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
---------------
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
----------------
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)
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