O + HCNO CNO- + H3O+

Initial (M) ---- 0.20 0 0

Change (M) ---- –6.5 x 10-3 + 6.5 x 10-3 + 6.5 x 10-3

Equilibrium (M) ---- 0.19 6.5 x 10-3 6.5 x 10-3

Now substitute into the equilibrium constant expression and evaluate Keq:

Ka = (0.19)

)10 x 5.6([HCNO]

]O[H][CNO 2-3

eq

eq+

3eq-

= = 2.2 x 10-4

15.35 To calculate concentrations at equilibrium from initial conditions, set up a concentration

table. For a gas phase reaction, concentrations must be expressed in atm. Let x = change in [H2]:

Reaction: H2 + Br2 2 HBr

Initial (atm) 0 0 10.0

Change (atm) + x + x –2x

Equilibrium (atm) x x 10.0 – 2x

Now substitute into the equilibrium constant expression and solve for x:

Chapter 15

483

Keq = ))((

)2 - (10.0)()(

)( 2

eqBreqH

2eqHBr

22xx

xpp

p= = 1.6 x 105

To simplify, assume that 2x << 10.0:

1.6 x 105 = 2

2

)((10.0)

x so

x2 = 510 x 1.6

100 = 6.25 x 10-4

x = 2.5 x 10-2 = ( pH2)eq = ( pBr2 )eq ;

(pHBr)eq = 10.0 – 2(2.5 x 10-2) = 10.0 atm 2(2.5 x 10-2) = 0.050 << 10.0, so the approximation is valid. 15.36 To calculate concentrations at equilibrium from initial conditions, set up a concentration

table. For a gas phase reaction, concentrations must be expressed in atm. Let x = change in [CO]:

Reaction: CO + Cl2 COCl2

Initial (atm) 0 0 0.250

Change (atm) + x + x –x

Equilibrium (atm) x x 0.250 – x


Keq = 2eqCleqCO

eqCOCl - 0.250)()(

)(

2

2

xx

ppp

= = 1.5 x 108; assume that x << 0.250:

1.5 x 108 = 2

0.250x

so

x2 = 810 x 1.5

0.250 = 1.67 x 10-9

x = 4.1 x 10-5 = ( pCO )eq = ( pCl2 )eq ;

eqCOCl )(2

p = 0.250 – 4.1 x 10-5 = 0.250 atm

4.1 x 10-5 << 0.250, so the approximation is valid. 15.37 To calculate concentrations at equilibrium from initial conditions, set up a concentration

table. For a gas phase reaction, concentrations must be expressed in atm. Let x = change in [CO2]:

Reaction: FeO(s) + CO(g) CO2 (g) + Fe(s)

Initial (M) excess 5.0 0 ----

Chapter 15

484

Change (M) ---- –x + x ----

Equilibrium (M) ---- 5.0 – x x ----


Keq = )- (5.0)(

)(

eqCO

eqCO 2

xx

pp

= = 0.403

x = (0.403)(5.0 – x) = 2.015 – 0.403 x so 1.403 x = 2.015 and x= 1.436

( pCO2

)eq = 1.4 atm

( pCO)eq = 5.0 – 1.4 = 3.6 atm. 15.38 To calculate concentrations at equilibrium from initial conditions, set up a concentration

table. For a gas phase reaction, concentrations must be expressed in atm. Use the ideal gas equation to calculate the initial pressure of Cl2 gas:

p =

nRTV

= mRT( MM )V

= (5.0 g)(0.08206 L atm / mol K )(1200 K)(70.90g / mol)(3.0 L)

= 2.3 atm

Let –x = change in [Cl2]:

Reaction: Cl2(g) 2Cl (g)

Initial (M) 2.3 0

Change (M) –x + 2x

Equilibrium (M) 2.3 – x 2x


Keq = )- (2.3

)(2)()( 2

eqCl

2eqCl

2x

xpp

= = 2.5 x 10-5; assume x << 2.3

4x2 = (2.5 x 10-5)(2.3) = 5.75 x 10-5 so

x2 = 1.44 x 10-5 and x = 3.8 x 10-3

( pCl )eq = 3.8 x 10-3 atm ( pCl2 )eq = 2.3 atm

0.0038 << 2.3, so the approximation is valid.

Chapter 15

485

15.39 To identify species in solution, first identify the nature of the solute. Strong acids, strong bases, and salts generate ions, while all other substances remain molecular: (a) Weak acid, major species are H2O and CH3CO2H; (b) salt, major species are H2O, NH4+ and Cl-; (c) salt, major species are H2O, K+ and Cl-; (d) salt, major species are H2O, Na+ and CH3CO2-; and (e) strong base, major species are H2O, Na+ and OH-.

15.40 To identify species in solution, first identify the nature of the solute. Strong acids, strong

bases, and salts generate ions, while all other substances remain molecular: (a) Weak acid, major species are H2O and HClO; (b) salt, major species are H2O, Ca2+ and Br-; (c) salt, major species are H2O, K+ and ClO-; (d) strong acid, major species are H2O, H3O+ and NO3-; and (e) weak acid, major species are H2O and HCN.

