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.

If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then:

• A. $\Delta H$ and $\Delta S$ both are $(+ve)$
• B. $\Delta H$ is $(-ve)$ and $\Delta S$ is $(+ve)$
• C. $\Delta H$ and $\Delta S$ both are $(-ve)$
• D. $\Delta H$ is $+ve$ and $\Delta S$ is $(-ve)$

Answer 1

Answer: (A)

$\Delta H$ and $\Delta S$ both are $(+ve)$

For endothermic reaction, $\Delta H=+ve$ as heat is absorbed during the reaction. As $\Delta G=\Delta H-T\Delta S$, is $+ve$ at low temperature if $\Delta S$ is positive but $T\Delta S<\Delta H$.
At high temperature, $T\Delta S>\Delta H$ and $\Delta G$ become $-ve$.
Thus at low temperature (freezing point), the process is non spontaneous (free energy change is positive) and at high temperature (boiling point), the process is spontaneous (free energy change is negative).