The reaction, $C O (g) + 3 H_{2} (g) \leftrightarrow C H_{4} (g) + H_{2} O (g)$ 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 $C H_{4}$ in the flask. Determine the concentration of $C H_{4}$ in the mixture.
The equilibrium constant, $K_{c}$ for the reaction at the given temperature is 3.90.

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Solution

Let the concentration of methane at equilibrium be x.
$\begin{matrix} C O_{(g)} & + & 3 H_{2 (g)} & \leftrightarrow & C H_{4 (g)} & + & H_{2} O_{(g)} A t e q u i l i b r i u m & \frac{0.3}{1} = 0.3 M & \frac{0.1}{1} = 0.1 M & X & \frac{0.02}{1} = 0.02 M \end{matrix}$
it is given that $K_{c} = 3.90$
Therefore,
$\frac{[C H_{4 (g)}] [H_{2} O_{(g)}]}{[C O_{(g)}] [H_{2 (g)}]^{3}} = K_{c} \Rightarrow \frac{x \times 0.02}{0.3 \times (0.1)^{3}} = 3.90 \Rightarrow x = \frac{3.90 \times 0.3 \times (0.1)^{3}}{0.02} \Rightarrow x = \frac{3.90 \times 0.3 \times (0.1)^{3}}{0.02} = \frac{0.00117}{0.02} = 0.0585 M = 5.85 \times 10^{- 2} M$
Hence, the concentration of $C H_{4}$ at equilibrium is $5.85 \times 10^{- 2} M$ .

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