15.41 The equilibria among major species depend on the nature of the species. The proton

transfer reaction of water always plays a role: H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) (a) weak acid equilibrium: CH3CO2H (aq) + H2O (l) CH3CO-(aq) + H3O+(aq) (b) ammonium ion is the weak conjugate acid of NH3: NH4+ (aq) + H2O (l) NH3 (aq) + H3O+(aq) (c) There are no equilibria other than the water equilibrium; (d) acetate ion is the weak conjugate base of acetic acid: CH3CO2-(aq) + H2O(l) CH3CO2H (aq) + OH-(aq) (e) There are no equilibria other than the water equilibrium. 15.42 The equilibria among major species depend on the nature of the species. The proton

transfer reaction of water always plays a role: H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) (a) weak acid equilibrium: HClO (aq) + H2O (l) ClO-(aq) + H3O+(aq) (b) There are no equilibria other than the water equilibrium; (c) ClO- ion is the conjugate base of HClO, so is a weak base: ClO-(aq) + H2O(l) HClO (aq) + OH-(aq) (d) There are no equilibria other than the water equilibrium; (e) weak acid equilibrium: HCN(aq) + H2O (l) CN-(aq) + H3O+(aq) 15.43 To identify species in solution, first identify the nature of the solute. Strong acids, strong

bases, and salts generate ions, while all other substances remain molecular:

Chapter 15

486

(a) major species are (CH3)2CO (acetone) and H2O; (b) salt, major species are H2O, K+, and Br-; (c) strong base, major species are H2O, Li+, and OH-; (d) strong acid, major species are H2O, H3O+, and HSO4-.

15.44 To identify species in solution, first identify the nature of the solute. Strong acids, strong

bases, and salts generate ions, while all other substances remain molecular: (a) salt, major species are H2O, Na+, and HCO3-; (b) major species are CH3OH (methanol) and H2O; (c) strong acid, major species are H2O, H3O+, and Br-; (d) weak acid, major species are C6H5CO2H (benzoic acid) and H2O.

15.45 Equilibrium constant expressions have the concentrations of the products in the

numerator and the concentrations of the reactants in the denominator with each concentration raised to the power of its stoichiometric coefficient. Remember to omit liquids and solids from the expressions. If the reaction deals with a weak acid (or base) reacting with water it will be related to Ka (or Kb if a base). If the reaction deals with a solid it will be related to Ksp.

(a) K =

[ClO2- ][H3O

+][HClO2]

= Ka;

(b) K = 33 ]][OH[Fe1

−+ = sp

1K

;

(c) K =

[HCN]

[CN-][H3O+]

=

1Ka

.

15.46 Equilibrium constant expressions have the concentrations of the products in the

numerator and the concentrations of the reactants in the denominator with each concentration raised to the power of its stoichiometric coefficient. Remember to omit liquids and solids from the expressions. If the reaction deals with a weak acid (or base) reacting with water it will be related Ka (or Kb if a base). If the reaction deals with a solid it will be related to Ksp. (a) K = [Ag+ ]2[SO4

2 − ]= Ksp ;

(b) K =

[H3O+ ][H2PO4

− ][H3PO4 ]

= a1

1K

;

(c) K =

[HCO2H]

[HCO2− ][H3O

+ ]=

1Ka

.

15.47 To identify the spectator ions present first determine the reaction (if any) that occurs

when the solutions are mixed. Whatever does not react will then be a spectator ion.

(a) CH3CO2H + OH- CH3CO2- + H2O; spectator ions : Na+;

Chapter 15

487

(b) 3 Ca2+ + 2 PO43− Ca3(PO4)2; spectator ions : Cl- and K+;

(c) H3O+ + OH- 2 H2O; spectator ions: K+ and NO3

- . 15.48 To identify the spectator ions present first determine the reaction (if any) that occurs

when the solutions are mixed. Whatever does not react will then be a spectator ion.

(a) Fe3+ + 3 OH- Fe(OH)3; spectator ions : Cl- and Na+; (b) NH3 + H3O

+ NH4+ + H2O; spectator ions : NO3

- ; (c) H3O

+ + CH3CO2- CH3CO2H + H2O; spectator ions: K+ and ClO 4

- . 15.49 Equilibrium constant expressions are determined by the stoichiometry of the overall

reaction, with pure liquids, solids, and solvent omitted from the expression. Equilibrium constant expressions have the concentrations of the products in the numerator and the concentration of the reactants in the denominator with each concentration raised to the power of its stoichiometric coefficient:

(a) Keq = eqOHeqCO )()(22

pp ; (b) Keq = 2eqN

3eqO

4eqNH

)()()(

2

23

ppp ; (c) Keq =

eqO2eqHC

2eqCHOCH

)()()(

242

3

ppp ;

(d) Keq = eq-2

42eq

+ ]SO[]Ag[ ; (e) Keq = eqNHeqSH )()(

32pp .

15.50 Equilibrium constant expressions are determined by the stoichiometry of the overall

reaction, with pure liquids, solids, and solvent omitted from the expression. Equilibrium constant expressions have the concentrations of the products in the numerator and the concentration of the reactants in the denominator with each concentration raised to the power of its stoichiometric coefficient:

(a) Keq = 5eqO

4eqNH

4eqNO

)()()(

23pp

p ; (b) Keq =

[CN- ]eq [H3O+ ]eq

[HCN]eq; (c) Keq =

1

[NH4+ ]eq

2 [SO42- ]eq

;

(d) Keq =

[OH-]eq[HSO4- ]eq

[SO42-]eq

; (e) Keq = 2eqO

3eqO

)()(

3

2

pp .

15.51 This problem describes an equilibrium reaction. We are asked to determine the

equilibrium pressures of all the gases. To calculate pressures at equilibrium from initial conditions, set up a concentration table, write the K expression, and solve for the pressures. For a gas phase reaction, concentrations must be expressed in atm. Use PV = nRT to calculate the initial pressure of phosgene:

P = atm 3.6L 2.55

K) 273 )(360 06mol)(0.082 (0.31 K molatm L

=+

Chapter 15

488

Reaction: COCl2 CO + Cl2

Initial (atm) 6.3 0 0

Change (atm) – x + x + x

Equilib(atm) 6.3 - x x x

Now substitute into the equilibrium constant expression and solve for x :

Keq = 8.3 x 10-4 = x

xp

pp– 6.3)(

)()()( 2

eqCOCl

eqCleqCO

2

2 = ; Assume that x << 6.3:

8.3 x 10-4 = 3.6

2x, so

x2 = (6.3)(8.3 x 10-4) = 5.23 x 10-3;

x = 7.2 x 10-2 atm = eqCleqCO )()(2

pp = ;

eqCOCl )(2

p = 6.3 – 7.2 x 10-2 = 6.2 atm;

7.23 x 10-2 is about 1.1% of 6.3, so x is << 6.3 and the approximation is valid. 15.52 This problem describes an equilibrium reaction. We are asked to determine the

equilibrium pressures of all the gases. To calculate pressures at equilibrium from initial conditions, set up a concentration table, write the K expression, and solve for the pressures.

