ch16

Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 16

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16.1 The reaction proceeds, converting cis–butene into trans–butene, but as the concentrationof 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 plotof concentration vs. time appears similar to Figure 16–2, with 0.75 bar for the finalpressure of trans–butene and 0.25 bar for the final pressure of cis–butene:

16.2 The reaction proceeds, converting trans–butene into cis–butene, but as the concentrationof cis–butene builds up, the reverse reaction becomes increasingly important until, whenthe concentration ratio is 1/3, the rates are equal. Then there is no net change. A plot ofconcentration vs. time appears similar to Figure 16–2, with 0.75 bar for the final pressureof trans–butene and 0.25 bar for the final pressure of cis–butene:

16.3 (a) According to the principle of reversibility, every elementary reaction can run in eitherdirection, so as the system approaches equilibrium, the reverse reactions becomeimportant:

Cl– (aq) + ClO2– (aq)

k−1⎯ →⎯⎯ ClO– (aq) + ClO– (aq)

Cl– (aq) + ClO3– (aq)

k−2⎯ →⎯⎯ ClO– (aq) + ClO2– (aq)

(b) The overall 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 overall 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 =k1k2k−1k−2


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16.4 (a) According to the principle of reversibility, every elementary reaction can run in eitherdirection, so the decomposition of H2CO is described by the reverse reactions:

H2CO k−3⎯ →⎯⎯ HCO + H

HCO k−2⎯ →⎯⎯ H + CO

2 Hk−1⎯ →⎯⎯ H2

(b) The overall reaction is H2 + CO ⎯ →⎯← ⎯⎯ H2CO . By inspection, the equilibrium

constant expression is Keq =[H2CO]eq

[H2]eq[CO]eq(c) The overall 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

16.5 A molecular picture of an elementary reaction shows the reactants, the products, and (ifnecessary) the intermediate collision complex.

16.6 A molecular picture of an elementary reaction shows the reactants, the products, and (ifnecessary) the intermediate collision complex.

16.7 To test the reversibility of a reaction, set up a system containing the products and observeif reactants form. Here, a solution containing Cl– and ClO3– ions should react to formsome ClO– ions. (If the reaction of ClO– to form Cl– and ClO3– goes virtually tocompletion, this experiment may not succeed.)


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16.8 To test the reversibility of a reaction, set up a system containing the products and observeif reactants form. Here, if a reactor is charged with formaldehyde gas (H2CO), it shouldbe possible to observe the formation of H2 and CO. (If the reaction between H2 and COto form H2CO goes virtually to completion, however, this experiment may not succeed.)

16.9 Equilibrium constant expressions can be written by inspection of the overallstoichiometry, remembering that pure liquids and solids do not appear in the expression.Equilibrium constant expressions contain product concentrations over reactantconcentrations raised to their stoichiometric coefficient:

(a) Keq =(pIF5 )eq

2

(pF2 )eq5 (b) Keq =

1

pO2( )eq5 (c) Keq = (pCO)eq

2

(d) Keq =1

(pCO)eq(pH2 )eq2 (e) Keq =

[H3O+]eq3 [PO4

3-]eq[H3PO4 ]eq

16.10 Equilibrium constant expressions can be written by inspection of the overallstoichiometry, remembering that pure liquids and solids do not appear in the expression.Equilibrium constant expressions contain product concentrations over reactantconcentrations raised to their coefficient:

(a) Keq =(pSO2 )eq

(pH2S)eq(pO2 )eq3 (b) Keq =

(pCO2 )eq3

(pCO)eq3 (c) Keq =

(pClF3 )eq2

(pF2 )eq3 (pCl2 )eq

(d) Keq =[NH4

+]eq[H3O

+]eq[NH3]eq(e)

Keq = 1(pH2 )eq

2 .

16.11 Equilibrium constant expressions can be written by inspection of the overallstoichiometry, remembering that pure liquids and solids do not appear in the expression:

(a) 2IF5 (g) ⎯ →⎯← ⎯⎯ I2 (s) + 5 F2 (g) Keq =(pF2 )eq

5

(pIF5 )eq2

(b) P4O10 (s) ⎯ →⎯← ⎯⎯ P4 (s) + 5 O2 (g) Keq = pO2( )eq5

(c) BaO (s) + 2 CO (g) ⎯ →⎯← ⎯⎯ BaCO3 (s) + C (s) Keq =1

(pCO)eq2

(d) CH3OH (l) ⎯ →⎯← ⎯⎯ CO (g) + 2 H2 (g) Keq = (pCO)eq(pH2 )eq2

(e) PO43– (aq) + 3 H3O+ (aq) ⎯ →⎯← ⎯⎯ H3PO4 (aq) + 3 H2O (l) Keq =

[H3PO4 ]eq[H3O

+]eq3 [PO4

3-]eq


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16.12 Equilibrium constant expressions can be written by inspection of the overallstoichiometry, 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) Keq =(pH2S)eq(pO2 )eq

3

(pSO2 )eq

(b) 2 Fe (s) + 3 CO2 (g) ⎯ →⎯← ⎯⎯ Fe2O3 (s) + 3 CO (g) Keq =(pCO)eq

3

(pCO2 )eq3

(c) 2 ClF3 (g) ⎯ →⎯← ⎯⎯ Cl2 (g) + 3 F2 (g) Keq =

(pF2 )eq3 (pCl2 )eq

(pClF3 )eq2

(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) Keq = (pH2 )eq

2

16.13 The standard states of gases are gases at 1 bar, the standard states of solutes are solutionsat 1 M, and the standard states of solvents and pure liquids and solids are unit molefractions, X = 1:(a) p = 1 bar for F2 and IF5, X = 1 for I2(b) p = 1 bar for O2, X = 1 for P4 and P4O10(c) p = 1 bar for CO, X = 1 for others(d) p = 1 bar for CO and H2, X = 1 for CH3OH(e) c = 1 M for H3PO4, H3O+, and PO43–, X = 1 for H2O

16.14 The standard states of gases are gases at 1 bar, the standard states of solutes are solutionsat 1 M, and the standard states of solvents and pure liquids and solids are unit molefractions, X = 1:(a) p = 1 bar for SO2, O2, and H2S, X = 1 for H2O(b) p = 1 bar for CO and CO2, X = 1 for Fe and Fe2O3(c) p = 1 bar for each gas(d) c = 1 M for H3O+ and NH4+, P = 1 bar for NH3, X = 1 for H2O(e) p = 1 bar for H2, X = 1 for others

16.15 Your views should be like those in Figure 16–4, showing two different size samples withequal sized portions of each highlighted to show that equal volumes contain equal numbersof atoms:


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16.16 Your views should be like those in Figure 16–4, showing two different size samples withequal sized portions of each highlighted to show that equal volumes contain equal numbersof atoms:

16.17 To determine an equilibrium constant at standard temperature from thermodynamic

tables, calculate ΔGoreaction from tabulated values for ΔGfo and then use Equation 16–3:

ΔGo = –RT ln Keq

The solubility reaction is LiCl (s) ⎯ →⎯← ⎯⎯ Li+ (aq) + Cl– (aq) Keq = Ksp

ΔGoreaction = [1 (–293.31 kJ/mol) + l (–131.0 kJ/mol)] – [1 (–384.4 kJ/mol)]= – 39.91 kJ/mol

ln Ksp = – –3.991 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = 16.11 Ksp = e16.11 = 9.9 x 106



ΔGo = –RT ln Keq

The solubility reaction is MgCl2 (s) ⎯ →⎯← ⎯⎯ Mg2+ (aq) + 2 Cl– (aq) Keq = Ksp

ΔGoreaction = [1 (–454.8 kJ/mol) + 2 (–131.0 kJ/mol)] – [1 (–591.8 kJ/mol)]= – 125 kJ/mol

ln Ksp = – –12.5 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = 50.45 Ksp = e50.45 = 8.2 x 1021



