$\Delta G$ is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that $\Delta G$ is a measure of free energy. Find the unit of $\Delta G$. If a reaction has positive enthalpy change and positive entropy change,under what condition will the reaction be spontaneous?

(N/A) Gibbs free energy is the thermodynamic quantity of a system,the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.
Mathematically,this result may be derived as follows:
The relationship between heat absorbed by a system $q$,the change in its internal energy $\Delta U$,and the work done by the system is given by the first law of thermodynamics:
$q = \Delta U + W_{\text{expansion}} + W_{\text{non-expansion}}$ $(i)$
Under constant pressure,the expansion work is $p \Delta V$. Since $\Delta H = \Delta U + p \Delta V$,we have:
$q = \Delta H + W_{\text{non-expansion}}$ $(ii)$
For a reversible process at constant temperature $T$,$\Delta S = \frac{q_{\text{rev}}}{T}$,so $q_{\text{rev}} = T \Delta S$ $(iii)$
Substituting $(iii)$ into $(ii)$:
$T \Delta S = \Delta H + W_{\text{non-expansion}}$
$\Delta H - T \Delta S = -W_{\text{non-expansion}}$ $(iv)$
Since $\Delta G = \Delta H - T \Delta S$,equation $(iv)$ becomes:
$\Delta G = -W_{\text{non-expansion}}$ $(v)$
Thus,$-\Delta G$ represents the maximum useful work (non-expansion work) obtainable from the system.
The unit of $\Delta G$ is the same as energy,which is joule $(J)$ or kilojoule $(kJ)$.
For a reaction with positive $\Delta H$ and positive $\Delta S$,the condition for spontaneity $(\Delta G < 0)$ is:
$\Delta H - T \Delta S < 0$
$T \Delta S > \Delta H$
$T > \frac{\Delta H}{\Delta S}$
Thus,the reaction is spontaneous at high temperatures.