Question

Free energy, $G=H-T\Delta S$, is a state function that indicates whether a reaction is spontaneous or non-spontaneous. If you think of $TS$ as the part of the system`s energy that is disordered already, then $(H-TS)$ is the part of the system`s energy that is still ordered and therefore free to cause spontaneous change by becoming disordered.
Also, $\Delta G=\Delta H-T\Delta S$

From the second law of Thermodynamics, a reaction is spontaneous if ${\Delta }_{total}S$ is positive, non-spontaneous if ${\Delta }_{total}S$ is negative, and at equilibrium if ${\Delta }_{total}S$ is zero.

Since, $-T\Delta S=\Delta G$ and since $\Delta G$ and $\Delta S$ have opposite signs, we can restate the Thermodynamic criterion for the spontaneity of a reaction carried out at a constant temperature and pressure.
If $\Delta G<0$, the reaction is spontaneous
If $\Delta G>0$, the reaction is non-spontaneous
If $\Delta G=0$, the reaction is at equilibrium

Read the above paragraph carefully and answer the following questions based on the above comprehension.

A particular reaction has a negative value for the free energy change. Then at ordinary temperature:

• A. It has a $(+ve)$ value for the entropy change and a $(-ve)$ change for the enthalpy change.
• B. It has a small $(+ve)$ value for enthalpy change
• C. It has a large $(+ve)$ value for the enthalpy change
• D. It has a large $(-ve)$ value for the entropy change

Answer 1

Answer: (A)

It has a $(+ve)$ value for the entropy change and a $(-ve)$ change for the enthalpy change.

$\Delta G=\Delta H-T\Delta S$ can be negative at ordinary temperature $T$ if $\Delta H$ is $-ve$ and $\Delta S$ is $+ve$ since it gives overall value of $\Delta G$ as negative.