Question

At 298 K and 1 atm pressure, the partial pressures in an equilibrium mixture of $N_2{O}_4$ and $N{O}_2$ are 0.7 and 0.3 atmosphere respectively. What will be the partial pressure of $N{O}_2$ when the gases are in equilibrium at 298 K and at a total (equilibrium) pressure of 10 atmospheres?

• A. 1.078 atm
• B. 2.23 atm
• C. 0.77 atm
• D. 1.60 atm

Answer 1

The correct option is A

1.078 atm

The temperature is maintained constant which means that $K_p$ does not change.

When the total pressure is 1 atmosphere,

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

Partial pressure at equlibrium 0.7 atm, 0.3 atm.

$K_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=\frac{0.3\times 0.3}{0.7}=0.1285$

Note: you can see that $K_p$ is small.

Now derive $K_p$ in terms of $\alpha$ and pressure P. We derived the value of $K_p$ from given partial pressures for equilibrium at 1 atm.

Let us assume there were initially n moles of $N_2{O}_4$ (g) and there were no products.

	$N_2{O}_4\left(g\right)$	$N{O}_2$(g)
t=0	1	0
t=eqm.	$(1-\alpha )$	$2\alpha$

At Equilibrium, total number of moles = (1 + $\alpha$)

Partial pressure of $N{O}_2$ (g) = mole fraction of $2N{O}_2$ (g) $\times$ Total pressure
=$\frac{2x}{1+x}\times 10$

Partial pressure of $N_2{O}_4$ (g) = mole fraction of $N_2{O}_4$ (g) x Total Pressure P
=$\frac{1-x}{1+x}\times 10$
Assume initially 1 mole of $N_2{O}_4$ is alone present. At equilibrium total pressure P, we have
$K_p$ = 0.1285 = $\frac{4P{\alpha }^2}{1-{\alpha }^2}\times 10$
(this should be derived)
For $\frac{{\alpha }^2}{1-{\alpha }^2}$ to be this small, a has to be very small as well. When $\alpha$ is that small, we can approximate 1 - ${\alpha }^2\approx 1$

Solving for a, we get $\alpha$ = 0.056

$P_{N{O}_2}$ = mole fraction of $N{O}_2$ $\times$ Total Pressure P
= $\frac{2\alpha }{1+\alpha }$ 10 atm = 1.078 atm.