CHAPTER 20 THERMODYNAMICS: ENTROPY, FREE ...

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CHAPTER 20 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND THE DIRECTION OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 20.1A Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than

that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, ∆S is positive.

Solution: a) PCl5(g). For substances with the same type of atoms and in the same physical state, entropy increases with increasing number of atoms per molecule because more types of molecular motion are available. b) BaCl2(s). Entropy increases with increasing atomic size. The Ba2+ ion and Cl– ion are larger than the Ca2+ ion and F– ion, respectively.

c) Br2(g). Entropy increases from solid → liquid → gas. 20.1B Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than

that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, ∆S is positive.

Solution: a) LiBr(aq). For substances with the same number of atoms and in the same physical state, entropy increases with increasing atomic size. LiBr has the lower molar mass of the two substances and, therefore, the lower entropy. b) Quartz. Quartz has a crystalline structure and the particles have less freedom (and lower entropy) in that structure than in glass, an amorphous solid.

c) Cyclohexane. Ethylcyclobutane has a side chain, which has more freedom of motion than the atoms in the ring. In cyclohexane, there are no side chains. Freedom of motion is restricted by the ring structure.

20.2A Plan: Predict the sign of rxnS∆ by comparing the randomness of the products with the randomness of the reactants.

Calculate rxnS∆ using Appendix B values and the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Solution: a) 4NO (g) → N2O(g) + N2O3(g)

The rxnS∆ is predicted to decrease ( rxnS∆ < 0) because four moles of random, gaseous product are transformed into two moles of random, gaseous product (the change in gas moles is –2).

rxnS∆ = ∑m productsS – ∑n reactantsS

rxnS∆ = [(1 mol N2O3)(Sº of N2O3) + (1 mol N2O)(Sº of N2O)] – [(4 mol NO)(Sº of NO)] rxnS∆ = [(1 mol N2O3)(314.7 J/mol•K) + (1 mol N2O)(219.7 J/mol•K)] − [(4 mol NO)(210.65 J/mol•K)] = – 308.2 J/K rxnS∆ < 0 as predicted.

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b) CH3OH(g) → CO(g) + 2H2(g) The change in gaseous moles is +2, so the sign of rxnS∆ is predicted to be greater than zero.

rxnS∆ = [(1 mol CO)(Sº of CO) + (2 mol H2)(Sº of H2)] − [(1 mol CH3OH)(Sº of CH3OH)] rxnS∆ = [(1 mol CO)(197.5 J/mol•K) + (2 mol H2)(130.6 J/mol•K)] − [(1 mol CH3OH)(238 J/mol•K)] = 220.7 = 221 J/K

rxnS∆ > 0 as predicted. 20.2B Plan: Predict the sign of rxnS∆ by comparing the randomness of the products with the randomness of the reactants.

Calculate rxnS∆ using Appendix B values and the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Solution: a) 2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)

The rxnS∆ is predicted to decrease ( rxnS∆ < 0) because the more random, gaseous reactant is transformed into a more ordered, liquid product.


rxnS∆ = [(1 mol Na2CO3)(Sº of Na2CO3) + (1 mol H2O)(Sº of H2O)] – [(2 mol NaOH)(Sº of NaOH) + (1 mol CO2)(Sº of CO2)] rxnS∆ = [(1 mol Na2CO3)(139 J/mol•K) + (1 mol H2O)(69.940 J/mol•K)] − [(2 mol NaOH)(64.454 J/mol•K) + (1 mol CO2)(213.7 J/mol•K)] = –133.668 = – 134 J/K rxnS∆ < 0 as predicted. b) 2Fe(s) + 3H2O(g) → Fe2O3(s) + 3H2(g)

The change in gaseous moles is zero, so the sign of rxnS∆ is difficult to predict. Iron(III) oxide has greater entropy than Fe because it is more complex, but this is offset by the greater molecular complexity of H2O versus H2.

rxnS∆ = [(1 mol Fe2O3)(Sº of Fe2O3) + (3 mol H2)(Sº of H2)] − [(2 mol Fe)(Sº of Fe) + (3 mol H2O)(Sº of H2O)] rxnS∆ = [(1 mol Fe2O3)(87.400 J/mol•K) + (3 mol H2)(130.6 J/mol•K)] − [(2 mol Fe)(27.3 J/mol•K) + (3 mol H2O)(188.72 J/mol•K)] = –141.56 = –141.6 J/K

The negative rxnS∆ shows that the greater entropy of H2O versus H2 does outweigh the greater entropy of Fe2O3 versus Fe.

20.3A Plan: Write the balanced equation for the reaction and calculate the rxnS∆ using Appendix B. Determine

the surr∆S by first finding rxn∆ H . Add surr∆S to rxnS∆ to verify that ∆Suniv is positive. Solution: P4(s) + 6Cl2(g) → 4PCl3(g) rxnS∆ = [(4 mol PCl3)(Sº of PCl3)] – [(1 mol P4)(Sº of P4) + (6 mol Cl2)(Sº of Cl2)]

rxnS∆ = [(4 mol PCl3)(312 J/mol•K)] – [(1 mol P4)(41.1 J/mol•K) + (6 mol Cl2) (223.0J/mol•K)]

rxnS∆ = –131J/K (The entropy change is expected to be negative because the change in gas moles is negative.)

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rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(4 mol PCl3)( fH∆ of PCl3)]

– [(1 mol P4)( fH∆ of P4) + (6 mol Cl2)( fH∆ of Cl2)]

rxnH∆ = [(4 mol PCl3)(–287 kJ/mol)] – [(1 mol P4)(0 kJ/mol) + (6 mol Cl2)(0 kJ/mol)] rxnH∆ = –1148 kJ

surr∆S = rxnHT

∆−

= 1148 kJ298 K

−− = 3.8523 kJ/K(103 J/1 kJ) = 3850 J/K

univS∆ = rxnS∆ + surr∆S = (–131 J/K) + (3850 J/K) = 3719 J/K

Because univS∆ is positive, the reaction is spontaneous at 298 K. 20.3B Plan: Write the balanced equation for the reaction and calculate the rxnS∆ using Appendix B. Determine

the surr∆S by first finding rxn∆ H . Add surr∆S to rxnS∆ to verify that ∆Suniv is positive. Solution: 2FeO(s) + 1/2O2(g) → Fe2O3(s) rxnS∆ = [(1 mol Fe2O3)(Sº of Fe2O3)] – [(2 mol FeO)(Sº of FeO) + (1/2 mol O2)(Sº of O2)]

rxnS∆ = [(1 mol Fe2O3)(87.400 J/mol•K)] – [(2 mol FeO)(60.75 J/mol•K) + (1/2 mol O2) (205.0J/mol•K)]

rxnS∆ = –136.6 J/K (entropy change is expected to be negative because gaseous reactant is converted to solid product). rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol Fe2O3)( fH∆ of Fe2O3)]

– [(2 mol FeO)( fH∆ of FeO) + (1/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol Fe2O3)(–825.5 kJ/mol)] – [(2 mol FeO)(–272.0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)] rxnH∆ = –281.5 kJ

surr∆S = rxnHT

∆−

= 281.5 kJ298 K

−− = 0.94463 kJ/K(103 J/1 kJ) = 944.63 J/K

univS∆ = rxnS∆ + surr∆S = (–136.6 J/K) + (944.63 J/K) = 808.03 = 808 J/K

Because univS∆ is positive, the reaction is spontaneous at 298 K. This process is also known as rusting. Common sense tells us that rusting occurs spontaneously. Although the entropy change of the system is negative, the increase in entropy of the surroundings is large enough to offset

rxnS∆ . 20.4A Plan: Calculate the rxnH∆ using fH∆ values from Appendix B. Calculate rxnS∆ from tabulated Sº values and then

use the relationship rxnG∆ = rxnH∆ – rxnT S∆ . Solution: rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(2 mol NOCl)( fH∆ of NOCl)] – [(2 mol NO)( fH∆ of NO) + (1 mol Cl2)( fH∆ of Cl2)]

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rxnH∆ = [(2 mol NOCl)(51.71 kJ/mol)] – [(2 mol NO)(90.29 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]

rxnH∆ = –77.16 kJ


rxnS∆ = [(2 mol NOCl)(Sº of NOCl)] – [(2 mol NO)(Sº of NO) + (1 mol Cl2)(Sº of Cl2)]

rxnS∆ = [(2 mol NOCl)(261.6 J/mol•K)] – [(2 mol NO)(210.65 J/mol•K) + (1 mol Cl2)(223.0 J/mol•K)]

rxnS∆ = –121.1 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –77.16 kJ – [(298 K)(–121.1 J/K)(1 kJ/103 J)] = –41.0722 = –41.1 kJ 20.4B Plan: Calculate the rxnH∆ using fH∆ values from Appendix B. Calculate rxnS∆ from tabulated Sº values and then

use the relationship rxnG∆ = rxnH∆ – rxnT S∆ . Solution: rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(2 mol NO2)( fH∆ of NO2)] – [(2 mol NO)( fH∆ of NO) + (1 mol O2)( fH∆ of O2)]

rxnH∆ = [(2 mol NO2)(33.2 kJ/mol)] – [(2 mol NO)(90.29 kJ/mol) + (1 mol O2)(0 kJ/mol)]

rxnH∆ = –114.18 = –114.2 kJ


rxnS∆ = [(2 mol NO2)(Sº of NO2)] – [(2 mol NO)(Sº of NO) + (1 mol O2)(Sº of O2)]

rxnS∆ = [(2 mol NO2)(239.9 J/mol•K)] – [(2 mol NO)(210.65 J/mol•K) + (1 mol O2)(205.0 J/mol•K)] rxnS∆ = –146.5 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –114.2 kJ – [(298 K)(–146.5 J/K)(1 kJ/103 J)] = –70.543 = –70.5 kJ 20.5A Plan: Use fG∆ values from Appendix B to calculate rxnG∆ using the relationship

rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: a) rxnG∆ = [(2 mol NOCl)( fG∆ of NOCl)] – [(2 mol NO)( fG∆ of NO) + (1 mol Cl2)( fG∆ of Cl2)]

rxnG∆ = [(2 mol NOCl)(66.07 kJ/mol)] – [(2 mol NO)(86.60 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]

rxnG∆ = –41.06 kJ

b) rxnG∆ = [(2 mol Fe)( fG∆ of Fe) + (3 mol H2O)( fG∆ of H2O)] –

[(3 mol H2)( fG∆ of H2) + (1 mol Fe2O3)( fG∆ of Fe2O3)]

rxnG∆ = [(2 mol Fe)(0 kJ/mol) + (3 mol H2O)(–228.60 kJ/mol)] – [(3 mol H2)(0 kJ/mol) + (1 mol Fe2O3)( –743.6 kJ/mol)]

rxnG∆ = 57.8 kJ 20.5B Plan: Use fG∆ values from Appendix B to calculate rxnG∆ using the relationship

rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: a) rxnG∆ = [(2 mol NO2)( fG∆ of NO2)] – [(2 mol NO)( fG∆ of NO) + (1 mol O2)( fG∆ of O2)]

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rxnG∆ = [(2 mol NO2)(51 kJ/mol)] – [(2 mol NO)(86.60 kJ/mol) + (1 mol O2)(0 kJ/mol)]

rxnG∆ = –71.2 = –71 kJ

b) rxnG∆ = [(2 mol CO)( fG∆ of CO)] – [(2 mol C)( fG∆ of C) + (1 mol O2)( fG∆ of O2)]

rxnG∆ = [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]

rxnG∆ = –274.4 kJ 20.6A Plan: Predict the sign of rxnS∆ by comparing the randomness of the products with the randomness of the

reactants. Use the relationship rxnG∆ = rxnH∆ – rxnT S∆ to answer b). Solution: a) The reaction is X2Y2(g) → X2(g) + Y2(g). Since there are more moles of gaseous product than there are of gaseous reactant, entropy increases and ∆S > 0. b) The reaction is only spontaneous above 325ºC or in other words, at high temperatures. In the relationship

rxnG∆ = rxnH∆ – rxnT S∆ , when ∆S > 0 so that – T∆S° is < 0, ∆G° will only be negative at high T if ∆H° > 0.

20.6B Plan: Predict the sign of rxnS∆ by comparing the randomness of the products with the randomness of the

reactants. Use the relationship rxnG∆ = rxnH∆ – rxnT S∆ to answer b). Solution: a) A solid forms a gas and a liquid, so ∆S > 0. A crystalline array breaks down, so ∆H > 0. b) For the reaction to occur spontaneously (∆G < 0), – T∆S° must be greater than ∆H, which would occur only at higher T.

20.7A Plan: Use the equation ∆G° = ∆H° – T∆S° to determine if the reaction is spontaneous (∆G° < 0). Then examine

the same equation to determine the effect of raising the temperature on the spontaneity of the reaction. Solution: a) rxnG∆ = rxnH∆ – rxnT S∆ = –192.7 kJ – [(298 K)(–308.2 J/K)(1 kJ/103 J)] = –100.8564 = –100.9 kJ Because ∆G < 0, the reaction is spontaneous at 298 K. b) As temperature increases, – T∆S° becomes more positive, so the reaction becomes less spontaneous at higher

temperatures. c) rxnG∆ = rxnH∆ – rxnT S∆ = –192.7 kJ – [(773 K)(–308.2 J/K)(1 kJ/103 J)] = 45.5386 = 45.5 kJ

20.7B Plan: Examine the equation ∆G° = ∆H° – T∆S° and determine which combination of enthalpy and entropy will

describe the given reaction. Solution: Two choices can already be eliminated: 1) When ∆H > 0 (endothermic reaction) and ∆S < 0 (entropy decreases), the reaction is always nonspontaneous, regardless of temperature, so this combination does not describe the reaction.

2) When ∆H < 0 (exothermic reaction) and ∆S > 0 (entropy increases), the reaction is always spontaneous, regardless of temperature, so this combination does not describe the reaction. Two combinations remain: 3) ∆H° > 0 and ∆S° > 0, or 4) ∆H° < 0 and ∆S° < 0. If the reaction becomes spontaneous at –40°C, this means that ∆G° becomes negative at lower temperatures. Case 3) becomes spontaneous at higher temperatures, when the –T∆S° term is larger than the positive enthalpy term. By process of elimination, Case 4) describes the reaction. At a lower temperature, the negative ∆H° becomes larger than the positive (–T∆S°) value, so ∆G° becomes negative.

20.8A Plan: To find the temperature at which the reaction becomes spontaneous, use

rxnG∆ = 0 = rxnH∆ – rxnT S∆ and solve for temperature.