ΔGo = –RT ln Keq

(a) ΔGoreaction = [1 (0 kJ/mol) + 1 (–394.4 kJ/mol)]– [1 (–137.2 kJ/mol) + 1 (–237.1 kJ/mol)] = – 20.1 kJ/mol

ln Keq = – –2.01 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = 8.11 Keq = e8.11 = 3.3 x 103


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(b) ΔGoreaction = 2 (–394.4 kJ/mol) – [1 (0 kJ/mol) + 2 (–137.2 kJ/mol)]= – 514.4 kJ/mol

ln Keq = – –5.144 x 105 J mol-1

(8.314 J mol-1 K-1)(298 K) = 207.6 Keq = e207.6 = 1 x 1090

(c) ΔGoreaction = [1 (–520.3 kJ/mol) + 2 (–137.2 kJ/mol)]– [1 (0 kJ/mol) + 1 (–1134.4 kJ/mol)] = 339.7 kJ/mol

ln Keq = – 3.397 x 105 J mol-1

(8.314 J mol-1 K-1)(298 K) = –137.1 Keq = e–137.1 = 3 x 10–60

(d) ΔGoreaction = [3 (–237.1 kJ/mol) + 1 (74.62 kJ/mol)]– [6 (0 kJ/mol) + 3 (–137.2 kJ/mol)] = –225.1 kJ/mol

ln Keq = – –2.251 x 105 J mol-1

(8.314 J mol-1 K-1)(298 K) = 90.86 Keq = e90.86 = 2.9 x 1039



ΔGo = –RT ln Keq

(a) ΔGoreaction = [3 (0 kJ/mol) + 1 (–137.2 kJ/mol)]– [1 (–50.5 kJ/mol) + 1 (–237.1 kJ/mol)] = 150.4 kJ/mol

ln Keq = – 1.504 x 105 J mol-1

(8.314 J mol-1 K-1)(298 K) = – 60.7 Keq = e–60.7 = 4 x 10–27

(b) ΔGoreaction = [4 (87.6 kJ/mol) + 6 (–237.1 kJ/mol)]–[4 (–16.4 kJ/mol) + 5 (0 kJ/mol)] = – 1006.6 kJ/mol

ln Keq = – –1.0066 x 106 J mol-1

(8.314 J mol-1 K-1)(298 K) = 406.3 Keq = e406.3 = 3 x 10176

(c) ΔGoreaction = [2 (–237.1 kJ/mol) + 1(0 kJ/mol)]– [1 (–515.8 kJ/mol) + 2 (0 kJ/mol)] = 41.6 kJ/mol

ln Keq = – 4.16 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = –16.8 Keq = e–16.8 = 5.1 x 10–8

(d) ΔGoreaction = [1 (–1015.4 kJ/mol) + 4 (0 kJ/mol)]– [3 (0 kJ/mol) + 4 (–237.1 kJ/mol)] = –67.0 kJ/mol

ln Keq = – –6.70 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = 27.0 Keq = e27.0 = 5.3 x 1011


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16.21 To estimate the equilibrium constant at a temperature different from 298 K, calculateΔHoreaction and ΔSoreaction at 298 K and then use Equations 14–10 and 16–3:

ΔGo = ΔHo– TΔSo ΔGo = –RT ln Keq

16.19 (b) ΔHoreaction = 2 (–393.5 kJ/mol) – [1 (0 kJ/mol) + 2 (–110.5 kJ/mol)]= – 566.0 kJ/mol

ΔSoreaction = 2 (213.8 J/mol K) – [1 (205.15 J/mol K) + 2 (197.7 J/mol K)]= – 173.0 J/mol K

ΔGoreaction, 250K ≅ (–566.0 kJ/mol) – (250 K)(–0.1730 kJ/mol K) = –522.8 kJ/mol

ln Keq = – –5.228 x 105 J mol-1

(8.314 J mol-1 K-1)(250 K) = 251.5 Keq = e251.5 = 2 x 10109

16.19 (c) ΔHoreaction = 1 (–548.0 kJ/mol) + 2 (–110.5 kJ/mol)– [1 (0 kJ/mol) + 1(–1213.0 kJ/mol)] = 444.0 kJ/mol

ΔSoreaction = [1 (72.1 J/mol K) + 2 (197.7 J/mol K)]–[1 (5.7 J/mol K) + 1 (112.1 J/mol K)] = 349.7 J/mol K

ΔGoreaction, 250K ≅ (444.0 kJ/mol) – (250 K)(0.3497 J/mol K) = 356.6 kJ/mol

ln Keq = – 356.6 x 103 J mol-1

(8.314 J mol-1 K-1)(250 K) = –171.6 Keq = e–171.6 = 3 x 10–75



16.20 (c) ΔHoreaction = [2 (–285.83 kJ/mol) + 1 (0 kJ/mol)]– [1 (–577.6 kJ/mol) + 2 (0 kJ/mol)] = 5.9 kJ/mol

ΔSoreaction = [2 (69.95 kJ/mol) + 1 (51.2 kJ/mol)]– [1 (49.0 J/mol K) + 2 (130.680 J/mol K)] = –119.3 J/mol K

ΔGoreaction, 350K ≅ (5.9 kJ/mol) – (350 K)(–0.1193 J/mol K) = 47.7 kJ/mol

ln Keq = – (47.7 kJ/mol)(103 J/kJ)(8.314 J/mol K)(350 K)

= – 16.4 Keq = e–16.4 = 7.5 x 10–8

16.20 (d) ΔHoreaction = [1 (–1118.4 kJ/mol) + 4 (0 kJ/mol)]– [3 (0 kJ/mol) + 4 (–285.83 kJ/mol)] = 24.9 kJ/mol

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

ΔGoreaction, 350K ≅ (24.9 kJ/mol) – (350 K)(0.3074 J/mol K) = –82.7 kJ/mol

ln Keq = – (–82.7 kJ/mol)(103 J/kJ)(8.314 J/mol K)(350 K)

= 28.4 Keq = e28.4 = 2.2 x 1012


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ΔGo = ΔHo– TΔSo ΔGo = –RT ln Keq16.19 (a)

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

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


ln Keq = – –2.46 x 104 J mol-1

(8.314 J mol-1 K-1)(395 K) = 7.49 Keq = e7.49 = 1.8 x 103

16.19 (d)ΔHoreaction = 1 (20.0 kJ/mol) + 3 (–241.83 kJ/mol)

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

–[3 (197.7 J/mol K) + 6 (130.680 J/mol K)] = –583.8 J/mol KΔGoreaction, 395K ≅ (–374.0 kJ/mol) – (395 K)(–0.5838 J/mol K) = –143.4 kJ/mol

ln Keq = – –1.434 x 105 J mol-1

(8.314 J mol-1 K-1)(395 K) = 43.67 Keq = e43.67 = 9.2 x 1018


ΔGo = ΔHo– TΔSo ΔGo = –RT ln Keq16.20 (c)

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

ΔSoreaction = [1 (51.2 J/mol K) + 2 (188.835 J/mol K)]– [1 (49.0 J/mol K) + 2 (130.680 J/mol K)] = 118.51 J/mol K

ΔGoreaction, 425K ≅ (93.9 kJ/mol) – (425 K)(0.18851 kJ/mol K) = 43.5 kJ/mol

ln Keq = – 4.35 x 104 J mol-1

(8.314 J mol-1 K-1)(425 K) = –12.3 Keq = e–12.3 = 4.5 x 10–6

16.20 (d)ΔHoreaction = 1 (–1118.4 kJ/mol) + 4 (0 kJ/mol)

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

–[3 (27.3 J/mol K) + 4 (188.835 J/mol K)] = –168.1 J/mol KΔGoreaction, 425K ≅ (–151.1 kJ/mol) – (425 K)(–0.1681 J/mol K) = –79.7 kJ/mol

ln Keq = – –7.97 x 104 J mol-1

(8.314 J mol-1 K-1)(425 K) = 22.6 Keq = e22.6 = 6.5 x 109


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16.25 Use Le Châtelier's principle to determine the effect on an equilibrium position caused byadding one reagent:In reactions (a), (b), and (d), CO is a reactant, so adding CO causes the reaction to go tothe right, forming products; in reaction (c), CO is a product, so adding CO causes thereaction to proceed to the left, forming reactants.

