One mole of `H_(2)O` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation,
`H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))`
Calculate the equilibrium constant for the reaction.

Explanation

` [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*04 xx0*04 )/(0*06 xx 0*06)= 0*444 .`