Question

At $450K,K_P=2.0\times {10}^{10}bar$ for the given reaction at equilibrium.
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
What is $K_c$, at this temperature?

• A. $7.4\times {10}^{11}Lmo{l}^{-1}$
• B. $7.4\times {10}^{-11}Lmo{l}^{-1}$
• C. $3.7\times {10}^{-11}Lmo{l}^{-1}$
• D. $3.7\times {10}^{11}Lmo{l}^{-1}$

Answer 1

The correct option is A $7.4\times {10}^{11}Lmo{l}^{-1}$

The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$.
For the reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$, the value of $\Delta n$ is $2+1-2=1$.
Substitute this value in the above expression.
$K_p=K_c(RT{)}^{\Delta n}=K_c(RT{)}^1$.
But $K_P=2.0\times {10}^{10}/bar,R=0.082\text{L atm /mol. K},T=450K$.
Substitute values in the above expression.
$2.0\times {10}^{10}/bar=K_c(0.082\times 450{)}^1$.
$K_c=7.4\times {10}^{11}Lmo{l}^{-1}$.