Question

In a reaction mixture containing $H_2,N_2$ and ${NH}_3$ at partial pressure of $1$ atm, $2$ atm and $3$ atm respectively, the value of $K_p$ at $725K$ is $4.28\times {10}^{-5}{atm}^{-2}$. In which direction the net reaction will go?

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

• A. Forward
• B. Backward
• C. No net reaction
• D. Direction cannot be predicted

Answer 1

Answer: (B)

Backward

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$Q_p=\frac{[P_{NH3}{]}^2}{\left[P_{N2}\right][P_{H2}{]}^3}$

$Q_p=\frac{3^2}{2\times 1^3}$

$Q_p=4.5$

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

$Q_p>K_p$

Thus, the reaction will favour in the backward direction.