Question

In the reaction $P+Q\to R+S$, there is no change in entropy. Enthalpy change for the reaction ($\Delta H$) is 12 kJ $mo{l}^{-1}$. Under the conditions, reaction will have negative value of free energy change:

• A. If $\Delta H$ is positive.
• B. If $\Delta H$ is negative.
• C. If $\Delta H$ is 24kJ $mo{l}^{-1}$.
• D. If temperature of reaction is high.

Answer 1

Answer: (B)

If $\Delta H$ is negative.

$\Delta G=\Delta H-T\Delta S\Delta S=0⟹Given\therefore \Delta G=\Delta H$

For negative value of $\Delta G$, $\Delta H$ should be negative.