A) 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g), ΔH∘=−844 kJ, ΔS∘=−165 J/K Express the Gibbs free energy in kilojoules to three significant digits.
B)
2POCl3(g)→2PCl3(g)+O2(g), ΔH∘=572 kJ, ΔS∘=179 J/K
Express the Gibbs free energy in kilojoules to three significant digits.
C)
At what temperature (if any) would the decomposition of POCl3 become spontaneous?
Express the temperature in kelvins to three significant digits. If there is no answer, enter none.
A)
ΔHo = -844.0 KJ/mol
ΔSo = -165 J/mol.K
= -0.165 KJ/mol.K
T = 298 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = -844.0 - 298.0 * -0.165
ΔGo = -794.83 KJ/mol
Answer: -795 KJ/mol
B)
ΔHo = 572.0 KJ/mol
ΔSo = 179 J/mol.K
= 0.179 KJ/mol.K
T = 298 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = 572.0 - 298.0 * 0.179
ΔGo = 518.658 KJ/mol
Answer: 519 KJ/mol
C)
ΔHo = 572.0 KJ/mol
ΔSo = 179 J/mol.K
= 0.179 KJ/mol.K
use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
572.0- T *0.179 < 0
T *0.179 > 572.0
T > 3195.5307 K
Answer: 3.20*10^3 K
A) 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g), ΔH∘=−844 kJ, ΔS∘=−165 J/K Express the Gibbs free energy in kilojoules to three significant...
For the values give for ΔH and ΔS, calculate ΔG foreach for each of the following reactions at 298 K. If the reactionis not spontaneous under standard conditions at 298 K, at whattemperature (if any) would the reaction become spontaneous? a) 2PbS(s)+3O2(g) ---> 2PbO(s)+2SO2(g) ΔH= -844kj ; ΔS = -165 J/K b)2POCl3(g)--->2PCl3(g)+O2(g) ΔH= 572kJ ; ΔS= 179 J/K
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