Question

At $450K,K_p=2.0\times 10^{10}$/ bar for the given reaction at equilibrium,
$2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$
The value of $K_C$ at this temperature will be

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

Answer 1

Answer: (B)

$2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$
$K_p=K_C(RT{)}^{\Delta n_g}$
$\because \Delta n_g=2-3=-1$
Given, $K_p=2.0\times 10^{10}{\text{bar}}^{-1}$
$R=0.0831L$ bar $K^{-1}mo{l}^{-1}$ and $T=450K$
$\therefore K_C=\frac{K_p}{(RT{)}^{\Delta n_g}}=\frac{K_p}{(RT{)}^{-1}}$
or $K_C=K_p\times RT$
$=2.0\times 10^{10}{\text{bar}}^{-1}\times 0.0831L$ bar $K^{-1}mo{l}^{-1}\times 450K$
$K_C=7.479\times 10^{11}Lmo{l}^{-1}$