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Standard XII

Chemistry

Derivation of Kp and Kc

The reaction,...

Question

The reaction,
$C O (g) + 3 H_{2} (g) ⇌ C H_{4} (g) + H_{2} O (g)$
is in 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 $H_{2}$ and $0.020 m o l$ 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.

4.85 \times 10^{- 3} M

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5.85 \times 10^{- 2} M

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3.85 \times 10^{- 2} M

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7.85 \times 10^{- 3} M

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Solution

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

$C o n c e n t r a t i o n = \frac{M o l e s}{v o l u m e}$
Hence, concentration at equilibrium: $C O = \frac{0.3}{1} = 0.3 M H_{2} = \frac{0.1}{1} = 0.1 M H_{2} O = \frac{0.02}{1} = 0.02 M$
Let concentration of $C H_{4}$ be $x M$
It is given that $K_{c} = 3.90$

$∴$ by definition

$\frac{[C H_{4 (g)}] [H_{2} O_{(g)}]}{[C O_{(g)}] {[H_{2 (g)}]}_{2 (g)}^{3}} = K_{c}$
$\Rightarrow \frac{x \times 0.02}{0.3 \times (0.1)^{3}} = 3.90$
$\Rightarrow x = \frac{3.90 \times 0.3 \times (0.1)^{3}}{0.02}$
$= \frac{0.00117}{0.02} = 0.0585 M$

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

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Similar questions

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C O (g) + 3 H_{2} (g) ⇌ C H_{4} (g) + H_{2} O (g)

is at equilibrium at

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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.
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Derivation of Kp and Kc

Standard XII Chemistry

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