Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\left(\because \mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\propto \mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e} \text{a consant P and T}\right)$
$\underset{\begin{array}{c}\text{I}\text{n}\text{i}\text{t}\text{i}\text{a}\text{l }\text{M}\text{o}\text{l}\text{e}\text{s}\\ \text{M}\text{o}\text{l}\text{e}\text{s }\text{a}\text{t }\text{e}\text{q}\text{u}\text{i}\text{l}\text{i}\text{b}\text{r}\text{i}\text{u}\text{m }\end{array}}{\mathrm{}}\underset{\begin{array}{c}\text{a}\\ \text{(}\text{a}\text{-}\text{x}\text{)}\end{array}}{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\rightleftharpoons \underset{\begin{array}{c}\text{0}\\ 2x\end{array}}{{\text{2NO}}_{\text{2}}\left(\mathrm{g}\right)}$
Given original volume$=\frac{75}{100}$of existing volume at equilibrium
or initial moles$\text{=}\left(\frac{\text{75}}{\text{100}}\right)\text{×moles at equilibrium}$
$\therefore \mathrm{}\mathrm{a}=\frac{75}{100}(\mathrm{a}+\mathrm{x})$
or$\mathrm{x}=\frac{25}{75}\mathrm{a}=0·33.33\mathrm{a}$
$\therefore \mathrm{D}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e} \mathrm{o}\mathrm{f} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}, \mathrm{\alpha }=\frac{\mathrm{x}}{\mathrm{a}}$
$=\frac{0.3333\text{a}}{\text{a}}=0.3333\text{or}33.33\text{\%}$