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 725 K. 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 725 K, 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

The correct option is B Backward

The given reaction is, $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$
$K_p=4.28\times {10}^{-5}at{m}^{-2}$ $Q_P=\frac{(P_{N{H}_3}{)}^2}{P_{N_2}\times (P_{H_2}{)}^3}=\frac{(3{)}^2}{\left(1\right)(2{)}^3}$
$Q_p=\frac{9}{8}at{m}^{-2}=1.125at{m}^{-2}$.
Since value of $Q_P$ is larger than $K_P$ $\left(4.28{10}^{-5}at{m}^{-2}\right)$, it indicates net reaction will proceed in backward direction.