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Solution: The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.

rxnG∆ = rxnH∆ – rxnT S∆ 0 = –192.7x103 J – (T)(–308.2 J/K) T = 625.2 K – 273.15 = 352.0°C

20.8B Plan: To find the temperature at which the reaction becomes spontaneous, use

rxnG∆ = 0 = rxnH∆ – rxnT S∆ and solve for temperature. rxnH∆ can be calculated from the individual fH∆ values of the reactants and products by using the relationship

rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ . rxnS∆ can be calculated from the individual S values of the

reactants and products by using the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Solution: CaO(s) + CO2(g) → CaCO3(s) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol CaCO3)( fH∆ of CaCO3)]

– [(1 mol CaO)( fH∆ of CaO) + (1 mol CO2)( fH∆ of CO2)]

rxnH∆ = [(1 mol CaCO3)( –1206.9 kJ/mol)] – [(1 mol CaO)(–635.1 kJ/mol) + (1 mol CO2)( –393.5 kJ/mol)] rxnH∆ = –178.3 kJ


rxnS∆ = [(1 mol CaCO3)(Sº of CaCO3)] – [(1 mol CaO)(Sº of CaO) + (1 mol CO2)(Sº of CO2)]

rxnS∆ = [(1mol CaCO3)(92.9 J/mol•K)] – [(1 mol CaO)(38.2 J/mol•K) + (1 mol CO2)(213.7 J/mol•K)] rxnS∆ = –159.0 J/K = – 0.159 kJ/K

rxnG∆ = 0 = rxnH∆ – rxnT S∆

rxnH∆ = rxnT S∆

= HTS

∆

∆

= –178.3 kJ–0.1590 kJ/K

= 1121.384 = 1121 K

The reaction becomes spontaneous at temperatures < 1121 K. 20.9A Plan: First find ∆G°, then calculate K from ∆G° = –RT ln K. Calculate G∆ using fG∆ values in the relationship

rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: 2C(graphite) + O2(g) 2CO(g) rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆

rxnG∆ = [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)( 0 kJ/mol)] = –274.4 kJ

ln K = GRT

∆−

= ( )( )

274.4 kJ/mol 1000 J8.314 J/mol•K 298 K 1 kJ

− −

= 110.7536 = 111

K = e111 = 1.6095x1048 = 1.6x1048

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20.9B Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. Solution: ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (2.22x10–15) = 8.35964x104 J = 83.6 kJ/mol 20.10A Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. Remember that each

particle represents 0.10 mol and that the volume is 1.0 L. A reaction that is proceeding to the right will have ∆G° < 0 and a reaction that is proceeding to the left will have ∆G° > 0. A reaction at equilibrium has ∆G° = 0. Solution: a) A(g) + 3B(g) AB3(g)

b) Qc = [AB3]

[A][B]3

Mixture 1: Qc = [AB3]

[A][B]3 =

[0.20][0.40][0.80]3

= 0.98


[A][B]3 =

[0.30][0.30][0.50]3

= 8


[A][B]3 =

[0.40][0.20][0.20]3

= 250

Mixture 2 is at equilibrium since Qc = Kc c) Qc < Kc for Mixture 1 and the reaction is proceeding right to reach equilibrium; thus ∆G° < 0. Mixture 2 is at equilibrium and ∆G° = 0. Mixture 3 proceeds to the left to reach equilibrium since Qc > Kc and ∆G° > 0. The ranking for most positive to most negative is 3 > 2 > 1.

20.10B Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. A reaction that

is proceeding to the right will have ∆G° < 0 and a reaction that is proceeding to the left will have ∆G° > 0. A reaction at equilibrium has ∆G° = 0. Solution: a) X2(g) + 2Y2(g) 2XY2(g)

Qc = 2

22

2 2

[XY ][X ][Y ]

Mixture 1: Qc = 2

22

2 2

[XY ][X ][Y ]

= 2

2[5]

[2][1] = 12.5

Mixture 2: Qc = 2

22

2 2

[XY ][X ][Y ]

= 2

2[4]

[2][2] = 2

Mixture 3: Qc = 2

22

2 2

[XY ][X ][Y ]

= 2

2[2]

[4][2] = 0.25

Mixture 2 is at equilibrium since Qc = Kc b) Qc > Kc for Mixture 1 and the reaction is proceeding left to reach equilibrium; thus ∆G° > 0. Mixture 2 is at equilibrium and ∆G° = 0. Mixture 3 proceeds to the right to reach equilibrium since Qc < Kc and ∆G° < 0. The ranking for most negative to most positive is 3 < 2 < 1.

c) Any reaction mixture moves spontaneously towards equilibrium so both changes have a negative ∆G°. 20.11A Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. The free energy of the

reaction under non-standard state conditions is calculated using ∆G = ∆G° + RT ln Q. Solution:

a) ln K = GRT

∆−

= ( )( )

33.5 kJ/mol 1000 J8.314 J/mol•K 298 K 1 kJ

− −

= 13.5213

K = e13.5213 = 7.4512x105 = 7.45x105

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b) Q = [C2H5Cl]

[C2H4][HCl] =

(1.5)(0.50)(1.0)

= 3.0

∆G = ∆G° + RT ln Q = (–33.5 kJ/mol) + (8.314 J/mol•K) (1 kJ/1000 J) (298 K) ln (3.0) = –30.7781 = –30.8 kJ/mol 20.11B Plan: Write a balanced equation for the dissociation of hypobromous acid in water. The free energy of the

reaction at standard state is calculated using ∆G° = –RT ln K. The free energy of the reaction under non-standard state conditions is calculated using ∆G = ∆G° + RT ln Q.

Solution: HBrO(aq) + H2O(l) BrO–(aq) + H3O+(aq) a) ∆G° = –RT ln K = – (8.314 J/mol•K)(298)ln (2.3x10–9) = 4.927979x104 = 4.9x104 J/mol = 49 kJ/mol

b) Q = [ ]

3H O BrO

HBrO

+ − =

[ ][ ]

46.0x10 0.10

0.20

−

∆G = ∆G° + RT ln Q = (4.927979x104 J/mol) + (8.314 J/mol•K) (298 K) ln [ ]

[ ]

46.0x10 0.10

0.20

−

= 2.9182399x104 = 2.9x104 J/mol = 29 kJ/mol The value of Ka is very small, so it makes sense that ∆G° is a positive number. The natural log of a negative exponent gives a negative number (ln 3.0 x 10–4), so the value of ∆G decreases with concentrations lower than the standard state 1 M values.

CHEMICAL CONNECTIONS BOXED READING PROBLEMS B20.1 Plan: Convert mass of glucose (1 g) to moles and use the ratio between moles of glucose and moles of ATP to

find the moles and then molecules of ATP formed. Do the same calculation with tristearin. Solution:

a) Molecules of ATP/g glucose =

( )231 mol glucose 36 mol ATP 6.022x10 molecules ATP1 g glucose

180.16 g glucose 1 mol glucose 1 mol ATP

= 1.20333x1023 = 1.203x1023 molecules ATP/g glucose

b) Molecules of ATP/g tristearin =

( )231 mol tristearin 458 mol ATP 6.022x10 molecules ATP1 g tristearin

897.50 g tristearin 1 mol tristearin 1 mol ATP

= 3.073065x1023 = 3.073x1023 molecules ATP/g tristearin B20.2 Plan: Add the two reactions to obtain the overall process; the values of the two reactions are then added to obtain for the overall reaction. Solution:

creatine phosphate → creatine + phosphate ∆G° = – 43.1 kJ/mol ADP + phosphate → ATP ∆G° = +30.5 kJ/mol creatine phosphate + ADP → creatine + ATP ∆G = –43.1 kJ/mol + 30.5 kJ/mol = –12.6 kJ/mol END–OF–CHAPTER PROBLEMS 20.1 Spontaneous processes proceed without outside intervention. The fact that a process is spontaneous does not mean that it will occur instantaneously (in an instant) or even at an observable rate. The rusting of iron is an example of a process that is spontaneous but very slow. The ignition of gasoline is an example of a process that is not spontaneous but very fast.

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20.2 A spontaneous process occurs by itself (possibly requiring an initial input of energy), whereas a nonspontaneous process requires a continuous supply of energy to make it happen. It is possible to cause a nonspontaneous process to occur, but the process stops once the energy source is removed. A reaction that is found to be nonspontaneous under one set of conditions may be spontaneous under a different set of conditions (different temperature, different concentrations). 20.3 a) The energy of the universe is constant. b) Energy cannot be created or destroyed. c) ∆Esystem = –∆Esurroundings The first law is concerned with balancing energy for a process but says nothing about whether the process can, in fact, occur. 20.4 Entropy is related to the freedom of movement of the particles. A system with greater freedom of movement has higher entropy. a) and b) Probability is so remote as to be virtually impossible. Both would require the simultaneous, coordinated movement of a large number of independent particles, so are very unlikely. 20.5 Vaporization is the change of a liquid substance to a gas so ∆Svaporization = Sgas – S liquid. Fusion is the change of a solid substance into a liquid so ∆S fusion = S liquid – Ssolid. Vaporization involves a greater change in volume than fusion. Thus, the transition from liquid to gas involves a greater entropy change than the transition from solid to liquid. 20.6 In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases (∆Ssurr > 0). In an endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases (∆Ssurr < 0). A chemical cold pack for injuries is an example of a spontaneous, endothermic chemical reaction as is the melting of ice cream at room temperature. 20.7 a) According to the third law the entropy is zero. b) Entropy will increase with temperature. c) The third law states that the entropy of a pure, perfectly crystalline element or compound may be taken as zero at zero Kelvin. Since the standard state temperature is 25°C and entropy increases with temperature, S° must be greater than zero for an element in its standard state. d) Since entropy values have a reference point (0 entropy at 0 K), actual entropy values can be determined, not just entropy changes. 20.8 Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Spontaneous, evaporation occurs because a few of the liquid molecules have enough energy to break away from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase. b) Spontaneous, a lion spontaneously chases an antelope without added force. This assumes that the lion has not just eaten. c) Spontaneous, an unstable substance decays spontaneously to a more stable substance. 20.9 a) The movement of Earth about the Sun is spontaneous. b) The movement of a boulder against gravity is nonspontaneous. c) The reaction of an active metal (sodium) with an active nonmetal (chlorine) is spontaneous. 20.10 Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. Solution: a) Spontaneous, with a small amount of energy input, methane will continue to burn without additional energy (the reaction itself provides the necessary energy) until it is used up. b) Spontaneous, the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so the reaction proceeds in the direction of dissolution. c) Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed.

20-10

20.11 a) If a satellite slows sufficiently, it will fall to Earth’s surface through a spontaneous process. b) Water is a very stable compound; its decomposition at 298 K and 1 atm is not spontaneous. c) The increase in prices tends to be spontaneous. 20.12 Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > S liquid > Ssolid. If the products of the process have more entropy than the reactants, ∆Ssys is positive. If the products of the process have less entropy than the reactants, ∆Ssys is negative. Solution: a) ∆Ssys positive, melting is the change in state from solid to liquid. The solid state of a particular substance always has lower entropy than the same substance in the liquid state. Entropy increases during melting.

b) ∆Ssys negative, the entropy of most salt solutions is greater than the entropy of the solvent and solute separately, so entropy decreases as a salt precipitates.

c) ∆Ssys negative, dew forms by the condensation of water vapor to liquid. Entropy of a substance in the gaseous state is greater than its entropy in the liquid state. Entropy decreases during condensation. 20.13 a) ∆Ssys positive b) ∆Ssys positive c) ∆Ssys negative 20.14 Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > S liquid > Ssolid. If the products of the process have more entropy than the reactants, ∆Ssys is positive. If the products of the process have less entropy than the reactants, ∆Ssys is negative. Solution: a) ∆Ssys positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have greater entropy than the liquid molecules. b) ∆Ssys positive, the process described is a change from solid to gas, an increase in possible energy states for the system. c) ∆Ssys positive, the perfume molecules have more possible locations in the larger volume of the room than inside the bottle. A system that has more possible arrangements has greater entropy. 20.15 a) ∆Ssys negative b) ∆Ssys negative c) ∆Ssys negative 20.16 Plan: ∆Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > S liquid > Ssolid; also, the greater the number of particles of a particular phase of matter, the higher the entropy. Solution: a) ∆Ssys negative, reaction involves a gaseous reactant and no gaseous products, so entropy decreases. The number of particles also decreases, indicating a decrease in entropy. b) ∆Ssys negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) ∆Ssys positive, when a solid salt dissolves in water, entropy generally increases since the entropy of the aqueous mixture has higher entropy than the solid. 20.17 a) ∆Ssys negative b) ∆Ssys negative c) ∆Ssys negative 20.18 Plan: ∆Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > S liquid > Ssolid; also, the greater the number of particles of a particular phase of matter, the higher the entropy. Solution: a) ∆Ssys positive, the reaction produces gaseous CO2 molecules that have greater entropy than the physical states of the reactants. b) ∆Ssys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system’s entropy decreases. c) ∆Ssys positive, the reaction produces a gas from a solid. 20.19 a) ∆Ssys negative b) ∆Ssys positive c) ∆Ssys negative

20-11

20.20 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, ∆S is positive. Solution: a) ∆Ssys positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the system increases. b) ∆Ssys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. c) ∆Ssys positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules. 20.21 a) ∆Ssys negative b) ∆Ssys positive c) ∆Ssys negative 20.22 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: a) Butane has the greater molar entropy because it has two additional C−H bonds that can vibrate and has greater rotational freedom around its bond. The presence of the double bond in 2-butene restricts rotation. b) Xe(g) has the greater molar entropy because entropy increases with atomic size. c) CH4(g) has the greater molar entropy because gases in general have greater entropy than liquids. 20.23 a) N2O4(g); it has greater molecular complexity. b) CH3OCH3(l); hydrogen bonding in CH3CH2OH would increase order. c) HBr(g); it has greater mass. 20.24 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: a) Ethanol, C2H5OH(l), is a more complex molecule than methanol, CH3OH, and has the greater molar entropy.

b) When a salt dissolves, there is an increase in the number of possible states for the ions. Thus, KClO3(aq) has the greater molar entropy.

c) K(s) has greater molar entropy because K(s) has greater mass than Na(s). 20.25 a) P4(g); it has greater molecular complexity. b) HNO3(aq); because S(solution) > S(pure). c) CuSO4 · 5H2O; it has greater molecular complexity. 20.26 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution:

a) Diamond < graphite < charcoal. Diamond has an ordered, three-dimensional crystalline shape, followed by graphite with an ordered two-dimensional structure, followed by the amorphous (disordered) structure of charcoal.

b) Ice < liquid water < water vapor. Entropy increases as a substance changes from solid to liquid to gas. c) O atoms < O2 < O3. Entropy increases with molecular complexity because there are more modes of movement (e.g., bond vibration) available to the complex molecules. 20.27 a) Ribose < glucose < sucrose; entropy increases with molecular complexity. b) CaCO3(s) < (CaO(s) + CO2(g)) < (Ca(s) + C(s) + 3/2O2(g)); entropy increases with moles of gas particles. c) SF4(g) < SF6(g) < S2F10(g); entropy increases with molecular complexity.

20-12

20.28 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. Solution: a) ClO4

–(aq) > ClO3–(aq) > ClO2

–(aq). The decreasing order of molar entropy follows the order of decreasing molecular complexity. b) NO2(g) > NO(g) > N2(g). N2 has lower molar entropy than NO because N2 consists of two of the same atoms while NO consists of two different atoms. NO2 has greater molar entropy than NO because NO2 consists of three atoms while NO consists of only two. c) Fe3O4(s) > Fe2O3(s) > Al2O3(s). Fe3O4 has greater molar entropy than Fe2O3 because Fe3O4 is more complex and more massive. Fe2O3 and Al2O3 contain the same number of atoms but Fe2O3 has greater molar entropy because iron atoms are more massive than aluminum atoms. 20.29 a) Ba(s) > Ca(s) > Mg(s); entropy decreases with lower mass. b) C6H14 > C6H12 > C6H6; entropy decreases with lower molecular complexity and lower molecular flexibility. c) PF2Cl3(g) > PF5(g) > PF3(g); entropy decreases with lower molecular complexity. 20.30 a) X2(g) + 3Y2(g) → 2XY3(g) b) ΔS < 0 since there are fewer moles of gas in the products than in the reactants. c) XY3 is the most complex molecule and thus will have the highest molar entropy. 20.31 A system at equilibrium does not spontaneously produce more products or more reactants. For either reaction

direction, the entropy change of the system is exactly offset by the entropy change of the surroundings. Therefore, for system at equilibrium, ∆Suniv = ∆Ssys + ∆Ssurr = 0. However, for a system moving to equilibrium, ∆Suniv > 0, because the second law states that for any spontaneous process, the entropy of the universe increases.

