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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 14 of 73
Marks: +1, -0
One mole of H2O\mathrm{H_2O} and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O(g)+CO(g)\mathrm{H_2O_{(g)}} + \mathrm{CO_{(g)}} ⇌ H2(g)+CO2(g)\mathrm{H_{2(g)}} + \mathrm{CO_{2(g)}}. Calculate the equilibrium constant for the reaction.
Solution:  
From the above reaction,
H2O(g)+CO(g)\mathrm{H_2O_{(g)}} + \mathrm{CO_{(g)}} ⇌ H2(g)+CO2(g)\mathrm{H_{2(g)}} + \mathrm{CO_{2(g)}}
Initial no. of moles1100At equilibrium0.60.60.40.4\begin{array}{lcccc} \text{Initial no. of moles} & 1 & 1 & 0 & 0 \\ \text{At equilibrium} & 0.6 & 0.6 & 0.4 & 0.4 \end{array}
∴ [H2O][\mathrm{H_2O}] = 0.610 mol L−1\frac{0.6}{10} \text{ mol L}^{-1} = 0.06 mol L−1\text{mol L}^{-1} ; [CO] = 0.610 mol L−1\frac{0.6}{10} \text{ mol L}^{-1} = 0.06 mol L−1\text{mol L}^{-1};
[H2][\mathrm{H_2}] = 0.410\frac{0.4}{10} = 0.04 mol L−1\text{mol L}^{-1} and [CO2][\mathrm{CO_2}] = 0.410\frac{0.4}{10} = 0.04 mol L−1\text{mol L}^{-1}
KcKc = [H2][CO2][H2O][CO]\frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} = 0.410×0.4100.610×0.610\frac{\frac{0.4}{10} \times \frac{0.4}{10}}{\frac{0.6}{10} \times \frac{0.6}{10}} = 0.44
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