Question

The degree of dissociation of $N_2{O}_4$ into ${NO}_2$ at one atmosphere and ${40}^{o}C$ is $0.310$. Calculate its $K_p$ at ${40}^{o}C$. Also report degree of dissociation at $10$ atmospheric pressure at the same temperature.

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$
Assume one mole of $N_2{O}_4$ is present initially.
0.310 moles of $N_2{O}_4$ will dissociate to form $2\times 0.310=0.620$ moles of $N{O}_2$.
$1-0.310=0.690$ moles of $N_2{O}_4$ will remain.
Total pressure is 1 atm.
The partial pressure of $N_2{O}_4=0.690\times 1atm=0.690atm$
The partial pressure of $N{O}_2=0.620\times 1atm=0.620atm$
$K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$
$K_p=\frac{(0.620atm{)}^2}{\left(0.690atm\right)}$
$K_p=0.557$
Total pressure is 10 atm.
Assume one mole of $N_2{O}_4$ is present initially.
x moles of $N_2{O}_4$ will dissociate to form $2x$ moles of $N{O}_2$.
$1-x$ moles of $N_2{O}_4$ will remain.
Total pressure is 10 atm.
The partial pressure of $N_2{O}_4=(1-x)\times 10atm=(10-10x)atm$
The partial pressure of $N{O}_2=2x\times 10atm=20xatm$
$K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$
$0.557=\frac{(20xatm{)}^2}{\left(\right(10-10x\left)atm\right)}$
$5.57-5.57x=400x^2$
$400x^2+5.57x-5.57=0$
This is quadratic equation with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-\left(5.57\right)\pm \sqrt{(5.57{)}^2-4(400\left)\right(-5.57)}}{2\left(400\right)}$
$x=\frac{-\left(5.57\right)\pm \sqrt{(5.57{)}^2-4(400\left)\right(-5.57)}}{800}$
$x=0.111$ or $x=-0.125$
The value $x=-0.125$ is discarded as the number of moles cannot be negative.
$x=0.111$
The degree of dissociation $=\frac{0.111}{1}=0.111$