Chemical And Ionic Equilibrium

Chemical And Ionic Equilibrium Result Question 54 — JEE PYQ Chapterwise

54. One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If $0.25$ per cent of nitrogen is converted to ammonia by the following reaction

$ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) \text {, then } $

calculate the equilibrium constant, $K_c$ in concentration units. What will be the value of $K_c$ for the following equilibrium?

$ \frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightleftharpoons \mathrm{NH}_3(g) $

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Solution:

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$ \left[\mathrm{N}_2\right]=\frac{0.75}{4},\left[\mathrm{H}_2\right]=\frac{2.25}{4},\left[\mathrm{NH}_3\right]=\frac{0.50}{4} $

$ \begin{aligned} &\begin{aligned} K_c & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{(0.50)^2}{(0.75)(2.25)^3} \times 16 \\ & =0.468 \mathrm{~L}^2 \mathrm{~mol}^{-2} \end{aligned}\\ &\text { Also for : }\\ &\begin{array}{r} \frac{1}{2} \mathrm{~N}_2+\frac{3}{2} \mathrm{H}_2 \rightleftharpoons \mathrm{NH}_3 \\ K_c^{\prime}=\sqrt{K_c}=0.68 \end{array} \end{aligned} $