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Question
The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below:
CO(g)+H2O(g)⇌CO2(g)+H2(g)
ΔHo300K=−41.16kJmol−1
ΔSo300K=−4.24×10−2kJmol−1
ΔHo1200K=−32.93kJmol−1
ΔSo1200K=−2.96×10−2kJmol−1
Calculate KP at each temperature and predict the direction of reaction at 300K and 1200K, when PCO=PCO2=PH2=PH2O=1 atm at initial state.
CO(g)+H2O(g)⇌CO2(g)+H2(g)
ΔHo300K=−41.16kJmol−1
ΔSo300K=−4.24×10−2kJmol−1
ΔHo1200K=−32.93kJmol−1
ΔSo1200K=−2.96×10−2kJmol−1
Calculate KP at each temperature and predict the direction of reaction at 300K and 1200K, when PCO=PCO2=PH2=PH2O=1 atm at initial state.
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Solution
At 300K
ΔGo=ΔHo−TΔSo
=41.16−300×(−4.24×10−2)=−28.44kJ
since, ΔGo is negative hence reaction is spontaneous in forward direction.
ΔGo=−2.303RTlogKP
−28.44=−2.303×8.314×10−3×300logKP
KP=893×104
At 1200K
ΔGo=ΔHo−TΔSo
2.59=−2.303×8.314×10−3×1200logKP
KP=0.77
ΔGo=ΔHo−TΔSo
=41.16−300×(−4.24×10−2)=−28.44kJ
since, ΔGo is negative hence reaction is spontaneous in forward direction.
ΔGo=−2.303RTlogKP
−28.44=−2.303×8.314×10−3×300logKP
KP=893×104
At 1200K
ΔGo=ΔHo−TΔSo
2.59=−2.303×8.314×10−3×1200logKP
KP=0.77
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