Which of the following are favourable conditions for spontaneous processes?

Options

A. `DeltaH (-), DeltaS(+)` any magnitude
B. `DeltaH (-), DeltaS(-) DeltaH lt TDeltaS`
C. `DeltaH(+), DeltaS(+) DeltaH lt T DeltaS`
D. `DeltaH(+), DeltaS(+) DeltaH gt T DeltaS`

Answer

To determine the favorable conditions for spontaneous processes, we need to analyze the Gibbs free energy change (ΔG) using the Gibbs-Helmholtz 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

For a process to be spontaneous, ΔG must be negative (ΔG < 0).

### Step-by-Step Solution:

1. **Identify the Conditions for Spontaneity**:
- A process is spontaneous if ΔG < 0.
- This can occur under different combinations of ΔH and ΔS.

2. **Analyze Each Case**:
- **Case 1**: ΔH < 0 (negative) and ΔS > 0 (positive)
- Here, ΔG = ΔH - TΔS will be negative because both terms contribute to a negative value. Thus, this process is spontaneous.

- **Case 2**: ΔH < 0 (negative) and ΔS < 0 (negative)
- In this case, ΔG = ΔH - TΔS. Since ΔH is negative and TΔS is also negative, the overall result can be positive if the magnitude of TΔS is greater than ΔH. Thus, this process is non-spontaneous.

- **Case 3**: ΔH > 0 (positive) and ΔS > 0 (positive)
- Here, ΔG = ΔH - TΔS. If TΔS is greater than ΔH, ΔG will be negative, making the process spontaneous.

- **Case 4**: ΔH > 0 (positive) and ΔS < 0 (negative)
- In this case, both ΔH and TΔS are positive, leading to ΔG being positive. Thus, this process is non-spontaneous.

3. **Conclusion**:
- The conditions favorable for spontaneous processes are:
- ΔH < 0 and ΔS > 0 (Spontaneous)
- ΔH > 0 and ΔS > 0 (Spontaneous if TΔS > ΔH)

### Summary of the Conditions:
- **Favorable for Spontaneity**:
- ΔH < 0 and ΔS > 0
- ΔH > 0 and ΔS > 0 (if TΔS > ΔH)