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, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)is4.28\times {10}^{-5}at{m}^{-2}at725K,$ 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

$Q_P=\frac{{\left(P_{N{H}_3}\right)}^2}{P_{N_2}\times {\left(P_{H_2}\right)}^3}=\frac{(3{)}^2}{\left(1\right)(2{)}^3}=\frac{9}{8}at{m}^{-2}=1.125at{m}^{-2}.$

Since value of $Q_p$ is larger than $K_p(4.28\times {10}^{-5}at{m}^{-2}),$ it indicates net reaction will proceed in backward direction.