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

Explanation

`K_(c )=([CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`
`:. 3.90=([CH_(4)][0.02])/((0.30)(0.10)^(3))`
(Molar concentration =Number of moles because volume of flask `=1 L`)
or `[CH_(4)]=0.0585 M=5.89xx10^(-2) M`