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Question
Calculate the equilibrium constant for the reaction H2(g)+CO2(g)⇌H2O(g)+CO(g) at 1395 K, if the equilibrium constants at 1395 K for following are:
2H2O(g)⇌2H2+O2(g); K1=2.1×10−13
2CO2(g)⇌2CO(g)+O2(g); K2=1.4×10−12.
2H2O(g)⇌2H2+O2(g); K1=2.1×10−13
2CO2(g)⇌2CO(g)+O2(g); K2=1.4×10−12.
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Solution
For 2H2O(g)⇌2H2(g)+O2(g); K1=[H2]2[O2][H2O]2....(1)
For 2CO2(g)⇌2CO(g)+O2(g) K2=[CO]2[O2][CO2]2...(2)
For CO2(g)+H2(g)⇌H2O(g)+CO(g) K=[H2O][CO][CO2][H2]...(3)
By dividing eq. (2) by eq. (1) :
K2K1=[CO]2[O2][CO2]2×[H2O]2[H2]2[O2]
K2K1=[CO]2[H2O]2[CO2]2[H2]2=K2 by Eq. (3)
or K=√(K2K1)=√(1.4×10−122.1×10−13)=2.58.
For 2CO2(g)⇌2CO(g)+O2(g) K2=[CO]2[O2][CO2]2...(2)
For CO2(g)+H2(g)⇌H2O(g)+CO(g) K=[H2O][CO][CO2][H2]...(3)
By dividing eq. (2) by eq. (1) :
K2K1=[CO]2[O2][CO2]2×[H2O]2[H2]2[O2]
K2K1=[CO]2[H2O]2[CO2]2[H2]2=K2 by Eq. (3)
or K=√(K2K1)=√(1.4×10−122.1×10−13)=2.58.
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