What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? `(R = 8.314 J "mol K"^(-))`

Options

A. 342`kj mol^(-1)`
B. `269kjmol^(-1)`
C. `34.7kj Mol^(-1)`
D. `15.1 kj Mol^(-1)`

Explanation

To find the activation energy (Ea) for the reaction, we can use the Arrhenius equation in the form of the logarithmic relationship between the rate constants at two different temperatures. The formula we will use is:

\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 1: Convert temperatures from Celsius to Kelvin
- Given temperatures:
- \( T_1 = 20^\circ C = 20 + 273 = 293 \, K \)
- \( T_2 = 35^\circ C = 35 + 273 = 308 \, K \)

### Step 2: Determine the rate constants
- The problem states that the rate doubles when the temperature increases from \( T_1 \) to \( T_2 \). Therefore:
- \( k_1 \) is the rate constant at \( T_1 \)
- \( k_2 = 2k_1 \)

### Step 3: Substitute values into the logarithmic equation
- Substitute \( k_2 \) into the equation:
\[ \log \frac{2k_1}{k_1} = \log 2 \]
Thus, we have:
\[ \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{308} \right) \]

### Step 4: Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \)
- Calculate \( \frac{1}{293} - \frac{1}{308} \):
\[ \frac{1}{293} \approx 0.003414 \, K^{-1} \]
\[ \frac{1}{308} \approx 0.003247 \, K^{-1} \]
\[ \frac{1}{293} - \frac{1}{308} \approx 0.003414 - 0.003247 = 0.000167 \, K^{-1} \]

### Step 5: Calculate \( \log 2 \)
- The value of \( \log 2 \) is approximately:
\[ \log 2 \approx 0.301 \]

### Step 6: Substitute and solve for \( E_a \)
- Now substitute the values into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \times 0.000167 \]
- Rearranging gives:
\[ E_a = 0.301 \times 2.303 \times 8.314 \times \frac{1}{0.000167} \]
- Calculate:
\[ E_a \approx 0.301 \times 2.303 \times 8.314 \times 5988.02 \approx 34673 \, J/mol \]

### Step 7: Convert Joules to Kilojoules
- Convert \( E_a \) from Joules to Kilojoules:
\[ E_a = 34673 \, J/mol \times \frac{1 \, kJ}{1000 \, J} \approx 34.673 \, kJ/mol \]

### Final Answer
- The activation energy \( E_a \) is approximately \( 34.7 \, kJ/mol \).