Let –x = change in [N2]:

Reaction: N2 + 3 H2 2 NH3

Initial (atm) 5.0 3.0 0

Change (atm) – x –3 x + 2x

Equilib(atm) 5.0 – x 3.0 – 3 x 2x


Keq = 2.81 x 10-5 = 3

2

3eqHeqN

2eqNH

)3– )(3.0– (5.0)(2

)()()(

22

3

xxx

ppp

= ; Assume that 3x << 3.0:

2.81 x 10-5 = 3

2

(5.0)(3.0)4x , so

4x2 = (5.0)(3.0)3(2.81 x 10-5) = 3.79 x 10-3; x = 3.08 x 10-2

( pNH3)eq = 2(3.08 x 10-2) = 6.2 x 10-2 atm;

Chapter 15

489

3(3.08 x 10-2) = 0.09, about 3% of 3.0, so 3 x is << 3.0 and the approximation is valid.

15.53 Equilibrium exists when the rates of evaporation and condensation are equal. A test tube has

a smaller surface area than a petri dish, so fewer molecules escape per unit time from the test tube. However, the rate of escape per unit surface area is the same for both samples, so equilibrium is established at the same pressure (molecules colliding per unit surface area) for both samples.

15.54 At equilibrium between a solid salt and solution, the rate at which ions leave the surface of

the solid equals the rate at which ions are captured from solution at the surface of the solid. Adding solid to a solution at equilibrium increases the total surface area, so the overall rate at which ions leave becomes greater. However, the rate at which ions are captured also becomes greater. The rate per unit surface area is unchanged, so the concentration remains the same.

15.55 At equilibrium, a molecular picture should show the presence of both reactants and

products, in relative amounts that are determined by the value of the equilibrium constant. Set up a "concentration" table to determine how many of each species are present at equilibrium:

Initial 12 12 0 0

Change – x – x + x + x

Equilibrium 12 – x 12 – x x x

Substitute in the equilibrium constant expression and solve for x:

Keq = 25 = 2

2

) - (12)(x

x ; Taking the square root of each side gives 5 = )– (12 x

x;

(60 – 5 x ) = x , 6 x = 60 and x = 10 The molecular picture should show (12 – 10) = 2 of each reactant and 10 of each product:

Chapter 15

490

15.56 At equilibrium, a molecular picture should show the presence of both reactants and

products, in relative amounts that are determined by the value of the equilibrium constant. Set up a "concentration" table to determine how many of each species are present at equilibrium:

Initial 12 12 0

Change – x – x + 2 x

Equilibrium 12 – x 12 – x 2 x

Substitute in the equilibrium constant expression and solve for x:

Keq = 4 = ) - )(12 - (12

)(2 2

xxx

; from which 4(12 – x)(12 – x) = 4x2

Cancel a factor of 4 and multiply out: 144 – 24 x + x 2 = x 2, or 24 x = 144 and x = 6;

The molecular picture should show (12 – 6) = 6 of each reactants, and 2(6) = 12 product molecules:

15.57 To calculate equilibrium pressures, set up a concentration table. In this problem, the

table can be short, because two of the three equilibrium pressures are provided:

Reaction: Br2 (g) + I2 (g) 2 IBr (g)

Equilib(atm) 0.512 0.327 x

Substitute into the equilibrium constant expression and solve for x :

322 =

[IBr]eq2

[Br2]eq[I2]eq =

)327.0)(512.0(

2x, from which

x2 = (322)(0.512)(0.327) = 53.91; x = [IBr]eq = 7.34 atm.

Chapter 15

491

15.58 To calculate equilibrium pressures, set up a concentration table. In this problem, the table can be short, because one equilibrium pressure is provided and the others must be stoichiometrically related:

Reaction: C (s) + 2 H2O (g) CO2 (g) + 2 H2 (g)

Equilib(atm) ----- 2.80 x 102 x 2x

Substitute into the equilibrium constant expression and solve for x:

0.38 = 2

2

2eq2

2eq2eq2

)280()2(

O][H][H][CO xx= =, from which

4x3 = (0.38)(280)2 = 2.98 x 104; x3 = 7.4 x 103, x = 19.5;

[CO2]eq = 20. atm and [H2]eq = 2(19.5) = 39 atm.

15.59 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ∆Gorxn from tabulated values for ∆Go

f; calculate ∆Horxn and ∆So

rxn at 298K and then use Equations 13-9 and 15-3 to estimate the equilibrium constant at a temperature different from 298 K:

∆Go = ∆Ho– T∆So ∆Go = –RT ln Keq at 298 K: ∆Goreaction = 1 mol(99.8 kJ/mol) – 2 mol(51.3 kJ/mol) = – 2.8 kJ/mol;

ln Keq = – K) )(298K mol J (8.314

mol J 10 x 2.8-1-1-

-13

= 1.1;

Keq = e1.1 = 3.0; At 525 K:

∆Horeaction = 1 mol(11.1 kJ/mol) – 2 mol(33.2 kJ/mol) = – 55.3 kJ/mol; ∆Soreaction = 1 mol(304.4 J/mol K) – 2 mol(240.1 J/mol K) = –175.8 J/mol K;

∆Goreaction = (–55.3 kJ/mol) – (525 K)

