Question

For the following equilibrium starting with $2$ moles $S{O}_2$ and $1mole{O}_2$ in $1L$ flask, $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$ Equilibrium mixture required $0.4moleMn{O}_4^{-}$ in acidic medium. Hence, $K_C$ is

• A. 2.0
• B. 0.4
• C. 1.6
• D. 2.6

Answer 1

Answer: (A)

$S{O}_2(g)+2Mn{O}_4^{-}$ to $M{n}^{2+}$

$5S{O}_2(g)+2Mn{O}_4^{-}(aq)+H^{+}⟶2M{n}^{2+}(aq)+5S{O}_4^{2-}(aq)$

$2Mn{O}_4^{-}=5S{O}_2$

$\therefore 0.4Mn{O}_4^{-}=\frac{5}{2}\times 0.4=1.0molS{O}_2$

Thus, $2-2x=1$

$2x=1$

$x=0.5$

$\left[S{O}_2\right]=\frac{2-2x}{1}=1mol{L}^{-1}$

$\left[O_2\right]=\frac{(1-x)}{1}=0.5mol{L}^{-1}$

$\left[S{O}_3\right]=\frac{2x}{1}=\frac{1}{1}mol{L}^{-1}$

$\therefore K_c=\frac{\left[S{O}_3\right]}{{\left[S{O}_2\right]}^2\left[O_2\right]}$

$=\frac{(1{)}^2}{(1{)}^2(0.5)}=2.0$