For `A rarr B, DeltaH= 3.5` kcal `mol^(-1), DeltaS= 10` cal `mol^(-1) K^(-1)`. Reaction is non spontaneous at

Options

A. 400K
B. `27^(@)C`
C. `77^(@)C`
D. 350K

Answer

To determine the temperature at which the reaction \( A \rightarrow B \) is non-spontaneous, we can use the Gibbs free energy equation:

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

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

### Step 1: Convert the given values
Given:
- \( \Delta H = 3.5 \, \text{kcal/mol} = 3.5 \times 10^3 \, \text{cal/mol} \)
- \( \Delta S = 10 \, \text{cal/mol K} \)

### Step 2: Set up the equation for non-spontaneity
For the reaction to be non-spontaneous, \( \Delta G \) must be greater than 0:

\[ \Delta H - T \Delta S > 0 \]

This can be rearranged to find the temperature:

\[ \Delta H > T \Delta S \]

\[ T < \frac{\Delta H}{\Delta S} \]

### Step 3: Calculate the critical temperature
Substituting the values we have:

\[ T < \frac{3.5 \times 10^3 \, \text{cal/mol}}{10 \, \text{cal/mol K}} = 350 \, \text{K} \]

### Step 4: Determine the non-spontaneous temperature range
The reaction is non-spontaneous at temperatures less than or equal to 350 K.

### Step 5: Analyze the options
Now we will analyze the options provided:
1. **400 K**: Greater than 350 K (spontaneous)
2. **27 °C**: \( 27 + 273 = 300 \, \text{K} \) (less than 350 K, non-spontaneous)
3. **77 °C**: \( 77 + 273 = 350 \, \text{K} \) (equal to 350 K, non-spontaneous)
4. **350 K**: Equal to 350 K (non-spontaneous)

### Conclusion
The reaction is non-spontaneous at:
- 27 °C (300 K)
- 77 °C (350 K)
- 350 K

Thus, the correct options are 27 °C, 77 °C, and 350 K.