If the value of ${\mathrm{\Delta G}}^0$ for the equilibrium reaction at $300\mathrm{ }\mathrm{K}$ is $57.44\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$, the equilibrium constant ${\mathrm{K}}_{\mathrm{c}}$ is

$$\begin{gathered} {\mathrm{\Delta G}}^0=-{\mathrm{RTlnK}}_{\mathrm{c}} \\ \Rightarrow 57.44=-8.314\times 300\times {\mathrm{lnK}}_{\mathrm{c}} \\ \Rightarrow \frac{57.44}{-8.314\times 300}={\mathrm{lnK}}_{\mathrm{c}} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{lnK}}_{\mathrm{c}}=-23.029 \\ \Rightarrow {\mathrm{K}}_{\mathrm{c}}={\mathrm{e}}^{-23.029} \\ \Rightarrow {\mathrm{K}}_{\mathrm{c}}=1.0\times {10}^{-10} \end{gathered}$$

Therefore, option (D) is correct.