Question

At 450 K, $K_p=2.0\times {10}^{10}/bar$ for the given reaction at equilibrium.
$2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\leftrightarrow 2S{O}_{3\left(g\right)}$
What is $K_c$ at this temperature?

Answer 1

For the given reaction,
$\Delta n=2-3=-1T=450KR=0.0831barLbar{K}^{-1}mo{l}^{-1}{K}_p=2.0\times {10}^{10}ba{r}^{-1}$
We know that,
$K_p=K_c\left(RT\right)\Delta n\Rightarrow 2.0\times {10}^{10}ba{r}^{-1}=K_c(0.831Lbar{K}^{-1}mo{l}^{-1}\times 450K{)}^{-1}\Rightarrow K_c=\frac{2.0\times {10}^{10}ba{r}^{-1}}{(0.0831Lbar{K}^{-1}mo{l}^{-1}\times 450K{)}^{-1}}=(2.0\times {10}^{10}ba{r}^{-1}\left)\right(0.0831Lbar{K}^{-1}mo{l}^{-1}\times 450K)=74.79\times {10}^{10}Lmo{l}^{-1}=7.48\times {10}^{11}Lmo{l}^{-1}=7.48\times {10}^{11}{M}^{-1}$