16.26 Use Le Châtelier's principle to determine the effect on an equilibrium position caused byadding one reagent. Note, however, that pure liquids and solids do not appear inequilibrium constant expressions, so changing their amounts do not affect the equilibriumposition. In all four reactions of Problem 16.20, H2O is a liquid reactant, so adding H2Ohas no effect on the equilibrium position.

16.27 Use Le Châtelier's principle to determine the effect on an equilibrium position caused bya 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.

16.28 Use Le Châtelier's principle to determine the effect on an equilibrium position caused bya 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 thetemperature shifts the equilibrium position in the direction of reactants, and theamount of SO3 decreases.

(b) Addition of O2 increases the concentration of one of the reactants, so more SO3 willform.

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

16.29 Use Le Châtelier's principle to determine what changes in conditions will drive anequilibrium in any given direction. This reaction can be driven to the left by removingSO2, removing Cl2, adding SO2Cl2, or increasing the temperature.

16.30 Use Le Châtelier's principle to determine what changes in conditions will drive anequilibrium in any given direction. This reaction can be driven to the left by removingPCl5, adding Cl2, adding PCl3, or decreasing the temperature.


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16.31 To calculate an equilibrium constant when amounts are available, we generally mustconvert amounts data into concentrations or pressures using stoichiometric reasoning,then complete an amounts table and substitute into the equilibrium constant expression.In this problem, however, the equilibrium constant expression contains p2 in thenumerator and denominator, so units cancel and amounts can be used directly. Set up anamounts table, using the fact that the change is 0.49 mol for CO:

Reaction: H2 + CO2 ⎯ →⎯← ⎯⎯ H2O + CO

Initial amount (mol) 1.00 1.00 0 0Change in amount (mol) –0.49 –0.49 + 0.49 + 0.49Equilibrium amount (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

16.32 To calculate an equilibrium constant from experimental data, complete a concentrationtable using concentration or pressures using stoichiometric reasoning. Then substituteinto the equilibrium constant expression. We are given initial pressures in bar. Use thefact that the change is 0.93 bar for Cl2:

Reaction: 4 HCl + O2 ⎯ →⎯← ⎯⎯ 2 Cl2 + 2 H2O

Initial pressure (bar) 2.30 1.00 0 0Change in pressure (bar) –1.86 –0.465 + 0.93 + 0.93Equilibrium pressure (bar) 0.44 0.535 0.93 0.93


Keq = (pH2O)eq

2 (pCl2 )eq2

(pHCl)eq4 (pO2 )eq

=(0.93)2(0.93)2

(0.44)4 (0.535) = 37

16.33 To calculate an equilibrium constant when experimental data concerning amounts areavailable, identify the reaction, convert amounts data into concentration usingstoichiometric reasoning, then complete a concentration table and substitute into theequilibrium constant expression. For solutes, concentrations must be in mol/L.

[C2H5CO2H]initial = 0.0500 mol

0.500 L = 0.100 M

To complete the concentration table, use stoichiometric reasoning and the fact that thechange is 1.15 x 10–3 M for H3O+ (The concentration of water is not needed, becausewater is the solvent):

Reaction: H2O + C2H5CO2H ⎯ →⎯← ⎯⎯ C2H5CO2– + H3O+

Initial conc. (M) 0.100 0 0Change in conc. (M) –1.15 x 10–3 + 1.15 x 10–3 +1.15 x 10–3

Equilibrium conc. (M) 0.099 1.15 x 10–3 1.15 x 10–3


477


Ka = [C2H5CO2

- ]eq[H3O+]eq[C2H5CO2H]eq

=(1.15 x 10-3)2

(0.099) = 1.3 x 10–5

16.34 To calculate an equilibrium constant when experimental data concerning concentrationsare available, identify the reaction, then complete a concentration table and substitute intothe equilibrium constant expression. To complete the concentration table, usestoichiometric reasoning and the fact that the change is 6.5 x 10–3 M for H3O+ (Theconcentration of water is not needed, because water is the solvent):

Reaction: H2O + HCNO ⎯ →⎯← ⎯⎯ CNO– + H3O+

Initial conc. (M) 0.20 0 0Change in conc. (M) –6.5 x 10–3 + 6.5 x 10–3 + 6.5 x 10–3

Equilibrium conc. (M) 0.19 6.5 x 10–3 6.5 x 10–3

Now substitute into the equilibrium constant expression and evaluate Keq:

Keq = Ka = [CNO-]eq[H3O+]eq

[HCNO]eq=

(6.5 x 10-3)2

(0.19) = 2.2 x 10–4

16.35 To calculate concentrations at equilibrium from initial conditions, set up a concentrationtable. For a gas phase reaction, concentrations must be expressed in bar. Let x = changein pH2 :

Reaction: H2 + Br2 ⎯ →⎯← ⎯⎯ 2 HBr

Initial pressure (bar) 0 0 10.0Change in pressure (bar) + x + x –2xEquilibrium pressure (bar) x x 10.0 – 2x

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

Keq = (pHBr )eq

2

(pH2 )eq(pBr2 )eq=

(10.0 - 2x)2

(x)(x) = 1.6 x 105

To simplify, assume that 2x << 10.0:

1.6 x 105 = (10.0)2

(x)2 so x2 =

1001.6 x 105 = 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 bar2(2.5 x 10–2) = 0.050 << 10.0, so the approximation is valid.


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16.36 To calculate concentrations at equilibrium from initial conditions, set up a concentrationtable. For a gas phase reaction, concentrations must be expressed in bar. Let x = changein pCO :

Reaction: CO + Cl2 ⎯ →⎯← ⎯⎯ COCl2

Initial pressure (bar) 0 0 0.250Change in pressure (bar) + x + x –xEquilibrium pressure (bar) x x 0.250 – x


Keq =

(pCOCl2 )eq(pCO)eq(pCl2 )eq

=0.250 - x

x2 = 1.5 x 108; assume that x << 0.250

1.5 x 108 = 0.250x2

so x2 = 0.250

1.5 x 108 = 1.67 x 10–9

x = 4.1 x 10–5 (pCO)eq = (pCl2 )eq = 4.1 x 10–5 bar

(pCOCl2 )eq = 0.250 – 4.1 x 10–5 = 0.250 bar

4.1 x 10–5 << 0.250, so the approximation is valid.

