Consider the reaction
\(\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}\)
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 \(P_{NH_3} \ll P_{total}\) at equilibrium)
\(\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}\)
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 \(P_{NH_3} \ll P_{total}\) at equilibrium)
Show Hint
- \(\frac{3^{3/2}K_P^{1/2}P^2}{4}\)
- \(\frac{K_P^{1/2}P^2}{4}\)
- \(\frac{3^{3/2}K_P^{1/2}P^2}{16}\)
- \(\frac{K_P^{1/2}P^2}{16}\)
The Correct Option is C
Solution and Explanation
Because \(P_{NH_3} \ll P\), the total pressure is essentially that of N2 and H2: \(P_{N_2} = \frac{P}{4}\), \(P_{H_2} = \frac{3P}{4}\).
For the given (formation) reaction: \(K_p = \frac{P_{NH_3}^2}{P_{N_2}P_{H_2}^3}\).
\(P_{NH_3}^2 = K_p\cdot\frac{P}{4}\cdot\left(\frac{3P}{4}\right)^3 = K_p\cdot\frac{27P^4}{256}\).
\(P_{NH_3} = K_p^{1/2}\,P^2\,\frac{3\sqrt{3}}{16} = \frac{3^{3/2}K_p^{1/2}P^2}{16}\).
Answer: (3) \(\frac{3^{3/2}K_P^{1/2}P^2}{16}\)
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