What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300 K to 400 K ? `(R="8.314 J mol"^(-1)"K"^(-1))`

Options

A. 68.8
B. 6.88
C. 34.4
D. 3.44

Explanation

To find the activation energy (EA) for a reaction where the rate constant doubles when the temperature is raised from 300 K to 400 K, we can use the Arrhenius equation in the form of the logarithmic relationship:

\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_A}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 1: Identify the values
- Given:
- \( K_2 = 2K_1 \) (since the rate constant doubles)
- \( T_1 = 300 \, \text{K} \)
- \( T_2 = 400 \, \text{K} \)
- \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)

### Step 2: Substitute the values into the equation
Since \( K_2 = 2K_1 \), we can express the logarithm as:

\[ \log \left( \frac{K_2}{K_1} \right) = \log(2) \]

Using the value of \( \log(2) \approx 0.301 \):

\[ 0.301 = \frac{E_A}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{400} \right) \]

### Step 3: Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \)
Calculating \( \frac{1}{300} - \frac{1}{400} \):

\[ \frac{1}{300} = 0.003333 \, \text{K}^{-1} \]
\[ \frac{1}{400} = 0.0025 \, \text{K}^{-1} \]
\[ \frac{1}{300} - \frac{1}{400} = 0.003333 - 0.0025 = 0.000833 \, \text{K}^{-1} \]

### Step 4: Substitute back into the equation
Now substituting this value back into the equation:

\[ 0.301 = \frac{E_A}{2.303 \times 8.314} \times 0.000833 \]

### Step 5: Solve for \( E_A \)
First, calculate \( 2.303 \times 8.314 \):

\[ 2.303 \times 8.314 \approx 19.1 \]

Now rearranging the equation to solve for \( E_A \):

\[ E_A = 0.301 \times 19.1 \times \frac{1}{0.000833} \]

Calculating \( \frac{1}{0.000833} \approx 1200 \):

\[ E_A \approx 0.301 \times 19.1 \times 1200 \]

Calculating \( 0.301 \times 19.1 \approx 5.75 \):

\[ E_A \approx 5.75 \times 1200 \approx 6900 \, \text{J/mol} \]

### Step 6: Convert to kJ/mol
To convert from J/mol to kJ/mol:

\[ E_A \approx \frac{6900}{1000} = 6.9 \, \text{kJ/mol} \]

### Final Answer
Thus, the activation energy \( E_A \) is approximately:

\[ E_A \approx 6.9 \, \text{kJ/mol} \]