Which of the following is correct ?

Options

A. `{:(DeltaH , DeltaS, " Nature of reaction ") , ((+),(-), "Nonspontaneous regardless of temperature "):}`
B. `{:(DeltaH , DeltaS, " Nature of reaction ") ,( (-),(+), "Spontaneous only at high temperature"):}`
C. `{:(DeltaH , DeltaS, " Nature of reaction "),((+),(+),"Spontaneous only at low temperature"):}`
D. `{:(DeltaH , DeltaS, " Nature of reaction ") ,((+),(-),"Spontaneous at all temperatures "):}`

Answer

To determine which of the given statements about the relationship between ΔH (enthalpy change) and ΔS (entropy change) is correct, we can analyze the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔG = Change in Gibbs free energy
- ΔH = Change in enthalpy
- ΔS = Change in entropy
- T = Temperature in Kelvin

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
- The spontaneity of a reaction is determined by the sign of ΔG. If ΔG < 0, the reaction is spontaneous. If ΔG > 0, the reaction is non-spontaneous.

2. **Case 1: ΔH is Positive and ΔS is Negative**:
- If ΔH > 0 (endothermic reaction) and ΔS < 0 (decrease in entropy), then:
\[ \Delta G = \Delta H - T \Delta S \]
- Here, ΔH is positive and TΔS is negative, making ΔG positive. Thus, the reaction is non-spontaneous at all temperatures.

3. **Case 2: ΔH is Negative and ΔS is Positive**:
- If ΔH < 0 (exothermic reaction) and ΔS > 0 (increase in entropy), then:
\[ \Delta G = \Delta H - T \Delta S \]
- Both terms contribute negatively, making ΔG negative. Thus, the reaction is spontaneous at all temperatures.

4. **Case 3: Both ΔH and ΔS are Positive**:
- If ΔH > 0 and ΔS > 0, then:
\[ \Delta G = \Delta H - T \Delta S \]
- At low temperatures, ΔG could be positive (non-spontaneous), but as temperature increases, the TΔS term becomes significant, potentially making ΔG negative. Thus, the reaction can become spontaneous at high temperatures.

5. **Case 4: ΔH is Positive and ΔS is Negative** (repeated for clarity):
- This case is similar to Case 1, confirming that the reaction is non-spontaneous at all temperatures.

### Conclusion:
From the analysis, we can conclude that:
- If ΔH is positive and ΔS is negative, the reaction is always non-spontaneous.
- If ΔH is negative and ΔS is positive, the reaction is always spontaneous.
- If both ΔH and ΔS are positive, the reaction may be spontaneous at high temperatures.

### Final Answer:
The correct statement is that if ΔH is negative and ΔS is positive, the reaction is spontaneous regardless of temperature.

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