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1) +4634.9 kJ/mol-rxn –371.2 kJ/mol-rxn –215.4 kJ/mol-rxn +371.2 kJ/mol-rxn –320

ID: 1006410 • Letter: 1

Question

1)

+4634.9 kJ/mol-rxn

–371.2 kJ/mol-rxn

–215.4 kJ/mol-rxn

+371.2 kJ/mol-rxn

–320.6 kJ/mol-rxn

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2)

QUESTION 15

What is ?rG° at 500.0 K for the following reaction?

Zn(s) + H2O(g) ? ZnO(s) + H2(g)

Substance

?fH°(kJ/mol-rxn) at 298 K

S° (J/K·mol-rxn) at 298 K

Zn(s)

0

41.6

H2O(g)

–241.8

188.7

ZnO(s)

–350.5

43.7

H2(g)

0

130.6

–80.7 kJ/mol-rxn

–92.0 kJ/mol-rxn

80.7 kJ/mol-rxn

–136.7 kJ/mol-rxn

92.0 kJ/mol-rxn

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3)

Balance the following half-reaction occurring in acidic solution.

NO3–(aq) ? NH4+aq)

NO3–(aq) + 10H+(aq) + 8e? ? NH4+(aq) + 3H2O(l)

NO3–(aq) + 10H+(aq) ? NH4+(aq) + 3H2O(l) + 10e?

NO3–(aq) + 3H2O(l) + 10e? ? NH4+(aq) + 10H+(aq)

NO3–(aq) + 8e? ? NH4+(aq) + 8H+(aq) + 3H2O(l)

NO3–(aq) + 10H+(aq)? NH4+(aq) + 3H2O(l)

+4634.9 kJ/mol-rxn

–371.2 kJ/mol-rxn

–215.4 kJ/mol-rxn

+371.2 kJ/mol-rxn

–320.6 kJ/mol-rxn

Explanation / Answer

1) Calculate rG° for the reaction at 25.0 °C

2 Na(s) + 2 H2O (l) 2 NaOH (aq) + H2 (g)

Given rH° = 366.6 kJ/mol-rxn

  rS° = 154.2 J/Kmol-rxn == 0.1542kJ/Kmol-rxn

T = 250C = 25 +273 = 298K

rG° = rH° T rS°

          = 366.6 kJ/mol-rxn – 298K ( 0.1542kJ/Kmol-rxn)

          = 366.6 kJ/mol-rxn + 45.95kJ/mol-rxn)

          = 320.65 kJ/mol-rxn

rG° for the reaction = 320.65 kJ/mol-rxn

2)

What is rG° at 500.0 K for the following reaction?

Zn(s) + H2O(g) ZnO(s) + H2(g)

Substance      fH°(kJ/mol-rxn) at 298 K         S° (J/K·mol-rxn) at 298 K

Zn(s)                                  0                                         41.6

H2O(g)                              –241.8                               188.7

ZnO(s)                               –350.5                               43.7

H2(g)                                 0                                         130.6

Given T = 500K

Step1. Calculate rH0

rH0 = –350.5    – (241.8) = 108.7kJ/mol-rxn

Step2. Calculate rS0

rS° = 43.7 +130.6 – 41.6 – 188.7

           = 55.4 J/K·mol-rxn

           = = 0.0554kJ/mol-rxn

rG° = rH° T rS°

        = 108.7 kJ/mol-rxn – 500K ( 0.0554kJ/Kmol-rxn)

          = 108.7 kJ/mol-rxn + 27.7kJ/mol-rxn)

           = 81 kJ/mol-rxn

   rG° for the reaction = 81 kJ/mol-rxn

3)

Balanced equation for the following half-reaction in acidic solution is

NO3–(aq) NH4+(aq)

NO3–(aq) + 10H+(aq) + 8e NH4+(aq) + 3H2O(l)