The rate of the chemical reaction doubles for an increase of `10K` in absolute temperature from `300K`. Calculate `E_(a)`.

Explanation

To solve the problem of calculating the activation energy \( E_a \) given that the rate of a chemical reaction doubles with a temperature increase of \( 10 \, K \) from \( 300 \, K \), we can use the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation relates the rate constant \( k \) to the activation energy \( E_a \) and temperature \( T \):
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = absolute temperature in Kelvin

### Step 2: Set Up the Problem
Given:
- Initial temperature \( T_1 = 300 \, K \)
- Final temperature \( T_2 = 310 \, K \)
- The rate doubles, so \( k_2 = 2k_1 \)

### Step 3: Use the Arrhenius Equation for Two Temperatures
Using the Arrhenius equation for both temperatures, we have:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \]

### Step 4: Substitute Known Values
Since \( k_2 = 2k_1 \):
\[ 2 = e^{\frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right)} \]

### Step 5: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln(2) = \frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right) \]

### Step 6: Calculate the Temperature Difference
Calculate \( \frac{1}{300} - \frac{1}{310} \):
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

### Step 7: Substitute Values into the Equation
Now substitute \( R = 8.314 \, J/mol·K \) and \( \ln(2) \approx 0.693 \):
\[ 0.693 = \frac{E_a}{8.314} \left( \frac{1}{9300} \right) \]

### Step 8: Solve for \( E_a \)
Rearranging gives:
\[ E_a = 0.693 \times 8.314 \times 9300 \]
Calculating:
\[ E_a \approx 0.693 \times 8.314 \times 9300 \approx 5330.4 \, J/mol \approx 53.3 \, kJ/mol \]

### Final Answer
The activation energy \( E_a \) is approximately \( 53.3 \, kJ/mol \).