The reaction, $C O (g) + 3 H_{2} (g) ⇌ C H_{4} (g) + H_{2} O (g)$ is at equilibrium at $1300 K$ in a one litre flask. The gaseous equilibrium mixture contains $0.30 m o l$ of $C O, 0.10 m o l$ of $H_{2}, 0.020 m o l$ 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 $3.90$ .

5.85 \times 10^{- 2} M

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5.85 \times 10^{- 3} M

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5.85 \times 10^{- 7} M

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5.85 \times 10^{- 1} M

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Solution

The correct option is A $5.85 \times 10^{- 2} M$
Let the concentration of $C H_{4}$ at equilibrium be $x M$ .
$C O_{(g)} + 3 H_{2 (g)} ⇋ C H_{4 (g)} + H_{2} O_{(g)}$

At equilibrium,
$[C O] = \frac{0.3}{1} = 0.3 M; [H_{2}] = \frac{0.1}{1} = 0.1 M; [C H_{4}] = x M; [H_{2} O] = \frac{0.02}{1} = 0.02 M$

We know,
$\frac{[C H_{4}] [H_{2} O]}{[C O] {[H_{2}]}_{2}^{3}} = K_{c}$
$∴ \frac{x \times 0.02}{0.3 \times (0.1)^{3}} = 3.90$
$\Rightarrow x = \frac{3.90 \times 0.3 \times (0.1)^{3}}{0.02}$
$= \frac{0.00117}{0.02}$
$= 0.0585 M$
$\Rightarrow x = 5.85 \times 10^{- 2} M$

Hence, the concentration of $C H_{4}$ at equilibrium is $5.85 \times 10^{- 2} M$ .

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