Part A:
ΔH∘rxn= 121 kJ ; ΔS∘rxn=− 246 J/K ; T= 291 K
Express your answer as an integer.
Part B:
ΔH∘rxn=− 121 kJ ; ΔS∘rxn= 246 J/K ; T= 291 K
Express your answer as an integer.
Part C:
ΔH∘rxn=− 121 kJ ; ΔS∘rxn=− 246 J/K ; T= 291 K
Express your answer as an integer.
Part D:
ΔH∘rxn=− 121 kJ ; ΔS∘rxn=− 246 J/K ; T= 600 K
Express your answer as an integer.
Part E:
Predict whether or not the reaction in part A will be spontaneous at the temperature indicated.
Part F:
Predict whether or not the reaction in part B will be spontaneous at the temperature indicated.
Part G:
Predict whether or not the reaction in part C will be spontaneous at the temperature indicated.
Part H:
Predict whether or not the reaction in part D will be spontaneous at the temperature indicated.
A)
ΔHo = 121 KJ
ΔSo = -246 J/K
= -0.246 KJ/K
T = 291 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = 121.0 - 291.0 * -0.246
ΔGo = 192.586 KJ
Answer: 193 KJ
B)
ΔHo = -121 KJ
ΔSo = 246 J/K
= 0.246 KJ/K
T = 291 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = -121.0 - 291.0 * 0.246
ΔGo = -192.586 KJ
Answer: -193 KJ
C)
ΔHo = -121 KJ
ΔSo = -246 J/K
= -0.246 KJ/K
T = 291 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = -121.0 - 291.0 * -0.246
ΔGo = -49.414 KJ
Answer: -49 KJ
D)
ΔHo = -121 KJ
ΔSo = -246 J/K
= -0.246 KJ/K
T = 600 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = -121.0 - 600.0 * -0.246
ΔGo = 26.6 KJ
Answer: 27 KJ
E)
Since ΔGo is positive, the reaction would be non spontaneous
Answer: non spontaneous
F)
Since ΔGo is negative, the reaction would be spontaneous
Answer: spontaneous
G)
Since ΔGo is negative, the reaction would be spontaneous
Answer: spontaneous
H)
Since ΔGo is positive, the reaction would be non spontaneous
Answer: non spontaneous
Part A: ΔH∘rxn= 121 kJ ; ΔS∘rxn=− 246 J/K ; T= 291 K Express your answer...
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