The reaction , CO(g) + 3 H_(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 is 3*90.

The reaction , CO(g) + 3 H_(2) (g) hArr CH_(4) (g) + H_(2)O (g),is at

1.	The reaction , CO(g) + 3 H_(2) (g) hArr CH_(4) (g) + H_(2)O (g),is at equilibrium at 1300 K in a 1 L flask. It also contains 030 mol of CO, 010 mol of H_(2) and 002 " 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 390.
Answer» Solution :` K_(c) = ([ CH_(4)][H_(2)O])/([CO][H_(2)]^(3))` ` :. 3* 90 = ([CH_(4)][H_(2)O])/([CO][H_(2)]^(3)) "" ("Molar conc = No. of moles because volume of flask = 1 L")` `[CH_(4)] = 00585 M = 585 xx 10^(-2) M`