J 1kJ10

K mol 1J 175.8- -3

= 37.0 kJ/mol

ln Keq = – K) )(525K mol J (8.314

mol J 10 x 3.701-1-

-14

= – 8.48;

Keq = e-8.48 = 2.1 x 10-4 Notice that the exothermic reaction has a smaller Keq at higher temperature. 15.60 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ∆Gorxn from tabulated values for

∆Gf

o; calculate ∆Horxn and ∆So

rxn at

Chapter 15

492

298 K and then use Equations 13-9 and 15-3 to estimate the equilibrium constant at a temperature different from 298 K:

∆Go = ∆Ho– T∆So ∆Go = –RT ln Keq

at 298 K: ∆Goreaction = 4 mol(87.6 kJ/mol)

– [2 mol(103.7 kJ/mol) + 1 mol(0 kJ/mol)] = 143.0 kJ/mol;

ln Keq = –K) )(298K mol J (8.314

mol J 10 x 143.01-1-

-13

= –57.7; Keq = e–57.7 = 8.7 x 10-26;

at 825 K: ∆Horeaction = 4 mol(91.3 kJ/mol) – [2 mol(81.6 kJ/mol) + 1 mol(0 kJ/mol)]

= 202.0 kJ/mol; ∆Soreaction = 4 mol(210.8 J/mol K)

– [2 mol(220.0 J/mol K) + 1 mol(205.152 J/mol K)] = 198.0 J/mol K;

∆Goreaction = (202.0 kJ/mol) – (825 K)

J 1kJ10

K mol 1J 198.0 -3

= 38.7 kJ/mol

ln Keq = – K) )(825K mol J (8.314

mol J 10 x 38.71-1-

-13

= –5.64;

Keq = e-5.64 = 3.6 x 10-3

Notice that the endothermic reaction has a larger Keq at higher temperature. 15.61 (a) The general weak base reaction is B + H2O BH+ + OH-, and B = (CH3)3N:

(CH3)3N + H2O (CH3)3NH+ + OH-, Kb =

[(CH3)3NH+ ]eq[OH- ]eq

[(CH3)3N]eq;

(b) The general weak acid reaction is HA + H2O A- + H3O+, and HA = HF:

HF + H2O F- + H3O+, Ka =

[F-]eq[H3O+ ]eq

[HF]eq;

(c) A solubility reaction involves solid dissolving to produce ions:

CaSO4 (s) Ca2+(aq) + SO42-(aq), Ksp = [Ca2+ ]eq[SO42- ]eq .

15.62 (a) This is a gas solubility equilibrium, CO(g) CO(aq): KH =

[CO]eq

( pCO )eq

(b) The general weak base reaction is B + H2O BH+ + OH-, and B = (C6H5)NH2:

(C6H5)NH2 + H2O (C6H5)NH3+ + OH-, Kb =

[(C6H5)NH3+ ]eq [OH-]eq

[(C6H5)NH2]eq;

(c) A solubility reaction involves solid dissolving to produce ions:

Ca(OH)2 (s) Ca2+(aq) + 2OH-(aq), Ksp = [Ca2+ ]eq[OH- ]eq

2 .

Chapter 15

493

15.63 To predict effects of changes on equilibrium position, apply Le Châtelier's principle: The

system will respond in the direction that reduces the effect of the change. The reaction in example 15-15 is: 2 NO (g) + O2 (g) 2 NO2 (g),

(a) NO2 is a product, so reducing its pressure causes the reaction to proceed to the right; (b) There are more gases on the reactant side, thus by doubling the volume the reaction

proceeds to the left; (c) Adding Ar does not change any of the pressures in the equilibrium expression, so this

change has no effect. 15.64 To predict effects of changes on equilibrium position, apply Le Châtelier's principle: The

system will respond in the direction that reduces the effect of the change. The reaction in example 15-13 is: HA + H2O A- + H3O+, where HA is benzoic acid and A- is the conjugate base, benzoate anion. (a) Adding sodium benzoate will increase the concentration of benzoate anion (A-) in the

solution, causing the reaction to proceed to the left; (b) Increasing the volume of the solution will reduce the concentrations of all solutes.

There are more solutes on the product side, thus the reaction will proceed to the right; (c) Adding NaCl does not change any of the concentrations in the equilibrium

expression, so this change has no effect. 15.65 To determine the volume that will contain a single molecule of SnH4 we first need to find

the equilibrium pressure of the gas. To calculate equilibrium pressures, set up a concentration table:

Reaction: Sn (s) + 2 H2 (g) SnH4 (g)

Initial (atm) ---- 200 0

Change (atm) ---- –2 x + x

Equilib(atm) ---- 200 – 2 x x

Because Keq is very small, we assume that 2x << 200:

1.07 x 10-33 =

[SnH4]eq

[H2]eq2

= 2(200)x

,

from which x = (1.07 x 10-33)(200)2 = 4.28 x 10-29

Thus [SnH4]eq = 4.28 x 10-29 atm (a very small pressure). To calculate the volume that would be expected to contain a single molecule, use the

ideal gas equation, rearranged to solve for V:

V = atm)10 x )(4.28mol10 x (6.022

K) )(298 0821.0(29-1-23

K molatm L

=pN

RT

A

= 9.49 x 105 L.

