Question

A reaction mixture containing $H_2,N_2$ and $N{H}_3$ has partial pressures $2$ atm, $1$ atm and $3$ atm respectively at $725K$. If the value of $K_p$ for the reaction, $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$ is $4.28\times {10}^{-5}at{m}^{-2}$ at $725K$, in which direction the net reaction will go?

• A. Forward
• B. Backward
• C. No net reaction
• D. Direction of reaction cannot be predicted

Answer 1

Answer: (B)

Backward

The reaction is $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}.$
The partial pressures of $H_2,N_2$ and $N{H}_3$ are $2$ atm, $1$ atm and $3$ atm respectively.
The expression for the reaction quotient is $Q_P=\frac{(P_{N{H}_3}{)}^2}{P_{N_2}\times (P_{H_2}{)}^3}=\frac{(3{)}^2}{\left(1\right)(2{)}^3}=\frac{9}{8}at{m}^{-2}=1.125at{m}^{-2}.$
The value of the equilibrium constant $K_p$ is $4.28\times {10}^{-5}at{m}^{-2}.$
The value of the reaction quotient is larger than the value of the equilibrium constant. Hence the net reaction will proceed in the backward direction.