Chapter 18 Homework Answers

Chapter 18 Homework Answers

18.22.

a. Since ∆G° = −RT lnK, as long as the temperature remains constant, the value of ∆G°

also remains constant.

b. In this case, ∆G = ∆G° + RT lnQ. Since the reaction quotient Q changes as the reaction proceeds, the value of ∆G will also change.

c. Since the value of K is large, the reaction is spontaneous.

d. No, the statement is not true. At equilibrium, ∆G is zero, but the value of ∆G° is constant, so it is not zero.

e. Since the value of K is large, at equilibrium the composition will be mostly B(g).

f. For the reverse reaction, Kr = 1/Kf, so ∆G° will have the same numerical value with the opposite sign and will be a constant. At equilibrium, ∆G will be zero, and the reaction mixture will still be mostly B(g). The reaction will not be spontaneous in this direction.

18.24.

a. Spontaneous. Sugar dissolves spontaneously in hot water.

b. Nonspontaneous. Rust does not spontaneously change to iron; rather, iron spontaneously rusts in air.

c. Spontaneous. The burning of butane in air is a spontaneous reaction.

d. Nonspontaneous. A pendulum, once stopped, will not spontaneously begin to move again.

e. Nonspontaneous. Water will not spontaneously decompose into its elements.

18.26.

a. Entropy increases; ∆S is positive; energy dispersal increases when the food coloring disperses throughout the water.

b. Entropy decreases; ∆S is negative; as a tree leafs out, energy dispersal decreases, and entropy decreases.

c. Entropy increases; ∆S is positive; as flowers wilt and stems decompose, energy dispersal increases and entropy increases.

d. Entropy decreases; ∆S is negative; as a liquid changes to a solid, there is a decrease in energy dispersal and a decrease of entropy.

e. Entropy increases; ∆S is positive; as a liquid changes to a vapor, energy dispersal increases and entropy increases.

18.34. At 25°C (298 K) and 1 atm (1.013 x 10⁵ Pa), the decrease in volume going from three moles to zero mol of gas is

−3 x 22.41 L x ^{298 K}

273 K = −73.386 L (−73.386 x 10⁻³ m³)

The enthalpy change for this reaction is two times the enthalpy of formation of H2O(l), which is −571.6 kJ. The work done on the chemical system by the atmosphere is

w = −P∆V = − (1.013 x 10⁵ Pa) x (−73.386 x 10⁻³ m³) = 7.434 x 10³ J = 7.434 kJ

∆U = qp + w = (−571.6 kJ) + 7.434 kJ = −564.16 = −564.2 kJ 18.38. First, determine the enthalpy change for the vaporization of 1.00 mol of CS2(l).

∆Hcond = −∆Hvap = −27.2 kJ/mol

The entropy change for this condensation at 25°C (298 K) is

∆S = ^∆Hcond T =

-2.72 x 10 J4

298 K = −91.27 J/K 18.40.

a. ∆S° is positive because there is an increase in moles of gas (∆ngas = +3) from one mole of liquid reactant forming three moles of gaseous products. (Entropy increases.) b. ∆S° is negative because there is a decrease in moles of gas (∆ngas = −1) from a liquid

reactant and three moles of gaseous reactant forming two moles of gaseous product and liquid product. (Entropy decreases.)

c. ∆S° is negative because there is a decrease in moles of gas (∆ngas = −5) from five moles of gaseous reactant and one mole of solid reactant forming one mole of solid product. (Entropy decreases.)

d. ∆S° is positive because there is an increase in moles of gas (∆ngas = +2) from a solid reactant forming two moles of gaseous product plus solid product. (Entropy increases.) 18.44.

CaCO₃(s) 2H⁺(aq) Ca²⁺(aq) H₂O(l) CO₂(g)

S°: 92.9 0 -53.1 69.95 213.7 J/K

+ → + +

∆S° = ΣnS°(products) − ΣmS°(reactants) =

[(−53.1 + 69.95 + 213.7) − (92.9)] J/K = 137.65 J/K S increases, as expected from the increase in moles of gas.

