One mole of `H_(2)O` and one mole of CO are taken in `10 L`vessel and heated to `725 K`. At equilibrium `40%` of water (by mass) reacts with `CO` according to the equation.
`H_(2)O(g) + CO(g) hArr H_(2)(g) + CO_(2)(g)`
Calculate the equilibrium constant for the reaction.

Explanation

At equilibrium,
`[H_(2)O]=(1-0.40)/10 mol L^(-1)=0.06 mol L^(-1)`
`[CO]=0.06 mol L^(-1)`,
`[H_(2)]=0.4/10 mol L^(-1)=0.04 mol L^(-1)`
`[CO_(2)]=0.04 mol L^(-1)`
`K=([H_(2)][CO_(2)])/([H_(2)O][CO])=(0.04xx0.04)/(0.06xx0.06)=0.444`