Question

For endothermic reaction when a change in entropy is negative, then the reaction is:
A.Not possible at any temperature
B.Possible at low temperature
C.Possible at all temperature
D.Possible at high temperature

Answer 1

Hint: Those chemical reactions in which there is absorption of heat energy by the reactants from the surroundings to form products are called endothermic reactions. In case of endothermic reactions, the enthalpy change is positive. From the enthalpy change and the entropy change, the change in free energy which decides the spontaneity of the chemical reaction can be determined.

Complete step by step answer:
The enthalpy change of an endothermic reaction is positive because the enthalpy of the products is greater than that of the reactants as there is absorption of heat. In other words, $\mathrm{\Delta }\text{H = + ve}$.
Entropy is the property of a system which measures the randomness or the amount of disorder of a system quantitatively. Greater the randomness of the system, greater is the entropy. It is a state function just like internal energy and enthalpy because the magnitude of its change also depends on the entropies of the system in the initial and final states. Now, according to the given question, change in entropy is negative. In other words, $\mathrm{\Delta }\text{S = - ve}$.
According to the Gibbs Helmholtz equation, 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 relation:
$\mathrm{\Delta }\text{G}=\mathrm{\Delta }\text{H - T}\mathrm{\Delta }\text{S}$
Here, T is the temperature of the system.
The Gibbs free energy concept is useful in determining the spontaneity or the feasibility of a process. The process will be spontaneous if $\mathrm{\Delta }\text{G}$ is negative and non-spontaneous when $\mathrm{\Delta }\text{G}$ is positive.
Now, for the given endothermic reaction, $\mathrm{\Delta }\text{H = + ve}$ and $\mathrm{\Delta }\text{S = - ve}$. From the Gibbs Helmholtz equation, we see that if $\mathrm{\Delta }\text{S = - ve}$, then the term $\text{T}\mathrm{\Delta }\text{S}$ is also negative.
Thus, we will have:
$$\begin{gathered} \mathrm{\Delta }\text{G = }\left(\text{ + ve}\right)\text{ - }\left(\text{ - ve}\right) \\ \Rightarrow \mathrm{\Delta }\text{G = + ve} \end{gathered}$$
Hence, free energy change is positive. Thus, the process is not spontaneous and is not possible at any temperature.

So, the correct option is A.

Note:
If both $\mathrm{\Delta }\text{H = + ve}$ and $\mathrm{\Delta }\text{S = + ve}$, then free energy change is positive when $\mathrm{\Delta }\text{H}$ is greater than $\text{T}\mathrm{\Delta }\text{S}$ and the reaction will be non-spontaneous. The free energy change is negative when $\mathrm{\Delta }\text{H}$ is less than $\text{T}\mathrm{\Delta }\text{S}$ and the reaction will be spontaneous.
If $\mathrm{\Delta }\text{H = - ve}$ and $\mathrm{\Delta }\text{S = - ve}$, then free energy change is negative when $\mathrm{\Delta }\text{H}$ is greater than $\text{T}\mathrm{\Delta }\text{S}$ and the reaction will be and the reaction will be spontaneous. The free energy change is positive when $\mathrm{\Delta }\text{H}$ is less than $\text{T}\mathrm{\Delta }\text{S}$ and the reaction will be non-spontaneous.