The reaction `CO_(g) + 3H_(2)(g) hArr CH_(4)(g) + H_(2)O(g)`
is at equilibrium at 1300 K in a 1L flask. It also contain 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 is 3.90.

Explanation

` " "K_c=([CH_4(g) ][H_2O (g)])/([CO(g)][H_2(g)]^(3))`
` therefore " " 3.90= ([CH_4(g)]xx0.02)/(0.30xx (0.10)^(3))`
` or " "[CH_4(g)]=(3.90 xx 0.30 xx (0.10 )^(3) )/(0.02 ) = 0.0585`
` " " = 5.85 xx 10 ^(-2) " mol " L^(-1)`