Question

At $450 K , K_{p}=2.0 \times 10^{10} $/ bar for the given reaction at equilibrium,
$2 SO _{2}(g)+ O _{2}(g) \rightleftharpoons 2 SO _{3}(g)$
The value of $K_{C}$ at this temperature will be

• A. $6.4 \times 10^{12}\, L\, mol ^{-1}$
• B. $7.479 \times 10^{11}\, L \,mol ^{-1}$
• C. $7.00 \times 10^{-11}\, L\, mol ^{-1}$
• D. $5.66 \times 10^{6} \,L\, mol ^{-1}$

Answer 1

Answer: (B)

$2 SO _{2}(g)+ O _{2}(g) \rightleftharpoons 2 SO _{3}(g)$
$K_{p}=K_{C}(R T)^{\Delta n_{g}}$
$\because \Delta n_{g}=2-3=-1$
Given, $ K_{p}=2.0 \times 10^{10}\, \text{bar} ^{-1}$
$R=0.0831 \, L$ bar $K ^{-1}\, mol ^{-1}$ and $T=450\, K$
$\therefore K_{C}=\frac{K_{p}}{(R T)^{\Delta n_{g}}}=\frac{K_{p}}{(R T)^{-1}}$
or $K_{C}=K_{p} \times R T$
$=2.0 \times 10^{10} \text{bar} ^{-1} \times 0.0831 \, L$ bar $K ^{-1}\, mol ^{-1} \times 450\, K$
$K_{C}=7.479 \times 10^{11} \, L\, mol ^{-1}$