For the formation of ammonia from its constit | vedclass

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(C) The chemical equation for the formation of ammonia is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$Initial moles: $N_2 = 1$,$H_2 = 3$,$NH_3 = 0$

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$\frac{4x^2V^2}{27(1-x)^4}$

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(C) The chemical equation for the formation of ammonia is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$Initial moles: $N_2 = 1$,$H_2 = 3$,$NH_3 = 0$At equilibrium moles: $N_2 = (1-x)$,$H_2 = (3-3x)$,$NH_3 = 2x$Concentrations at equilibrium (in volume $V$):$[N_2] = \frac{1-x}{V}$$[H_2] = \frac{3(1-x)}{V}$$[NH_3] = \frac{2x}{V}$The equilibrium constant $K_C$ is given by:$K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}$$K_C = \frac{(\frac{2x}{V})^2}{(\frac{1-x}{V}) 	imes (\frac{3(1-x)}{V})^3}$$K_C = \frac{\frac{4x^2}{V^2}}{(\frac{1-x}{V}) 	imes \frac{27(1-x)^3}{V^3}}$$K_C = \frac{4x^2}{V^2} 	imes \frac{V^4}{27(1-x)^4}$$K_C = \frac{4x^2V^2}{27(1-x)^4}$

For the formation of ammonia from its constituent elements ($1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$) in a closed vessel of volume $V \ L$,the value of $K_C$ is [units of $K_C = mol^{-2} \ L^2$].

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