20.32 Plan: Since entropy is a state function, the entropy changes can be found by summing the entropies of the products and subtracting the sum of the entropies of the reactants. Solution: rxnS∆ = [(2 mol HClO)(Sº of HClO)] – [(1 mol H2O)(Sº of H2O) + (1 mol Cl2O)(Sº of Cl2O)]

Rearranging this expression to solve for Sº of Cl2O gives: Sº of Cl2O = 2(Sº of HClO) – Sº of H2O – rxnS∆ 20.33 Plan: To calculate the standard entropy change, use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS .

To predict the sign of entropy recall that in general Sgas > S liquid > Ssolid, and entropy increases as the number of particles of a particular phase of matter increases, and with increasing atomic size and molecular complexity.

Solution: a) Prediction: ∆S° negative because number of moles of (∆n) gas decreases.

∆S° = [(1 mol N2O)(S° of N2O) + (1 mol NO2)(S° of NO2)] – [(3 mol NO)(S° of NO)] ∆S° = [(1 mol)(219.7 J/mol•K) + (1 mol)(239.9 J/mol•K)] – [(3 mol)(210.65 J/mol•K)] ∆S° = –172.35 = –172.4 J/K b) Prediction: Sign difficult to predict because ∆n = 0, but possibly ∆S° positive because water vapor has greater complexity than H2 gas. ∆S° = [(2 mol Fe)(S° of Fe) + (3 mol H2O)(S° of H2O)] – [(3 mol H2)(S° of H2) + (1 mol Fe2O3)(S° of Fe2O3)] ∆S° = [(2 mol)(27.3 J/mol•K) + (3 mol)(188.72 J/mol•K)] – [(3 mol)(130.6 J/mol•K) + (1 mol)(87.400 J/mol•K)] ∆S° = 141.56 = 141.6 J/K

c) Prediction: ∆S° negative because a gaseous reactant forms a solid product and also because the number of moles of gas (∆n) decreases. ∆S° = [(1 mol P4O10)(S° of P4O10)] – [(1 mol P4)(S°of P4) + (5 mol O2)(S° of O)]

∆S° = [(1 mol)(229 J/mol•K)] – [(1 mol)(41.1 J/mol•K) + (5 mol)(205.0 J/mol•K)] ∆S° = –837.1 = –837 J/K

20-13

20.34 a) 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g) ∆S° negative ∆S° = [(2 mol HNO3)(S° of HNO3) + (1 mol NO)(S° of NO)] – [(3 mol NO2)(S° of NO2) +( 1 mol H2O)(S° of H2O)] ∆S° = [(2 mol)(155.6 J/K•mol) + (1 mol)(210.65 J/K•mol)] – [(3 mol)(239.9 J/K•mol) + (1 mol)(69.940 J/K•mol)] ∆S° = –267.79 = –267.8 J/K b) N2(g) + 3F2(g) → 2NF3(g) ∆S° negative ∆S° = [(2 mol NF3)(S° of NF3] – [(1 mol N2)(S° of N2) + (3 mol F2)(S° of F2)] ∆S° = [(2 mol)(260.6 J/K•mol)] – [(1 mol)(191.5 J/K•mol) + (3 mol)(202.7 J/K•mol)] ∆S° = –278.4 J/K c) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g) ∆S° positive ∆S° = [(6 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)] – [(1 mol C6H12O6)(S° of C6H12O6) + (6 mol O2)(S° of O2)] ∆S° = [(6 mol)(213.7 J/K•mol) + (6 mol H2O)(188.72 J/K•mol)] – [(1 mol)(212.1 J/K•mol) + (6 mol)(205.0 J/K•mol)] ∆S° = 972.42 = 972.4 J/K 20.35 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship

rxnS∆ = ∑m productsS – ∑n reactantsS . To predict the sign of entropy recall that in general Sgas > S liquid > Ssolid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity.

Solution: The balanced combustion reaction is: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) ∆S° = [(4 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)] – [(2 mol C2H6)(S° of C2H6) + (7 mol O2)(S° of O2)] ∆S° = [(4 mol)(213.7 J/mol•K) + (6 mol)(188.72 J/mol•K)] – [(2 mol)(229.5 J/mol•K) + (7 mol)(205.0 J/mol•K)] ∆S° = 93.12 = 93.1 J/K

The entropy value is not per mole of C2H6 but per two moles. Divide the calculated value by two to obtain entropy per mole of C2H6.

Yes, the positive sign of ∆S° is expected because there is a net increase in the number of gas molecules from nine moles as reactants to ten moles as products. 20.36 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆S° = [(1 mol CO2)(S° of CO2) + (2 mol H2O)(S° of H2O)] – [(1 mol CH4)(S° of CH4) + (2 mol O2)(S° of O2)] ∆S° = [(1 mol)(213.7 J/K•mol) + (2 mol)(69.940 J/K•mol)] – [(1 mol)(186.1 J/K•mol) + (2 mol)(205.0 J/K•mol)] ∆S° = –242.52 = –242.5 J/K Yes, a decrease in the number of moles of gas should result in a negative ∆S° value. 20.37 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . To predict the sign of entropy recall that in general Sgas > S liquid > Ssolid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity. Solution: The balanced chemical equation for the described reaction is: 2NO(g) + 5H2(g) → 2NH3(g) + 2H2O(g) Because the number of moles of gas decreases, i.e., ∆n = 4 – 7 = –3, the entropy is expected to decrease. ∆S° = [(2 mol NH3)(S° of NH3) + (2 mol H2O)(S° of H2O)] – [(2 mol NO)(S° of NO) + (5 mol H2)(S° of H2)] ∆S° = [(2 mol)(193 J/mol•K) + (2 mol)(188.72 J/mol•K)] – [(2 mol)(210.65 J/mol•K) + (5 mol)(130.6 J/mol•K)] ∆S° = –310.86 = –311 J/K Yes, the calculated entropy matches the predicted decrease.

20-14

20.38 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) ∆S° = [(4 mol NO2)(S° of NO2) + (6 mol H2O)(S° of H2O)] – [(4 mol NH3)(S° of NH3) + (7 mol O2)(S° of O2)] ∆S° = [(4 mol)(239.9 J/K•mol) + (6 mol)(188.72 J/K•mol)] – [(4 mol)(193 J/K•mol) + (7 mol)(205.0 J/K•mol)] ∆S° = –115.08 = –115 J/K Yes, a loss of one mole of a gas should result in a small negative ∆S° value. 20.39 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ Use ΔHo of the reaction to calculate ΔS of the surroundings.

surr∆S = rxnHT

∆−

Add ΔS of the surroundings and ΔSo of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature.

univS∆ = rxnS∆ + surr∆S Solution: a) The reaction for forming Cu2O from copper metal and oxygen gas is: 2Cu(s) + 1/2O2(g) → Cu2O(s) ∆S° = [(1 mol Cu2O)(S° of Cu2O)] – [(2 mol Cu)(S° of Cu) + (1/2 mol O2)(S° of O2)] ∆S° = [(1 mol)(93.1 J/mol•K)] – [(2 mol)(33.1 J/mol•K) + (1/2 mol)(205.0 J/mol•K)] ∆S° = –75.6 J/K b) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol Cu2O)( fH∆ of Cu2O)]

– [(2 mol Cu)( fH∆ of Cu) + (1/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol Cu2O)(–168.6 kJ/mol)] – [(2 mol Cu)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)] rxnH∆ = –168.6 kJ

surr∆S = rxnHT

∆−

= −−168.6 kJ298 K

= 0.56577 kJ/K(103 J/1 kJ) = 565.77 J/K

univS∆ = rxnS∆ + surr∆S = (–75.6 J/K) + (565.77 J/K) = 490.17 = 490. J/K

Because univS∆ is positive, the reaction is spontaneous at 298 K. 20.40 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ Use ΔHo of the reaction to calculate ΔS of the surroundings.

surr∆S = rxnHT

∆−


univS∆ = rxnS∆ + surr∆S

20-15

Solution: a) One mole of hydrogen iodide is formed from its elements in their standard states according to the following equation: 1/2H2(g) + 1/2I2(s) → HI(g)

∆S° = [(1 mol HI)(S° of HI)] – [(1/2 mol H2)(S° of H2) + (1/2 mol I2)(S° of I2)] ∆S° = [(1 mol)(206.33 J/K•mol)] – [(1/2 mol)(130.6 J/K•mol) + (1/2 mol)(116.14 J/K•mol)] ∆S°= 82.96 = 83.0 J/K b) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol HI)( fH∆ of HI)]

– [(1/2 mol H2)( fH∆ of H2) + (1/2 mol I2)( fH∆ of I2)]

rxnH∆ = [(1 mol HI)(25.9 kJ/mol)] – [(1/2 mol H2)(0 kJ/mol) + (1/2 mol I2)(0 kJ/mol)] rxnH∆ = 25.9 kJ

surr∆S = rxnHT

∆−

= − 25.9 kJ298 K

= – 0.0869128 kJ/K(103 J/1 kJ) = –86.9128 J/K

univS∆ = rxnS∆ + surr∆S = (83.0 J/K) + (–86.9128 J/K) = –3.9128 = –3.9 J/K

Because univS∆ is negative, the reaction is not spontaneous at 298 K. 20.41 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ Use ΔHo of the reaction to calculate ΔS of the surroundings.

surr∆S = rxnHT

∆−


univS∆ = rxnS∆ + surr∆S Solution: a) One mole of methanol is formed from its elements in their standard states according to the following equation: C(g) + 2H2(g) + 1/2O2(g) → CH3OH(l) ∆S° = [(1 mol CH3OH)(S° of CH3OH)] – [(1 mol C)(S° of C) + (2 mol H2)(S° of H2) + (1/2 mol O2)(S° of O2)] ∆S° = [(1 mol)(127 J/mol•K)] – [(1 mol)(5.686 J/mol•K) + (2 mol)(130.6 J/mol•K) + (1/2 mol)(205.0 J/mol•K)] ∆S° = –242.386 = –242 J/K b) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol CH3OH)( fH∆ of CH3OH)]

– [(1 mol C)( fH∆ of C) + (2 mol H2)( fH∆ of H2) + (1/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol CH3OH)(–238.6 kJ/mol)] – [(1 mol C)(0 kJ/mol) + (2 mol H2)( 0 kJ/mol) + (1/2 mol O2)( 0 kJ/mol)] rxnH∆ = –238.6 kJ

surr∆S = rxnHT

∆−

= −−238.6 kJ298 K

= 0.80067 kJ/K(103 J/1 kJ) = 800.67 J/K

20-16

univS∆ = rxnS∆ + surr∆S = (–242 J/K) + (800.67 J/K) = 558.67 = 559 J/K

Because univS∆ is positive, the reaction is spontaneous at 298 K. 20.42 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ Use ΔHo of the reaction to calculate ΔS of the surroundings.

surr∆S = rxnHT

∆−


univS∆ = rxnS∆ + surr∆S Solution:

a) One mole of dinitrogen oxide is formed from its elements in their standard states according to the following equation: N2(g) + 1/2O2(g) → N2O(g)

∆S° = [(1 mol N2O)(S° of N2O] – [(1 mol N2)(S° of N2) + (1/2 mol O2)(S° of O2)] ∆S° = [(1 mol)(219.7 J/K•mol)] – [(1 mol)(191.5 J/K•mol) + (1/2 mol)(205.0 J/K•mol)] ∆S° = –214.775 = –74.3 J/K b) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol N2O)( fH∆ of N2O)]

– [(1 mol N2)( fH∆ of N2) + (1/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol N2O)(82.05 kJ/mol)] – [(1 mol N2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)] rxnH∆ = 82.05 kJ

surr∆S = rxnHT

∆−

= − 82.05 kJ298 K

= −0.27534 kJ/K(103 J/1 kJ) = −275.34 J/K

univS∆ = rxnS∆ + surr∆S = (–74.3 J/K) + (−275.34J/K) = –349.64 = –340.6 J/K

Because univS∆ is negative, the reaction is not spontaneous at 298 K. 20.43 SO2(g) + Ca(OH)2(s) → CaSO3(s) + H2O(l) ∆S° = [(1 mol CaSO3)(S° of CaSO3) + (1 mol H2O)(S° of H2O)] – [(1 mol SO2)(S° of SO2) + (1 mol Ca(OH)2)(S° of Ca(OH)2)] ∆S° = [1 mol)(101.4 J/K•mol) + (1 mol)(69.940 J/K•mol)] – [(1 mol)(248.1 J/K•mol) + (1 mol)(83.39 J/K•mol)] ∆S° = –160.15 = –160.2 J/K 20.44 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Solution: Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the products. C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(g)

20-17

∆S° = [(2 mol CO2)(S° of CO2) + (1 mol H2O)(S° of H2O)] – [(1 mol C2H2)(S° of C2H2) + (5/2 mol O2)(S° of O2)] ∆S° = [(2 mol)(213.7 J/mol•K) + (1 mol)(188.72 J/mol•K)] – [(1 mol)(200.85 J/mol•K) + (5/2 mol)(205.0 J/mol•K)] ∆S° = –97.23 = –97.2 J/K 20.45 Reaction spontaneity may now be predicted from the value of only one variable (∆Gsys) rather than two

(∆Ssys and ∆Ssurr). 20.46 A spontaneous process has ∆Suniv > 0. Since the Kelvin temperature is always positive, ∆Gsys must be negative (∆Gsys < 0) for a spontaneous process. 20.47 a) ∆G = ∆H – T∆S. Since T∆S > ∆H for an endothermic reaction to be spontaneous, the reaction is more likely

to be spontaneous at higher temperatures. b) The change depicted is the phase change of a solid converting to a gas (sublimation).

1. Energy must be absorbed to overcome intermolecular forces to convert a substance in the solid phase to the gas phase. This is an endothermic process and ∆H is positive.

2. Since gases have higher entropy values than solids, the process results in an increase in entropy and ∆S is positive.

3. This is an endothermic process so the surroundings lose energy to the system. ∆Ssurr is negative. 4. ∆G = ∆H – T∆S. Both ∆H and ∆S are positive. At low temperature, the ∆H term will predominate and ∆G will be positive; at high temperatures, the T∆S term will predominate and ∆G will be negative.

20.48 Plan: Examine the provided diagrams. Determine whether bonds are being formed or broken in order to determine

the sign of ΔH. Determine the relative number of particles before and after the reaction in order to determine the sign of ΔS.

Solution: a) The sign of ΔH is negative. Bonds are being formed, so energy is released and ΔH is negative. b) The sign of ΔS is negative. There are fewer particles in the system after the reaction, so entropy decreases.

c) The sign of ΔSsurr is positive. surr∆S = rxnHT

∆−

Since ΔHrxn is negative, ΔSsurr will be positive.

d) Because both ΔH and ΔS are negative, this reaction will become more spontaneous (ΔG will become more negative) as temperature decreases.

20.49 ∆H° is positive and ∆S° is positive. The reaction is endothermic (∆H° > 0) and requires a lot of heat from its surroundings to be spontaneous. The removal of heat from the surroundings results in ∆S° < 0. The only way an endothermic reaction can proceed spontaneously is if ∆S° > 0, effectively offsetting the decrease in surroundings entropy. In summary, the values of ∆H° and ∆S° are both positive for this reaction. Melting is an example.

