The rate of a reaction becomes 4 times when temperature is raised from 293 K to 313 K. The activation energy for such reaction would be

Options

A. 50.855 kJ `" mol"^(-1)`
B. 52.849 kJ `" mol"^(-1)`
C. 54.855 kJ `" mol"^(-1)`
D. 56.855 kJ `" mol"^(-1)`

Explanation

To find the activation energy (Ea) of the reaction given that the rate increases 4 times when the temperature is raised from 293 K to 313 K, we can use the Arrhenius equation in its logarithmic form:

### Step 1: Write down the Arrhenius equation in logarithmic form
The Arrhenius equation in logarithmic form is given by:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
where:
- \( k_1 \) is the rate constant at temperature \( T_1 \)
- \( k_2 \) is the rate constant at temperature \( T_2 \)
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( E_a \) is the activation energy

### Step 2: Identify the values
From the problem:
- The rate becomes 4 times, which means \( k_2 = 4 k_1 \).
- Therefore, \( \frac{k_2}{k_1} = 4 \).
- The temperatures are \( T_1 = 293 \, \text{K} \) and \( T_2 = 313 \, \text{K} \).

### Step 3: Substitute the values into the equation
Substituting the values into the logarithmic form:
\[ \log 4 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{313} \right) \]

### Step 4: Calculate \( \log 4 \)
Calculating \( \log 4 \):
\[ \log 4 \approx 0.602 \]

### Step 5: Calculate \( \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \)
Calculating \( \left( \frac{1}{293} - \frac{1}{313} \right) \):
\[ \frac{1}{293} \approx 0.003412 \, \text{K}^{-1} \]
\[ \frac{1}{313} \approx 0.003194 \, \text{K}^{-1} \]
\[ \frac{1}{293} - \frac{1}{313} \approx 0.003412 - 0.003194 = 0.000218 \, \text{K}^{-1} \]

### Step 6: Substitute back into the equation
Now substituting back into the equation:
\[ 0.602 = \frac{E_a}{2.303 \times 8.314} \times 0.000218 \]

### Step 7: Solve for \( E_a \)
Rearranging to solve for \( E_a \):
\[ E_a = 0.602 \times 2.303 \times 8.314 \times \frac{1}{0.000218} \]

Calculating:
\[ E_a = 0.602 \times 2.303 \times 8.314 \times 4587.156 \]
\[ E_a \approx 52.849 \, \text{kJ/mol} \]

### Final Answer
Thus, the activation energy \( E_a \) for the reaction is approximately:
\[ E_a \approx 52.849 \, \text{kJ/mol} \]