18.46. The reaction, standard enthalpy changes, and standard entropies are as follows:

C₃H₈(g) 5O₂(g) 3CO₂(g) 4H₂O(g)

∆H_f^°: -104.7 0 3 x (-393.5) 4 x (-241.8) kJ

S^°: 270.2 5 x 205.0 3 x 213.7 4 x 188.7 J/K

+ → +

Calculate ∆H° and ∆S° for the reaction.

∆H° = Σn∆Hf°(products) − Σm∆Hf°(reactants) =

[3 x (−393.5) + 4 x (−241.8) − (−104.7)] kJ = −2043.0 kJ

∆S° = ΣnS°(products) − ΣmS°(reactants) =

[(3 x 213.7 + 4 x 188.7) − (270.2 + 5 x 205.0)] J/K = 100.7 (0.1007 kJ/K)

∆G° = ∆H° − T∆S° = −2043.0 kJ − (298 K)(0.1007 kJ/K) = −2073.00 =

−2073.0 kJ 18.52.

a. Spontaneous reaction b. Spontaneous reaction c. Nonspontaneous reaction d. Nonspontaneous reaction

e. Equilibrium mixture; significant amounts of both 18.58. Calculate ∆G° per 1 mol Zn(s) using the given ∆Gf° values.

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

∆G_f^°: 0 0 -147.0 0 kJ

∆G° = [(−147.0) − 0] kJ = −147.0 kJ/mol Zn

−147.0 kJ/mol Zn x (3.65 g ÷ 65.39 g/mol Zn) = −8.205 = −8.21 kJ

Maximum work equals ∆G° equals −8.21 kJ. Because maximum work is stipulated, no entropy is produced.

18.60.

a. K = Kp = ³

2 CH OH 2 CO H

P P P

b. K = ¹

Ksp = _{+ 2 2-}

4

1 [ Ag ] [ CrO ] c. K = ²

2+

CO + 2

[ Ca ] [ H ]

P

18.70. First, calculate ∆H° and ∆S° using the given ∆Hf° and S° values.

2HgO(s) → 2Hg(g ) + O₂(g)

∆H_f°:

∆S°:

2 x (-90.79) 2 x 70.27

2 x 61.38 2 x 174.9

0 kJ 205.0 J/K

∆H° = [(2 x 61.38) − 2(−90.79)] kJ = 304.34 kJ

∆S° = [(2 x 174.9 + 205.0) − 2(70.27)] J/K = 414.26 J/K (0.41426 kJ/K) Substitute these values into ∆G° = ∆H° − T∆S°, let ∆G° = 0, and rearrange to solve for T.

T = ^{∆ °}

∆ ° H

S = ^{304.34 kJ}

0.41426 kJ/K = 734.65 = 734.7 K

18.76. When the liquid evaporates, it absorbs heat: ∆Hvap = 29.1 kJ/mol, or 2.91 x 10⁴ J/mol, at 56°C (329 K). The entropy change is

∆S = ^∆Hvap T =

2.91 x 10 J/g4

329 K = 88.45 = 88.4 J/(K^•mol) 18.78.

a. Entropy decreases; ∆S° is negative because there is a decrease in the moles of gas (∆ngas = −2) from four moles of gaseous reactant forming two moles of gaseous products.

b. Entropy increases; ∆S° is certainly positive because there is an increase in the moles of gas (∆ngas = +2) from a solid reactant forming two moles of gas.

c. Entropy decreases; ∆S° is negative because there is a decrease in the moles of gas (∆ngas = −3) from three moles of gaseous reactant forming a liquid product.

d. Entropy increases; ∆S° is certainly positive because there is an increase in moles of gas (∆ngas = +1) from solid and liquid reactants forming one mole of gaseous product.

