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(C) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.The expression for the equilibrium c

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$417$

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(C) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.The expression for the equilibrium constant $K_c$ is: $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$.Substituting the given equilibrium concentrations:$K_c = \frac{(1.5 	imes 10^{-2})^2}{(2 	imes 10^{-2}) 	imes (3 	imes 10^{-2})^3}$.$K_c = \frac{2.25 	imes 10^{-4}}{(2 	imes 10^{-2}) 	imes (27 	imes 10^{-6})}$.$K_c = \frac{2.25 	imes 10^{-4}}{54 	imes 10^{-8}}$.$K_c = \frac{2.25}{54} 	imes 10^4 = 0.04166 	imes 10^4 = 416.66 \approx 417$.

The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

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