Chemical and Ionic Equilibrium - Result Question 63
####7. Consider the reaction,
$$ N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) $$
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 $p _{NH _3} \ll<p _{\text {total }}$ 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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Answer:
Correct Answer: 7. (d)
Solution:
$$ \begin{aligned} & N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) \ & \text { At equilibrium: } p _{N _2}=P, \quad p _{H _2}=3 P, \quad p _{NH _3}=2 P \ & \Rightarrow \quad p _{\text {(total) }}=p _{N _2}+p _{H _2}+p _{NH _3} \simeq p _{N _2}+p _{H _2} \ & =p+3 p=4 p \ & {\left[\because P _{\text {(total) }} \gg p _{NH _3}\right]} \ & \text { Now, } \quad K _p=\frac{p^{2} NH _3}{p _{N _2} \times p _{H _2}^{3}}=\frac{p _{NH _3}^{2}}{p \times(3 p)^{3}} \ & =\frac{p _{NH _3}^{2}}{27 \times p^{4}}=\frac{p _{NH _3}^{2}}{27 \times\left(\frac{P}{4}\right)^{4}} \ & K _p=\frac{p _{NH _3}^{2} \times 4^{4}}{3^{2} \times 3 \times P^{4}} \ & \Rightarrow \quad p _{NH _3}^{2}=\frac{3^{2} \times 3 \times P^{4} \times K _p}{4^{4}} \ & \Rightarrow \quad p _{NH _3}=\frac{3 \times 3^{1 / 2} \times P^{2} \times K _p^{1 / 2}}{4^{2}}=\frac{3^{3 / 2} \times P^{2} \times K _p^{1 / 2}}{16} \end{aligned} $$
