Which of the following conditions is not favourable for a spontaneous reaction?

Options

A. `DeltaH = + ve, TDeltaS = +ve and TDeltaS gt DeltaH`
B. `DeltaH = +ve, T DeltaS = +ve and TDeltaS lt DeltaH`
C. `DeltaH = -ve, T DeltaS = +ve and TDeltaS gt DeltaH`
D. `DeltaH = -ve, T DeltaS = -ve and TDeltaS lt DeltaH`

Answer

To determine which of the given conditions is not favorable for a spontaneous reaction, we can use the Gibbs-Helmholtz equation:

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

Where:
- \(\Delta G\) = change in Gibbs free energy
- \(\Delta H\) = change in enthalpy
- \(T\) = temperature (in Kelvin)
- \(\Delta S\) = change in entropy

For a reaction to be spontaneous, \(\Delta G\) must be negative. Let's analyze each condition step by step.

### Step 1: Analyze Condition A
- **Given**: \(\Delta H > 0\) (positive) and \(T \Delta S > 0\) (positive), with \(|T \Delta S| > |\Delta H|\).
- **Analysis**: Since \(|T \Delta S|\) is greater than \(|\Delta H|\), the negative term \(T \Delta S\) dominates. Therefore, \(\Delta G\) will be negative, indicating that the reaction is spontaneous.

### Step 2: Analyze Condition B
- **Given**: \(\Delta H > 0\) (positive) and \(T \Delta S > 0\) (positive), with \(|\Delta H| > |T \Delta S|\).
- **Analysis**: Here, the positive term \(\Delta H\) dominates. Thus, \(\Delta G\) will be positive, indicating that the reaction is non-spontaneous.

### Step 3: Analyze Condition C
- **Given**: \(\Delta H < 0\) (negative) and \(T \Delta S > 0\) (positive), with \(|T \Delta S| > |\Delta H|\).
- **Analysis**: The negative term \(\Delta H\) and the positive term \(T \Delta S\) combine to make \(\Delta G\) even more negative, indicating that the reaction is spontaneous.

### Step 4: Analyze Condition D
- **Given**: \(\Delta H < 0\) (negative) and \(T \Delta S < 0\) (negative), with \(|\Delta H| > |T \Delta S|\).
- **Analysis**: In this case, the negative term \(\Delta H\) dominates, leading to a positive \(\Delta G\). Therefore, the reaction is also spontaneous.

### Conclusion
From the analysis, the condition that is not favorable for a spontaneous reaction is **Condition B**, where both \(\Delta H\) and \(T \Delta S\) are positive, but \(|\Delta H|\) is greater than \(|T \Delta S|\).

### Final Answer
**Condition B is not favorable for a spontaneous reaction.**

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