Question

For a chemical reaction, $\Delta {\text{G}}$ will always be negative if
A.$\Delta {\text{H}}$ and ${\text{T}}\Delta {\text{S}}$ both are positive
B.$\Delta {\text{H}}$ and ${\text{T}}\Delta {\text{S}}$ both are negative
C.$\Delta {\text{H}}$ is negative and ${\text{T}}\Delta {\text{S}}$ is positive
D.$\Delta {\text{H}}$ is positive and ${\text{T}}\Delta {\text{S}}$ is negative

Answer 1

The change in the free energy of a system is related to the change in enthalpy as well as the change in entropy by the Gibbs Helmholtz equation:
$\Delta {\text{G}} = \Delta {\text{H - T}}\Delta {\text{S}}$
Here, T denotes the temperature of the system.
If $\Delta {\text{G}}$ is negative, the reaction process will be spontaneous.

Complete step by step answer:
When both $\Delta {\text{H}}$ and ${\text{T}}\Delta {\text{S}}$ are positive, it indicates that the energy factor (change in enthalpy) opposes the process but the randomness factor (the entropy factor) favours the process. Then we have the following possibilities.
If the energy factor is greater than the randomness factor, i.e., $\Delta {\text{H}}$ is greater than ${\text{T}}\Delta {\text{S}}$ , then the change in free energy will be positive, i.e., $\Delta {\text{G}}$ will be positive. Hence, the process is non – spontaneous.
If the energy factor is less than the randomness factor, i.e., $\Delta {\text{H}}$ is less than ${\text{T}}\Delta {\text{S}}$ , then the change in free energy will be negative, i.e., $\Delta {\text{G}}$ will be negative. Hence, the process is spontaneous.
If $\Delta {\text{H}}$ is equal to ${\text{T}}\Delta {\text{S}}$ , then $\Delta {\text{G}}$ will be zero and the process is in equilibrium.
Thus, option A is wrong.
When $\Delta {\text{H}}$ and ${\text{T}}\Delta {\text{S}}$ are negative, it indicates that the energy factor favours the process but the randomness factor opposes it. Then,
If $\Delta {\text{H}}$ is greater than ${\text{T}}\Delta {\text{S}}$ , then the change in free energy will be negative, i.e., $\Delta {\text{G}}$ will be negative. Hence, the process is spontaneous.
If $\Delta {\text{H}}$ is less than ${\text{T}}\Delta {\text{S}}$ , then $\Delta {\text{G}}$ will be positive. Hence, the process is nonspontaneous.
If $\Delta {\text{H}}$ is equal to ${\text{T}}\Delta {\text{S}}$ , then $\Delta {\text{G}}$ will be zero and the process is in equilibrium.
Thus, option B is wrong.
When $\Delta {\text{H}}$ is negative and ${\text{T}}\Delta {\text{S}}$ is positive, it indicates that both the energy factor and the randomness factor favours the process. Thus, the process is highly spontaneous and $\Delta {\text{G}}$ will be highly negative at all temperatures. Thus, option C is correct.
When $\Delta {\text{H}}$ is positive and ${\text{T}}\Delta {\text{S}}$ is negative, it indicates that both the energy factor and the randomness factor oppose the process. Thus, the process is highly non-spontaneous and $\Delta {\text{G}}$ will be highly positive at all temperatures. Thus, option D is wrong.

Hence option C is correct.

Note:
In case of endothermic reactions, there is absorption of heat energy by the reactants from the surroundings and so the change in enthalpy for endothermic reactions is always positive. Hence, endothermic reactions will be spontaneous if the randomness factor is positive and greater than the energy factor.
In case of exothermic reactions, there is release of heat. And so change in enthalpy for exothermic reactions is always negative. Hence, exothermic reactions will be highly spontaneous if the randomness factor is also positive.