Chapter 15

494

15.66 To calculate concentrations at equilibrium from initial conditions, set up a concentration table. For a gas phase reaction, concentrations must be expressed in atm. Let –x = change in [H2]:

Reaction: H2 + F2 2 HF


Change (atm) – x – x + 2x

Equilib(atm) 3.00 – x 3.00 – x 2x


Keq = 115 = )+ 6.00 - (9.00

4)- )(3.00- (3.00

)(2)()(

)(2

22

eqFeqH

2eqHF

22xx

xxx

xpp

p==

4x2 = (115)(9.00 – 6.00x + x2) = 1035 – 690 x + 115 x2 0 = 111 x2 – 690 x + 1035

− b ± b2 − 4ac2a

= 690 ± (690)2 − 4(111)(1035)2(111)

=

690 ± 16560222

= 3.69 or 2.53

The initial pressures are 3.00 atm, so a change of 3.69 atm is impossible; thus x = 2.53

( pH2)eq = ( pF2

)eq = 3.00 – 2.53 = 0.47 atm;

( pHF )eq = 2(2.53) = 5.06 atm. 15.67 To calculate an equilibrium constant from initial and equilibrium conditions, set up a

concentration table:

Reaction: CCl4 (g) 2 Cl2 (g) + C (s)

Initial (atm) 1.00 0 ----

Change (atm) – x + 2x ----

Equilib(atm) 1.00 – x 2x ----

The problem gives the total equilibrium pressure, which is the sum of partial pressures: 1.35 atm = (1.00 – x) + 2 x = 1.00 + x, from which x = (1.35 – 1.00) = 0.35; Substitute into the equilibrium constant expression and calculate K:

Keq = 65.070.0

)35.000.1()]35.0)(2[(

)()( 22

eqCCl

2eqCl

4

2 =−

=pp

= 0.75.

Chapter 15

495

15.68 When the volume of a gas system changes, all partial pressures change proportionally. This generates a new set of "initial" conditions that form the starting point for a new concentration table. When volume doubles, pressure is cut in half:

pCCl4 = 265.0

= 0.325 atm;

pCl2 = 270.0

= 0.35 atm. To return to equilibrium, the system responds in the direction

that increases pressure, so pCl2 increases.

Reaction: CCl4 (g) 2 Cl2 (g) + C (s)

Initial (atm) 0.325 0.35 ----

Change (atm) – x + 2x ----

Equilib(atm) 0.325 – x 0.35 + 2x ----

Substitute into the equilibrium constant expression and solve for x:

0.75 = )325.0()235.0( 2

xx

−+

, from which (0.75)(0.325 – x) = (0.35 + 2x)2

0.24375 – 0.75 x = 0.1225 + 1.40 x + 4 x2, or 4 x2 + 2.15 x – 0.12125 = 0

x =

− b ± b2 − 4ac2a

= − 2.15 ± (2.15)2 − 4(4)(− 0.12125)2(4)

=

− 2.15 ± 4.6225 + 1.948

;

x = 0.0515; (pCCl4)eq = (0.325 – 0.0515) = 0.27 atm; (pCl2)eq = [0.35 + 2(0.0515)] = 0.45 atm.

15.69 (a) The equilibrium constant expression for a reaction can be written by inspection of the

stoichiometry of the reaction, omitting pure liquids, solids, and solvents:

Keq =[CO3

2- ]eq

( pCO2)eq [OH-]eq

2

(b) Use Le Châtelier's Principle to predict the effect of changes on a system at equilibrium. Dissolving Na2CO3 leads to an increase in the concentration of CO32-, so the equilibrium shifts to the left and the pressure of CO2 increases.

(c) At first glance, it may appear that HCl will not affect this equilibrium, but recall that HCl is a strong acid which generates H3O+ in solution. This, in turn, will react with OH-, reducing the concentration of a reactant. Again the equilibrium shifts to the left and the pressure of CO2 increases.

15.70 To calculate concentrations at equilibrium from initial conditions, set up a concentration

table. For a gas phase reaction, concentrations must be expressed in atm.

Chapter 15

496

Use the ideal gas equation to calculate the initial pressure of PCl5 gas:

p = L) )(3.00mol g (208.22

K) )(395 g)(0.08206 (2.001-

K molatm L

==(MM)V

mRTV

nRT = 0.1038 atm

Let x = change in pCl2:

Reaction: PCl5(g) PCl3(g) + Cl2(g)


Change (atm) – x + x + x

Equilib(atm) 0.1038 – x x x


Keq = )– (0.1038

)()(

)()( 2

eqPCl

eqCleqPCl

5

23

xx

ppp

= = 1.83 x 10-3;

x2 = (1.83 x 10-3)(0.1038 – x) = 1.90 x 10-4 – (1.83 x 10-3)x x2 + (1.83 x 10-3)x – 1.90 x 10-4 = 0

x = )72.3(2

)10 x 90.1(4)10 x 83.1()10 x 83.1(2

4 -42-3-32 +±−=−±−a

acbb

x = 2

)10 x 63.7()10 x 83.1( -4-3 ±− =

2)10 x 76.2()10 x 83.1( -2-3 +−

= 1.29 x 10-2

( pPCl3 )eq = ( pCl2 )eq = 1.29 x 10-2 atm, and

( pPCl5 )eq = 0.1038 – 0.0129 = 0.0909 atm. 15.71 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ∆Goreaction from tabulated values for ofG∆ ; to estimate the equilibrium

constant at a temperature different from 298 K, calculate ∆Horeaction and ∆Soreaction at 298 K and then use Equations 13-9 and 15-3:

∆Go = ∆Ho – T∆So ∆Go = –RT ln Keq (a) ∆Goreaction = 1 mol(–210.7 kJ/mol)

– [1 mol(31.8 kJ/mol) + 1 mol(–178.6 kJ/mol)] = – 63.9 kJ/mol;

ln Keq = –K) )(298K mol J (8.314

mol J 10 x 6.39-1-1-

-14

= 25.8;

Keq = e25.8 = 1.6 x 1011; (b) When the equilibrium pressure is 1.00 atm, Keq = 1, ln(Keq) = 0 and ∆Go

rxn = 0:

0 = ∆Ho – T∆So, so T∆So =∆Ho and T = o

o

SH

∆∆

∆Horeaction = 1 mol(–265.4 kJ/mol)

