Pure $N_2O_4(g)$ is introduced in a closed container. At $27^oC$ and $1 \;atm$, this gas is dissociated to $20\%$ forming $NO_2(g)$.
Calculate the equilibrium constant $K_p$.
Let $n$ be the number of moles of $N_2O_4(g)$ before dissociation, then $0.2n$ moles will be dissociated at equilibrium, and there will remain $n$ $-$ $0.2n$ $=$ $0.8n$ moles.
According to the reaction's equation we have:
$N_2O_4$ $2NO_2$,
$0.4n$ moles $NO_2$ will have appeared.
The total pressure at equilibrium is therefore
$ P$ $ =$ $\frac{(0.8n+0.4n)\cdot RT}{V}$ $=$ $\frac{1.2n\cdot RT}{V}$ $=$ $1 (1)$
(the total pressure is given!).
The partial pressure of $NO_2$ at equilibrium is
$ p_{NO_2}$ $=$ $\frac{(n-2)\cdot RT}{V} (2)$.
The partial pressure of $N_2O_4$ at equilibrium is
$ p_{N_2O_4} =\frac{0,4n\cdot RT}{V} (3)$.
From (1): $\frac{n\cdot RT}{V}$ $=$ $\frac{1}{1,2}$
Introducing in (2) and (3), we get:
$p_{NO_2}$ $=$ $0.4\frac{1}{1.2}=\frac{1}{3}$ atm
$p_{N_2O_4}$ $=$ $0.8\frac{1}{1.2}$ $=$ $\frac{2}{3}$ atm
$K_p$ $=$ $\frac{(\frac{1}{3})^2}{\frac{2}{3}}$ $=$ $\frac{1}{6}$ atm
$2NO_2$,
$0.4n$ moles $NO_2$ will have appeared.
The total pressure at equilibrium is therefore
$ P$ $ =$ $\frac{(0.8n+0.4n)\cdot RT}{V}$ $=$ $\frac{1.2n\cdot RT}{V}$ $=$ $1 (1)$
(the total pressure is given!).
The partial pressure of $NO_2$ at equilibrium is
$ p_{NO_2}$ $=$ $\frac{(n-2)\cdot RT}{V} (2)$.
The partial pressure of $N_2O_4$ at equilibrium is
$ p_{N_2O_4} =\frac{0,4n\cdot RT}{V} (3)$.
From (1): $\frac{n\cdot RT}{V}$ $=$ $\frac{1}{1,2}$
Introducing in (2) and (3), we get:
$p_{NO_2}$ $=$ $0.4\frac{1}{1.2}=\frac{1}{3}$ atm
$p_{N_2O_4}$ $=$ $0.8\frac{1}{1.2}$ $=$ $\frac{2}{3}$ atm
$K_p$ $=$ $\frac{(\frac{1}{3})^2}{\frac{2}{3}}$ $=$ $\frac{1}{6}$ atm