For the formation of NH_3 from N_2 and H_2 at | vedclass

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(B) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.For the reverse reaction,the equatio

Vedclass · Accepted Answer

$2.81 	imes 10^{-3}$

Vedclass · Answer

(B) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.For the reverse reaction,the equation is: $2NH_3(g) ightleftharpoons N_2(g) + 3H_2(g)$.The equilibrium constant $K_C$ for the reverse reaction is given by the expression: $K_C = \frac{[N_2][H_2]^3}{[NH_3]^2}$.Substituting the given equilibrium concentrations:$K_C = \frac{(1.5 	imes 10^{-2}) 	imes (3.0 	imes 10^{-2})^3}{(1.2 	imes 10^{-2})^2}$.$K_C = \frac{(1.5 	imes 10^{-2}) 	imes (27.0 	imes 10^{-6})}{1.44 	imes 10^{-4}}$.$K_C = \frac{40.5 	imes 10^{-8}}{1.44 	imes 10^{-4}} = 28.125 	imes 10^{-4} = 2.81 	imes 10^{-3}$.

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

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