Chapter 15

497

– [1 mol(61.4 kJ/mol) + 1 mol(–224.3 kJ/mol)] = – 102.5 kJ/mol; ∆Soreaction = 1 mol(191.6 J/mol K)

– [1 mol(175.0 J/mol K) + 1 mol(146.0 J/mol K)] = – 129.4 J/mol K;

T =

−

J 129.4K mol 1

kJ 1J10

mol 1kJ 102.5- 3

= 792 K

(c) ∆Goreaction, 1050K = (–102.5 kJ/mol) – (1050 K)

− −

J 1kJ 10

K mol 1J 129.4 3

= 33.4 kJ/mol

ln Keq = – K) )(1050K mol J (8.314

mol J 10 x 3.341-1-

-14

= –3.83;

Keq = e-3.83 = 2.2 x 10-2

Notice that Keq decreases as T increases for this exothermic reaction. 15.72 To relate an equilibrium constant to rate constants for elementary steps, set forward and

reverse reaction rates equal to each other, rearrange to obtain equilibrium expressions, and then multiply the expressions:

H2SO4 SO3 + H2O Rate constants: k1, k-1

k1[H2SO4]eq = k-1[SO3]eq[H2O]eq, so

k1k− 1

=[SO3]eq [H2O]eq

[H2SO4]eq

SO3 + C6H6 C6H6SO3 Rate constants: k2, k-2

k2[SO3]eq[C6H6]eq = k-2[C6H6SO3]eq, so

k2k− 2

=[C6H6SO3]eq

[SO3]eq[C6H6]eq

C6H6SO3 + H2O C6H5SO3- + H3O+ Rate constants: k3, k-3 k3[C6H6SO3]eq[H2O]eq = k-3[H3O+]eq[C6H5SO3-]eq so

k3k− 3

=[H3O+ ]eq [C6H5SO3

- ]eq

[H2O]eq[C6H6SO3]eq

Multiply these three expressions and cancel concentration terms:

k1k2k3k− 1k− 2k− 3

=[H3O+ ]eq [C6H5SO3

- ]eq

[C6H6]eq [H2SO4]eq= Keq.

15.73 To find an equilibrium total pressure, it is necessary to calculate equilibrium partial

pressures of all gaseous participants. First determine the initial pressures of the gases, using the ideal gas equation:

p = L 1.00

K) 273)(102006mol)(0.082 (0.500 K molatm L +=

VnRT = 53.1 atm

Chapter 15

498

Set up a concentration table, write the equilibrium expression and solve for the pressure:

Reaction: C(s) + CO2 (g) 2 CO (g)

Initial (atm) ---- 53.1 53.1

Change (atm) ---- – x + 2 x

Equilib(atm) ---- 53.1 - x 53.1 + 2 x

Substitute into the equilibrium constant expression and solve for x :

167.5 = )– (53.1)2 + (53.1 2

xx

(167.5)(53.1 – x ) = (53.1 + 2 x )2 8894 – 167.5 x = 2820 + 212.4 x + 4 x 2; 4 x 2 + 379.9 x – 6074 = 0

x = )4(2

)6074)(4(4)9.379(9.3792

4 22 −−±−=−±−a

acbb = 8

4.4919.379 ±− = 13.9;

Rule out the negative value, which would give a negative pressure; ( pCO2

)eq = 53.1 – 13.9 = 39.2 atm;

( pCO )eq = 53.1 + 2(13.9) = 80.9 atm; Ptotal = 39.2 + 80.9 = 1.20 x 102 atm. 15.74 To calculate an equilibrium constant from initial and equilibrium conditions, set up a


Reaction: 2 Ef (g) + 3 N2 (g) 2 EfN3 (g)


Change (atm) – 2 x – 3 x + 2 x

Equilib(atm) 0.75 – 2 x 1.00 – 3 x 2 x

The problem gives the total equilibrium pressure, which is the sum of partial pressures: 0.85 = (0.75 – 2 x ) + (1.00 – 3 x ) + 2 x = 1.75 – 3 x , from which

3 x = (1.75 – 0.85) = 0.90; Thus, x = 0.30; pEf = 0.75 – 2(0.30) = 0.15 atm pN2 = 1.00 – 3(0.30) = 0.10 atm pEfN3 = 2(0.30) = 0.60 atm

Chapter 15

499

Ptotal = 0.15 + 0.10 + 0.60 = 0.85 atm which matches the experimental pressure

Substitute into the equilibrium constant expression and calculate Keq:

Keq =

[EfN3]eq2

[Ef ]eq2 [N2]eq

3 = (0.60)2

(0.15)2(0.10)3 = 1.6 x 104.

15.75 This molecular picture illustrates starting conditions and equilibrium conditions. The

symbols indicate BG3 as the starting material and G2 and GB as products. Count numbers of symbols to determine initial and equilibrium concentrations. Set up a concentration table using numbers of symbols:

Initial 15 0 0

Change -12 +12 +12

Equilibrium 3 12 12

(a) From the amounts of change, the stoichiometry is 1:1, so the net reaction is

; (b) Use numbers of symbols at equilibrium to calculate the equilibrium constant:

Keq =

[GB]eq[G2]eq

[BG3]eq= (12)(12)

(3)= 48 .

15.76 (a) Follow the five-step procedure for solving equilibrium problems: 1.) Species are acetic acid (A) and its dimer (A2) 2.) The reaction is 2 A(g) A2(g)

3.) Keq =

( pA2 )eq

( pA )eq2

=3.72;

4.) The total pressure at equilibrium is given, Ptotal = 0.75 atm. Let x = pA 2

; then pA = 0.75 – x

5.) Substitute into the equilibrium constant expression and solve for x :

3.72 = 2) - (0.75 xx

Solve for x using the quadratic equation. x = (3.72)(0.75 – x )2 = (3.72)( x 2 – 1.50 x + 0.563) = 3.72 x 2 – 5.58 x + 2.093; 0 = 3.72 x 2 – 6.58 x + 2.093;

Chapter 15

500

x =

− b ± b2 − 4ac2a

= 6.58 ± (6.58)2 − 4(3.72)(2.093)2(3.72)

=

6.58 ± 43.296 − 31.1447.44

x =

6.58 ±3.497.44

= 1.35 or 0.415;

Rule out 1.35, which would give a negative pressure for the monomer; ( pA2

)eq = 0.415 atm

(b) The reaction to form dimer is exothermic, so raising the temperature will shift the equilibrium position to the left. The equilibrium constant is lower at 200 °C.

