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, $N_2+3H_2\rightleftharpoons 2N{H}_3$ is $4.28\times {10}^{-5}{atm}^{-2}$ 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 quotient

$Q_p=\frac{P_{N{H}_3}^2}{P_{N_2}{P}_{H_2}^3}$

$Q_P=\frac{3^2}{1\times 2^2}=2.25at{m}^{-2}$

But $K_p=4.28\times {10}^{-5}{atm}^{-2}$

Hence, $Q_P>K_P$

The partial pressure of ammonia is higher than the equilibrium partial pressure and the partial pressures of nitrogen and hydrogen are lower than the equilibrium partial pressures.

Hence, the net reaction will shift in a backward direction.