20.50 For a given substance, the entropy changes greatly from one phase to another, e.g., from liquid to gas. However, the entropy changes little within a phase. As long as the substance does not change phase, the value of ∆S° is relatively unaffected by temperature. 20.51 Plan: ∆G° can be calculated with the relationship ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: a) ∆G° = [(2 mol MgO)( fG∆ of MgO)] – [(2 mol Mg)( fG∆ of Mg) + (1 mol O2)( fG∆ of O2)]

Both Mg(s) and O2(g) are the standard-state forms of their respective elements, so their fG∆ values are zero. ∆G° = [(2 mol)(–569.0 kJ/mol)] – [(2 mol)(0) + (1 mol)(0)] = –1138.0 kJ b) ∆G° = [(2 mol CO2)( fG∆ of CO2) + (4 mol H2O)( fG∆ of H2O)]

– [(2 mol CH3OH)( fG∆ of CH3OH) + (3 mol O2)( fG∆ of O2)]

20-18

∆G° = [(2 mol)(–394.4 kJ/mol) + (4 mol)(–228.60 kJ/mol)] − [(2 mol)(–161.9 kJ/mol) + (3 mol)(0)] ∆G° = –1379.4 kJ c) ∆G° = [(1 mol BaCO3)( fG∆ of BaCO3)] – [(1 mol BaO)( fG∆ of BaO) + (1 mol CO2)( fG∆ of CO2)]

∆G° = [(1 mol)(–1139 kJ/mol)] – [(1 mol)(–520.4 kJ/mol) + (1 mol)(–394.4 kJ/mol)] ∆G° = –224.2 = –224 Kj 20.52 a) H2(g) + I2(s) → 2HI(g) ∆G° = [(2 mol HI)( fG∆ of HI] – [(1 mol H2)( fG∆ of H2) + (1 mol I2)( fG∆ of I2)] ∆G° = [(2 mol)(1.3 kJ/mol)] – [(1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)] ∆G° = 2.6 kJ b) MnO2(s) + 2CO(g) → Mn(s) + 2CO2(g) ∆G° = [(1 mol Mn)( fG∆ of Mn) + (2 mol CO2)( fG∆ of CO2)]

– [(1 mol MnO2)( fG∆ of MnO2) + (2 mol CO)( fG∆ of CO)] ∆G° = [(1 mol)(0 kJ/mol) + (2 mol)(–394.4 kJ/mol)] – [(1 mol)(–466.1 kJ/mol) + (2 mol)(–137.2 kJ/mol)] ∆G° = –48.3 kJ c) NH4Cl(s) → NH3(g) + HCl(g) ∆G° = [(1 mol NH3)( fG∆ of NH3) + (1 mol HCl)( fG∆ of HCl)] – [(1 mol NH4Cl)( fG∆ of NH4Cl)] ∆G° = [(1 mol)(–16 kJ/mol) + (1 mol)(–95.30 kJ/mol)] – [1 mol)(–203.0 kJ/mol)] ∆G° = 91.7 = 92 kJ 20.53 Plan: rxnH∆ can be calculated from the individual fH∆ values of the reactants and products by using the

relationship rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ . rxnS∆ can be calculated from the individual S values

of the reactants and products by using the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Once rxnH∆ and

rxnS∆ are known, ∆G° can be calculated with the relationship rxnG∆ = rxnH∆ – rxnT S∆ . rxnS∆ values in J/K

must be converted to units of kJ/K to match the units of rxnH∆ . Solution: a) rxnH∆ = [(2 mol MgO)( fH∆ of MgO)] – [(2 mol Mg)( fH∆ of Mg) + (1 mol O2)( fH∆ of O2)]

rxnH∆ = [(2 mol)(–601.2 kJ/mol)] –[(2 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]

rxnH∆ = –1202.4 kJ

rxnS∆ = [(2 mol MgO)( S of MgO)] – [(2 mol Mg)( S of Mg) + (1 mol O2)( S of O2)]

rxnS∆ = [(2 mol)(26.9 J/mol•K)] – [(2 mol)(32.69 J/mol•K) + (1 mol)(205.0 J/mol•K)]

rxnS∆ = –216.58 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –1202.4 kJ – [(298 K)(–216.58 J/K)(1 kJ/103 J)] = –1137.859 = –1138 kJ

b) rxnH∆ = [(2 mol CO2)( fH∆ of CO2) + (4 mol H2O)( fH∆ of H2O)]

– [(2 mol CH3OH)( fH∆ of CH3OH) + (3 mol O2)( fH∆ of O2)]

rxnH∆ = [(2 mol)(–393.5 kJ/mol) + (4 mol)(–241.826 kJ/mol)] – [(2 mol)(–201.2 kJ/mol) + (3 mol)(0 kJ/mol)]

rxnH∆ = –1351.904 kJ

rxnS∆ = [(2 mol CO2)( S of CO2) + (4 mol H2O)( S of H2O)]

– [(2 mol CH3OH)( S of CH3OH) + (3 mol O2)( S of O2)]

rxnS∆ = [(2 mol)(213.7 J/mol•K) + (4 mol)(188.72 J/mol•K)]

20-19

– [(2 mol)(238 J/mol•K) + (3 mol)(205.0 J/mol•K)] = 91.28 J/K rxnG∆ = rxnH∆ – rxnT S∆ = –1351.904 kJ – [(298 K)(91.28 J/K)(1 kJ/103 J)] = –1379.105 = –1379 kJ

c) rxnH∆ = [(1 mol BaCO3)( fH∆ of BaCO3)] – [(1 mol BaO)( fH∆ of BaO) + (1 mol CO2)( fH∆ of CO2)]

rxnH∆ = [(1 mol)(–1219 kJ/mol)] – [(1 mol)(–548.1 kJ/mol) + (1 mol)(–393.5 kJ/mol)]

rxnH∆ = –277.4 kJ

rxnS∆ = [(1 mol BaCO3)( S of BaCO3)] – [(1 mol BaO)( S of BaO) + (1 mol CO2)( S of CO2)]

rxnS∆ = [(1 mol)(112 J/mol•K)] – [(1 mol)(72.07 J/mol•K) + (1 mol)(213.7 J/mol•K)]

rxnS∆ = –173.77 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –277.4 kJ – [(298 K)(–173.77 J/K)(1 kJ/103 J)] = –225.6265 = –226 kJ 20.54 a) rxnH∆ = [(2 mol HI)(25.9 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol I2)(0 kJ/mol)]

rxnH∆ = 51.8 kJ

rxnS∆ = [(2 mol HI)(206.33 J/K•mol)] – [(1 mol H2)(130.6 J/K•mol) + (1 mol I2)(116.14 J/K•mol)]

rxnS∆ = 165.92 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = 51.8 kJ – [(298 K)(165.92 J/K)(1 kJ/103 J)] = 2.3558 = 2.4 kJ

b) rxnH∆ = [(1 mol Mn)(0 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)] – [(1 mol MnO2)(–520.9 kJ/mol) + (2 mol CO)(–110.5 kJ/mol)] rxnH∆ = –45.1 kJ

rxnS∆ = [(1 mol Mn)(31.8 J/K•mol) + (2 mol CO2)(213.7 J/K•mol)] – [(1 mol MnO2)(53.1 J/K•mol) + (2 mol CO)(197.5 J/K•mol)] rxnS∆ = 11.1 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –45.1 kJ – [(298 K (11.1 J/K)(1 kJ/103 J)] = –48.4078 = –48.4 kJ

c) rxnH∆ = [(1 mol NH3)(–45.9 kJ/mol) + (1 mol HCl)(–92.3 kJ/mol)] – [(1 mol NH4Cl)(–314.4 kJ/mol)]

rxnH∆ = 176.2 kJ

rxnS∆ = [(1 mol NH3)(193 J/K•mol) + (1 mol HCl)(186.79 J/K•mol)] – [(1 mol NH4Cl)(94.6 J/K•mol)]

rxnS∆ = 285.19 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = 176.2 kJ – [(298 K)(285.19 J/K)(1 kJ/103 J)] = 91.213 = 91.2 kJ 20.55 Plan: rxnG∆ can be calculated with the relationship ∑m f (products)G∆ – ∑n f (reactants)G∆ . Alternatively, rxnG∆

can be calculated with the relationship rxnG∆ = rxnH∆ – rxnT S∆ . Entropy decreases (is negative) when there are fewer moles of gaseous products than there are of gaseous reactants. Solution: a) Entropy decreases ( oΔS negative) because the number of moles of gas decreases from reactants (1 1/2 mol) to products (1 mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then releases energy (exothermic, oΔH negative) which is typical of all combustion reactions.

b) Method 1: Calculate rxnG∆ from fG∆ values of products and reactants.

rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆

rxnG∆ = [(1 mol CO2)( fG∆ of CO2)] – [(1 mol CO)( fG∆ of CO) + (1/2 mol)( fG∆ of O2)]

20-20

rxnG∆ = [(1 mol)(–394.4 kJ/mol)] – [(1 mol)(–137.2 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –257.2 kJ

Method 2: Calculate rxnG∆ from rxnH∆ and rxnS∆ at 298 K (the degree superscript indicates a reaction at standard state, given in the Appendix at 25°C). rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol CO2)( fH∆ of CO2)] – [(1 mol CO)( fH∆ of CO) + (1/2 mol)( fH∆ of O2)]

rxnH∆ = [(1 mol)(–393.5 kJ/mol)] – [(1 mol)(–110.5 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –283.0 kJ


rxnS∆ = [(1 mol CO2)( S of CO2)] – [(1 mol CO)( S of CO) + (1/2 mol)( S of O2)]

rxnS∆ = [(1mol)(213.7 J/mol•K)] – [(1mol)(197.5 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]

rxnS∆ = –86.3 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = (–283.0 kJ) – [(298 K)(–86.3 J/K)(1 kJ/103 J)] = –257.2826 = –257.3 kJ 20.56 C4H10(g) + 13/2O2(g) → 4CO2(g) + 5H2O(g) a) An increase in the number of moles of gas should result in a positive ∆S° value. The combustion of C4H10(g) will result in a release of energy or a negative ∆H° value. b) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(4 mol CO2)(–393.5 kJ/mol) + (5 mol H2O)(–241.826 kJ/mol)] – [(1 mol C4H10)(–126 kJ/mol) + (13/2 mol O2)(0 kJ/mol)] rxnH∆ = –2657.13 kJ


rxnS∆ = [(4 mol CO2)(213.7 J/K•mol) + (5 mol H2O)(188.72 J/K•mol)] – [(1 mol C4H10)(310 J/K•mol) + (13/2 mol O2)(205.0 J/K•mol)]

rxnS∆ = 155.9 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = –2657.13 kJ – [(298 K)(155.9 J/K)(1 kJ/103 J)] = –2703.588 = –2704 kJ


rxnG∆ = [(4 mol CO2)(–394.4 kJ/mol) + (5 mol H2O)(–228.60 kJ/mol)] – [(1 mol C4H10)(–16.7 kJ/mol) + (13/2 mol O2)(0 kJ/mol)] rxnG∆ = –2703.9 kJ 20.57 Plan: Use the relationship rxnG∆ = rxnH∆ – rxnT S∆ to find rxnS∆ , knowing rxnH∆ and rxnG∆ . This

relationship is also used to find rxnG∆ at a different temperature. Solution: Reaction is Xe(g) + 3F2(g) → XeF6(g) a) rxnG∆ = rxnH∆ – rxnT S∆

∆S° = H G

T∆ ° − ∆ °

= ( )402 kJ/mol 280. kJ/mol

298 K− − −

= –0.40939597 = –0.409 kJ/mol•K

b) rxnG∆ = rxnH∆ – rxnT S∆ = (–402 kJ/mol) – [(500. K)(–0.40939597 kJ/mol•K)] = –197.302 = –197 kJ/mol

20.58 a) ∆S° = H G

T∆ ° − ∆ °

= ( ) ( )220. kJ/mol 206 kJ/mol

298 K− − −

= –0.046979865 = –0.047 kJ/mol•K

20-21

b) rxnG∆ = rxnH∆ – rxnT S∆ = –220. kJ/mol – (450. K)(– 0.046979865 kJ/K•mol) = –198.859 = –199 kJ/mol

20.59 Plan: rxnH∆ can be calculated from the individual fH∆ values of the reactants and products by using the




must be converted to units of kJ/K to match the units of rxnH∆ . The temperature at which a reaction becomes spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at equilibrium) and solving for T.

Solution: a) rxnH∆ = [(1 mol CO)( fH∆ of CO) + (2 mol H2)( fH∆ of H2)] – [(1 mol CH3OH)( fH∆ of CH3OH)]

rxnH∆ = [(1 mol)(–110.5 kJ/mol) + (2 mol)(0 kJ/mol)] – [(1 mol)(–201.2 kJ/mol)]

rxnH∆ = 90.7 kJ

rxnS∆ = [(1 mol CO)( S of CO) + (2 mol H2)( S of H2)] – [(1 mol CH3OH)( S of CH3OH)]

rxnS∆ = [(1 mol)(197.5 J/mol•K) + (2 mol)(130.6 J/mol•K)] – [(1 mol)(238 J/mol•K)]

rxnS∆ = 220.7 = 221 J/K

b) rxnG∆ = rxnH∆ – rxnT S∆ T1 = 28 + 273 = 301 K ∆G° = 90.7 kJ – [(301 K)(220.7 J/K)(1 kJ/103 J)] = 24.2693 = 24.3 kJ

T2 = 128 + 273 = 401 K ∆G° = 90.7 kJ – [(401 K)(220.7 J/K)(1 kJ/103 J)] = 2.1993 = 2.2 kJ T3 = 228 + 273 = 501 K ∆G° = 90.7 kJ – [(501 K)(220.7 J/K)(1 kJ/103 J)] = –19.8707 = –19.9 kJ

c) For the substances in their standard states, the reaction is nonspontaneous at 28°C, near equilibrium at 128°C, and spontaneous at 228°C. Reactions with positive values of rxnH∆ and rxnS∆ become spontaneous at high temperatures. d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.

rxnG∆ = rxnH∆ – rxnT S∆ 0 = 90.7 x 103 J – (T)(221 J/K) T = 410. K At temperatures above 410. K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the reaction becomes spontaneous above this temperature.)





must be converted to units of kJ/K to match the units of rxnH∆ . The temperature at which a reaction becomes spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at equilibrium) and solving for T.