16.37 To calculate concentrations at equilibrium from initial conditions, set up a concentrationtable. For a gas phase reaction, concentrations must be expressed in bar. Let x = changein pCO2 :

Reaction: FeO (s) + CO (g) ⎯ →⎯← ⎯⎯ CO2 (g) + Fe (s)

Initial pressure (bar) excess 5.0 0 ––––Change in pressure (bar) –––– –x + x ––––Equilibrium pressure (bar) –––– 5.0 – x x ––––Now substitute into the equilibrium constant expression and solve for x:

Keq = (pCO2 )eq(pCO)eq

=x

(5.0 -x) = 0.403

x = (0.403)(5.0 – x) = 2.015 – 0.403 x so 1.403 x = 2.015x = 1.436 (pCO2 )eq = 1.4 bar (pCO)eq = 5.0 – 1.4 = 3.6 bar

16.38 This problem describes an equilibrium reaction. We are asked to determine theequilibrium pressures of all the gases. To calculate pressures at equilibrium from initialconditions, set up a concentration table, write the K expression, and solve for thepressures:Let –x = change in [N2]:

Reaction: N2 + 3 H2 ⎯ →⎯← ⎯⎯ 2 NH3

Initial pressure (bar) 5.0 3.0 0Change in pressure (bar) – x –3 x + 2xEquilibrium pressure (bar) 5.0 – x 3.0 – 3 x 2x


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Keq = 2.81 x 10–5 = (pNH3 )eq

2

(pN2 )eq(pH2 )eq3 =

(2x)2

(5.0 –x)(3.0 –3x)3 ; Assume that 3x << 3.0:

2.81 x 10–5 = 4x2

(5.0)(3.0)34x2 = (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 bar

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

16.39 To identify species in solution, first identify the nature of the solute. Strong acids, strongbases, 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 areH2O, Na+ and CH3CO2–; and (e) strong base, major species are H2O, Na+ and OH–.

16.40 To identify species in solution, first identify the nature of the solute. Strong acids, strongbases, 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 areH2O, H3O+ and NO3–; and (e) weak acid, major species are H2O and HCN.

16.41 The equilibria among major species depend on the nature of the species. The protontransfer 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.

16.42 The equilibria among major species depend on the nature of the species. The protontransfer 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.


480

(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)

16.43 To identify species in solution, first identify the nature of the solute. Strong acids, strongbases, and salts generate ions, while all other substances remain molecular:(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–; and (d) strong acid, majorspecies are H2O, H3O+, and HSO4–.

16.44 To identify species in solution, first identify the nature of the solute. Strong acids, strongbases, 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–; and(d) weak acid, major species are C6H5CO2H (benzoic acid) and H2O.

16.45 Equilibrium constant expressions have the concentrations of the products in thenumerator and the concentrations of the reactants in the denominator with eachconcentration raised to the power of its stoichiometric coefficient. Remember to omitliquids and solids from the expressions. If the reaction involves a weak acid (or base)reacting with water, Keq will be related to Ka (or Kb if a base). If the reaction involves asolid, Keq will be related to Ksp:

(a) Keq = [ClO2

- ][H3O+]

[HClO2]= Ka (b) Keq =

1[Fe3+ ][OH− ]3

= 1Ksp

(c) Keq = [HCN]

[CN-][H3O+]

= 1Ka

16.46 Equilibrium constant expressions have the concentrations of the products in thenumerator and the concentrations of the reactants in the denominator with eachconcentration raised to the power of its stoichiometric coefficient. Remember to omitliquids and solids from the expressions. If the reaction involves a weak acid (or base)reacting with water, Keq will be related to Ka (or Kb if a base). If the reaction involves asolid, Keq will be related to Ksp:

(a) Keq = [Ag+]2[SO42-]= Ksp

(b) Keq = [H3O

+][H2PO4- ]

[H3PO4 ]=

1Ka1

(c) Keq = [HCO2H]

[HCO2- ][H3O

+]=

1Ka


481

16.47 To identify the spectator ions, first determine the reaction (if any) that occurs when thesolutions are mixed. Species that do not participate in the reaction are spectator ions:

(a) CH3CO2H + OH– ⎯ →⎯← ⎯⎯ CH3CO2– + H2O; spectator ion: Na+

(b) 3 Ca2+ + 2 PO43- ⎯ →⎯← ⎯⎯ Ca3(PO4)2; spectator ions: Cl– and K+

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

16.48 To identify the spectator ions, first determine the reaction (if any) that occurs when thesolutions are mixed. Species that do not participate in the reaction are spectator ions:

(a) Fe3+ + 3 OH– ⎯ →⎯← ⎯⎯ Fe(OH)3; spectator ions: Cl– and Na+

(b) NH3 + H3O+ ⎯ →⎯← ⎯⎯ NH4+ + H2O; spectator ion: NO3

-

(c) H3O+ + CH3CO2– ⎯ →⎯← ⎯⎯ CH3CO2H + H2O; spectator ions: K+ and ClO4

-

16.49 Equilibrium constant expressions are determined by the stoichiometry of the overallreaction, with pure liquids, solids, and solvent omitted from the expression. Equilibriumconstant expressions have the concentrations of the products in the numerator and theconcentrations of the reactants in the denominator, with each concentration raised to thepower of its stoichiometric coefficient:

(a) Keq = (pCO2 )eq(pH2O)eq (b) Keq = (pNH3 )eq

4 (pO2 )eq3

(pN2 )eq2

(c) Keq = (pCH3CHO)eq

2

(pC2H4 )eq2 (pO2 )eq

(d) Keq =[Ag+]eq2 [SO4

2-]eq

(e) Keq = (pH2S)eq(pNH3 )eq

16.50 Equilibrium constant expressions are determined by the stoichiometry of the overallreaction, with pure liquids, solids, and solvent omitted from the expression. Equilibriumconstant expressions have the concentrations of the products in the numerator and theconcentrations of the reactants in the denominator, with each concentration raised to thepower of its stoichiometric coefficient:

(a) Keq = (pNO)eq

4

(pNH3 )eq4 (pO2 )eq

5 (b) Keq = [CN-]eq[H3O

+]eq[HCN]eq

(c) Keq = 1

[NH4+]eq2 [SO4

2-]eq(d) Keq =

[OH-]eq[HSO4- ]eq

[SO42- ]eq

(e) Keq = (pO2 )eq

3

(pO3 )eq2


482

16.51 This problem describes an equilibrium reaction. We are asked to determine theequilibrium pressures of all the gases. To calculate pressures at equilibrium from initialconditions, set up a concentration table, write the Keq expression, and solve for thepressures. For a gas phase reaction, concentrations must be expressed in bar.p1T1

=p2T2

p2 = p1T2T1

= 3.00 bar( )(273 + 352) K

298 K( ) = 6.29 bar

Reaction: COCl2 ⎯ →⎯← ⎯⎯ CO + Cl2

Initial pressure (bar) 6.29 0 0Change in pressure (bar) – x + x + xEquilibrium pressure (bar) 6.29 – x x x

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

Keq = 8.3 x 10–4 = (pCO)eq(pCl2 )eq

(pCOCl2 )eq=

(x)2

6.29 –x; Assume that x << 6.29:

8.3 x 10–4 = x2

6.29x2 = (6.29)(8.3 x 10–4) = 5.22 x 10–3

x = 7.2 x 10–2 bar = (pCO)eq = (pCl2 )eq (pCOCl2 )eq = 6.29 – 7.2 x 10–2 = 6.22 bar

7.2 x 10–2 is 1.1% of 6.29, so x is << 6.29 and the approximation is valid.

16.52 To calculate concentrations at equilibrium from initial conditions, set up a concentrationtable. For a gas phase reaction, concentrations must be expressed in bar.We know the pressure of Cl2 gas at 298 K, from which we can calculate the initialpressure at 1200 K:p1T1

=p2T2

p2 = p1T2T1

= 0.57 bar( ) 1200 K( )

298 K( ) = 2.3 bar

Let –x = change in pCl2 :

Reaction: Cl2 (g) ⎯ →⎯← ⎯⎯ 2Cl (g)

Initial pressure (bar) 2.3 0Change in pressure (bar) –x + 2xEquilibrium pressure (bar) 2.3 – x 2x


Keq = (pCl)eq

2

(pCl2 )eq=

(2x)2

(2.3 -x) = 2.5 x 10–5; assume x << 2.3

4x2 = (2.5 x 10–5)(2.3) = 5.75 x 10–5 x2 = 1.44 x 10–5

x = 3.8 x 10–3 (pCl)eq = 3.8 x 10–3 bar (pCl2 )eq = 2.3 bar

0.0038 << 2.3, so the approximation is valid.


