Standard entropy of A, B and C are 30, 60 and 100 `JK^-1` `mol^-1` respectively. For the reaction `(2A + 5Brarr 6C)`, If the AH of the reaction is 300 kJ then the temperature at which the reaction will become spontaneous is

Options

A. 1150 K
B. 1240 K
C. 1000 K
D. 1260 K

Answer

To determine the temperature at which the reaction \(2A + 5B \rightarrow 6C\) becomes spontaneous, we can follow these steps:

### Step 1: Write down the given data
- Standard entropies:
- \(S^\circ_A = 30 \, \text{JK}^{-1} \text{mol}^{-1}\)
- \(S^\circ_B = 60 \, \text{JK}^{-1} \text{mol}^{-1}\)
- \(S^\circ_C = 100 \, \text{JK}^{-1} \text{mol}^{-1}\)
- Enthalpy change of the reaction:
- \(\Delta H = 300 \, \text{kJ} = 300,000 \, \text{J}\)

### Step 2: Calculate the change in entropy (\(\Delta S\)) for the reaction
Using the formula:
\[ \Delta S = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]
For the reaction \(2A + 5B \rightarrow 6C\):
\[ \Delta S = [6 \times S^\circ_C] - [2 \times S^\circ_A + 5 \times S^\circ_B] \]
Substituting the values:
\[ \Delta S = [6 \times 100] - [2 \times 30 + 5 \times 60] \]
Calculating each term:
\[ \Delta S = 600 - [60 + 300] = 600 - 360 = 240 \, \text{J K}^{-1} \]

### Step 3: Convert \(\Delta S\) to kJ
Since \(\Delta H\) is in kJ, we convert \(\Delta S\) to kJ:
\[ \Delta S = 240 \, \text{J K}^{-1} = 0.240 \, \text{kJ K}^{-1} \]

### Step 4: Use the Gibbs free energy equation
The condition for spontaneity is:
\[ \Delta G < 0 \]
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
For spontaneity:
\[ \Delta H - T \Delta S < 0 \]
Rearranging gives:
\[ T \Delta S > \Delta H \]
\[ T > \frac{\Delta H}{\Delta S} \]

### Step 5: Substitute the values
Substituting the known values:
\[ T > \frac{300 \, \text{kJ}}{0.240 \, \text{kJ K}^{-1}} = \frac{300}{0.240} = 1250 \, \text{K} \]

### Step 6: Conclusion
The temperature at which the reaction will become spontaneous must be greater than \(1250 \, \text{K}\).