For a certain reaction at $300 \mathrm{K}, \mathrm{K}=10,$ then $\mathrm{\Delta G}°$ for the same reaction is – ________ $\times {10}^{-1} \mathrm{kJ} {\mathrm{mol}}^{-1}.$ $$ (Given $\mathrm{R}=8.314{\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}$)

The standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one  mole of that substance from its component element available at their standard states or the most stable form of the element which is at $25 °$C and $100$ kPa. Its symbol is ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{G}˚$.

$\mathrm{\Delta G}°=-\mathrm{RT\ell nK}$

$=-8.314\times 300\ln \left(10\right)$

$=5744.14 \mathrm{J}/\mathrm{mole}$

$=57.44\times {10}^{-1} \mathrm{kJ}/\mathrm{mole}$