Question

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

The correct option is A Forward

$N_2+{3H}_2\rightleftharpoons 2{NH}_3$

at eqn, 1atm 2atm 3atm

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

$Q_p=\frac{{\left(P_{{NH}_3}\right)}^2}{\left(P_{N_2}\right){\left(P_{H_2}\right)}^3}=\frac{{\left(3\right)}^2}{1\times {\left(2\right)}^3}=\frac{3\times 3}{2\times 2\times 2}=\frac{9}{8}=1.125{atm}^{-2}$

Since, $Q_p>>K_p$

Reaction will move in forward direction.