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 x 10^-2 M,[ H_2]=3 x 10^-2 M and [ NH_3]=1.5 x 10^-2 M. Equilibrium constant for the reaction is ________.

K_ C=frac[ NH_3]^2[ N_2][ H_2]^3 \ K_ C=frac(1.5 x 10^-2)^2(2 x 10^-2) x (3 x 10^-2)^3 \ K_ C=417 \

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The following concentrations were observed at 500 K for... | (Chemistry) | DivineJEE

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Chemistry

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Chemistry

The following concentrations were observed at $500 \mathrm{~K}$ for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$ . At equilibrium ; $\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M}$ and $\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M}$ . Equilibrium constant for the reaction is ________.

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Explanation

$\begin{aligned} & \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\ & \mathrm{~K}_{\mathrm{C}}=\frac{\left(1.5 \times 10^{-2}\right)^2}{\left(2 \times 10^{-2}\right) \times\left(3 \times 10^{-2}\right)^3} \\ & \mathrm{~K}_{\mathrm{C}}=417 \end{aligned}$