Form the given data of equilibrium constants of the following reactions:
`CuO(s)+H_(2)(g)hArrCu(s)+H_(2)O(g),K=67`
`CuO(s)+CO(g)hArrCu(s)+CO_(2)(g),K=490`
Calculate the equilibrium constant of the reaction,
`CO_(2)(g)+H_(2)(g)hArrCu(s)+CO_(2)(g)+H_(2)O(g)`

Explanation

`Cu(s)+H_(2)(g)(g)hArrCuO(s)+H_(2)O(g),K_(1)=67`
Now reversing the second reaction,
`Cu(s)+CO_(2)(g)hArrCuO(s)+CO(g),`
`K_(2)=1/490`
Adding the two reactions, we get,
`Co_(2)(g)+H_(2)hArrCO(g)+H_(2)O(g)`
for which `K=K_(1).K_(2)=67xx1/490=0.137`