Question

For the following equilibrium starting with $2\,$ moles $SO _{2}$ and $1\, mole O _{2}$ in $1\, L$ flask, $2 SO _{2}( g )+ O _{2}( g ) \rightleftharpoons 2 SO _{3}( g )$ Equilibrium mixture required $0.4\, mole\, MnO _{4}^{-}$ in acidic medium. Hence, $K _{ C }$ is

• A. 2.0
• B. 0.4
• C. 1.6
• D. 2.6

Answer 1

Answer: (A)

$SO _{2}( g )+2 MnO _{4}^{-}$ to $Mn ^{2+}$

$5 SO _{2}( g )+2 MnO _{4}^{-}( aq )+ H ^{+} \longrightarrow 2 Mn ^{2+}( aq )+5 SO _{4}^{2-}( aq )$

$2 MnO _{4}^{-}=5 SO _{2}$

$\therefore 0.4 MnO _{4}^{-}=\frac{5}{2} \times 0.4=1.0\, mol\, SO _{2}$

Thus, $2-2 x=1$

$2 x=1$

$x=0.5$

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

$\left[O_{2}\right]=\frac{(1-x)}{1}=0.5\, mol\, L^{-1}$

$\left[S O_{3}\right]=\frac{2 x}{1}=\frac{1}{1} mol\, L ^{-1}$

$\therefore K_{c}=\frac{\left[S O_{3}\right]}{\left[S O_{2}\right]^{2}\left[O_{2}\right]}$

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