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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium3 Marks
Question
The reaction, $\text{CO (g) + 3H}_2\text{ (g)}\rightleftharpoons\text{CH}_4\text{ (g) + }\text{H}_2\text{O (g)}$ is at equilibrium at 1300K in a 1L flask. It also contain 0.30 mol of $\mathrm{CO}, 0.10 \mathrm{~molof} \mathrm{~H}_2$ and 0.02 mol of $\mathrm{H}_2 \mathrm{O}$ and an unknown amount of $\mathrm{CH}_4$ in the flask. Determine the concentration of $\mathrm{CH}_4$ in the mixture. The equilibrium constant, $\mathrm{K}_{\mathrm{c}}$ for the reaction at the given temperature is 3.90 .
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Answer

Let the concentration of methane at equilibrium be x.
  $\text{CO}_\text{(g)}$ $+$ $\text{3H}_{2\text{(g)}}$ $\leftrightarrow$ $\text{CH}_{4\text{(g)}}$ $+$ $\text{H}_2\text{O}_\text{(g)}$
At equilibrium $\frac{0.3}{1}=0.3\text{M}$   $\frac{0.1}{1}=0.1\text{M}$   $\text{x}$   $\frac{0.02}{1}=0.02\text{M}$
It is given that $\text{K}_\text{c}=3.90.$
Therefore,
$\frac{[\text{CH}_{4\text{(g)}}][\text{H}_2\text{O}_\text{(g)}]}{[\text{CO}_\text{(g)}][\text{H}_{2\text{(g)}}]}=\text{K}_\text{c}$
$\Rightarrow\frac{\text{x}\times0.02}{0.3\times(0.1)^3}=3.90$
$\Rightarrow\text{x}=\frac{3.90\times0.3\times(0.1)^3}{0.02}$
$=\frac{0.00117}{0.02}$
$=0.0585\text{M}$
$=5.85\times10^{-2}\text{M}$
Hence, the concentration of $CH_4$ at equilibrium is $5.85\times10^{-2}\text{M}$

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