Question

A reaction mixture containing $H_2,N_2$ and $N{H}_3$ has partial pressure 2 atm, 1atm and 3 atm respectively at 725 K. If the value of K for the reaction. $N_2+3H_2\rightleftharpoons 2N{H}_3$ is $4.28\times {10}^{-5}$

atm${}^{-2}$ 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

$Q_p=\frac{p_{N{H}_3}^2}{p_{N_2}{p}_{H_2}^3}$ $=\frac{3^2}{1\times 2^3}=1.125at{m}^{-1}$

Since it is grater that $K_p$ the reaction will go in backward direction to achieve $K_P$