Under which of the following sets of conditions, `triangleG` is always negative ?

Options

A. `triangleH=-ve and triangleS=+ve`
B. `triangleH=-ve and triangleS=-ve`
C. `triangleH=+ve and triangleS=+ve`
D. `triangleH=+ve and triangleS=-ve`

Answer

To determine under which conditions the change in Gibbs free energy (ΔG) is always negative, we will analyze the relationship between ΔG, enthalpy (ΔH), temperature (T), and entropy (ΔS) using the Gibbs-Helmholtz equation:

### Step-by-Step Solution:

1. **Understanding the Gibbs Free Energy Equation:**
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature (in Kelvin)
- ΔS = change in entropy

2. **Identifying Conditions for ΔG < 0:**
For a reaction to be spontaneous (ΔG < 0), we need to analyze the signs of ΔH and ΔS under different conditions.

3. **Condition 1: ΔH < 0 and ΔS > 0**
- If ΔH is negative (exothermic reaction) and ΔS is positive (increase in disorder), then:
\[ \Delta G = \Delta H - T \Delta S \]
Here, both terms contribute to making ΔG negative. Thus, this condition guarantees that ΔG is always negative.

4. **Condition 2: ΔH < 0 and ΔS < 0**
- If ΔH is negative and ΔS is negative, then:
\[ \Delta G = \Delta H - T \Delta S \]
At low temperatures, ΔG can be negative, but at high temperatures, the negative term from TΔS can dominate, making ΔG positive. Therefore, this condition does not guarantee ΔG < 0.

5. **Condition 3: ΔH > 0 and ΔS > 0**
- If ΔH is positive and ΔS is positive, then:
\[ \Delta G = \Delta H - T \Delta S \]
At high temperatures, TΔS can dominate and make ΔG negative, but at low temperatures, ΔG can be positive. Thus, this condition does not guarantee ΔG < 0.

6. **Condition 4: ΔH > 0 and ΔS < 0**
- If ΔH is positive and ΔS is negative, then:
\[ \Delta G = \Delta H - T \Delta S \]
Here, both terms are positive, so ΔG will always be positive. This condition does not guarantee ΔG < 0.

### Conclusion:
The only condition under which ΔG is always negative is when ΔH is negative and ΔS is positive. Therefore, the correct answer is:

**ΔH < 0 and ΔS > 0.**