Question

An equilibrium mixture at 300K contains $N_2{O}_4$ and $N{O}_2$ at 0.28 and 1.1 atm respectively. If the volume of the container is doubled and the new equilibrium pressures of $N{O}_2$ is $x\times {10}^{-2}$ atm, then $x-60$ is: (nearest integer value)

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$
$0.28$ $1.1$

$K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{(1.1{)}^2}{0.28}=4.32atm$

lf the volume is doubled, the pressure becomes half and the reaction proceeds in the forward direction.

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

Now, the pressure at equilibrium $(\frac{0.28}{2}-p\left)\right(\frac{1.1}{2}+2p)$, where, reactant $N_2{O}_4$ equivalent to pressure P is used up and $N{O}_2$ equivalent to pressure 2P is formed.

$K_p=\frac{(\frac{1.1}{2}+2P{)}^2}{(\frac{0.28}{2}-P)}=\frac{(0.55+2P{)}^2}{(0.14-P)}=4.32$, neglecting $P^2$ term,

$P=0.045atm$

$P_{N_2{O}_4}=0.14-0.045=0.095atm$

$p_{N{O}_2}=0.55+2\times 0.045=0.64atm=64\times {10}^{-2}atm$

Hence, $x$ is 64

So, $64-60=4$.