Calculate standard molar entropy ${\mathrm{O}}_2\left(\mathrm{g}\right)$from the following thermodynamic data at $300 \mathrm{K}$ :

$\begin{array}{l}4\mathrm{Cr}\left(s\right)+3{\mathrm{O}}_2\left(g\right)\to 2{\mathrm{Cr}}_2{\mathrm{O}}_3\left(s\right),\mathrm{\Delta }{G}^{\circ }=-2093.4 \mathrm{kJ}/\mathrm{mo}1\\ \mathrm{\Delta }{H}_f^{\circ }{\mathrm{Cr}}_2{\mathrm{O}}_3\left(s\right)=-1129.05 \mathrm{kJ}/\mathrm{mole}\\ S_m^{\circ }\mathrm{Cr}\left(s\right)=24 \mathrm{J}/\mathrm{K} \mathrm{mole}\\ S_m^{\circ }{\mathrm{Cr}}_2{\mathrm{O}}_3\left(s\right)=81 \mathrm{J}/\mathrm{K} \mathrm{mole}\end{array}$

$\begin{array}{l}4\mathrm{Cr}\left(s\right)+3{\mathrm{O}}_2⟶2{\mathrm{Cr}}_2{\mathrm{O}}_3\left(s\right)\mathrm{\Delta }{G}^{\circ }=-2093.4 \mathrm{kJ}/\mathrm{mole}\\ \mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }\\ \begin{array}{rl}\mathrm{\Delta }{H}^{\circ }& =2\mathrm{\Delta }{H}_f^{\circ }{\mathrm{Cr}}_2{\mathrm{O}}_3=2\times (-1129.05) \mathrm{kJ}\\ \mathrm{\Delta }{S}^{\circ }& =2S_m^{\circ }{\mathrm{Cr}}_2{\mathrm{O}}_3\left(s\right)-4S_m^{\circ }\mathrm{Cr}\left(s\right)-3S_m^{\circ }{\mathrm{O}}_2\\ & =\left(2\times 81-4\times 24-3S_m^{\circ }{\mathrm{O}}_2\right) \mathrm{J}/\mathrm{K}\end{array}\end{array}$
Substituting the values
$\begin{array}{l}-2093.4=-(2\times 1129.05)\\ -300\left(81\times 2-4\times 24-3S_m^{\circ }{\mathrm{O}}_2\right){10}^{-3}\\ S_m^{\circ }{\mathrm{O}}_2=205 \mathrm{J}/\mathrm{K}\text{ mole }\end{array}$