Solution: a) N2(g) + O2(g) 2NO(g)

rxnH∆ = [(2 mol NO)(90.29 kJ/mol)] – [(1 mol N2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]

rxnH∆ = 180.58 kJ

rxnS∆ = [(2 mol NO)(210.65 J/K•mol)] – [(1 mol N2)(191.5 J/K•mol) + (1 mol O2)(205.0 J/K•mol)]

20-22

rxnS∆ = 24.8 J/K b) ∆G°373 = ∆H° – ((273 + 100.)K) (∆S°) = 180.58 kJ – [(373 K)(24.8 J/K)(1 kJ/103 J)] = 171.3296 = 171.33 kJ ∆G°2833 = ∆H° – ((273 + 2560.)K) (∆S°) = 180.58 kJ – [(2833 K)(24.8 J/K)(1 kJ/103 J)] = 110.3216 = 110.3 kJ ∆G°3813 = ∆H° – ((273 + 3540.)K) (∆S°) = 180.58 kJ – [(3813 K)(24.8 J/K)(1 kJ/103 J)] = 86.0176 = 86.0 kJ c) The values of ∆G became smaller at higher temperatures. The reaction is not spontaneous at any of these temperatures; however, the reaction becomes less nonspontaneous as the temperature increases.

d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0.

rxnG∆ = rxnH∆ – rxnT S∆ 0 = 180.58 x 103 J – (T)(24.8 J/K) T = 7280 K At temperatures above 7280 K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the reaction becomes spontaneous above this temperature.)





must be converted to units of kJ/K to match the units of rxnH∆ . To find the temperature at which the reaction

becomes spontaneous, use rxnG∆ = 0 = rxnH∆ – rxnT S∆ and solve for temperature. Solution: a) The reaction for this process is H2(g) + 1/2O2(g) → H2O(g). The coefficients are written this way (instead of 2H2(g) + O2(g) → 2H2O(g)) because the problem specifies thermodynamic values “per (1) mol H2,” not per 2 mol H2. rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol H2O)( fH∆ of H2O)] – [(1 mol H2)( fH∆ of H2) + (1/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol H2O)(–241.826 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

rxnH∆ = –241.826 kJ


rxnS∆ = [(1 mol H2O)( S of H2O)] – [(1 mol H2)( S of H2) + (1/2 mol O2)( S of O2)]

rxnS∆ = [(1 mol)(188.72 J/mol•K)] – [(1 mol)(130.6 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]

rxnS∆ = –44.38 = –44.4 J/K = –0.0444 kJ/K

rxnG∆ = rxnH∆ – rxnT S∆

rxnG∆ = –241.826 kJ – [(298 K)( –0.0444 kJ/K)]

rxnG∆ = –228.6 kJ b) Because ∆H < 0 and ∆S < 0, the reaction will become nonspontaneous at higher temperatures because the positive (–T∆S) term becomes larger than the negative ∆H term.

20-23

c) The reaction becomes spontaneous below the temperature where rxnG∆ = 0


rxnH∆ = rxnT S∆

= HTS

∆

∆

= –241.826 kJ–0.0444 kJ/K

= 5446.53 = 5.45x103 K

20.62 C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(2 mol C2H5OH)(–277.63 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)] – [1 mol C6H12O6)(–1273.3 kJ/mol)] rxnH∆ = –68.96 kJ = –69.0 kJ


rxnS∆ = [(2 mol C2H5OH)(161 J/K•mol) + (2 mol CO2)(213.7 J/K•mol)] – [(1 mol C6H12O6)(212.1 K•mol)]

rxnS∆ = 537.3 J/K = 537 J/K

rxnG∆ = rxnH∆ – rxnT S∆

rxnG∆ = –68.96 kJ – [(298 K)( 0.5373 kJ/K)]

rxnG∆ = –229.0754 kJ/mol = –229.1 kJ/mol No, a reaction with a negative value for ∆H and a positive value for ∆S is spontaneous at all temperatures. 20.63 a) An equilibrium constant that is much less than one indicates that very little product is made to reach

equilibrium. The reaction, thus, is not spontaneous in the forward direction and ∆G° is a relatively large positive value. b) A large negative ∆G° indicates that the reaction is quite spontaneous and goes almost to completion. At equilibrium, much more product is present than reactant so K > 1. Q depends on initial conditions, not equilibrium conditions, so its value cannot be predicted from ∆G°.

20.64 For a spontaneous process, ∆G is the maximum useful work obtainable from the system. In reality, the actual amount of useful work is less due to energy lost as heat. If the process is run in a slower or more controlled fashion, the actual amount of available work approaches ∆G. 20.65 a) Point x represents the difference between Greactants and Gproducts or ∆G°, the standard free energy change

for the reaction. b) Scene A corresponds to Point 1 on the graph. This point corresponds to the pure substances, not a mixture. c) Scene C corresponds to Point 2 on the graph. Point 2 represents equilibrium; for this reaction, products dominate at equilibrium (the minimum in the curve is close to the XY side of the graph).

20.66 The standard free energy change, ∆G°, occurs when all components of the system are in their standard states. Standard state is defined as 1 atm for gases, 1 M for solutes, and pure solids and liquids. Standard state does not specify a temperature because standard state can occur at any temperature. ∆G° = ∆G when all concentrations equal 1 M and all partial pressures equal 1 atm. This occurs because the value of Q = 1 and ln Q = 0 in the equation ∆G = ∆G° + RT ln Q. 20.67 Plan: For each reaction, first find ∆G°, then calculate K from ∆G° = –RT ln K. Calculate G∆ using fG∆ values

in the relationship rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: a) MgCO3(s) Mg2+(aq) + CO3

2–(aq)

20-24


rxnG∆ = [(1 mol Mg2+)(–456.01 kJ/mol) + (1 mol CO32–)(–528.10 kJ/mol)]

– [(1 mol MgCO3)(–1028 kJ/mol)] = 43.89 kJ

ln K = GRT

∆−

= ( ) ( )

343.89 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –17.7149

K = e–17.7149 = 2.0254274x10–8 = 2.0x10–8 b) H2(g) + O2(g) H2O2(l)

rxnG∆ = [(1 mol H2O2)(–120.4 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)] = –120.4 kJ/mol

ln K = GRT

∆−

= ( ) ( )

3120.4 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

− −

= 48.59596

K = e48.59596 = 1.2733777x1021 = 1.27x1021 20.68 Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. Solution: a) ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (6.57 x 10173) = –9.91598 x105 J/mol = –9.92x105 J/mol b) ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (4.46 x 10-15) = 8.18680 x 104 J/mol = 8.19x104 J/mol 20.69 Plan: For each reaction, first find ∆G°, then calculate K from ∆G° = –RT ln K. Calculate G∆ using fG∆ values

in the relationship rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution:

a) rxnG∆ = [(1 mol NaCN)( fG∆ of NaCN) + (1 mol H2O)( fG∆ of H2O)]

– [(1 mol HCN)( fG∆ of HCN) + (1 mol NaOH)( fG∆ of NaOH)] NaCN(aq) and NaOH(aq) are not listed in Appendix B. Converting the equation to net ionic form will simplify the problem: HCN(aq) + OH–(aq) CN–(aq) + H2O(l) rxnG∆ = [(1 mol CN–)( fG∆ of CN–) + (1 mol H2O)( fG∆ of H2O)]

– [(1 mol HCN)( fG∆ of HCN) + (1 mol OH–)( fG∆ of OH–)]

rxnG∆ = [(1 mol)(166 kJ/mol) + (1 mol)(–237.192 kJ/mol)] – [(1 mol)(112 kJ/mol) + (1 mol)(–157.30 kJ/mol)]

rxnG∆ = –25.892 kJ

ln K = GRT

∆−

= ( ) ( )

325.892 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

− −

= 10.45055

K = e10.45055 = 3.4563x104 = 3.46x104 b) SrSO4(s) Sr2+(aq) + SO4

2–(aq) rxnG∆ = [(1 mol Sr2+)(–557.3 kJ/mol) + (1 mol SO4

2–)(–741.99 kJ/mol)] – [(1 mol SrSO4)(–1334 kJ/mol)] = 34.71 kJ

ln K = GRT

∆−

= ( ) ( )

334.71 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –14.00968

K = e–14.00968 = 8.23518x10–7 = 8.2x10–7 20.70 Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. Solution: a) ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (1.58 x 107) = –4.10670 x104 J/mol = –4.11x104 J/mol b) ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (3.25 x 1037) = –2.13999 x 105 J/mol = –2.14x105 J/mol

20-25

20.71 Plan: At the normal boiling point, defined as the temperature at which the vapor pressure of the liquid equals 1 atm, the phase change from liquid to gas is at equilibrium. For a system at equilibrium, the change in Gibbs free energy is zero. Since the gas is at 1 atm and the liquid assumed to be pure, the system is at standard state and ∆G° = 0. The temperature at which this occurs can be found from rxnG∆ = 0 = rxnH∆ – rxnT S∆ . rxnH∆ can be

calculated from the individual fH∆ values of the reactants and products by using the relationship

rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆ . rxnS∆ can be calculated from the individual S values of the

reactants and products by using the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . Solution: Br2(l) Br2(g) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol Br2)( fH∆ of Br2(g))] – [(1 mol Br2)( fH∆ of Br2(l))]

rxnH∆ = [(1 mol)(30.91 kJ/mol)] – [(1 mol)(0 kJ/mol)] = 30.91 kJ


rxnS∆ = [(1 mol Br2)( S of Br2(g))] – [(1 mol Br2)( S of Br2(l))]

rxnS∆ = [(1 mol)(245.38 J/K•mol)] – [(1 mol)(152.23 J/K•mol)] = 93.15 J/K = 0.09315 kJ/K


rxnH∆ = rxnT S∆

= HTS

∆

∆

= 30.91 kJ0.09315 kJ/K

= 331.830 = 331.8 K

20.72 S(rhombic) S(monoclinic) rxnH∆ = [1 mol S(monoclinic)(0.30 kJ/mol)] – [1 mol S(rhombic)(0 kJ/mol)] = 0.30 kJ

rxnS∆ = [1 mol S(monoclinic)(32.6 J/K•mol)] – [1 mol S(rhombic)(31.9 J/K•mol)] = 0.7 J/K = 0.0007 kJ/K


rxnH∆ = rxnT S∆

= HTS

∆

∆

= 30.91 kJ0.09315 kJ/K

= 428.571 = 4x102 K

20.73 Plan: Write the balanced equation. First find ∆G°, then calculate K from ∆G° = –RT ln K. Calculate G∆ using

fG∆ values in the relationship rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution:

The solubility reaction for Ag2S is Ag2S(s) + H2O(l) 2Ag+(aq) + HS–(aq) + OH–(aq) rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆

rxnG∆ = [(2 mol Ag+)( fG∆ of Ag+) + (1 mol HS–)( fG∆ of HS–) + (1 mol OH–)( fG∆ of OH–]

− [(1 mol Ag2S)( fG∆ of Ag2S) + (1 mol H2O)( fG∆ of H2O)]

rxnG∆ = [(2 mol)(77.111 kJ/mol) + (1 mol)(12.6 kJ/mol) + (1 mol)(–157.30 kJ/mol)] − [(1 mol)(–40.3 kJ/mol) + (1 mol)(–237.192 kJ/mol)] rxnG∆ = 287.014 kJ

20-26

ln K = GRT

∆−

= ( ) ( )

3287.014 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –115.8448675

K = e–115.8448675 = 4.8889241x10–51 = 4.89x10–51 20.74 CaF2(s) Ca2+(aq) + 2F–(aq) rxnG∆ = [(1 mol Ca2+)(–553.04 kJ/mol) + (2 mol F–)(–276.5 kJ/mol)] – [(1 mol CaF2)(–1162 kJ/mol)]

rxnG∆ = 55.96 kJ

ln K = GRT

∆−

= ( ) ( )

355.96 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –22.586629

K = e–22.586629 = 1.5514995x10–10 = 1.55x10–10 20.75 Plan: First find ∆G°, then calculate K from ∆G° = –RT ln K. Calculate G∆ using fG∆ values in the relationship

rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆ . Recognize that I2(s), not I2(g), is the standard state for iodine. Solution:


rxnG∆ = [(2 mol ICl)( fG∆ of ICl)] – [(1 mol I2)( fG∆ of I2) + (1 mol Cl2)( fG∆ of Cl2)]

rxnG∆ = [(2 mol)(–6.075 kJ/mol)] – [(1 mol)(19.38 kJ/mol) + (1 mol)(0 kJ/mol)]

rxnG∆ = –31.53 kJ

ln Kp = GRT

∆−

= ( ) ( )

331.53 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

− −

= 12.726169

Kp = e12.726169 = 3.3643794x105 = 3.36x105 20.76 CaCO3(s) CaO(s) + CO2(g) Kp =

2COP

rxnG∆ = [(1 mol CaO)(–603.5 kJ/mol) + (1 mol CO2)(–394.4 kJ/mol)] – [(1 mol CaCO3)(–1128.8 kJ/mol)]

rxnG∆ = 130.9 kJ

ln Kp = GRT

∆−

= ( ) ( )

3130.9 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –52.83398

Kp = e–52.83398 = 1.1336892x10–23 = 1.13x10–23 atm = 2COP

20.77 Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. Solution: ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (1.7x10–5) = 2.72094x104 J/mol = 2.7x104 J/mol The large positive ∆G° indicates that it would not be possible to prepare a solution with the concentrations of lead and chloride ions at the standard-state concentration of 1 M. A Q calculation using 1 M solutions will confirm this:

PbCl2(s) Pb2+(aq) + 2Cl−(aq) Q = [Pb2+][Cl–]2 = (1 M)(1 M)2 = 1 Since Q > Ksp, it is impossible to prepare a standard-state solution of Pb2+(aq) and Cl–(aq).

20-27

20.78 ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (3.0x10–2) = 8.6877x103 J/mol = 8.7x103 J/mol The large positive ∆G° indicates that it would not be possible to prepare a solution with the concentrations of zinc and fluoride ions at the standard-state concentration of 1 M. A Q calculation using 1 M solutions will confirm this:

Q = [Zn2+][F–]2 = (1 M)(1 M)2 = 1 Since Q > Ksp, it is impossible to prepare a standard state solution of ZnF2. 20.79 Plan: The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. ∆G is found by using the relationship ∆G = ∆G° + RT ln Q. Solution: a) ∆G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (9.1x10–6) = 2.875776x104 = 2.9x104 J/mol b) Since rxnG∆ is positive, the reaction direction as written is nonspontaneous. The reverse direction, formation of reactants, is spontaneous, so the reaction proceeds to the left. c) Calculate the value for Q and then use it to find ∆G.

Q =

2 22 2

23 22

Fe Hg

Fe Hg

+ +

+ +

= [ ] [ ][ ] [ ]

2 2

2

0.010 0.025

0.20 0.010 = 1.5625x10–4

∆G = ∆G° + RT ln Q = 2.875776x104 J/mol + (8.314 J/mol•K)(298) ln (1.5625x10–4) = 7.044187x103 = 7.0x103 J/mol Because ∆G298 > 0 and Q > K, the reaction proceeds to the left to reach equilibrium. 20.80 a) ∆G° = –RT ln K = –(8.314 J/mol•K)((273 + 25) K) ln (5.6x108) = –4.9906841x104 = –5.0x104 J/mol b) Since rxnG∆ is negative, the reaction direction as written is spontaneous. The reaction proceeds to the right. c) Calculate the value for Q and then use it to find ∆G.

Q = 2

3 6

623

Ni(NH )

Ni NH

+

+

= [ ]

[ ][ ]6

0.010

0.0010 0.0050 = 6.4x1014

∆G = ∆G° + RT ln Q = –4.9906841x104 J/mol + (8.314 J/mol•K)(298) ln (6.4x1014) = 3.4559756x104 = 3.5x104 J/mol Because ∆G298 > 0 and Q > K, the reaction proceeds to the left to equilibrium. 20.81 Plan: Start by writing the balanced equation for the reaction. Use the balanced equation to write the reaction

quotient expression for the reaction. For part a, count the number of each type of particle present in the individual scenes. Using the given information that each particle represents 0.10 mol and the volume of the container is 0.10 L, calculate the molarity of each species in each individual scene. Use these concentrations and the reaction quotient expression to calculate the value of the reaction quotient for each scene to determine which scene is at equilibrium. For part b, the scene that is at equilibrium has ΔG = 0. Use the values of Q and the properties of logarithms to determine the relative values of ΔG for the other scenes.