483

16.53 At equilibrium, a molecular picture should show the presence of both reactants andproducts, in relative amounts that are determined by the value of the equilibriumconstant. Set up a "concentration" table to determine how many of each species arepresent at equilibrium:

Initial 12 12 0 0Change – x – x + x + xEquilibrium 12 – x 12 – x x x

Substitute in the equilibrium constant expression and solve for x:

Keq = 25 = (x)2

(12 - x)2 ; Taking the square root of each side gives 5 = x

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

16.54 At equilibrium, a molecular picture should show the presence of both reactants andproducts, in relative amounts that are determined by the value of the equilibriumconstant. Set up a "concentration" table to determine how many of each species arepresent at equilibrium:

Initial 12 12 0Change – x – x + 2 xEquilibrium 12 – x 12 – x 2 x

Substitute in the equilibrium constant expression and solve for x:

Keq = 4 = (2x)2

(12 - x)(12 - x); Taking the square root of each side gives 2 =

2x(12 – x)

2(12 – x ) = 2 x 4 x = 24 and x = 6The molecular picture should show (12 – 6) = 6 of each reactants, and 2(6) = 12 productmolecules:


484

16.55 At equilibrium between a solute and solution, the rate at which molecules leave the surface ofthe solute equals the rate at which molecules are captured from solution at the surface of thesolute. Adding more solute to a solution that is saturated increases the total surface area, sothe overall rate at which molecules enter the solution becomes greater. However, the rate atwhich molecules are captured also becomes greater. The rate per unit surface area isunchanged, so the concentration of chloroform in the aqueous solution remains the same.

16.56 At equilibrium between a solid salt and solution, the rate at which ions leave the surface ofthe 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 atwhich ions leave becomes greater. However, the rate at which ions are captured alsobecomes greater. The rate per unit surface area is unchanged, so the concentration of sodiumchloride in the aqueous solution remains the same.

16.57 To calculate equilibrium pressures, set up a concentration table. In this problem, thetable can be short, because two of the three equilibrium pressures are provided:

Reaction: Br2 (g) + I2 (g) ⎯ →⎯← ⎯⎯ 2 IBr (g)

Equilibrium pressure (bar) 0.512 0.327 xSubstitute into the equilibrium constant expression and solve for x :

322 = (pIBr )eq

2

(pBr2 )eq(pI2 )eq =

x2

(0.512)(0.327)x2 = (322)(0.512)(0.327) = 53.91

x = 7.34 (pIBr )eq = 7.34 bar

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

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

Equilibrium pressure (bar) ––––– 2.80 x 102 x 2xSubstitute into the equilibrium constant expression and solve for x:

0.38 = (pCO2 )eq(pH2 )eq

2

(pH2O)eq2 =

x(2x)2

(280)2

4x3 = (0.38)(280)2 = 2.98 x 104

x3 = 7.4 x 103, x = 19.5(pCO2 )eq = 20. bar (pH2 )eq = 2(19.5) = 39 bar


485

16.59 To determine an equilibrium constant at standard temperature from thermodynamictables, calculate ΔGoreaction from tabulated values for ΔGo

f. To estimate the equilibrium

constant at a temperature different from 298 K, calculate ΔHoreaction and ΔSoreaction at298K and then use Equations 14–10 and 16–3:

ΔGo = ΔHo– TΔSo ΔGo = –RT ln KeqAt 298 K:

ΔGoreaction = 1 (99.8 kJ/mol) – 2 (51.3 kJ/mol) = – 2.8 kJ/mol

ln Keq = – –2.8 x 103 J mol-1

(8.314 J mol-1 K-1)(298 K) = 1.1

Keq = e1.1 = 3.0At 525 K:

ΔHoreaction = 1 (11.1 kJ/mol) – 2 (33.2 kJ/mol) = – 55.3 kJ/molΔSoreaction = 1 (304.4 J/mol K) – 2 (240.1 J/mol K) = –175.8 J/mol K

ΔGoreaction ≅ (–55.3 kJ/mol) – (525 K)–175.8 J1 mol K

⎛⎝⎜

⎞⎠⎟

10-3 kJ1 J

⎛

⎝⎜

⎞

⎠⎟ = 37.0 kJ/mol

ln Keq = – 3.70 x 104 J mol-1

(8.314 J mol-1 K-1)(525 K) = – 8.48 Keq = e–8.48 = 2.1 x 10–4

Notice that the exothermic reaction has a smaller Keq at higher temperature.

16.60 To determine an equilibrium constant at standard temperature from thermodynamictables, calculate ΔGoreaction from tabulated values for ΔGo

f. To estimate the equilibrium

constant at a temperature different from 298 K, calculate ΔHoreaction and ΔSoreaction at298K and then use Equations 14–10 and 16–3:

ΔGo = ΔHo– TΔSo ΔGo = –RT ln KeqAt 298 K:

ΔGoreaction = 4 (87.6 kJ/mol) – [2 (103.7 kJ/mol) + 1 (0 kJ/mol)]= 143.0 kJ/mol

ln Keq = –143.0 x 103 J mol-1

(8.314 J mol-1 K-1)(298 K) = –57.7; Keq = e–57.7 = 8.7 x 10–26

At 825 K:ΔHoreaction = 4 (91.3 kJ/mol) – [2 (81.6 kJ/mol) + 1 (0 kJ/mol)] = 202.0 kJ/molΔSoreaction = 4 (210.8 J/mol K)

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

ΔGoreaction ≅ (202.0 kJ/mol) – (825 K)198.0 J1 mol K

⎛⎝⎜

⎞⎠⎟

10-3 kJ1 J

⎛

⎝⎜

⎞

⎠⎟ = 38.7 kJ/mol

ln Keq = – 38.7 x 103 J mol-1

(8.314 J mol-1 K-1)(825 K) = –5.64 Keq = e–5.64 = 3.6 x 10–3

Notice that the endothermic reaction has a larger Keq at higher temperature.


486

16.61 (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

16.62 (a) The general weak acid reaction is HA + H2O ⎯ →⎯← ⎯⎯ A– + H3O+, and HA = Cl3CCO2H:

Cl3CCO2H + H2O Cl3CCO2– + H3O+ Ka =

[Cl3CCO2-]eq[H3O

+]eq[Cl3CCO2H]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-]eq2

16.63 To predict effects of changes on equilibrium position, apply Le Châtelier's principle: Thesystem will respond in the direction that reduces the effect of the change. The reaction in

Example 16–14 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 moles of gas on the reactant side, doubling the volume causes thereaction to proceed to the left. (c) Adding Ar does not change any of the pressures in theequilibrium expression, so this change has no effect.

16.64 To predict effects of changes on equilibrium position, apply Le Châtelier's principle: Thesystem will respond in the direction that reduces the effect of the change. The reaction in

Example 16–12 is: HA + H2O ⎯ →⎯← ⎯⎯ A– + H3O+, where HA is benzoic acid and A– isthe conjugate base, benzoate anion:(a) Adding sodium benzoate will increase the concentration of benzoate anion (A–) in thesolution, causing the reaction to proceed to the left. Increasing the volume of the solutionwill reduce the concentrations of all solutes. (b) There are more moles of solutes on theproduct side, so the reaction will proceed to the right. (c) Adding NaCl does not changeany concentrations in the equilibrium expression, so this change has no effect.


