Question

Starting with the thermodynamic relationship $G = H + TS$ ,derive the following relationship $\Delta G = \mathop { - T\Delta S}\nolimits_{Total}$

Answer 1

It is the total entropy change $$\Delta H = \mathop { - T\Delta S}\nolimits_{surr} ........\left( 2 \right)$$$\mathop {\Delta S}\nolimits_{Total}$that decides the spontaneity of process but most of the chemical reactions fall into category of either closed system or open system. Therefore , for most of the chemical reactions there are changes in both enthalpy and entropy. The quantity that relates these two are called Gibbs energy and denoted by G. G is an extensive property .It is a state function. It has units of energy that are joules.

Complete answer:
$\;G = H - TS$
on differentiating $\Delta G = \Delta H - T\left( {\Delta S} \right) - \left( {\Delta T} \right)S$
the reaction occurs at constant temperature, ΔT=0
$\Delta G = \Delta H - T\left( {\Delta S} \right)......\left( 1 \right)$
We also know that entropy changes the surroundings.
$\mathop {\Delta S}\nolimits_{surrounding} = T - \Delta H$
$\Delta H = \mathop { - T\Delta S}\nolimits_{surr} ......\left( 2 \right)$
Substituting (2) in (1)
$\Delta G = - T.\mathop {\Delta S}\nolimits_{Surr} - T\Delta S\Delta G = - T\left( {\mathop {\Delta S}\nolimits_{Surr} + \Delta S} \right)$we know, $\left( {\mathop {\Delta s}\nolimits_{Surr} + \Delta S} \right) = \mathop {\Delta S}\nolimits_{TOTAL}$
⇒$\Delta G = \mathop { - T(\Delta S)}\nolimits_{total}$
Thus, the relationship $\Delta G = \mathop { - T(\Delta S)}\nolimits_{total}$is derived.

Note:
Knowledge of the sign and magnitude of free energy change of a chemical reactions allows:-
Prediction of the spontaneity of a chemical reaction
Prediction of the useful work that could be extracted from it.
If change in G is negative then the process is spontaneous .
If change in G is positive then the process is nonspontaneous
Entropy is defined as the degree of randomness in a state