Question

Gibbs-Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as
$(\Delta G{)}_{PT}=\Delta H-T\Delta S$
The magnitude of $\Delta H$ does not change much with the change in temperature but the entropy factor $T\Delta S$ changes appreciably.
Thus, spontaneity of a process depends very much on temperature.

For the reaction at ${25}^{\circ }C$, $X_2{O}_4\left(l\right)⟶2X{O}_2$
$\Delta H=2.0Kcal$ and $\Delta S=20cal{K}^{-1}$ the reaction would be:

• A. spontaneous
• B. at equilibrium
• C. unpredictable
• D. non-spontaneous

Answer 1

The correct option is A spontaneous

a. The relationship between the Gibbs free energy change $\Delta G$, the enthalpy change $\Delta H$ and the entropy change $\Delta S$ is as shown.
$\Delta G=\Delta H-T\Delta S$ But $\Delta H=2.0Kcal=2000.0cal$ $\Delta S=20cal{K}^{-1}$ $T=298K$
Substitute values in the above expression
$\Delta G=2000.0-298\times 20cal{K}^{-1}$ $=-3960.0$ $=-ve$,
hence spontaneous.
Note: For a spontaneous process, the free energy change is negative. For equilibrium process, the free energy change is zero and for non spontaneous process, free energy change is positive.