487



The solubility reaction is LiCl (s) ⎯ →⎯← ⎯⎯ Li+ (aq) + Cl– (aq)

ΔHoreaction = [1 (–167.1 kJ/mol) + 1 (–278.47 kJ/mol)]– [1 (–408.6 kJ/mol)] = – 36.97 kJ/mol

ΔSoreaction = [1 (56.5 J/mol K) + 1 (12.2 J/mol K)]– [1 (59.3 J/mol K)] = 9.4 J/mol K


ln Keq = – –4.05 x 104 J mol-1

(8.314 J mol-1 K-1)(373 K) = 13.06 Keq = e13.06 = 4.7 x 105

From Problem 16.17, at 298 K, Keq = 9.9 x 106; LiCl is less soluble at high temperature.


ΔGo = ΔHo – TΔSo ΔGo = –RT ln Keq

The solubility reaction is MgCl2 (s) ⎯ →⎯← ⎯⎯ Mg2+ (aq) + 2 Cl– (aq)

ΔHoreaction = [2 (–167.1 kJ/mol) + 1 (–467.0 kJ/mol)]– [1 (–641.3 kJ/mol)] = – 159.9 kJ/mol

ΔSoreaction = [2 (56.5 J/mol K) + 1 (–137 J/mol K)]– [1 (89.6 J/mol K)] = – 113.6 J/mol K


ln Keq = – –1.175 x 105 J mol-1

(8.314 J mol-1 K-1)(373 K) = 37.9

Keq = e37.9 = 2.9 x 1016; from Problem 16.18, at 298 K, Keq = 8.2 x 1021

By this estimate, MgCl2 is less soluble at high temperature than at low.

16.67 To calculate an equilibrium constant from initial and equilibrium conditions, set up aconcentration table:

Reaction: CCl4 (g) ⎯ →⎯← ⎯⎯ 2 Cl2 (g) + C (s)

Initial pressure (bar) 1.00 0 ––––Change in pressure (bar) – x + 2x ––––Equilibrium pressure (bar) 1.00 – x 2x ––––

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

Keq = (pCl2 )eq

2

(pCCl4 )eq=[(2)(0.35)]2

(1.00 − 0.35)=0.702

0.65 = 0.75


488

16.68 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 newconcentration table. When volume doubles, pressure is cut in half:

pCCl4 = 0.652

= 0.325 bar pCl2 = 0.702

= 0.35 bar

To return to equilibrium, pCl2 increases.

Reaction: CCl4 (g) ⎯ →⎯← ⎯⎯ 2 Cl2 (g) + C (s)

Initial pressure (bar) 0.325 0.35 ––––Change in pressure (bar) – x + 2x ––––Equilibrium pressure (bar) 0.325 – x 0.35 + 2x ––––

Substitute into the equilibrium constant expression and solve for x:

0.75 = (0.35 + 2x)2

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

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

x = −b ± b2 − 4ac

2a=−2.15 ± (2.15)2 − 4(4)(−0.12125)

2(4) = 0.0515

(pCCl4 )eq= (0.325 – 0.0515) = 0.27 bar (pCl2 )eq= [0.35 + 2(0.0515)] = 0.45 bar

ptotal = 0.72 bar. This pressure is larger than the pressure that would result fromexpansion without reaction, in conformity with Le Châtelier's Principle.


tables, calculate ΔGoreaction from tabulated values for ΔGfo ; to estimate the equilibrium

constant at a temperature different from 298 K, calculate ΔHoreaction and ΔSoreaction at298 K and then use Equations 14–10 and 16–3:

ΔGo = ΔHo – TΔSo ΔGo = –RT ln Keq

(a) ΔGoreaction = 1 (–210.7 kJ/mol)– [1 (31.8 kJ/mol) + 1 (–178.6 kJ/mol)] = – 63.9 kJ/mol

ln Keq = ––6.39 x 104 J mol-1

(8.314 J mol-1 K-1)(298 K) = 25.8 Keq = e25.8 = 1.6 x 1011

(b) When the equilibrium pressure is 1.00 bar, Keq = 1, ln(Keq) = 0 and ΔGoreaction = 0

0 = ΔHo – TΔSo TΔSo =ΔHo and T =

ΔH o

ΔSo

ΔHoreaction = 1 (–265.4 kJ/mol)– [1 (61.4 kJ/mol) + 1 (–224.3 kJ/mol)] = – 102.5 kJ/mol

ΔSoreaction = 1 (191.6 J/mol K)– [1 (175.0 J/mol K) + 1 (146.0 J/mol K)] = – 129.4 J/mol K

T = –102.5 kJ

1 mol⎛⎝⎜

⎞⎠⎟

103 J1 kJ

⎛

⎝⎜

⎞

⎠⎟

1 mol K−129.4 J

⎛⎝⎜

⎞⎠⎟

= 792 K


489

(c) ΔGoreaction, 1050 K ≅ (–102.5 kJ/mol) – (1050 K)−129.4 J1 mol K

⎛⎝⎜

⎞⎠⎟

10−3 kJ1 J

⎛

⎝⎜

⎞

⎠⎟ = 33.4 kJ/mol

ln Keq = – 3.34 x 104 J mol-1

(8.314 J mol-1 K-1)(1050 K) = –3.83

Keq = e–3.83 = 2.2 x 10–2

Notice that Keq decreases as T increases for this exothermic reaction.

16.70 To calculate concentrations at equilibrium from initial conditions, set up a concentrationtable. For a gas phase reaction, concentrations must be expressed in bar.Use the ideal gas equation to calculate the initial pressure of PCl5 gas:

p =

nRT

V=

mRT

MM V=

(2.00 g)(0.08206 L atmmol K )(390 K)

(208.22 g mol-1)(3.00 L) = 0.1025 atm

Convert from atm to bar:

0.1025 atm 1.013 bar

1 atm⎛⎝⎜

⎞⎠⎟ = 0.1038 bar

Let x = change in pCl2 :

Reaction: PCl5 (g) ⎯ →⎯← ⎯⎯ PCl3 (g) + Cl2 (g)

Initial pressure (bar) 0.1038 0 0Change in pressure (bar) – x + x + xEquilibrium pressure (bar) 0.1038 – x x x


Keq = (pPCl3 )eq(pCl2 )eq

(pPCl5 )eq=

(x)2

(0.1038 –x) = 1.83 x 10–3

x2 = (1.83 x 10–3)(0.1038 – x) = 1.90 x 10–4 – (1.83 x 10–3)xx2 + (1.83 x 10–3)x – 1.90 x 10–4 = 0

x = −b ± b2 − 4ac

2a=−(1.83 x 10-3) ± (1.83 x 10-3)2 + 4(1.90 x 10-4 )

2(3.72)

x = −(1.83 x 10-3) ± (7.63 x 10-4 )

2 =

−(1.83 x 10-3) + (2.76 x 10-2 )2

= 1.29 x 10–2

(pPCl3 )eq = (pCl2 )eq = 1.29 x 10–2 bar (pPCl5 )eq = 0.1038 – 0.0129 = 0.0909 bar

16.71 (a) The equilibrium constant expression for a reaction can be written by inspection of thestoichiometry of the reaction, omitting pure liquids, solids, and solvents:

Keq =[CO3

2-]eq(pCO2 )eq[OH

-]eq2


490

(b) Use Le Châtelier's Principle to predict the effect of changes on a system atequilibrium. 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 thatHCl is a strong acid which generates H3O+ in solution. This, in turn, will react withOH–, reducing the concentration of a reactant. Again the equilibrium shifts to the leftand the pressure of CO2 increases.