15.77 (a) To calculate an equilibrium constant from initial and equilibrium conditions, set up a


Reaction: C (s) + CO2 (g) 2 CO (g)

Initial (atm) --- 0.464 0

Change (atm) --- – x + 2x

Equilibrium (atm) --- 0.464 – x 2x

The problem gives the equilibrium pressure, which is the sum of partial pressures: 0.746 atm = (0.464 – x) + 2 x = 0.464 + x, from which

x = (0.746 – 0.464) = 0.282;

Substitute into the equilibrium constant expression and calculate K:

Keq = 182.0

2564.0)282.0464.0(

2)]282.0)(2[(

eq)(

2eq)(

2CO

CO

=−

=p

p = 1.75

(b) If the container is compressed to one third of its initial volume, then the pressure of

CO2 should triple by the ideal gas law. pCO2 = 3(0.464 atm) = 1.392 atm

Reaction: C (s) + CO2 (g) 2 CO (g)

Initial (atm) --- 1.392 0

Change (atm) --- – x + 2x

Equilibrium (atm) --- 1.392 – x 2x

Write the equilibrium expression and solve for the pressures of the gases:

Chapter 15

501

Keq = 1.75 = x

xx

xpp

−=

−=

392.14

392.1)2(

)()( 22

eq2CO

2eqCO

x = 0.592 atm pCO = 2x = 2(0.592 atm) = 1.184 atm; pCO2 = 1.392 atm - x = 1.392 atm – 0.592 atm = 0.800 atm

The equilibrium pressure is the sum of the partial pressures:

Peq = 1.184 atm + 0.800 atm = 1.984 atm 15.78 (a) To calculate an equilibrium constant when experimental data concerning amounts are

available, convert amounts data into concentration using stoichiometric reasoning, then complete an amounts table and substitute into the equilibrium constant expression.

Generally, pressures in atmospheres are required for gas-phase reactions, but in this problem, the equilibrium constant expression contains p2 in the numerator and denominator, so units cancel and amounts can be used directly. Set up a concentration table, using stoichiometric reasoning and the fact that the change is 0.182 mol for I2:

Reaction: 2 HI H2 + I2

Initial (mol) 1.00 0 0

Change (mol) -2(0.182) + 0.182 + 0.182

Equilibrium (mol) 0.636 0.182 0.182

Now substitute into the equilibrium constant expression and evaluate Keq:

Keq =

( pH2)eq ( pI2

)eq

( pHI )eq2 =

(nH2)eq (nI2 )eq

(nHI )eq2 = (0.182)2

(0.636)2 = 0.0819.

(b) To estimate the Keq using thermodynamical values we want to use the equation:

∆Go = -RTlnKeq ∆Gf

o for H2, I2, and HI are 0.00, 19.3, and 1.7 kJ/mol respectively. ∆Greaction

o = 1 mol(19.3 kJ/mol)+ 1 mol(0.0 kJ/mol) – 2 mol(1.7 kJ/mol) = 15.9 kJ/mol

ln Keq = 13.2K) )(898.15Kmol J (8.314

mol J 159001-1-

-1o

−=−=∆−RTG

Keq = e-2.13 = 0.12 (c) Given an initial amount of 1.00 mol HI, use the ideal gas law to determine the pressure of HI initially:

pHI = L 1.00

K) 273)(62506mol)(0.082 (1.00 K molatm L +=

VnRT = 73.7 atm

Chapter 15

502

Construct a concentration table, write the equilibrium expression, and solve for the pressures.

Reaction: 2 HI H2 + I2


Change (atm) -2x + x + x

Equilibrium (atm) 73.7 - 2x x x

Using the Keq expression calculate the pressure of H2 formed.

Keq = 2

2

2eqHI

eqIeqH

)2-(73.7)(

)(

)()(22

xx

p

pp= = 0.12; use the quadratic equation to solve for x

x = 15.1 pH2 = pI2 = 15.1 atm pHI = 73.7 – 2(15.1) = 43.5 atm

15.79 (a) Gas pressures and solubilities are connected through the equilibrium described by

Henry's law: KH =

[CO2(aq )]eq

( pCO2)eq

, from which [CO2(aq)]eq = KH ( pCO2)eq ;

From Table 15-1, KH (25 °C) = 3.4 x 10-2;

[CO2(aq)]eq = (3.4 x 10-2)(1.10) = 0.0374 M (remember that in all equilibrium constant expressions, solute concentrations are molarities)

Complete the calculation using standard stoichiometric methods:

n = VM =

−

L 1mol 0.0374

mL 1L10

mL 2253

= 0.00842 mol;

m = n MM = 0.00842 mol

mol 1g 44.01

= 0.37 g.

(b) From part (a), [CO2(aq)]eq = 3.74 x 10-2 M;

From Table 15-1, KH (0.0 °C) = 7.8 x 10-2;

( pCO2)eq = 2

2

10 x 7.810 x 3.74

−

−

= 0.48 atm.

