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Question
Calculate the equation constant for the reaction: H2(g)+CO2(g)⇌H2O(g)+CO at 1395 K, if the equilibrium constants at 1395 K for the following are
2H2O(g)⇌2H2(g)+O2(g) K1= 2.1 × 10−13
2CO2(g)⇌2CO(g)+O2(g) K2= 1.4 × 10−12
2H2O(g)⇌2H2(g)+O2(g) K1= 2.1 × 10−13
2CO2(g)⇌2CO(g)+O2(g) K2= 1.4 × 10−12
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Solution
2CO2(g)⇌2CO(g)+O2(g);K2=1.4×10−12ii⟵CO2(g)⇌CO(g)+12O2(g);K12=(1.4×10−12)122H2O(g)⇌2H2(g)+O2(g);K1=2.1×10−13i⟵H2(g)+12O2(g)⇌H2O(g);K11=+(2.1×10−13)12H2(g)⇌H2(g);K
K=K11×K21=+(2.1×10−13)12×(1.4×10−12)12
K=1.71×10−12Therefore, the equilibrium canstant for the given reaction is 1.71×10−12.
ii⟵CO2(g)⇌CO(g)+12O2(g);K12=(1.4×10−12)12
2H2O(g)⇌2H2(g)+O2(g);K1=2.1×10−13
i⟵H2(g)+12O2(g)⇌H2O(g);K11=+(2.1×10−13)12
H2(g)⇌H2(g);K
K=K11×K21=+(2.1×10−13)12×(1.4×10−12)12
K=1.71×10−12
Therefore, the equilibrium canstant for the given reaction is 1.71×10−12.
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