Question

At constant pressure, addition of helium to the reaction system :
${\left(\text{N}\right)}_{\text{2}}{\left(\text{(g) + 3H}\right)}_{\text{2}}\text{(g) }\rightleftharpoons {\left(\text{2NH}\right)}_{\text{3}}\text{(g)}$

• A. Favours the formation of ammonia
• B. Reduces the formation of ammonia
• C. Does not affect the position of equilibrium
• D. Reduces the dissociation of ammonia.

Answer 1

Answer: (B)

$K_p=\frac{{\left(n^2\right)}_{N{H}_3}}{n_{N_2}\times n_{H_2}^3}\times {\left(\frac{P}{\Sigma n}\right)}^{\Delta n}$
At constant pressure, with the introduction inert gas $\Sigma n$ increases. Since $\Delta n=-2$ , therefore on increasing $\Sigma n{\left(\frac{P}{\Sigma n}\right)}^{\Delta n}$ increases, but $K_p$ remain constant, hence no. of
moles of $N_2\left(g\right)$ & $H_2\left(g\right)$ are increased and backward reaction is favoured.