7.34     The reaction,

                  CO(g)+3H_{2}(g)\rightleftharpoons 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_2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Answers (1)
M manish

Given that,
Total volume = 1L
0.3 mol of CO, 0.10 mol of dihydrogen(H_2)and the 0.02 mol of water(H_2O)
the equilibrium constant = 3.90

Let x be the concentration of methane at equilibrium. The given reaction is-

                                        CO(g)+3H_{2}(g)\rightleftharpoons CH_{4}(g)+H_{2}O(g)
At equilibrium,                    0.3 mol/L         0.1 mol/L        x           0.02 mol/L

Therefore,

K_c = \frac{0.02\times x}{(0.3)(0.1)}=3.90
        \Rightarrow x = \frac{(3.9)(0.3)(0.1)^3}{(0.02)}
                   =5.85\times 10^{-2}

Thus the concentration of methane at equilibrium is =5.85\times 10^{-2}

Exams
Articles
Questions