Chemical and Ionic Equilibrium 1 Question 7
7. Consider the reaction,
$$ \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$
The equilibrium constant of the above reaction is $K_{p}$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $p_{\mathrm{NH}{3}} \ll<p{\text {total }}$ at equilibrium)
(2019 Main, 11 Jan I)
(a) $\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{4}$
(b) $\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{16}$
(c) $\frac{K_{p}^{1 / 2} P^{2}}{16}$
(d) $\frac{K_{p}{ }^{1 / 2} P^{2}}{4}$
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Solution:
$$ \begin{aligned} & \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}{3}(g) \ & \text { At equilibrium: } p{\mathrm{N}{2}}=P, \quad p{\mathrm{H}{2}}=3 P, \quad p{\mathrm{NH}{3}}=2 P \ & \Rightarrow \quad p{\text {(total) }}=p_{\mathrm{N}{2}}+p{\mathrm{H}{2}}+p{\mathrm{NH}{3}} \simeq p{\mathrm{N}{2}}+p{\mathrm{H}{2}} \ & =p+3 p=4 p \ & {\left[\because P{\text {(total) }} \gg p_{\mathrm{NH}{3}}\right]} \ & \text { Now, } \quad K{p}=\frac{p^{2} \mathrm{NH}{3}}{p{\mathrm{N}{2}} \times p{\mathrm{H}{2}}^{3}}=\frac{p{\mathrm{NH}{3}}^{2}}{p \times(3 p)^{3}} \ & =\frac{p{\mathrm{NH}{3}}^{2}}{27 \times p^{4}}=\frac{p{\mathrm{NH}{3}}^{2}}{27 \times\left(\frac{P}{4}\right)^{4}} \ & K{p}=\frac{p_{\mathrm{NH}{3}}^{2} \times 4^{4}}{3^{2} \times 3 \times P^{4}} \ & \Rightarrow \quad p{\mathrm{NH}{3}}^{2}=\frac{3^{2} \times 3 \times P^{4} \times K{p}}{4^{4}} \ & \Rightarrow \quad p_{\mathrm{NH}{3}}=\frac{3 \times 3^{1 / 2} \times P^{2} \times K{p}^{1 / 2}}{4^{2}}=\frac{3^{3 / 2} \times P^{2} \times K_{p}^{1 / 2}}{16} \end{aligned} $$
