Question

During a process the enthalpy change of the system, \( \Delta H \) is positive and entropy change of the system, \( \Delta S \) is negative. The process will be:

• A. spontaneous at all temperatures
• B. spontaneous when \( \Delta H \) is negative and \( \Delta S \) is positive
• C. spontaneous when \( \Delta H \) is positive and \( \Delta S \) is negative
• D. non-spontaneous at all temperatures

Hint:

To determine the spontaneity of a process, check the signs of \( \Delta H \) and \( \Delta S \). If \( \Delta H \) is positive and \( \Delta S \) is negative, the process will always be non-spontaneous.

Answer 1

The Correct Option is D

The spontaneity of a process is determined by the Gibbs free energy change, \( \Delta G \), which is given by the equation: \[ \Delta G = \Delta H - T \Delta S \] For a process to be spontaneous, \( \Delta G \) must be negative.
In this case:
- \( \Delta H \) is positive, which means the system absorbs heat.
- \( \Delta S \) is negative, indicating a decrease in entropy.
Thus, \( \Delta G = \Delta H - T \Delta S \) will always be positive (since both \( \Delta H>0 \) and \( \Delta S<0 \)). Therefore, the process will always be non-spontaneous at all temperatures.