$\Delta G$ is 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?
Gibbs free energy is that 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 results 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 equation of the first law of thermodynamics, therefore,
$q=\Delta U+W_{\text {expansion }}+W_{\text {non-expansion }}$ .... (i)
Under constant pressure condition, the expansion work is given by $p \Delta V$. $$ \begin{aligned} \therefore \quad q & =\Delta U+p \Delta V+W_{\text {non expansion }} \quad (\because \Delta U+p \Delta V=\Delta H) \\ & =\Delta H+W_{\text {non expansion }} \quad \ldots \text { (ii) } \end{aligned}$$
For a reversible change taking place at constant temperature,
$$\Delta S=\frac{q_{\mathrm{rev}}}{T} \text { or } q_{\mathrm{rev}}=T \Delta \mathrm{S}\quad \text{.... (iii)}$$
Substituting the value of $q$ from Eq. (iii) in Eq. (ii), we get
$$\begin{aligned} T \Delta S & =\Delta H+W_{\text {non expansion }} \\ \text{or}\quad \Delta H-T \Delta S & =-W_{\text {non-expansion }}\quad \text{... (iv)} \end{aligned}$$
For a change taking place under conditions of constant temperature and pressure,
$$\Delta G=\Delta H-T \Delta S$$
Substituting this value in equation (iv), we get
$$\Delta G=-W_{\text {non expansion }}\quad \text{... (v)}$$
Thus, free energy change can be taken as a measure of work other than the work of expansion. For most changes, the work of expansion can not be converted to other useful work, whereas the non-expansion work is convertible to useful work.
Rearranging equation (v), it may write as
$\Delta G=\Delta H-T \Delta S$
If $\Delta H=$ positive and $\Delta S=$ positive, then
$\Delta G$ will be negative i.e., process will be spontaneous only when $T \Delta S>\Delta H$ in magnitude, which will be so when temperature is high.
Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $\left(p_i, V_i\right)$ to $\left(p_f, V_f\right)$. With the help of a $p V$ plot compare the work done in the above case with that carried out against a constant pressure $p_f$.
(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $\left(p_i V_i\right)$ to $\left(p_f, V_f\right)$. Reversible work is represented by the combined areas $A B C$ and $B C V_i V_f$.
(ii) Work against constant pressure, $p_f$ is represented by the area $B C V_i V_t$. Work (i) $>$ work (ii)