15.80 This problem describes an equilibrium reaction pertinent to smog. Use standard

procedures to determine the partial pressures of each pollutant. Begin by converting all the initial pressures to the same units. Because K is so large, take the reaction to completion and then determine the equilibrium pressures:

Chapter 15

503

Reaction: O3 + NO → O2 + NO2

Initial (ppm) 6.5 x 10-3 1.5 2.1 x 105 0

Change (ppm) -6.5 x 10-3 -6.5 x 10-3 +6.5 x 10-3 +6.5 x 10-3

Completion (ppm) 0 1.5 2.1 x 105 6.5 x 10-3

Use these values in a concentration table, write the equilibrium expression, and solve for the equilibrium pressures:

Reaction: O3 + NO O2 + NO2

Initial (ppm) 0 1.5 2.1 x 105 6.5 x 10-3

Change (ppm) + x +x - x - x

Equilibrium (ppm) x 1.5 + x 2.1 x 105 6.5 x 10-3 - x

The equilibrium expression is:

Keq = 6.0 x 1034 = )x1.5(

)10 x )(6.52.1x10( 35

NOO

NOO

3

22

+−=

−

xx

pppp

; assume x<<6.5 x 10-3

x = 1.5 x 10-32 atm = pO3; assumption is valid pNO = 1.5 ppm

pNO2 = 6.5 x 10-3 ppm pO2 = 2.1 x 105 ppm 15.81 The molecular picture of the equilibrium conditions shows 4 O2, 5 SO3, and 5 SO2

molecules. From this, calculate the "equilibrium constant" for the reaction:

Keq =[SO3]eq

2

[SO2]eq2 [O2]eq

= (5)2

(5)2(4)= 0.25

(a) To determine what happens under the new conditions, compare amounts of molecules with equilibrium amounts.

Equilibrium conditions: 4 O2, 5 SO3, and 5 SO2 molecules. The new picture shows 4 O2, 3 SO3, and 5 SO2 molecules. The amounts of reactants are the same as before, but there are fewer product molecules,

so Q < Keq and the reaction will proceed to the right, forming additional SO3.

(b) To determine the effect of temperature, compare amounts of molecules at equilibrium for the two temperatures.

Equilibrium conditions at 1100 K: 4 O2, 5 SO3, and 5 SO2 molecules.

Chapter 15

504

Equilibrium conditions at 1300 K: 5 O2, 3 SO3, and 7 SO2 molecules. The equilibrium shifts to the left, forming additional reactant molecules, when

temperature increases. According to Le Châtelier's principle, the reaction shifts in the endothermic direction (absorbs heat) when temperature increases. Therefore, the forward reaction is exothermic.

(c) To determine what happens under the new conditions, compare amounts of molecules

with equilibrium amounts. Equilibrium conditions: 4 O2, 5 SO3, and 5 SO2 molecules. The new picture shows 4 O2, 9 SO3, and 9 SO2 molecules. Qualitative reasoning is insufficient to decide what happens, because the number of

molecules of a reactant and a product both have increased. Compare Q for the second set of conditions with Keq, calculated from the equilibrium conditions:

Keq =[SO3]eq

2

[SO2]eq2 [O2]eq

= (5)2

(5)2(4)= 0.25 and

Q =[SO3]2

[SO2]2[O2]= (9)2

(9)2(4)= 0.25 ;

The two values are identical, so the new conditions represent equilibrium, and no change will occur.

15.82 (a) This problem describes an equilibrium reaction pertinent to sulfuric acid synthesis.

Use standard procedures to determine the partial pressures of each gas. Because K is large, take the reaction to completion and then determine the equilibrium pressures. Since information is given about both reactants this is a limiting reactant problem.

Divide the pressures by the stoichiometric coefficients to determine the limiting reactant:

SO2: 2350.0 = 0.175 atm (LR)

O2: 0.762 atm Use a concentration table to determine the pressures at completion:

Reaction: 2 SO2 + O2 → 2 SO3

Initial (atm) 0.350 0.762 0

Change (atm) -0.350 –½(0.350) + 0.350

Completion (atm) 0 0.587 0.350

Use these values in a new concentration table, write the equilibrium expression, and solve for the equilibrium pressures:

Reaction: 2 SO2 + O2 2 SO3

Initial (atm) 0 0.587 0.350

Chapter 15

505

Change (atm) +2x +x –2x

Equilibrium (atm) 2x 0.587 + x 0.350–2x

The equilibrium expression is:

Keq = 5.60 x 104 = )587.0()2(

)2350.0(2

2

O2SO

2SO

22

3

xxx

ppp

+−= ; assume 2x<<0.350

5.60 x 104 = )587.0()2(

)350.0(2

2

x; rearrange to get:

x2 = )587.0)(10 x 60.5(4

)350.0(4

2

= 9.32 x 10-7 from which

x = 9.65 x 10-4 Now solve for the partial pressures of the three gases:

pSO2 = 2(9.65 x 10-4) = 1.93 x 10-3 atm pO2 = 0.587 + 9.65 x 10-4 = 0.588 atm pSO3 = 0.350 – 2(9.65 x 10-4) = 0.348 atm (b) To estimate the Keq using thermodynamical values we want to use the equation:

∆G = -RTlnKeq ∆Gf

o for SO2, O2, and SO3 are –300.1, 0.00, and –371.1 kJ/mol respectively. ∆Goreaction = 2 mol(-371.1 kJ/mol)

– [1 mol(0.000 kJ/mol) + 2 mol(-300.1 kJ/mol)] = -142.0 kJ/mol

ln Keq = K) )(298Kmol J (8.314

mol J 10 x 1.42-1-1-

-15o

−=∆−RTG

= 57.3

Keq = e57.3 = 7.7 x 1024 Based on this result, the reaction should be run at lower temperature to improve the yield. (c) Although the reaction is thermodynamically more favorable at lower temperatures, the reaction is kinetically very slow. Therefore, it must be run at higher temperatures (700K) to speed up the reaction while still being favorable thermodynamically.

Chapter 15 471 15.1 The reaction proceeds, converting cis ...

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