Question

At a certain temperature, $K_C$ for ${SO}_2\left(g\right)+{NO}_2\left(g\right)\rightleftharpoons {SO}_3\left(g\right)+NO\left(g\right)$ is 16. If we take one mole of each of all the four gases in one litre container, what would be the equilibrium concentration of $NO$ and ${NO}_2$?

• A. $NO=3.6M$
  ${NO}_2=10M$
• B. $NO=2M$
  ${NO}_2=2.4M$
• C. $NO=1.6M$
  ${NO}_2=1.2M$
• D. $NO=1.6M$
  ${NO}_2=0.4M$

Answer 1

The correct option is D $NO=1.6M$

${NO}_2=0.4M$
${SO}_2\left(g\right)+{NO}_2\left(g\right)\rightleftharpoons {SO}_3\left(g\right)+NO\left(g\right)$
1 1 1 1 Moles at $t=0$
(1-x) (1-x) (1+x) (1+x) (Moles at equilibrium)

$\therefore K_C=\frac{\left(\frac{1+x}{V}\right)\left(\frac{1+x}{V}\right)}{\left(\frac{1-x}{V}\right)\left(\frac{1-x}{V}\right)}=\frac{{(1+x)}^2}{{(1-x)}^2}$
$\therefore \frac{{(1+x)}^2}{{(1-x)}^2}=16$ $(\because K_C=16)$
$\therefore \frac{(1+x)}{(1-x)}=4$ or $x=\frac{3}{5}=0.6$
$\therefore \left[{SO}_2\right]=\left[{NO}_2\right]=1-x=1-0.6=0.4M$
$\left[{SO}_3\right]=\left[NO\right]=1+x=1+0.6=1.6M$