Question

What is the nature of reaction at $298\, K$, if the entropy change and enthalpy change for a chemical reaction are $7.4\, cal \,K ^{-1}$ and $-2.5 \times 10^{3}\, cal$ respectively?

• A. Reversible
• B. Spontaneous
• C. Non-Spontaneous
• D. Irreversible

Answer 1

Answer: (B)

Given, $T=298 \,K$

Entropy change $(\Delta S)=7.4$ cal $K ^{-1}$

Enthalpy change $(\Delta H)=-2.5 \times 10^{3}$ cal

As we know that,

$ \Delta G=\Delta H-T \Delta S $

$ \therefore \Delta G=-2.5 \times 10^{3}-(298 \times 7.4) $

$ \rightarrow \Delta G=-2.5 \times 10^{3}-2.2 \times 10^{3} cal $

$ \Delta G=-4.7 \times 10^{3} cal $

Since, the value of $\Delta G$ is negative and thus, the reaction is spontaneous in nature.