Question

Consider the reaction,
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
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 $P_{N{H}_3}<<P_{total}$ at equilibrium)

• A. $\frac{3^{\frac{3}{2}}{K}_p^{\frac{1}{2}}{P}^2}{4}$
• B. $\frac{3^{\frac{3}{2}}{K}_p^{\frac{1}{2}}{P}^2}{16}$
• C. $\frac{K_p^{\frac{1}{2}}{P}^2}{16}$
• D. $\frac{K_p^{\frac{1}{2}}{P}^2}{4}$

Answer 1

The correct option is B $\frac{3^{\frac{3}{2}}{K}_p^{\frac{1}{2}}{P}^2}{16}$

$\begin{array}{cccccc}& 2N{H}_3& \rightleftharpoons & N_2& +& 3H_2\\ t=0& P_1& & 0& & 0\\ t=eq& P_{N{H}_3}& & \frac{P_1}{2}& & \frac{3P_1}{2}\end{array}$

$P=P_{total}=\frac{P_1}{2}+\frac{3P_1}{2}=2P_1$

$\therefore P_1=\frac{P}{2}$

$k_{eq}=\frac{\left(P_{N_2}\right)\left(P_{H_2}\right)}{(P_{N{H}_3}{)}^2}\Rightarrow \frac{1}{k_p}=\frac{\left(\frac{P_1}{2}\right){\left(\frac{3P_1}{2}\right)}^3}{(P_{N{H}_3}{)}^2}$

$\therefore \frac{(P_{N{H}_3}{)}^2}{K_P}=\frac{P^4}{4^4}\times 3^3$

$\therefore P_{N{H}_3}^2=K_p\frac{P^4}{4^4}\times 3^3$

$P_{N{H}_3}={\left(K_p\frac{P^4}{4^4}{3}^3\right)}^{\frac{1}{2}}=\frac{(K_p{)}^{\frac{1}{2}}{3}^{\frac{3}{2}}{P}^2}{16}$