$\text{ For the reaction, }{X}_2{\mathrm{O}}_{4\left(l\right)}⟶2{\mathrm{XO}}_{2\left(g\right)}$

$\Delta U=2.1\mathrm{kcal},\Delta S=20{\mathrm{calK}}^{-1}\text{at }300 \mathrm{K}$

$\text{ Hence, }\Delta G\text{ is }$

$\Delta H=\Delta U+\Delta n_{g}RT$

$\text{ Given, }\Delta U=2.1\mathrm{kcal},\Delta n_g=2$

$R=2\times {10}^{-3}\mathrm{kcal},T=300 \mathrm{K}$

$\therefore \Delta H=2.1+2\times 2\times {10}^{-3}\times 300=3.3\mathrm{kcal}$

$\text{ Again, }\Delta G=\Delta H-T\Delta S$

$\text{ Given, }\Delta S=20\times {10}^{-3}{\mathrm{kcalK}}^{-1}$

$\text{ On putting the value of }\Delta H\text{ in the equation, we get }$

$\Delta G=3.3-300\times 20\times {10}^{-3}$

$=3.3-6\times {10}^3\times {10}^{-3}=-2.7\mathrm{kcal}$