At 450 K, ${\mathrm{K}}_{\mathrm{p}}= 2.0\times {10}^{10} \mathrm{bar}$ for the given reaction at equilibrium,

$2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right)$

What is ${\mathrm{K}}_{\mathrm{C}}$ at this temperature?

$2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right); {\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{C}}(\mathrm{RT}{)}^{\mathrm{\Delta n}}$

${\mathrm{\Delta n}}_{ }=2-3=-1$

${\mathrm{K}}_{\mathrm{p}}=2.0\times {10}^{10}$

$\mathrm{R}=0.0831 \mathrm{L}\text{ bar }$$\text{ and }\mathrm{T}=450 \mathrm{K}$

${\mathrm{K}}_{\mathrm{C}}=\frac{{\mathrm{K}}_{\mathrm{p}}}{(\mathrm{RT}{)}^{\mathrm{\Delta n}}} = \frac{{\mathrm{K}}^{ }\mathrm{p}}{(\mathrm{RT}{)}^{-1}}$

$\text{ or }{\mathrm{K}}_{\mathrm{C}}={\mathrm{K}}_{\mathrm{p}}\times \mathrm{RT}$

$=2.0\times {10}^{10}\times 0.0831 \times 450$

${\mathrm{K}}_{\mathrm{C}}=7.479\times {10}^{11} \mathrm{L}{ \mathrm{mol}}^{-1}$