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

`CaO(s)+H_(2)(g) Leftrightarrow Co(s)+H_(2)O(g)" "K_(1)=67…….(1)`
Now reversing the second reaction,
`Co(s)+CO_2(g) Leftrightarrow CoO(s)+CO(g)" "K_2=1/(490)….(2)` (Eqn. (5))
Adding the two reactions, we get
`CO_(2)(g)+H_(2)(g) Leftrightarrow Co(g)+H_(2)O(g)`
for which `K=K_1.K_2 =67 xx (1)/(490) =0.137`