For the reaction, $${X_2}{O_4}\left( l \right) \to 2X{O_2}\left( g \right),$$     $$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$$       at $$300\,K.$$  Hence, $$\Delta G$$  is

Solution :
The change in Gibbs free energy is given by
$$\Delta G = \Delta H - T\Delta S$$
where, $$\Delta H = $$  change enthalpy of the reaction
            $$\Delta S = $$  change entropy of the reaction
Thus, in order to determine $$\Delta G,$$ the values of $$\Delta H$$ must be known. The value of $$\Delta H$$ can be calculated by using equation
$$\Delta H = \Delta U + \Delta {n_g}RT\,\,\,...{\text{(i)}}$$
where, $$\Delta U = $$  change in internal energy using
$$\Delta {n_g} =$$  number of moles of gaseous products $$ - $$ number of moles of gaseous reactants
$$= 2 - 0 = 2$$
$$R = {\text{gas constant}} = 2\,cal$$
given, $$\Delta U = 2.1\,kcal$$
                  $$ = 2.1 \times {10^3}cal\,\left[ {\because \,\,1\,kcal = {{10}^3}\,cal} \right]$$
By putting the values in eq. (i) we get,
$$\eqalign{ & \therefore \,\,\Delta H = \left( {2.1 \times {{10}^3}} \right) + \left( {2 \times 2 \times 300} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3300\,cal \cr & {\text{Hence,}}\,\Delta G = \Delta H - T\Delta S \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta G = \left( {3300} \right) - \left( {300 \times 20} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta G = - 2700\,cal \cr & \therefore \,\,\,\,\,\,\,\,\,\,\Delta G = - 2.7\,kcal \cr} $$