For the reaction: NH_{3(g)} rightleftharpoons | vedclass

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(A) For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alph

Vedclass · Accepted Answer

$\alpha = \left( 1 + \frac{3 \sqrt{3} P^o}{4 K_p} \right)^{ - \frac{1}{2}}$

Vedclass · Answer

(A) For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alpha), \frac{\alpha}{2}, \frac{3\alpha}{2}$Total moles at equilibrium: $(1 - \alpha) + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$Partial pressures: $p_{NH_3} = \frac{1-\alpha}{1+\alpha} P^o, p_{N_2} = \frac{\alpha/2}{1+\alpha} P^o, p_{H_2} = \frac{3\alpha/2}{1+\alpha} P^o$$K_p = \frac{(p_{N_2})^{1/2} (p_{H_2})^{3/2}}{p_{NH_3}} = \frac{(\frac{\alpha/2}{1+\alpha} P^o)^{1/2} (\frac{3\alpha/2}{1+\alpha} P^o)^{3/2}}{\frac{1-\alpha}{1+\alpha} P^o}$$K_p = \frac{(\alpha/2)^{1/2} (3\alpha/2)^{3/2}}{(1-\alpha)} \cdot \frac{(P^o)^{1/2+3/2-1}}{(1+\alpha)^{1/2+3/2-1}} = \frac{\alpha^{1/2} \cdot 3^{3/2} \cdot \alpha^{3/2}}{2^{1/2} \cdot 2^{3/2} \cdot (1-\alpha)} \cdot \frac{P^o}{1+\alpha}$$K_p = \frac{3 \sqrt{3} \alpha^2}{4(1-\alpha^2)} P^o$$\frac{1-\alpha^2}{\alpha^2} = \frac{3 \sqrt{3} P^o}{4 K_p} \implies \frac{1}{\alpha^2} - 1 = \frac{3 \sqrt{3} P^o}{4 K_p}$$\frac{1}{\alpha^2} = 1 + \frac{3 \sqrt{3} P^o}{4 K_p} \implies \alpha = \left( 1 + \frac{3 \sqrt{3} P^o}{4 K_p} \right)^{-1/2}$

For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_p$. The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as:

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