Question

The equilibrium constant $K_p$ for the reaction
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is $4.5$
What would be the average molar mass (in g/mol) of an equilibrium mixture of $N_2{O}_4$ and $N{O}_2$ formed by the dissociation of pure $N_2{O}_4$ at a total pressure of $2$ atm?

• A. $69$
• B. $57.5$
• C. $80.5$
• D. $85.5$

Answer 1

The correct option is B $57.5$

Let initial mole of $N_2{O}_4\left(g\right)$ is one
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

initial moles	$1$	$0$
At equilibrium	$1-\alpha$	$2x$

Total number of moles at equilibrium $=1+\alpha$
$P_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }\times P$
$P_{N{O}_2}=\frac{2\alpha }{(1+\alpha )}\times P$
Hence,
$K_P=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{4{\alpha }^2}{(1-{\alpha }^2)}\times 2$
$\therefore 4.5=\frac{4{\alpha }^2}{(1-{\alpha }^2)\times 2}$
$\Rightarrow \alpha =0.6$
Mole fraction of $N_2{O}_4$:
$X_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }=0.25$
$\therefore X_{N{O}_2}=0.75$
Average molar mass of mixture
$=0.25\times 92+0.75\times 46=57.5$