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Question

The equilibrium constant for the folowing reactions have been measured at 823.
CoO(s)+H2(g)Co(s)+H2O(g);K=67
CoO(s)+CO(g)Co(s)+CO2(g);K=490
The value of equilibrium constant of the reaction following reaction is
2CO2(g)+2H2(g)2CO(g)+2H2O(g)

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Solution

2nd reaction is reversed so K12=1K2
CoO(s)+H2(g)Co(s)+H2O(g)K1=67
CO2(g)+Co(s)CoO(s)+CO(g)K12=1490
CO2+H2CO+H2OK=K1×K12
=67490=0.136
as reaction is doubled, K is squared K1=K2=0.0186

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