# 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.

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 $=5.85\times 10^{-2}$

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