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

Number of moles of water originally present =1 mol
percentage of water reacted =40 %
Number of moles of water reacted `=(1xx40)/(100) =0.4 mol`
number of moles of water left `=(1 -0.4) = 0.6 mol`
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus , the molar conc.`//`litre per litre of the reactants and products before the reaction and the equilibrium point are as follows:
`{:(,H_(2)O(g)+,CO(g)hArr,H_(2)(g)+,CO_(2)(g)),("Initial moles"//"litre",(1)/(10),(1)/(10),0,0),("Moles"//"litre at equilibrium point",(1-0.4)/(10)=(0.6)/(10)(1-0.4)/(10)=,(0.6)/(10),(0.4)/(10),(0.4)/(10)):}`
Applying law of chemical equilibrium
Equilibrium constant `(K_(c)) =[[H_(2)(g)][CO_(2)(g)]]/[[H_(2)O(g)][CO(g)]]=(((0.4)/(10)mol L^(-1))xx((0.4)/(10) mol L^(-1)))/(((0.6)/(10) mol L^(-1))xx((0.6)/(10) mol L^(-1)))= (0.16)/(0.36) =0.44`