Question

For a reaction at $25{}^{\circ }\text{C}$ enthalpy change $\left(\Delta H\right)$ and entropy change $\left(\Delta S\right)$ are $-11.7\times {10}^3{\text{J mol}}^{-1}$ and $-105{\text{J mol}}^{-1}{\text{K}}^{-1}$ respectively. The reaction is:

• A. Spontaneous
• B. Non-spontaneous
• C. Instantaneous
• D. None of these

Answer 1

The correct option is B Non-spontaneous

As we know,
$\Delta G=\Delta H-T\Delta S$
$\therefore \Delta G=-11700-298\times (-105)=+19590\text{J}=19.59\text{kJ}$
Thus, reaction is non-spontaneous.