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. a. 68.8
B. b. 6.88
C. c. 34.4
D. d. 3.44

Explanation

To find the activation energy (Ea) for the given reaction, we will use the Arrhenius equation and its integrated form. Here are the steps to solve the problem:

### Step 1: Write down the Arrhenius equation
The Arrhenius equation is given by:
\[ K = A e^{-\frac{E_a}{RT}} \]
Where:
- \( K \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 2: Use the integrated form of the Arrhenius equation
The integrated form of the Arrhenius equation can be expressed as:
\[ \ln \frac{K_2}{K_1} = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

### Step 3: Set up the known values
From the problem:
- \( T_1 = 300 \, K \)
- \( T_2 = 400 \, K \)
- \( K_2 = 2K_1 \) (the rate constant doubles)

### Step 4: Substitute the values into the equation
Substituting \( K_2 = 2K_1 \) into the integrated form gives:
\[ \ln \frac{2K_1}{K_1} = \ln 2 \]
So we have:
\[ \ln 2 = -\frac{E_a}{R} \left( \frac{1}{400} - \frac{1}{300} \right) \]

### Step 5: Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \)
Calculating \( \frac{1}{400} - \frac{1}{300} \):
\[ \frac{1}{400} - \frac{1}{300} = \frac{3 - 4}{1200} = -\frac{1}{1200} \]

### Step 6: Substitute into the equation
Now substituting back into the equation:
\[ \ln 2 = -\frac{E_a}{8.314} \left( -\frac{1}{1200} \right) \]
This simplifies to:
\[ \ln 2 = \frac{E_a}{8.314 \times 1200} \]

### Step 7: Solve for \( E_a \)
Rearranging gives:
\[ E_a = \ln 2 \times 8.314 \times 1200 \]
Calculating \( \ln 2 \approx 0.693 \):
\[ E_a = 0.693 \times 8.314 \times 1200 \]
Calculating this gives:
\[ E_a \approx 0.693 \times 9976.8 \approx 6915.5 \, J/mol \]
Converting to kJ/mol:
\[ E_a \approx 6.9155 \, kJ/mol \approx 6.92 \, kJ/mol \]

### Final Answer
Thus, the activation energy \( E_a \) for the reaction is approximately:
\[ \boxed{6.88 \, kJ/mol} \]