Question

$N_2{O}_4$ is 25% dissociated at ${37}^{o}C$ and one atmosphere pressure. Calculate $K_p$ and the percentage dissociation at $0.1$ atmosphere and ${37}^{o}C$.

• A. 0.166 atm and 37% respecively
• B. 0.266 atm and 87% respectively
• C. 0.316 atm and 47% respectively
• D. 0.316 atm and 37% respectively

Answer 1

The correct option is B 0.266 atm and 87% respectively

Let initially 1 mole of $N_2{O}_4$ be present. Total pressure is 1 atm.

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

Initial moles	$1$	$0$
Equilibrium moles	$0.75$	$0.5$

The expression for the equilibrium constant is:

$K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{{0.4}^2}{0.6}=0.266$.

Now total pressure is 0.1 atm.

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

Equilibrium moles are $1-xand2x$

Mole fraction are $\frac{1-x}{1+x}$, $\frac{2x}{1+x}$

The expression for the equilibrium constant is $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$. $0.266=\frac{(\frac{2x}{1+x}{)}^2}{\left(\frac{1-x}{1+x}\right)\times 0.1}$. Thus $x=0.87$. Hence, the degree of dissociation is 87%.