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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 34 of 73
Marks: +1, -0
The reaction, CO(g)+3H2(g)\mathrm{CO}_{(g)} + 3\mathrm{H}_{2(g)} ⇌ CH4(g)+H2O(g)\mathrm{CH}_{4(g)} + \mathrm{H_2O}_{(g)} is at equilibrium at 1300 K in a 1 L flask. It also contains 0.30 mol of CO, 0.10 mol of H2\mathrm{H}_2 and 0.02 mol of H2O\mathrm{H_2O} and an unknown amount of CH4\mathrm{CH}_4 in the flask. Determine the concentration of CH4\mathrm{CH}_4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Solution:  
CO(g)+3H2(g)\mathrm{CO}_{(g)} + 3\mathrm{H}_{2(g)} ⇌ CH4(g)+H2O(g)\mathrm{CH}_{4(g)} + \mathrm{H_2O}_{(g)}
Applying law of chemical equilibrium,
KcK_c = [CH4][H2O][CO][H2]3\frac{[\mathrm{CH}_4][\mathrm{H_2O}]}{[\mathrm{CO}][\mathrm{H}_2]^3} ⇒ 3.90 = [CH4]×0.020.30×(0.1)3\frac{[\mathrm{CH}_4] \times 0.02}{0.30 \times (0.1)^3}
[CH4][\mathrm{CH}_4] = 3.9×0.3×0.1×0.1×0.10.02\frac{3.9 \times 0.3 \times 0.1 \times 0.1 \times 0.1}{0.02} = 39×32×10−3\frac{39 \times 3}{2} \times 10^{-3} = 58.5 × 10−310^{-3} = 5.85 × 10−2 mol / L10^{-2} \text{ mol / L}
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