Question

The degree of dissociation of $N_2{O}_4$ into $N{O}_2$ at $1$ atmosphere and ${40}^{\circ }C$ is $0.25$. Calculate its $K_p$ at ${40}^{\circ }C$.

Answer 1

Consider the following chemical reaction.

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$att=0,10$

$atequil.,1-\alpha 2\alpha$

Here,

Total moles of gas $=1+\alpha$

If $P$ is the total pressure, then partial pressures are:

$P_{N_2{O}_4}=\frac{P(1-\alpha )}{(1+\alpha )}$

$P_{N{O}_2}=\frac{P\left(2\alpha \right)}{(1+\alpha )}$

Therefore,

$K_P=\frac{{\left(P_{N{O}_2}\right)}^2}{P_{N_2{O}_4}}$

$K_P=\frac{{\left(\frac{P\left(2\alpha \right)}{1+\alpha }\right)}^2}{\frac{P(1-\alpha )}{1+\alpha }}$

$K_P=\frac{P{\left(2\alpha \right)}^2}{(1-\alpha )(1+\alpha )}$

$K_P=\frac{1atm{(2\times 0.25)}^2}{(1-0.25)(1+0.25)}$

$K_P=\frac{0.25}{0.75\times 1.25}atm$

$K_P=0.267atm$

Hence, this is the required result.