Question

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`.

Answer 1

`CO(g) +3H_(2)(g)hArr CH_(4)(g) +H_(2)O(g)`
According to available data,
`K_(c) =[[CH_(4)]xx[H_(2)O]]/[[CO]xx[H_(2)]^(3)] " or " 3.90 = [[CH_(4)][0.02]]/[[0.30]xx [0.1]^(3)]`
`[CH_(4)] = ((3.9) xx (0.30) xx(0.001))/((0.02)) =5.85 xx 10^(-2) M`