Solution: a) The balanced equation for the reaction is: 2A(g) A2(g) The corresponding reaction quotient expression is:

Q = [A2][A]2

Number of A Particles Number of A2 Particles Scene 1 3 3 Scene 2 5 2 Scene 3 1 4

Scene 1:

20-28

Molarity of A: (3 particles)�0.10 mol

particle �

0.10 L = 3.0 M

Molarity of A2: (3 particles)�0.10 mol

particle �

0.10 L = 3.0 M

Q = (3.0 M)

(3.0 M)2 = 0.33

Scene 2:

Molarity of A: (5 particles)�0.10 mol

particle �

0.10 L = 5.0 M


particle �

0.10 L = 2.0 M

Q = (2.0 M)

(5.0 M)2 = 0.080

Scene 3:

Molarity of A: (1 particle)�0.10 mol

particle �

0.10 L = 1.0 M


particle �

0.10 L = 4.0 M

Q = (4.0 M)

(1.0 M)2 = 4.0

The reaction quotient for scene 1 is equal to the equilibrium constant (K = 0.33), so scene 1 is at equilibrium. b) ∆G = ∆G° + RT ln Q Scene 1 is at equilibrium, so its ΔG = 0. Scene 2 has Q = 0.08. Since Q < 1, ln Q is negative, which makes ΔG more negative than it is for the equilibrium scene. Scene 3 has Q = 4.0. Since Q > 1, ln Q is positive and ΔG is positive for scene 3. Scene 3 has the most positive ΔG, followed by Scene 1 (ΔG = 0) and Scene 2 (ΔG is negative).

20.82 Plan: Start by writing the balanced equation for the reaction. Use the balanced equation to write the reaction

quotient expression for the reaction. For part a, count the number of each type of particle present in the individual scenes. Using the given information that each particle represents 0.10 atm, calculate the partial pressure of each species in each individual scene. Use these partial pressures and the reaction quotient expression to calculate the value of the reaction quotient for each scene to determine which scene is at equilibrium. For part b, the scene that is at equilibrium has ΔG = 0. Use the values of Q and the properties of logarithms to determine the relative values of ΔG for the other scenes.

Solution: a) The balanced equation for the reaction is: X(g) + Y2(g) XY(g) + Y(g) The corresponding reaction quotient expression is:

Q = [XY][Y][X][Y2]

Number of X

Particles Number of Y2

Particles Number of XY

Particles Number of Y

Particles Scene 1 1 1 4 2 Scene 2 3 2 2 2 Scene 3 2 1 3 3

Scene 1:

20-29

Partial Pressure of X: (1 particle)(0.10 atm/particle) = 0.10 atm Partial Pressure of Y2: (1 particle)(0.10 atm/particle) = 0.10 atm Partial Pressure of XY: (4 particles)(0.10 atm/particle) = 0.40 atm Partial Pressure of Y: (2 particles)(0.10 atm/particle) = 0.20 atm

Q = (0.40 atm)(0.20 atm)(0.10 atm)(0.10 atm)

= 8.0

Scene 2: Partial Pressure of X: (3 particles)(0.10 atm/particle) = 0.30 atm

Partial Pressure of Y2: (2 particles)(0.10 atm/particle) = 0.20 atm Partial Pressure of XY: (2 particles)(0.10 atm/particle) = 0.20 atm Partial Pressure of Y: (2 particles)(0.10 atm/particle) = 0.20 atm

Q = (0.20 atm)(0.20 atm)(0.30 atm)(0.20 atm)

= 0.67

Scene 3: Partial Pressure of X: (2 particles)(0.10 atm/particle) = 0.20 atm

Partial Pressure of Y2: (1 particle)(0.10 atm/particle) = 0.10 atm Partial Pressure of XY: (3 particles)(0.10 atm/particle) = 0.30 atm Partial Pressure of Y: (3 particles)(0.10 atm/particle) = 0.30 atm

Q = (0.30 atm)(0.30 atm)(0.20 atm)(0.10 atm)

= 4.5

The reaction quotient for scene 3 is equal to the equilibrium constant (K = 4.5), so scene 3 is at equilibrium. b) ∆G = ∆G° + RT ln Q Scene 3 is at equilibrium, so its ΔG = 0. Scene 2 has Q = 0.67. Since Q < 1, ln Q is negative, which makes ΔG more negative than it is for the equilibrium scene. Scene 1 has Q = 8.0. Since Q > 1, ln Q is positive and ΔG is positive for scene 1. Scene 1 has the most positive ΔG, followed by Scene 3 (ΔG = 0) and Scene 2 (ΔG is negative).

20.83 Plan: To decide when production of ozone is favored, both the signs of rxnH∆ and rxnS∆ for ozone are needed.

The values of fH∆ and S can be used. Once rxnH∆ and rxnS∆ are known, ∆G° can be calculated with the

relationship rxnG∆ = rxnH∆ – rxnT S∆ . rxnS∆ values in J/K must be converted to units of kJ/K to match the

units of rxnH∆ . ∆G is found by using the relationship ∆G = ∆G° + RT ln Q. Solution: a) Formation of O3 from O2 : 3O2(g) 2O3(g) or per mole of ozone: 3/2O2(g) O3(g). rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol O3)( fH∆ of O3)] – [(3/2 mol O2)( fH∆ of O2)]

rxnH∆ = [(1 mol)(143 kJ/mol] – [(3/2 mol)(0 kJ/mol)] = 143 kJ


rxnS∆ = [(1 mol O3)( S of O3)] – [(3/2 mol O2)( S of O2)]

rxnS∆ = [(1 mol)(238.82 J/mol•K] – [(3/2 mol)(205.0 J/mol•K)] = –68.68 J/K = –0.06868 kJ/K

The positive sign for rxnH∆ and the negative sign for rxnS∆ indicates the formation of ozone is favored at no temperature. The reaction is nonspontaneous at all temperatures.

b) rxnG∆ = rxnH∆ – rxnT S∆

20-30

rxnG∆ = 143 kJ – [(298 K)(–0.06868 kJ/K)] = 163.46664 = 163 kJ for the formation of one mole of O3. c) Calculate the value for Q and then use to find ∆G.

Q = [ ]

33

22

O

O

= [ ]

7

32

5x10 atm

0.21 atm

− = 5.195664x10–6

∆G = ∆G° + RT ln Q = 163 kJ/mol + (8.314 J/mol•K)(298)(1 kJ/103 J) ln (5.195664x10–6) = 132.85368 = 1x102 kJ/mol 20.84 BaSO4(s) Ba2+(aq) + SO4

2–(aq) The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K.

ln Kp = GRT

∆−

= ( ) ( )

359.1 kJ/mol 10 J8.314 J/mol•K (273 37)K 1 kJ

− +

= –22.930618

K = e–22.930618 = 1.099915x10–10 = 1.10x10–10 Ksp = [Ba2+][SO4

2–] = 1.099915x10–10 = S2

S = 101.099915x10− = 1.0487683x10–5 = 1.05x10–5 M Ba2+ 20.85 a) diamond → graphite rxnH∆ = [(1 mol graphite)(0 kJ/mol)] – [(1 mol diamond)(1.896 kJ/mol)] = –1.896 kJ/mol

rxnS∆ = [(1 mol graphite)(5.686 J/K•mol)] – [(1 mol diamond)(2.439 kJ/mol)] = 3.247 J/K

rxnG∆ = [(1 mol graphite)(0 kJ/mol)] – [(1 mol diamond)(2.866 kJ/mol)] = –2.866 kJ/mol b) Since ∆G° is negative, the reaction diamond → graphite is spontaneous at room temperature. However, this does not give any information about the rate of reaction, which is very slow. Therefore, diamonds are not forever, but they are for a very long time. c) graphite → diamond For this process, the signs of ∆H and ∆S are, like the reaction, reversed. A process with ∆H positive and ∆S negative is nonspontaneous at all temperatures. Thus, something other than a change in temperature is necessary. That is why diamonds also require a change in pressure. d) Graphite cannot be converted to diamond spontaneously at 1 atm. At all temperatures ∆G° > 0 (nonspontaneous). 20.86

∆S rxn ∆Hrxn ∆Grxn Comment (a) + − − Spontaneous (b) (+) 0 − Spontaneous (c) − + (+) Not spontaneous (d) 0 (−) − Spontaneous (e) (−) 0 + Not spontaneous (f) + + (−) T∆S > ∆H

a) The reaction is always spontaneous when ∆Grxn < 0, so there is no need to look at the other values other than to check the answer. b) Because ∆Grxn = ∆H – T∆S = –T∆S, ∆S must be positive for ∆Grxn to be negative. c) The reaction is always nonspontaneous when ∆Grxn > 0, so there is no need to look at the other values other than to check the answer. d) Because ∆Grxn = ∆H – T∆S = ∆H, ∆H must be negative for ∆Grxn to be negative. e) Because ∆Grxn = ∆H – T∆S = –T∆S, ∆S must be negative for ∆Grxn to be positive. f) Because T∆S > ∆H, the subtraction of a larger positive term causes ∆Grxn to be negative. 20.87 ∆S° > 0 (Four aqueous species become seven aqueous species. The increase in the number of species increases the entropy.)

20-31

∆G° < 0 (The reaction has K > 1.) The reaction proceeds due to an increase in entropy. 20.88 At the freezing point, the system is at equilibrium, so ∆G° = 0. ∆G° = 0 = ∆H° – T∆S° ∆H° = T∆S° ∆S° = ∆H°/T = (–2.39 kJ/mol)/((273.2 + 63.7)K) = 7.0940932x10–3 kJ/mol•K ∆S° = (0.200 mol)(7.0940932x10–3 kJ/mol•K) = 1.4188x10–3 = 1.42x10–3 kJ/K 20.89 a) False, all spontaneous reactions occur without outside intervention. b) True, a reaction cannot be spontaneous in both directions. c) False, all exothermic processes release heat. d) True e) False, if a process increases the freedom of motion of the particles of a system, the entropy of the system increases. f) False, the energy of the universe is constant; the entropy of the universe increases toward a maximum. g) False, all systems with their surroundings increase the freedom of motion spontaneously. h) True 20.90 Plan: Write the equilibrium expression for the reaction. The equilibrium constant, K, is related to ∆G° through the equation ∆G° = –RT ln K. Once K is known, the ratio of the two species is known. Solution:

a) For the reaction, K = [ ][ ][ ][ ]

2

2

Hb•CO OHb•O CO

since the problem states that [O2] = [CO]; the K expression simplifies to:

K = [ ][ ]2

Hb•COHb•O

ln Kp = GRT

∆−

= ( ) ( )

314 kJ/mol 10 J8.314 J/mol•K (273 37)K 1 kJ

− − +

= 5.431956979

K = e5.431956979 = 228.596 = 2.3x102 = [ ][ ]2

Hb•COHb•O

b) By increasing the concentration of oxygen, the equilibrium can be shifted in the direction of Hb•O2. Administer oxygen-rich air to counteract the CO poisoning.

20.91 a) MgCO3(s) MgO(s) + CO2(g) b) Locate the thermodynamic values. rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol MgO)(–601.2 kJ/mol) + (1 mol CO2)(–393.5 kJ/mol)] – [(1 mol MgCO3)(–1112 kJ/mol)] = 117.3 kJ rxnS∆ = ∑m productsS – ∑n reactantsS

rxnS∆ = [(1 mol MgO)(26.9 J/K•mol) + (1 mol CO2)(213.7 J/K•mol)] – [(1 mol MgCO3)(65.86 J/K•mol)] = 174.74 J/K rxnG∆ = rxnH∆ – rxnT S∆ = 117.3 kJ – (298 K)(174.74 J/K)(1 kJ/103 J) = 65.22748 = 65.2 kJ c) rxnG∆ = 0 = rxnH∆ – rxnT S∆

rxnH∆ = rxnT S∆

20-32

= HTS

∆

∆

= 117.3 kJ0.17474 kJ/K

= 671 K

d) Kp = 2COP and ∆G° = –RT ln K are necessary.

ln K = GRT

∆−

= ( ) ( )

365.22748 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –26.327178

K = e–26.327178 = 3.6834253x10–12 = 3.68x10–12 atm =2COP

e) This is similar to part d) except a new ∆G° must be determined at 1200 K. rxnG∆ = rxnH∆ – rxnT S∆ = 117.3 kJ – (1200 K)(174.74 J/K)(1 kJ/103 J) = –92.388 kJ

ln K = GRT

∆−

= ( ) ( )

392.388 kJ/mol 10 J8.314 J/mol•K 1200 K 1 kJ

− −

= 9.26028

K = e9.26028 = 1.0512x104 = 1.05x104 atm =2COP

20.92 Plan: Sum the two reactions to yield an overall reaction. Use the relationship rxnG∆ = rxnH∆ – rxnT S∆ to

calculate rxnG∆ . rxnH∆ and rxnS∆ will have to be calculated first. Solution: UO2(s) + 4HF(g) → UF4(s) + 2H2O(g) UF4(s) + F2(g) → UF6(s) UO2(s) + 4 HF(g) + F2(g) → UF6(g) + 2H2O(g) (overall process) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol UF6)( fH∆ of UF6) + (2 mol H2O)( fH∆ of H2O)]

– [(1 mol UO2)( fH∆ of UO2) + (4 mol HF)( fH∆ of HF) + (1 mol F2)( fH∆ of F2)]

rxnH∆ = [(1 mol)(–2197 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(1 mol)(–1085 kJ/mol) + (4 mol)(–273 kJ/mol) + (1 mol)(0 kJ/mol)] rxnH∆ = –503.652 kJ


rxnS∆ = [(1 mol UF6)( S of UF6) + (2 mol H2O)( S of H2O)]

– [(1 mol UO2)( S of UO2) + (4 mol HF)( S of HF) + (1 mol F2)( S of F2)]

rxnS∆ = [(1 mol)(225 J/mol•K) + (2 mol)(188.72 J/mol•K)] – [(1 mol)(77.0 J/mol•K) + (4 mol)(173.67 J/mol•K) + (1 mol)(202.7 J/mol•K)] rxnS∆ = –371.94 J/K

rxnG∆ = rxnH∆ – rxnT S∆ = (–503.652 kJ) – ((273 + 85)K)(–371.94 J/K)(kJ/103 J) = –370.497 = –370. kJ 20.93 CO(g) + 2H2(g) → CH3OH(l) a) rxnH∆ = [(1 mol CH3OH)(–238.6 kJ/mol)] – [(1 mol CO)(–110.5 kJ/mol) + (2 mol H2)(0 kJ/mol)] = –128.1 kJ rxnS∆ = [(1 mol CH3OH)(127 J/K•mol)] – [(1 mol CO)(197.5 J/K•mol) + (2 mol H2)(130.6 J/K•mol)] = –331.7 J/K rxnG∆ = rxnH∆ – rxnT S∆ = –128.1 kJ – [(298 K)(–331.7 J/K)(1 kJ/103 J)] = –29.2534 = –29.2 kJ The negative value of ∆G° indicates that the reaction is spontaneous (feasible). b) ∆H° negative and ∆S° negative means the reaction is favored at low temperature.

