The value of rate constant for a second order reaction is `6.7 xx 10^(-5) mol^(-) L s^(-)` at 298 K and `1.64 xx 10^(-4) mol^(-) L s^(-) ` at 313 K. Find the Arrhenius frequency factor A and activation energy of the reaction.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
We have two rate constants for a second-order reaction:
- \( k_1 = 6.7 \times 10^{-5} \, \text{mol}^{-1} \text{L s}^{-1} \) at \( T_1 = 298 \, \text{K} \)
- \( k_2 = 1.64 \times 10^{-4} \, \text{mol}^{-1} \text{L s}^{-1} \) at \( T_2 = 313 \, \text{K} \)

### Step 2: Use the Arrhenius equation
The Arrhenius equation relates the rate constant \( k \) to the temperature \( T \) and the activation energy \( E_a \):
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
- \( A \) = Arrhenius frequency factor
- \( R \) = Universal gas constant \( = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)

### Step 3: Use the logarithmic form of the Arrhenius equation
We can use the logarithmic form to find the activation energy:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ \log \frac{1.64 \times 10^{-4}}{6.7 \times 10^{-5}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{313} \right) \]

### Step 5: Calculate the left-hand side
Calculating the left-hand side:
\[ \log \frac{1.64 \times 10^{-4}}{6.7 \times 10^{-5}} = \log 2.447 \approx 0.388 \]

### Step 6: Calculate the right-hand side
Calculate \( \frac{1}{298} - \frac{1}{313} \):
\[ \frac{1}{298} - \frac{1}{313} = \frac{313 - 298}{298 \times 313} = \frac{15}{298 \times 313} \approx 0.000158 \]

### Step 7: Rearranging to find \( E_a \)
Rearranging the equation to solve for \( E_a \):
\[ E_a = \frac{0.388 \times 2.303 \times 8.314}{0.000158} \]

### Step 8: Calculate \( E_a \)
Calculating \( E_a \):
\[ E_a \approx \frac{0.388 \times 2.303 \times 8.314}{0.000158} \approx 119.06 \, \text{kJ/mol} \]

### Step 9: Find the Arrhenius frequency factor \( A \)
Using the equation:
\[ \log k_1 = \log A - \frac{E_a}{2.303RT_1} \]
Rearranging gives:
\[ \log A = \log k_1 + \frac{E_a}{2.303RT_1} \]

### Step 10: Substitute values to find \( A \)
Substituting the values:
\[ \log A = \log(6.7 \times 10^{-5}) + \frac{119060}{2.303 \times 8.314 \times 298} \]

### Step 11: Calculate \( \log A \)
Calculating \( \log(6.7 \times 10^{-5}) \approx -4.173 \):
\[ \log A \approx -4.173 + \frac{119060}{2.303 \times 8.314 \times 298} \approx -4.173 + 3.937 \]
\[ \log A \approx -0.236 \]

### Step 12: Find \( A \)
Taking the antilog:
\[ A \approx 10^{-0.236} \approx 0.576 \, \text{mol}^{-1} \text{L s}^{-1} \]

### Final Results
- Activation Energy \( E_a \approx 119.06 \, \text{kJ/mol} \)
- Arrhenius Frequency Factor \( A \approx 0.576 \, \text{mol}^{-1} \text{L s}^{-1} \)