18.84. Calculate ∆G° using the ∆Gf° values from Appendix C.

N₂(g) + CH₄(g) → HCN(g) + NH₃(g)

∆G_f°: 0 -50.80 124.7 -16.40 kJ

∆G° = Σn∆Gf°(products) − Σm∆Gf°(reactants) =

[(−16.40 + 124.7) − (−50.80)] kJ = 159.10 = 159.1 kJ Because ∆G° is positive, the reaction is nonspontaneous as written, at 25°C.

18.88. First, calculate ∆G° using the values in Appendix C.

BaSO₄(s) Ba₂+(aq) + SO₄^2-(aq)

∆G_f°: -1362.3 -560.7 -744.6 kJ

Hence, ∆G° for the reaction is

∆G° = [−560.7 + (−744.6) − (−1362.3)] kJ = 57.0 kJ

Now, substitute numerical values into the equation relating ln K and ∆G°.

ln K = ^{-∆ °G} RT =

57.0 x 103

- 8.31 x 298 = −23.017 K = Ksp = e^−23.017 = 1.00 x 10⁻¹⁰ = 1 x 10⁻¹⁰ 18.96.

a. ∆G° = [3(−228.6) − (−763.1)] kJ = 77.3 kJ Since ∆G° is positive, K will be less than 1.

b. ∆H° = [3(−241.8) − (−839.9)] kJ = 114.5 kJ

∆G° = ∆H° − T∆S°

77.3 kJ = 114.5 kJ − (298 K) ∆S°

∆S° = 0.1248 kJ/K = 125 J/K c. T = ^∆

∆ H

S =

114.5 x 10 J3

124.8 J/K = 917.4 = 917 K

d. The driving force is the change in entropy. At high temperatures, the term T∆S becomes very important.

18.104.

a. ∆H° = [−626.2 − (−608.8)] kJ = −17.4 kJ = −17.4 x 10³ J

∆S° = [139.8 − 232.2] J/K = −92.4 J/K

∆G° = ∆H° − T∆S°

∆G° = −17.4 x 10³ J − (298 K)(−92.4 J/K) = 10.135 x 10³ J

Now, substitute numerical values into the equation relating ln K and ∆G°.

ln K = ^{-∆ °G} RT =

10.135 x 103

- 8.31 x 298 = −4.092 K = e^−4.092 = 0.01669 = 0.017

b. The change in entropy is negative, greater order, so this causes H2SO3 to be a weak acid. The enthalpy change favors the acid strength of H2SO3, and the entropy is a very important term.

18.118. For the formation of HI, assume ∆H and ∆S are constant over the temperature range from 25°C to 205°C, and calculate the value of each to use to calculate K at 205°C. Start by calculating ∆H° and ∆S° at 25°C using ∆Hf° and S° values. From Appendix C, we have

H₂(g) I₂(g) 2HI(g)

0 62.42 2 x 26.36 kJ

130.6 260.6 2 x 206.5 J/K

∆H_f°:

S°:

+

Calculate ∆H° and ∆S° from these values.

∆H° = [2(26.36) − 62.42] kJ = −9.70 kJ

∆S° = [2(206.5) − 130.6 − 260.6)] J/K = 21.80 J/K = 0.02180 kJ/K Substitute ∆H°, ∆S° ( = −0.02180 kJ/K), and T (478 K) into the equation for ∆GT°.

∆GT° = ∆H° − T∆S° = −9.70 kJ − (478 K)(0.02180 kJ/K) = −20.120 kJ

= −20.120 x 10³ J

Now, substitute numerical values into the equation relating ln K and ∆G° ( = ∆GT°).

ln K = ^{-∆ °G} RT =

-20.120 x 103

- 8.31 x 478 = 5.0653 K = e^5.0653 = 158.43

Letting [HI] = 2x, [H2] = 0.500 mol − x, and [I2] = 1.00 mol − x, substitute into the equilibrium expression:

K =

2 2

[HI]

[H ][I ] =

(2 )2

(0.500 - )(1.00 - ) x

x x = 158.43