Question

1 mole $N_2$ and 3 mol $H_2$ are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$. The equilibrium constant $K_p$ for dissociation of $N{H}_3$ is:

• A. $\frac{1}{0.5}\times (1.5{)}^{3}at{m}^{-2}$
• B. $0.5\times (1.5{)}^{3}at{m}^2$
• C. $\frac{0.5\times (1.5{)}^3}{3\times 3}at{m}^2$
• D. $\frac{3\times 3}{0.5\times (1.5{)}^3}at{m}^{-2}$

Answer 1

The correct option is B $0.5\times (1.5{)}^{3}at{m}^2$

Initial partial pressures of reactants can be found out using $p_{component}$ = X*$p_{total}$, where X is the mole fraction of the component.

Therefore, $p_{N_2}=\frac{1}{4}\ast 4=1$,

similarly, $p_{H_2}=3$.

On attainment of equilibrium, if $p_0$ is the drop in pressure for $N_2$,
then we have the following equilibrium,

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

$(1-p_0)$ $(3-3p_0)$ $\left(2p_0\right)$

Total pressure is $4-2p_0=3$.

Hence, $p_0=0.5$

$K_p=\frac{p_{N{H}_3}^2}{p_{H_2}^3\times p_{N_2}}$
= $\frac{1}{(1.5{)}^3\times 0.5}at{m}^{-2}$

This is for formation of $N{H}_3$.

For its dissociation, $K_p=(1.5{)}^3\times 0.5at{m}^2$

Therefore, the correct option is $B$