Question

At ${35}^{o}C$ and $1$ atmospheric pressure, $N_2{O}_4$ is $27.2$% dissociated into ${NO}_2$. What is the value of $K_p$ under these conditions?

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$
Let n be the initial number of moles of $N_2{O}_4$. 27.2% of these or $\frac{\text{27.2}}{\text{100}}\times \text{n}=\text{0.272n}$ moles will dissociate to form
$2\times 0.272n=0.544n$ moles of $N_2{O}_4$
$n-0.272n=0.728n$ moles of $N_2{O}_4$ will remain.
Total number of moles $=0.728n+0.544n=1.272n$
Mole fraction of $N_2{O}_4=\frac{0.728n}{1.272n}=0.572$
Total pressure $=1$ atm.
Partial pressure of $N_2{O}_4=0.572\times 1=0.572$ atm.
Partial pressure of $N{O}_2=1-0.572=0.428$ atm.
$K_p=\frac{P_{NO2}^2}{P_{N2O4}}$
$K_p=\frac{(0.428{)}^2}{0.572}$
$K_p=0.3195$ atm