Question

Identify from following the correct set of thermodynamic conditions for the reaction to be spontaneous below equilibrium temperature.

• A. \( \Delta H<0 \) and \( \Delta S<0 \)
• B. \( \Delta H>0 \) and \( \Delta S>0 \)
• C. \( \Delta H<0 \) and \( \Delta S>0 \)
• D. \( \Delta H>0 \) and \( \Delta S<0 \)

Hint:

$\Delta H (-)$, $\Delta S (+)$ $\rightarrow$ Spontaneous always.
$\Delta H (-)$, $\Delta S (-)$ $\rightarrow$ Spontaneous at low $T$.
$\Delta H (+)$, $\Delta S (+)$ $\rightarrow$ Spontaneous at high $T$.

Answer 1

The Correct Option is A

Step 1: Gibbs Free Energy Equation
$\Delta G = \Delta H - T \Delta S$. For spontaneity, $\Delta G<0$.
Step 2: Condition Analysis
If both $\Delta H$ and $\Delta S$ are negative ($\Delta H<0, \Delta S<0$), the equation becomes $\Delta G = (-\text{Value}) - T(-\text{Value}) = -\Delta H + T\Delta S$.
Step 3: Temperature Dependence
To keep $\Delta G$ negative, the magnitude of $|\Delta H|$ must be greater than $|T\Delta S|$. This happens at **low temperatures** (below equilibrium temperature $T = \Delta H / \Delta S$).
Step 4: Conclusion
Exothermic reactions with decreasing entropy are spontaneous only at low temperatures.
Final Answer:(A)