16.72 To relate an equilibrium constant to rate constants for elementary steps, set forward andreverse 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 k1k−1

=[SO3]eq[H2O]eq[H2SO4 ]eq

SO3 + C6H6 ⎯ →⎯← ⎯⎯ C6H6SO3 Rate constants: k2, k–2

k2[SO3]eq[C6H6]eq = k–2[C6H6SO3]eqk2k−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 k3k−3

=[H3O

+]eq[C6H5SO3- ]eq

[H2O]eq[C6H6SO3]eqMultiply these three expressions and cancel concentration terms:

k1k2k3k−1k−2k−3

=[H3O

+]eq[C6H5SO3- ]eq

[C6H6]eq[H2SO4 ]eq= Keq

16.73 To find an equilibrium total pressure, it is necessary to calculate equilibrium partialpressures of all gaseous participants. First determine the initial pressures of the gases,using the ideal gas equation:

p =

nRT

V=

(0.494 mol)(0.08206 L atmmol K )(1020 + 273 K)

1.00 L = 52.4 atm

Convert from atm to bar: 52.4 atm1.013 bar

1 atm⎛⎝⎜

⎞⎠⎟ = 53.1 bar

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

Reaction: C (s) + CO2 (g) ⎯ →⎯← ⎯⎯ 2 CO (g)

Initial pressure (bar) –––– 53.1 53.1Change in pressure (bar) –––– – x + 2 xEquilibrium pressure (bar) –––– 53.1 – x 53.1 + 2 x


491

Substitute into the equilibrium constant expression and solve for x :

167.5 = (53.1 + 2x)2

(53.1 – x)(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 = −b ± b2 − 4ac

2a=−379.9 ± (379.9)2 − 4(4)(−6074)

2(4) =

−379.9 ± 491.48

= 13.9

Rule out the negative value, which would give a negative pressure:(pCO2 )eq = 53.1 – 13.9 = 39.2 bar (pCO)eq = 53.1 + 2(13.9) = 80.9 bar

Ptotal = 39.2 + 80.9 = 1.20 x 102 bar

16.74 To calculate an equilibrium constant from initial and equilibrium conditions, set up aconcentration table:

Reaction: 2 Ef (g) + 3 N2 (g) ⎯ →⎯← ⎯⎯ 2 EfN3 (g)

Initial pressure (bar) 0.75 1.00 0Change in pressure (bar) – 2 x – 3 x + 2 xEquilibrium pressure (bar) 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

3 x = (1.75 – 0.85) = 0.90 and x = 0.30pEf = 0.75 – 2(0.30) = 0.15 barpN2 = 1.00 – 3(0.30) = 0.10 bar

pEfN3 = 2(0.30) = 0.60 bar

Ptotal = 0.15 + 0.10 + 0.60 = 0.85 bar, which matches the experimental pressure.Substitute into the equilibrium constant expression and calculate Keq:

Keq = [EfN3]eq

2

[Ef]eq2 [N2 ]eq

3 =(0.60)2

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

16.75 To determine the volume that will contain a single molecule of SnH4 we first need to findthe equilibrium pressure of the gas. To calculate equilibrium pressures, set up aconcentration table:

Reaction: Sn (s) + 2 H2 (g) ⎯ →⎯← ⎯⎯ SnH4 (g)

Initial pressure (bar) –––– 200 0Change in pressure (bar) –––– –2 x + xEquilibrium pressure (bar) –––– 200 – 2 x x

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

1.1 x 10–33 = [SnH4 ]eq[H2 ]eq

2 = x

(200)2x = (1.1 x 10–33)(200)2 = 4.4 x 10–29

[SnH4]eq = 4.4 x 10–29 bar (a very small pressure)


492

To calculate the volume that would be expected to contain a single molecule, usepV = nRT # = nNA

V#=

RTpNA

First, convert the pressure from bar to atm:

p = (4.4 x 10–29 bar)1 atm

1.013 bar⎛⎝⎜

⎞⎠⎟ = 4.3 x 10–29 atm

V/1 molecule =

RT

pNA=

(0.08206 L atmmol K )(298 K)

(6.022 x 1023 mol-1)(4.3 x 10-29 bar) = 9.4 x 105 L

16.76 (a) Follow the five–step procedure for solving equilibrium problems:1.) The species are acetic acid (A) and its dimer (A2)

2.) The reaction is 2 A (g) ⎯ →⎯← ⎯⎯ A2 (g)

3.) Keq = (pA2 )eq(pA)eq

2 =3.72

4.) The total pressure at equilibrium is given, Ptotal = 0.75 bar.Let x = pA2 and pA = 0.75 – x

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

3.72 = x

(0.75 - x)2

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 + 2.0930 = 3.72 x 2 – 6.58 x + 2.093

x = −b ± b2 − 4ac

2a=6.58 ± (6.58)2 − 4(3.72)(2.093)

2(3.72) = 6.58 ± 43.296 − 31.144

7.44

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 bar

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


493

16.77 This molecular picture illustrates starting conditions and equilibrium conditions. Thesymbols indicate BG3 as the starting material and G2 and GB as products. Count numbersof symbols to determine initial and equilibrium concentrations. Set up a concentrationtable using numbers of symbols:

ReactionInitial 15 0 0Change –12 +12 +12Equilibrium 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 =[BG]eq[G2 ]eq[BG3]eq

=(12)(12)(3)

= 48

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

Reaction: H2 + F2 ⎯ →⎯← ⎯⎯ 2 HF

Initial pressure (bar) 3.00 3.00 0Change in pressure (bar) – x – x + 2xEquilibrium pressure (bar) 3.00 – x 3.00 – x 2x


Keq = 115 = (pHF )eq

2

(pH2 )eq(pF2 )eq=

(2x)2

(3.00 -x)(3.00 -x)=

4x2

(9.00 - 6.00x +x2 )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 ± 16560

222 = 3.69 or 2.53

The initial pressures are 3.00 bar, so a change of 3.69 bar is impossible; thus, x = 2.53(pH2 )eq = (pF2 )eq = 3.00 – 2.53 = 0.47 bar

(pHF )eq = 2(2.53) = 5.06 bar


494

16.79 (a) To calculate an equilibrium constant from initial and equilibrium conditions, set up aconcentration table:


Initial pressure (bar) ––– 0.464 0Change in pressure (bar) ––– – x + 2xEquilibrium pressure (bar) ––– 0.464 – x 2xThe problem gives the equilibrium pressure, which is the sum of partial pressures:0.746 bar = (0.464 – x) + 2 x = 0.464 + x, from which x = (0.746 – 0.464) = 0.282Substitute into the equilibrium constant expression and calculate K:

Keq = (pCO)eq

2

(pCO2 )eq=[(2)(0.282)]2

(0.464 − 0.282)=0.5642

0.182 = 1.75

(b) Applying the ideal gas equation, the pressure of CO2 will be multiplied by 3 if thecontainer is compressed to one third of its initial volume:pCO2 = 3(0.464 bar) = 1.392 bar

Set up a new concentration table using this as the initial pressure:


Initial pressure (bar) ––– 1.392 0Change in pressure (bar) ––– – x + 2xEquilibrium pressure(bar)

––– 1.392 – x 2x

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

Keq = 1.75 = (pCO)eq

2

(pCO2)eq=

(2x)2

1.392 − x=

4x2

1.392 − xx = 0.592 pCO = 2x = 2(0.592 bar) = 1.184 barpCO2 = 1.392 bar – x = 1.392 bar – 0.592 bar = 0.800 bar

The total pressure at equilibrium is the sum of the partial pressures:Peq = 1.184 bar + 0.800 bar = 1.98 bar (final result rounds to 3 significant figures)

