Question:
For a reaction to be spontaneous, the required conditions are:
For a reaction to be spontaneous, the required conditions are:
Show Hint
Use
\[
\Delta G=\Delta H-T\Delta S
\]
to determine spontaneity.
\[
\Delta G\lt 0 \Rightarrow \text{Spontaneous}
\]
\[
\Delta G\gt 0 \Rightarrow \text{Non-spontaneous}
\]
For \(\Delta H\gt 0\) and \(\Delta S\gt 0\), spontaneity is favored at high temperature.
Updated On: Jun 26, 2026
- \(\Delta_rH^\circ=-ve,\ \Delta_rS^\circ=-ve,\) at High \(T\)
- \(\Delta_rH^\circ=+ve,\ \Delta_rS^\circ=+ve,\) at High \(T\)
- \(\Delta_rH^\circ=+ve,\ \Delta_rS^\circ=+ve,\) at Low \(T\)
- \(\Delta_rH^\circ=+ve,\ \Delta_rS^\circ=-ve,\) at all \(T\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Recall the spontaneity criterion.
A reaction is spontaneous when Gibbs free energy change is negative: \[ \Delta G=\Delta H-T\Delta S \] For spontaneity, \[ \Delta G\lt 0 \]
Step 2: Analyze Option (1).
Given: \[ \Delta H\lt 0,\qquad \Delta S\lt 0 \] Then, \[ \Delta G=\Delta H+T|\Delta S| \] At high temperature, the positive term \[ T|\Delta S| \] becomes large and may make \(\Delta G\) positive.
Hence, spontaneity is favored at low temperature, not high temperature.
Therefore, Option (1) is incorrect.
Step 3: Analyze Option (2).
Given: \[ \Delta H\gt 0,\qquad \Delta S\gt 0 \] Then, \[ \Delta G=\Delta H-T\Delta S \] At high temperature, the term \[ T\Delta S \] becomes sufficiently large and can exceed \(\Delta H\).
Thus, \[ \Delta G\lt 0 \] and the reaction becomes spontaneous.
Therefore, Option (2) is correct.
Step 4: Analyze Options (3) and (4).
For Option (3): \[ \Delta H\gt 0,\qquad \Delta S\gt 0 \] At low temperature, \[ T\Delta S \] is small, so \(\Delta G\) remains positive. Hence, not spontaneous.
For Option (4): \[ \Delta H\gt 0,\qquad \Delta S\lt 0 \] Then, \[ \Delta G=\Delta H+T|\Delta S| \] which is always positive. Therefore, the reaction is non-spontaneous at all temperatures.
Step 5: Final conclusion.
The reaction is spontaneous under the condition: \[ \boxed{\Delta_rH^\circ=+ve,\ \Delta_rS^\circ=+ve,\ \text{at High }T} \] Hence, the correct option is \[ \boxed{(2)} \]
A reaction is spontaneous when Gibbs free energy change is negative: \[ \Delta G=\Delta H-T\Delta S \] For spontaneity, \[ \Delta G\lt 0 \]
Step 2: Analyze Option (1).
Given: \[ \Delta H\lt 0,\qquad \Delta S\lt 0 \] Then, \[ \Delta G=\Delta H+T|\Delta S| \] At high temperature, the positive term \[ T|\Delta S| \] becomes large and may make \(\Delta G\) positive.
Hence, spontaneity is favored at low temperature, not high temperature.
Therefore, Option (1) is incorrect.
Step 3: Analyze Option (2).
Given: \[ \Delta H\gt 0,\qquad \Delta S\gt 0 \] Then, \[ \Delta G=\Delta H-T\Delta S \] At high temperature, the term \[ T\Delta S \] becomes sufficiently large and can exceed \(\Delta H\).
Thus, \[ \Delta G\lt 0 \] and the reaction becomes spontaneous.
Therefore, Option (2) is correct.
Step 4: Analyze Options (3) and (4).
For Option (3): \[ \Delta H\gt 0,\qquad \Delta S\gt 0 \] At low temperature, \[ T\Delta S \] is small, so \(\Delta G\) remains positive. Hence, not spontaneous.
For Option (4): \[ \Delta H\gt 0,\qquad \Delta S\lt 0 \] Then, \[ \Delta G=\Delta H+T|\Delta S| \] which is always positive. Therefore, the reaction is non-spontaneous at all temperatures.
Step 5: Final conclusion.
The reaction is spontaneous under the condition: \[ \boxed{\Delta_rH^\circ=+ve,\ \Delta_rS^\circ=+ve,\ \text{at High }T} \] Hence, the correct option is \[ \boxed{(2)} \]
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