$S{O_2} + N{O_2} \rightleftharpoons S{O_3} + NO$
If we take one mole of each of four gases in one $L$ container. What would be equilibrium concentration of $NO$ and $NO_2$ respectively
111 Moles at $t=0$
$(1-x) \quad(1+x) \quad(1+x)$ (Moles at equilibrium)
$\therefore \mathrm{K}_{\mathrm{C}}=\frac{\left(\frac{1+\mathrm{x}}{\mathrm{V}}\right)\left(\frac{1+\mathrm{x}}{\mathrm{V}}\right)}{\left(\frac{1-\mathrm{x}}{\mathrm{V}}\right)\left(\frac{1-\mathrm{x}}{\mathrm{V}}\right)}=\frac{(1+\mathrm{x})^{2}}{(1-\mathrm{x})^{2}}$
$\therefore \frac{(1+\mathrm{x})^{2}}{(1-\mathrm{x})^{2}}=16 \quad\left(\because \mathrm{K}_{\mathrm{C}}=16\right)$
$\therefore \frac{(1+x)}{(1-x)}=4 \quad$ or $x=\frac{3}{5}=0.6$
$\therefore\left[\mathrm{SO}_{2}\right]=\left[\mathrm{NO}_{2}\right]=1-\mathrm{x}=1-0.6=0.4 \mathrm{M}$
$\left[\mathrm{SO}_{3}\right]=[\mathrm{NO}]=1+\mathrm{x}=1+0.6=1.6 \mathrm{M}$
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Assume that there is no other way of consuming stored energy.Given : The enthalpy of evaporation of water is $45\,kJ\,mol ^{-1}$
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$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$
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