16.80 (a) To calculate an equilibrium constant when experimental data concerning amounts areavailable, convert amounts data into concentration using stoichiometric reasoning,then complete an amounts table and substitute into the equilibrium constantexpression. Generally, pressures in bar are required for gas–phase reactions, but inthis problem, the equilibrium constant expression contains p2 in the numerator anddenominator, so units cancel and amounts can be used directly. Set up aconcentration table, using the change of 0.182 mol for I2:

Reaction: 2 HI ⎯ →⎯← ⎯⎯ H2 + I2

Initial amount (mol) 1.00 0 0Change in amount (mol) –2(0.182) + 0.182 + 0.182Equilibrium amount (mol) 0.636 0.182 0.182


495

Now substitute into the equilibrium constant expression and evaluate Keq:

Keq = (pH2 )eq(pI2 )eq

(pHI)eq2 =

(nH2 )eq(nI2 )eq(nHI)eq

2 =(0.182)2

(0.636)2 = 0.0819

(b) To estimate the equilibrium constant at a temperature different from 298 K, calculateΔHoreaction and ΔSoreaction at 298 K and then use Equations 14–10 and 16–3:


ΔHoreaction = [1 (0 kJ/mol) + 1 (62.4 kJ/mol)] – 2 (26.5 kJ/mol) = 9.4 kJ/molΔSoreaction = [1 (130.680 J/mol K) + 1 (260.7 J/mol K)] – 2 (206.6 J/mol K)

= –21.82 J/mol KΔGoreaction, 625 ˚C ≅ 9.4 kJ/mol) – (625 + 273 K)(–0.02182 kJ/mol K) = 29.0 kJ/mol

ln Keq = – 29.0 x103J mol-1

(8.314 J mol-1 K-1)(898 K) = –3.88

Keq = e–3.88 = 0.021

(c) As in part (a), we can make use of the fact that both numerator and denominator of theequilibrium constant expression contain p2, so units cancel and we can work directlywith moles. Construct a concentration table, write the equilibrium expression, andsolve for the pressures.

Reaction: 2 HI ⎯ →⎯← ⎯⎯ H2 + I2

Initial amount (mol) 1.00 0 0Change in amount (mol) –2x + x + xEquilibrium amount (mol) 1.00 – 2x x x

Use the Keq expression to calculate the amount of H2 formed:

Keq = (pH2 )eq(pI2 )eq

(pHI)eq2 =

(x)2

(1.00-2x)2 = 0.021

Take the square root of both sides:x

1.00 − 2x( ) = 0.145 so x = 0.145 – 0.290 x

1.290 x = 0.145 from which x = 0.112 mol and (1.00 – 2x) = 0.776 molUse the ideal gas equation to calculate the partial pressures, which we leave in atm:

pH2 = pI2 =

nRT

V=

(0.112 mol)(0.08206 L atmmol K )(898 K)

1.00 L = 8.25 atm

pHI =

nRT

V=

(0.776 mol)(0.08206 L atmmol K )(898 K)

1.00 L = 57.2 atm


496

16.81 This problem describes an equilibrium reaction pertinent to smog. Use standardprocedures to determine the partial pressures of each pollutant. The equilibrium constantexpression is dimensionless because it has p2 in both numerator and denominator, so wecan use any convenient units. Begin by converting all the initial pressures to the sameunits, ppm. Because Keq is so large, take the reaction to completion and then determinethe equilibrium pressures:

Reaction: O3 + NO → O2 + NO2

Initial (ppm) 6.5 x 10–3 1.5 2.1 x 105 0Change (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 as the initial values in a new concentration table, write the equilibriumexpression, 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 – xEquilibrium (ppm) x 1.5 + x 2.1 x 105 6.5 x 10–3 – xThe equilibrium expression is:

Keq = 6.0 x 1034 =

pO2 pNO2

pO3 pNO=

(2.1 x 105 )(6.5 x 10−3 − x)x(1.5 + x)

; assume x << 6.5 x 10–3

x = 1.5 x 10–32 (assumption is valid)Convert ppm to bar, using the fact that 1 ppm = 1 part in 106 and standard atmosphericpressure is 1.01325 bar:pO3 = (1.5 x 10–32 ppm)(1 atm)(10–6/ppm)(1.01325 bar) = 1.5 x 10–38 bar

pNO = (1.5 ppm)(1 atm)(10–6/ppm)(1.01325 bar) = 1.5 x 10–6 bar

pNO2 = (6.5 x 10–3 ppm)(1 atm)(10–6/ppm)(1.01325 bar) = 6.6 x 10–9 bar

pO2 = (2.1 x 105 ppm)(1 atm)(10–6/ppm)(1.01325 bar) = 0.21 bar

16.82 (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 islarge, 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 limitingreactant:

O2: 0.762 bar SO2: 0.3502

= 0.175 bar (LR)

Use a concentration table to determine the pressures at completion:Reaction: 2 SO2 + O2 → 2 SO3

Initial (bar) 0.350 0.762 0Change (bar) –0.350 – (0.350)/2 + 0.350Completion (bar) 0 0.587 0.350


497

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

Reaction: 2 SO2 + O2 ⎯ →⎯← ⎯⎯ 2 SO3

Initial (bar) 0 0.587 0.350Change (bar) +2x +x –2xEquilibrium (bar) 2x 0.587 + x 0.350–2x

The equilibrium expression is:

Keq = 5.60 x 104 =

pSO32

pSO22 pO2

=(0.350 − 2x)2

(2x)2(0.587 + x); assume 2x<<0.350

5.60 x 104 = (0.350)2

(2x)2(0.587); rearrange to get:

x2 = (0.350)2

4(5.60 x 104 )(0.587) = 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 bar

pO2 = 0.587 + 9.65 x 10–4 = 0.588 bar

pSO3 = 0.350 – 2(9.65 x 10–4) = 0.348 bar

(b) To estimate Keq using thermodynamic values, use the equation:

ΔGo = –RT ln Keq

ΔGfo for SO2, O2, and SO3 are –300.1, 0.00, and –371.1 kJ/mol respectively.

ΔGoreaction = 2 (–371.1 kJ/mol)– [1 (0.000 kJ/mol) + 2 (–300.1 kJ/mol)] = –142.0 kJ/mol

ln Keq = −ΔGo

RT= −

–1.42 x 105 J mol-1

(8.314 J mol-1K-1)(298 K)= 57.3

Keq = e57.3 = 7.7 x 1024

Based on this result, the reaction should be run at lower temperature to improve theyield.

(c) Although the reaction is thermodynamically more favorable at lower temperatures, thereaction is kinetically very slow. Therefore, it must be run at higher temperatures(700 K) to speed up the reaction while still being thermodynamically favorable.


498

16.83 The molecular picture of the equilibrium conditions shows 4 O2, 5 SO3, and 5 SO2molecules. 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 moleculeswith equilibrium amounts.Equilibrium conditions show 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 productmolecules, so Q < Keq and the reaction will proceed to the right, forming additionalSO3.

(b) To determine the effect of temperature, compare amounts of molecules at equilibriumfor the two temperatures.Equilibrium conditions at 1100 K: 4 O2, 5 SO3, and 5 SO2 moleculesEquilibrium conditions at 1300 K: 5 O2, 3 SO3, and 7 SO2 moleculesThe equilibrium shifts to the left, forming additional reactant molecules, whentemperature increases. According to Le Châtelier's principle, the reaction shifts in theendothermic direction (absorbs heat) when temperature increases. Therefore, theforward reaction is exothermic.

(c) To determine what happens under the new conditions, compare amounts of moleculeswith equilibrium amounts.Equilibrium conditions show 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 ofmolecules of a reactant and a product both have increased. Compare Q for the secondset 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 nochange will occur.

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