Question

The degree of dissociation of $N_2{O}_4$ into $N{O}_2$ at $1.5$ atmosphere and ${40}^{\circ }C$ is $0.25$. Calculate its $K_P$ at ${40}^{\circ }C$. Also report degree of dissociation at $10$ atmospheric pressure at same temperature.

Answer 1

$N_2{O}_4\leftrightharpoons 2N{O}_{2}initially10atequlibrium1-\alpha 2\alpha$

Total moles of gas = 1+$\alpha$

If P is the total pressure, then partial pressure are :-

From Roult's law we have,

Partial pressure of the gas-

$p\left(g\right)=P_T\times x$

$P_{N_2{O}_4}=\frac{P(1-\alpha )}{(1+\alpha )}{P}_{N{O}_2}=\frac{P\left(2\alpha \right)}{(1+\alpha )}{K}_P=\frac{{\left(P_{N{O}_2}\right)}^2}{P_{N_2{O}_4}}=\frac{{\left[\frac{P\left(2\alpha \right)}{(1+\alpha )}\right]}^2}{\left[\frac{P(1-\alpha )}{(1+\alpha )}\right]}=\frac{{\left(P\right(2\alpha \left)\right)}^2}{P(1-{\alpha }^2)}=\frac{1atm{(2\times (0.25\left)\right)}^2}{1-{0.25}^2}=\frac{1atm\times \left(0.25\right)}{0.9375}=0.26atm$

$\therefore K_P$ at 40℃ is 0.26 atm.