Question

$K_p$ for $N_2{O}_4\rightleftharpoons 2N{O}_2$ at equilibrium pressure of $3P_o$ is 0.5 $P_o$. On halving the volume of container equilibrium is distributed. Find the new equilibrium pressure.

Answer 1

Given reaction is :-

$N_2{O}_4\rightleftharpoons 2{NO}_2;K_P=0.5P_0$

Initial pressure $P$ $0$

At eqm $P-p$ $2p$

Now, given that, equilibrium pressure $=3P_0$

$\Rightarrow P-p+2p=3P_0$

$\Rightarrow P+p=3P_0\to \left(I\right)$

So, $K_P=\frac{{\left(P{NO}_2\right)}^2}{{PN}_2{O}_4};{PNO}_2,{PN}_2{O}_4$ are partial pressure of ${NO}_2\&N_2{O}_4$ at eqm

$\Rightarrow 0.5P_0=\frac{{\left(2p\right)}^2}{P+p}=\frac{{4p}^2}{P+p}\to \left(II\right)$

By $\left(I\right)$ & $\left(II\right)$ $\Rightarrow \frac{{4p}^2}{3P_0}=0.5P_0$

$\Rightarrow 4p^2=1.5{P_0}^2$

$\Rightarrow P^2=\frac{1.5}{4}{P_0}^2\Rightarrow P=0.6P_0$

So, ${PNO}_2=1.2P_0\&{PN}_2{O}_4=3P_0-0.6P_0=2.4P_0$

Now, when volume gets halved pressure will get doubled because by Boyle's law $\left(P\alpha \frac{1}{V}\right)$.

$\Rightarrow {PNO}_2=2.4P_0\&{PN}_2{O}_4=4.8P_0$

$\Rightarrow 2p=2.4P_0\&P.p=4.8P_0$

$\Rightarrow p=1.2P_0\Rightarrow P=6P_0$

so, equilibrium pressure $=P+p=6P_0+1.2P_0=7.2P_0$