Question

The equilibrium constant $K_P$ for the reaction
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2$ (g) 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 $N_2{O}_4\left(g\right)$ be 1

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$initialmoles10$
$atequilibrium1-\alpha 2\alpha$
$\therefore$ 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, $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{4{\alpha }^2}{(1-{\alpha }^2)}\times P$
$\Rightarrow 4.5=\frac{4{\alpha }^2}{(1-{\alpha }^2)}\times 2$
$\Rightarrow \alpha =0.6$

Mole fraction of $N_2{O}_4$,
${\chi }_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }=0.25$

So average molar mass of mixture,
$=0.25\times 92+0.75\times 46=57.5$

$\therefore {\chi }_{N{O}_2}=1-0.25=0.75$