At a given temperature, Ke is 4 for the reaction H_(2(g)) + CO_(2(g)) hArr H_(2)O_((g)) + CO_((g)). Initially 0.6 moles each of H_(2) and CO_(2) are taken in 1lit flask. The equilibrium concentration of H_(2) O_((g)) is

At a given temperature, Ke is 4 for the reaction H_(2(g)) + CO_(2(g))

1.	At a given temperature, Ke is 4 for the reaction H_(2(g)) + CO_(2(g)) hArr H_(2)O_((g)) + CO_((g)). Initially 0.6 moles each of H_(2) and CO_(2) are taken in 1lit flask. The equilibrium concentration of H_(2) O_((g)) is
Answer» 0.4M 0.46M 0.2M 0.8M Solution : `K_(c)=(X^(2))/((0.6-x)^(2))=4 implies (x)/(0.6-x)=2` x=1.2-3x, 3x=1.2, `x=0.4=[H_(2)O]`