The reaction `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)` is at equilibrium at 1300K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of `H_(2)` and 0.02 mol of `H_2O` and an unknown amount of `CH_(4)` in the flask. Determine the concentration on `CH_(4)` in the mixture. the equilibrium constant, `K_(c)` for the reaction at the given temperature is 3.90.

Explanation

Let the concentration of `CH_(4)` in the equilibrium mixture `=x" "mol*L^(-1)`
`K_(c)=([CH_(4)][H_(2)O])/([CO][H_(2)]^(3))=(x xx 0.02)/(0.30xx(0.1)^(3))=3.9`
`therefore x=0.0585`
So, in the equilibrium mixture, `[CH_(4)]=0.0585 mol*L^(-1)`