20-33

c) CH3OH(g) + 1/2O2(g) → CH2O(g) + H2O(g) rxnH∆ = [(1 mol CH2O)(–116 kJ/mol) + (1 mol H2O)(–241.826 kJ/mol)] − [(1 mol CH3OH)(–201.2 kJ/mol) + (1/2 mol O2)(0 kJ/mol)] = –156.626 kJ rxnS∆ = [(1 mol CH2O)(219 J/K•mol) + (1 mol H2O)(188.72 J/K•mol)] − [(1 mol CH3OH)(238 J/K•mol) + (1/2 mol O2)(205.0 J/K•mol)] = 67.22 J/K rxnG∆ = rxnH∆ – rxnT S∆ = –156.626 kJ – [((273 + 100.)K)(67.22 J/K)(1 kJ/103 J)] = –181.699 = –182 kJ 20.94 Plan: rxnG∆ can be calculated with the relationship ∑m f (products)G∆ – ∑n f (reactants)G∆ . ∆G is found by using

the relationship ∆G = ∆G° + RT ln Q. Solution: a) 2N2O5(g) + 6F2(g) → 4NF3(g) + 5O2(g) b) rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆

rxnG∆ = [(4 mol NF3)( fG∆ of NF3) + (5 mol O2)( rxnG∆ of O2)]

– [(2 mol N2O5)( rxnG∆ of N2O5)) + (6 mol F2) ( rxnG∆ of F2)]

rxnG∆ = [(4 mol)(–83.3 kJ/mol)) + (5 mol)(0 kJ/mol)] − [(2 mol)(118 kJ/mol) + (6 mol)(0 kJ/mol)]

rxnG∆ = –569.2 = –569 kJ c) Calculate the value for Q and then use to find ∆G.

Q = [ ][ ]

4 53 2

2 62 5 2

NF O

N O F

= [ ] [ ]

[ ] [ ]

4 5

2 6

0.25atm 0.50atm

0.20atm 0.20atm = 47.6837

∆G = ∆G° + RT ln Q = –569.2 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(298) ln (47.6837) = –559.625 = –5.60x102 kJ/mol 20.95 a) rxnS∆ = [(2 mol NO)(210.65 J/K•mol) + (1 mol Br2)(245.38 J/K•mol)] – [(2 mol NOBr)(272.6 J/mol•K)]

rxnS∆ = 121.48 = 121.5 J/K

b) rxnG∆ = –RT ln K = –(8.314 J/mol•K)(373 K) ln (0.42) = 2690.225 = 2.7x103 J/mol

c) rxnG∆ = rxnH∆ – rxnT S∆

rxnH∆ = rxnG∆ + rxnT S∆ = 2690.225 J/mol + (373)(121.48 J/K) = 4.8002265x104 = 4.80x104 J/mol

d) rxnH∆ = [(2 mol NO)(90.29 kJ/mol) + (1 mol Br2)(30.91 kJ/mol)] – [(2 mol NOBr)( fH∆ of NOBr)]

(2 mol NOBr)( fH∆ of NOBr)] = [(2 mol)(90.29 kJ) + (1 mol)(30.91 kJ)] – (4.8002265x104 J)(1 kJ/103 J))

fH∆ of NOBr = 81.7438675 = 81.7 kJ/mol

e) rxnG∆ = rxnH∆ – rxnT S∆ = 4.8002265x104 J/mol – (298 K)(121.48 J/mol•K) = 1.1801225x104 = 1.18x104 J/mol f) rxnG∆ = [(2 mol NO)(86.60 kJ/mol) + (1 mol Br2)(3.13 kJ/mol)] – [(2 mol NOBr(g)( fG∆ of NOBr)]

(2 mol NOBr(g)( fG∆ of NOBr) = [(2 mol)(86.60 kJ) + (1 mol)(3.13 kJ)] – (1.1801225x104 J)(1 kJ/103 J)

fG∆ of NOBr = 82.264 = 82.3 kJ/mol 20.96 Plan: rxnH∆ can be calculated from the individual fH∆ values of the reactants and products by using the

20-34


of the reactants and products by using the relationship rxnS∆ = ∑m productsS – ∑n reactantsS . rxnG∆ can be

calculated with the relationship ∑m f (products)G∆ – ∑n f (reactants)G∆ . Solution: The reaction for the hydrogenation of ethene is C2H4(g) + H2(g) → C2H6(g). rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol C2H6)( fH∆ of C2H6)] – [(1 mol C2H4)( fH∆ of C2H4) + (1 mol H2)( fH∆ of H2)] = [(1 mol)(–84.667 kJ/mol)] – [(1 mol)(52.47 kJ/mol) + (1 mol)(0 kJ/mol)] = –137.137 = –137.14 kJ rxnS∆ = ∑m productsS – ∑n reactantsS

rxnS∆ = [(1 mol C2H6)( S of C2H6)] – [(1 mol C2H4)( S of C2H4)) + (1 mol H2)( S of H2)] = [(1 mol)(229.5 J/mol•K)] – [(1 mol)(219.22 J/mol•K) + (1 mol)(130.6 J/mol•K)] = –120.32 = –120.3 J/K rxnG∆ = ∑m f (products)G∆ – ∑n f (reactants)G∆

rxnG∆ = [(1 mol C2H6)( fG∆ of C2H6))] – [(1 mol C2H4)( fG∆ of C2H4) + (1 mol H2)( fG∆ of H2)] = (1 mol)(–32.89 kJ/mol)] – [(1 mol)(68.36 kJ/mol) + (1 mol)(0 kJ/mol)] = –101.25 kJ 20.97 a) C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g) rxnH∆ = [(1 mol H2)( fH∆ of H2) + (1 mol C6H5CHCH2)( fH∆ of C6H5CHCH2)]

– [(1 mol C6H5CH2CH3)( fH∆ of C6H5CH2CH3)]

rxnH∆ = [(1 mol)(0 kJ/mol) + (1 mol)(103.8 kJ/mol)] – [(1 mol)( –12.5 kJ/mol)] = 116.3 kJ

rxnG∆ = [(1 mol H2)( fG∆ of H2) + (1 mol C6H5CHCH2)( fG∆ of C6H5CHCH2)]

– [(1 mol C6H5CH2CH3)( fG∆ of C6H5CH2CH3)]

rxnG∆ = [(1 mol)(0 kJ/mol) + (1 mol)(202.5 kJ/mol)] – [(1 mol)( 119.7 kJ/mol)] = 82.8 kJ

rxnS∆ = [(1 mol H2)( S of H2) + (1 mol C6H5CHCH2)( S of C6H5CHCH2)]

– [(1 mol C6H5CH2CH3)( S of C6H5CH2CH3)]

rxnS∆ = [(1 mol)(130.6 J/mol•K) + (1 mol)(238 J/mol•K)] – [(1 mol)( 255 J/mol•K)] = 113.6 = 114 J/K

b) rxnG∆ = 0 = rxnH∆ – rxnT S∆

rxnH∆ = rxnT S∆

= HTS

∆

∆

= 116.3 kJ0.114 kJ/K

= 1020.1754 = 1020 K = 747°C

c) rxnG∆ = rxnH∆ – rxnT S∆ = 116.3 kJ – (600°C + 273)(114 J/K)(1 kJ/103 J) = 16.778 = 16.8 kJ/mol

ln K = GRT

∆−

= ( ) ( )

316.778 kJ/mol 10 J8.314 J/mol•K 873 K 1 kJ

−

= –2.311617

K = e–2.311617 = 9.9101x10–2 = 9.9x10–2 d) Calculate the value for Q and then use it to find ∆G. C6H5CH2CH3(g) → H2(g) + C6H5CHCH2(g) Initial 1.0 0 0 React – 0.50 +0.50 +0.50 (50% conversion) Remaining 0.50 0.50 0.50

20-35

Mole fraction of each gas = 2

amount of gas 0.5 = H + styrene + ethylbenzene + steam 0.5 + 0.5 + 0.5 + 5

= 0.076923

Partial pressure of each gas = mole fraction x total pressure = 0.076923 x 1.3 atm = 0.10 atm

Q = [ ]2 6 5 2

6 5 2 2

H C H CHCHC H CH CH

= [ ][ ]

[ ]0.10atm 0.10atm

0.10atm = 0.10

∆G = ∆G° + RT ln Q = 16.778 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(873) ln (0.10) = 0.0655565 = 0.07 kJ/mol 20.98 a) The formation reaction for propylene is 3C(s) + 3H2(g) → CH3CH=CH2(g)

fS = [(1 mol CH3CH=CH2)( S of CH3CH=CH2)] – [(3 mol C)( S of C) + (3 mol H2)( S of H2)] = [(1 mol)(267.1 J/mol•K)] – [(3 mol)(5.686 J/mol•K) + (3 mol)(130.6 J/mol•K)] = –141.758 = –141.8 J/K

b) fG∆ = fH∆ – fT S∆

fG∆ = 20.4 kJ/mol – (298 K)(–141.758 J/K)(1 kJ/103 J) = 62.643884 = 62.6 kJ/mol c) The dehydrogenation reaction is CH3CH2CH3(g) → CH3CH=CH2(g) + H2(g)

rxnH∆ = [(1 mol CH3CH=CH2)( fH∆ of CH3CH=CH2) + (1 mol H2)( fH∆ of H2)]

– [(1 mol CH3CH2CH3) ( fH∆ of CH3CH2CH3)] = [(1 mol)(20.4 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)( –105 kJ/mol)] = 125.4 = 125 kJ

rxnG∆ = [(1 mol CH3CH=CH2)( fG∆ of CH3CH=CH2) + (1 mol H2)( fG∆ of H2)]

– [(1 mol CH3CH2CH3)( fG∆ of CH3CH2CH3)] = [(1 mol )(62.643884 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)( –24.5 kJ/mol)] = 87.143884 = 87.1 kJ

d) Find K from rxnG∆ at this elevated temperature. rxnS∆ must be found first:

rxnS∆ = [(1 mol CH3CH=CH2)( S of CH3CH=CH2) + (1 mol H2)( S of H2)]

– [(1 mol CH3CH2CH3)( S of CH3CH2CH3)] = [(1 mol)(267.1 J/mol•K) + (1 mol)(130.6 J/mol•K)] – [(1 mol)(269.9 J/mol•K)] = 127.8 J/K

fG∆ = fH∆ – fT S∆

fG∆ = 125.4 kJ – (853 K)( 127.8 J/K)(1 kJ/103 J) = 16.3866 kJ

ln K = GRT

∆−

= ( ) ( )

316.3866 kJ/mol 10 J8.314 J/mol•K 853 K 1 kJ

−

= –2.3106267

K = e–2.3106267 = 0.099199 = 0.0992

K = ( )( )

( )3 2 2

3 2 3

CH CH CH H

CH CH CH

P P

P=

x= pressure of CH3CH=CH2 and the pressure of H2 and 1.00 – x = pressure of CH3CH2CH3

0.099199 = ( )( )

( )x x

1.00 x−

x2 + 0.099199x – 0.099199 = 0 Solving the quadratic equation:

x = 2b b 4ac

2a− ± −

x = ( ) ( )( )

( )

2(0.099199) 0.099199 4 1 0.099199

2 1

− ± − −

x = 0.2692408 = 0.269 atm The theoretical yield of propylene is 27%.

20-36

e) If hydrogen could escape through the reactor walls, the reaction would be shifted to the right, improving the yield. f) rxnH∆ = rxnT S∆

= HTS

∆

∆

= 125.4 kJ0.1278 kJ/K

= 981.22066 = 981 K = 708°C

20.99 The reactions are (from text): ATP4– + H2O ADP3– + HPO4

2– + H+ ∆G° =–30.5 kJ Glucose + HPO4

2– + H+ [glucose phosphate]– + H2O ∆G° = 13.8 kJ Glucose + ATP4– [glucose phosphate]– + ADP3– ∆G° =–16.7 kJ ∆G° = –RT ln K T = (273 + 25) K = 298 K

a) ln K = GRT

∆−

= ( ) ( )

330.5 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

− −

= 12.3104

K = e12.3104 = 2.2199x105 = 2.22x105

b) ln K = GRT

∆−

= ( ) ( )

313.8 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

−

= –5.569969

K = e–5.569969 = 3.8105985x10–3 = 3.81x10–3

c) ln K = GRT

∆−

= ( ) ( )

316.7 kJ/mol 10 J8.314 J/mol•K 298 K 1 kJ

− −

= 6.74047

K = e6.74047 = 8.45958x102 = 8.46x102 d) Repeat the calculations at the new temperature. T = (273 + 37) = 310. K

ln K = GRT

∆−

= ( ) ( )

330.5 kJ/mol 10 J8.314 J/mol•K 310. K 1 kJ

− −

= 11.8339

K = e11.8339 = 1.37847x105 = 1.38x105

ln K = GRT

∆−

= ( ) ( )

313.8 kJ/mol 10 J8.314 J/mol•K 310. K 1 kJ

−

= –5.3543576

K = e–5.3543576 = 4.7275x10–3 = 4.73x10–3

ln K = GRT

∆−

= ( ) ( )

316.7 kJ/mol 10 J8.314 J/mol•K 310. K 1 kJ

− −

= 6.4795

K = e6.4795 = 6.51645x102 = 6.52x102 20.100 a) ATP4– + H2O ADP3– + HPO4

2– + H+ ∆G° = –30.5 kJ

ln K = GRT

∆−

= ( ) ( )

330.5 kJ/mol 10 J8.314 J/mol•K 310. K 1 kJ

− −

= 11.8339

K = e11.8339 = 1.37847x105 = 1.38x105 b) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) rxnG∆ = [(6 mol CO2)( fG∆ of CO2) + (6 mol H2O)( fG∆ of H2O)]

– [(1 mol C6H12O6 )( fG∆ of C6H12O6) + (6 mol O2)( fG∆ of O2)]

rxnG∆ = [(6 mol)(–394.4 kJ/mol) + (6 mol)(–237.192 kJ/mol)] − [(1 mol)(–910.56 kJ/mol) + (6 mol)(0 kJ/mol)]

rxnG∆ = –2878.992 = –2879.0 kJ

c) 1 mol ATP 2878.992 kJ

30.5 kJ 1 mol glucose − −

= 94.39318 = 94.4 mol ATP/mol glucose

20-37

d) The actual yield is 36 moles ATP (this is assumed to be an exact value). The percent yield is:

36 mol ATP x 100%

94.39318 mol ATP

= 38.1383 = 38.1%

20.101 SO3(g) + SCl2(l) → SOCl2(l) + SO2(g) rxnG∆ = –75.2 kJ

a) rxnH∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

rxnH∆ = [(1 mol SOCl2)( fH∆ of SOCl2) + (1 mol SO2)( fH∆ of SO2)]

– [(1 mol SO3)( fH∆ of SO3) + (1 mol SCl2 )( fH∆ of SCl2)]

rxnH∆ = [(1 mol)(–245.6 kJ/mol) + (1 mol)(–296.8 kJ/mol)] – [(1 mol)(–396 kJ/mol) + (1 mol)(–50.0 kJ/mol)]

rxnH∆ = –96.4 kJ

rxnG∆ = –75.2 kJ/mol = rxnH∆ – rxnT S∆

–75.2 kJ/mol = –96.4 kJ/mol – rxnT S∆

75.2 kJ/mol – 96.4 kJ/mol = –21.2 kJ/mol = rxnT S∆

(–21.2 kJ/mol)/(298 K) = rxnS∆

rxnS∆ = (–0.0711409 kJ/mol•K)(103 J/1 kJ) = –71.1409 J/mol•K

rxnS∆ = [(1 mol SOCl2)( S of SOCl2) + (1 mol SO2)( S of SO2)]

– [(1 mol SO3)( S of SO3) + (1 mol SCl2)( S of SCl2)]

–71.1409 J/mol•K = [(1 mol)( S of SOCl2) + (1 mol)(248.1 J/mol•K)] – [(1 mol)(256.7 J/mol•K)) + (1 mol)(184 J/mol•K)]

–71.1409 J/mol•K = [(1 mol)( S of SOCl2) + 248.1 J/mol•K] – [(440.7 J/mol•K)]

–71.1409 J/mol•K – 248.1 J/mol•K + 440.7 J/mol•K = 1 mol( S of SOCl2)

121.4591 J/mol•K = 1 mol( S of SOCl2)

S of SOCl2 = 121 J/mol•K

b) rxnG∆ =0 = rxnH∆ – rxnT S∆

rxnH∆ = rxnT S∆

= HTS

∆

∆

= 96.4 kJ0.0711409 kJ/K

−−

= 1355.057 = 1.4x103 K

20.102 Oxidation of iron: 4Fe(s) + 3O2(g) → 2Fe2O3(s) The change in Gibbs free energy for this reaction will be twice fG∆ of Fe2O3 since the reaction forms the oxide from its elements and is reported per mole. fG∆ = [(2 mol Fe2O3)(–743.6 kJ/mol)] – [(4 mol Fe)(0 kJ/mol) + (3 mol O2)(0 kJ/mol)] = –1487.2 kJ Oxidation of aluminum: 4Al(s) + 3O2(g) → 2Al2O3(s) The change in Gibbs free energy for this reaction will be twice fG∆ of Al2O3 since the reaction forms the oxide from its elements and is reported per mole. fG∆ = [(2 mol Al2O3)(–1582 kJ/mol)] – [(4 mol Al)(0 kJ/mol) + (3 mol O2)(0 kJ/mol)] = –3164 kJ With ∆G° < 0, both reactions are spontaneous at 25°C.

