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

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(A) Initial moles: $NH_3 = 1$,$N_2 = 0$,$H_2 = 0$At equilibrium: $NH_3 = (1-\alpha)$,$N_2 = \frac{\alpha}{2}$,$H_2 = \frac{3\alpha}{2}$Total moles at

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

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(A) Initial moles: $NH_3 = 1$,$N_2 = 0$,$H_2 = 0$At equilibrium: $NH_3 = (1-\alpha)$,$N_2 = \frac{\alpha}{2}$,$H_2 = \frac{3\alpha}{2}$Total moles at equilibrium: $n_{total} = (1-\alpha) + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$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)(1+\alpha)} P^o = \frac{\alpha^{1/2} \cdot 3^{3/2} \cdot \alpha^{3/2}}{2^{1/2} \cdot 2^{3/2} (1-\alpha^2)} P^o = \frac{3\sqrt{3} \alpha^2}{4(1-\alpha^2)} P^o$Rearranging for $\alpha$: $\frac{4 K_P}{3\sqrt{3} P^o} = \frac{\alpha^2}{1-\alpha^2} \implies \frac{3\sqrt{3} P^o}{4 K_P} = \frac{1-\alpha^2}{\alpha^2} = \frac{1}{\alpha^2} - 1$$\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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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: