${A_2}\left( g \right) + {B_2}\left( g \right) \rightleftharpoons {C_2}\left( g \right) + {D_2}\left( g \right)$
If we take $1\ mole$ of each of the four gases in a $10\ litre$ container, what would be equilibrium concentration of $A_2(g)$?
$\therefore \mathrm{Q}_{\mathrm{c}}>\mathrm{K}_{\mathrm{c}}$ so reaction will proceed in backward direction
$\mathrm{A}_{2}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{2}(\mathrm{g})+\mathrm{D}_{2}(\mathrm{g})$
conc. eqm. $\frac{1+x}{10} \quad \frac{1+x}{10} \quad \frac{1-x}{10} \quad \frac{1-x}{10}$
$0.25=\frac{\left(\frac{1-\mathrm{x}}{10}\right)^{2}}{\left(\frac{1+\mathrm{x}}{10}\right)^{2}} \Rightarrow \mathrm{x}=0.333$
$\left[\mathrm{A}_{2}(\mathrm{g})\right]=\frac{1+\mathrm{x}}{10}=\frac{1.333}{10}=0.133$
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(Round off to the Nearest Integer).