20-38

20.103 a) H2(g) + I2(g) 2HI(g)

Kc = [ ]

[ ][ ]

2

2 2

HIH I

= [ ]

[ ][ ]

20.100.010 0.020

= 50 Kc > 1

b) Kp = Kc(RT)Δngas

Kp = 50[(0.0821 Latm/molK)(733 K)]0 Kp = 50 = Kc c) ∆G° = –RT ln K = –(8.314 J/mol•K)(733 K) ln (50) = 2.38405x104 = 2.4x104 J/mol

d) Kc = [ ]

[ ][ ]

2

2 2

HIH I

= [ ]

[ ][ ]

20.100.020 0.010

= 50

The value of Kc is 50 in this situation as in a) so ∆G° does not change.

20.104 Plan: Use the relationship ∆G ̥̥° = –RT ln K to calculate ∆G°. Use the relationship ∆G = ∆G° + RT ln Q to calculate ∆G. Solution: The given equilibrium is: G6P F6P K = 0.510 at 298 K a) ∆G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (0.510) = 1.6682596x103 = 1.67x103 J/mol b) Q = [F6P]/[G6P] = 10.0 ∆G = ∆G° + RT ln Q ∆G = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln 10.0 ∆G = 7.3730799x103 = 7.37x103 J/mol c) Repeat the calculation in part b) with Q = 0.100 ∆G = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln 0.100 ∆G = –4.0365607x103 = –4.04x103 J/mol d) ∆G = ∆G° + RT ln Q (–2.50 kJ/mol)(103 J/1 kJ) = 1.6682596x103 J/mol + (8.314 J/mol•K)(298 K) ln Q (–4.1682596x103 J/mol) = (8.314 J/mol•K)(298 K) ln Q ln Q = (–4.1682596x103 J/mol)/[(8.314 J/mol•K)(298 K)] = –1.68239696 Q = 0.18592778 = 0.19 20.105 a) Graph D depicts how Gsys changes for the chemical reaction. G decreases as the reaction proceeds from either pure reactant or pure product until it reaches the minimum at equilibrium. Beyond that in either direction the reaction is nonspontaneous. b) Graph A depicts how Gsys changes as ice melts at 1°C. 1°C is higher than the melting point of water, therefore the system is not at equilibrium. Melting is spontaneous at 1°C and 1 atm, and the Gsys will decrease until the system reaches equilibrium. 20.106 The equilibrium may be represented as: double helix 2 random coils. a) ∆S° is positive. The formation of a larger number of components increases randomness. b) ∆G = ∆H – [T∆S] Energy is required to overcome the hydrogen bonds and the dispersion forces between the strands. Thus, ∆H is positive. The Kelvin temperature is, of course, positive. (+) – [(+) (+)] ∆G is positive when T∆S < ∆H. c) ∆G = 0 (at equilibrium) ∆G = 0 = ∆H – T∆S ∆H = T∆S T = ∆H/∆S

20-39

20.107 a) Respiration: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Fermentation: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) Ethanol oxidation: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) b) rxnG∆ = [(6 mol CO2 )( fG∆ of CO2) +(6 mol H2O)( fG∆ of H2O)]

– [(1 mol C6H12O6)( fG∆ of C6H12O6) + (6 mol O2 )( fG∆ of O2)]

rxnG∆ = [(6 mol)(–394.4 kJ/mol) + (6 mol)(–237.192 kJ/mol)] – [(1 mol)(–910.56 kJ/mol) + (6 mol)(0 kJ/mol)]

rxnG∆ = –2878.992 kJ for one mole of C6H12O6

rxnG∆ /g of glucose = ( ) 1 mol glucose 2878.992 kJ1.00 g glucose180.16 g glucose 1 mol glucose

−

= –15.980195 = –16.0 kJ/g

c) rxnG∆ = [(2 mol CO2)( fG∆ of CO2) + (2 mol C2H5OH)( fG∆ of C2H5OH)]

– [(1 mol C6H12O6)( fG∆ of C6H12O6)]

rxnG∆ = [(2 mol)(–394.4 kJ/mol) + (2 mol)(–174.8 kJ/mol)] – [(1 mol)(–910.56 kJ/mol)]

rxnG∆ = –227.84 kJ for one mole of C6H12O6

rxnG∆ /g of glucose = ( ) 1 mol glucose 227.84 kJ1.00 g glucose180.16 g glucose 1 mol glucose

−

= –1.264653641 = –1.26 kJ/g

d) rxnG∆ = [(2 mol CO2)( fG∆ of CO2) + (3 mol H2O)( fG∆ of H2O)]

– [(1 mol C2H5OH)( fG∆ of C2H5OH) + (3 mol O2)( fG∆ of O2)]

rxnG∆ = [(2 mol)(–394.4 kJ/mol) + (3 mol)(–237.192 kJ/mol)] – [(1 mol)(–174.8 kJ/mol) + (3 mol)(0 kJ/mol)]

rxnG∆ = –1325.576 kJ for one mole of C2H5OH

rxnG∆ = ( ) 2 5

2 5

2 mol C H OH1 mol glucose 1325.576 kJ1.00 g glucose180.16 g glucose 1 mol glucose 1 mol C H OH

−

= –14.71554 = –14.7 kJ/g

20.108 Plan: First calculate H∆ and S∆ . According to the relationship ln K = GRT

∆−

, since ln Kp = 0 when Kp = 1.00,

G∆ = 0. Use the relationship G∆ = 0 = H∆ – T S∆ to find the temperature. For part b), calculate G∆ at

the higher temperature with the relationship G∆ = H∆ – T S∆ and then calculate K with ln K = GRT

∆−

.

Solution: a) H∆ = ∑m f (products)H∆ – ∑n f (reactants)H∆

H∆ = [(2 mol NH3)( fH∆ of NH3)] – [(1 mol N2)( fH∆ of N2) + (3 mol H2)( fH∆ of H2)]

H∆ = [(2 mol)(–45.9 kJ/mol)] – [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)] = –91.8 kJ

S∆ = ∑m productsS – ∑n reactantsS

S∆ = [(2 mol NH3)( S of NH3)] – [(1 mol N2)( S of N2) + (3 mol H2)( S of H2)]

S∆ = [(2 mol)(193 J/mol•K)] – [(1 mol)(191.50 J/mol•K) + (3 mol)(130.6 J/mol•K)]

S∆ = –197.3 J/K

ln K = GRT

∆−

Since ln Kp = 0 when Kp = 1.00, G∆ = 0.

G∆ = 0 = H∆ – T S∆

20-40

H∆ = T S∆

= HTS

∆

∆

= 391.8 kJ 10 J

197.3 J/K 1 kJ − −

= 465.281 = 465 K

b) G∆ = H∆ – T S∆ = (–91.8 kJ)(103 J/1 kJ) – (673 K)(–197.3 J/K)= 4.09829x104 J

ln K = GRT

∆−

= ( ) ( )

44.09829x10 J/mol8.314 J/mol•K 673 K

−

= –7.32449

K = e–7.32449 = 6.591958x10–4 = 6.59x10–4 c) The reaction rate is higher at the higher temperature. The time required (kinetics) overshadows the lower yield (thermodynamics).

20.109 a) At equilibrium, all three free energies are equal. b) Kyanite has the lowest enthalpy. Lowering the temperature will shift the equilibrium in the exothermic direction and favor the substance with the lowest enthalpy. c) Sillimanite has the highest entropy. At equilibrium, the substance with the highest enthalpy (opposite to kyanite) must have the highest entropy. d) Andalusite has the lowest density. Decreasing pressure shifts the equilibrium toward the substance with more volume or less density. 20.110 a) 2CH4(g) + 1/2O2(g) → C2H2(g) + 2H2(g) + H2O(g) H∆ = [(1 mol C2H2)( fH∆ of C2H2) + (2 mol H2)( fH∆ of H2) + (1 mol H2O)( fH∆ of H2O)]

– [(2 mol CH4)( fH∆ of CH4) + (1/2 mol O2)( fH∆ of O2)]

H∆ = [(1 mol)(227 kJ/mol) + (2 mol)(0 kJ/mol) + (1 mol)(–241.826 kJ/mol)] – [(2 mol)(–74.87 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

H∆ = 134.914 = 135 kJ

S∆ = [(1 mol C2H2)( S of C2H2) + (2 mol H2)( S of H2) + (1 mol H2O)( S of H2O)]

– [(2 mol CH4)( S of CH4) + (1/2 mol O2)( S of O2)]

S∆ = [(1 mol)(200.85 J/mol•K) + (2 mol)(130.6 J/mol•K) + (1 mol)(188.72 J/mol•K)] – [(2 mol)(186.1 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]

S∆ = 176.07 = 176.1 J/K

G∆ = 0 = H∆ – T S∆

H∆ = T S∆

= HTS

∆

∆

= 3134.914 kJ 10 J

176.07 J/K 1 kJ

= 766.25206 = 766 K

b) 2C(s) + H2(g) → C2H2(g) H∆ = [(1 mol C2H2)( fH∆ of C2H2)] – [(2 mol C)( fH∆ of C) + (1 mol H2)( fH∆ of H2)]

H∆ = [(1 mol)(227 kJ/mol)] – [(2 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]

H∆ = 227 kJ

S∆ = [(1 mol C2H2)( S of C2H2)] – [(2 mol C)( S of C) + (1 mol H2)( S of H2)]

S∆ = [(1 mol)(200.85 J/mol•K)] – [(2 mol)(5.686 J/mol•K) + (1 mol)(130.6 J/mol•K)]

S∆ = 58.878 = 58.9 J/K

G∆ = 0 = H∆ – T S∆

20-41

H∆ = T S∆

= HTS

∆

∆

== 3227 kJ 10 J

58.878 J/K 1 kJ

= 3855.43 = 3860 K

c) Considering the reverse reaction of its formation, the acetylene is produced under conditions at which acetylene is unstable and can decompose back into its elements. It must be quickly cooled to a temperature where its thermal decomposition rate is slow. 20.111 a) CH4(g) + H2O(g) CO(g) + 3H2(g) steam re-forming

H∆ = [(1 mol CO)( fH∆ of CO) + (3 mol H2)( fH∆ of H2)]

– [(1 mol CH4)( fH∆ of CH4) + (1 mol H2O)( fH∆ of H2O)]

H∆ = [(1 mol)(–110.5 kJ/mol) + (3 mol)(0)] – [(1 mol)(–74.87 kJ/mol) + (1 mol)(–241.826 kJ/mol)] = 206.196 = 206.2 kJ

S∆ = [(1 mol CO)( S of CO) + (3 mol H2)( S of H2)] – [(1 mol CH4)( S of CH4) + (1 mol H2O)( S of H2O)]

S∆ = [(1 mol)(197.5 J/mol•K) + (3 mol)(130.6 J/mol•K)] – [(1 mol)(186.1 J/mol•K) + (1 mol)(188.72 J/mol•K)]

S∆ = 214.48 = 214.5 J/K

G∆ = H∆ – T S∆ = (206.2 kJ)(103 J/1 kJ) – (1273.K)(214.5 J/K)

G∆ = –6.6858x104 = –66.86 kJ CO(g) + H2O(g) CO2(g) + H2(g) water-gas shift reaction

H∆ = [(1 mol CO2)( fH∆ of CO2) + (1 mol H2)( fH∆ of H2)]

– [(1 mol CO)( fH∆ of CH4) + (1 mol H2O)( fH∆ of H2O)]

H∆ = [(1 mol)(–393.5 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)(–110.5 kJ/mol) + (1 mol)(–241.826 kJ/mol)]

H∆ = –41.174 = –41.2 kJ

S∆ = [(1 mol CO2)( S of CO2) + (1 mol H2)( S of H2)] – [(1 mol CO)( S of CO) + (1 mol H2O)( S of H2O)]

S∆ = [(1 mol)(213.7 J/mol•K) + (1 mol)(130.6 J/mol•K)] – [(1 mol)(197.5 J/mol•K) + (1 mol)(188.72 J/mol•K)]

S∆ = – 41.92 = –41.9 J/K

G∆ = H∆ – T S∆ = (–41.2 kJ)(103 J/1 kJ) – (1273 K)(–41.9 J/K)

G∆ = 1.21387x104 = 12.1 kJ b) 2CH4(g) + 3H2O(g) CO2(g) + CO(g) + 7H2(g)

Since this reaction is two steam re-forming reactions and one water-gas shift reaction, G∆ = 2( G∆ for the steam

Re-forming reaction) + G∆ for the water-gas shift reaction:

G∆ = 2(–66.86 kJ) + 12.1 kJ = –121.62 = –121.6 kJ

Since G∆ is negative, the reaction is spontaneous at this temperature. c) At 98 atm and 50% conversion, the partial pressures of the reactants and products are as follows (from the ratio of the coefficients: CH4, 14 atm; H2O, 21 atm; CO2, 7.0 atm; CO, 7.0 atm; H2, 49 atm. Calculate Q:

Q = ( )( )( )

( ) ( )

72 2

2 34 2

CO CO H

CH H O =

( )( )( )( ) ( )

7

2 3

7.0 7.0 49

14 21 = 1.8309x107

20-42

∆G = ∆G° + RT ln Q = –121.6 kJ/mol + (1 kJ/103 J) (8.314 J/mol•K)(1273 K) ln (1.8309x107) = 55.391 = 55.4 kJ/mol

The reaction is not spontaneous at this point. d) At 98 atm and 90% conversion, the partial pressures of the reactants and products are as follows: CH4, 2.3 atm; H2O, 3.4 atm; CO2, 10.3 atm; CO, 10.3 atm; H2, 71.8 atm. Calculate Q:

Q = ( )( )( )

( ) ( )

72 2

2 34 2

CO CO H

CH H O =

( )( )( )( ) ( )

7

2 3

10.3 10.3 71.8

2.3 3.4 = 5.019x1012

∆G = ∆G° + RT ln Q = –121.6 kJ/mol + (1 kJ/103 J)(8.314 J/mol•K)(1273 K) ln (5.019x1012) = 187.91 = 188 kJ/mol

The reaction is not spontaneous at this point.

CHAPTER 20 THERMODYNAMICS: ENTROPY, FREE ...

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