The activation energy for a reaction that doubles the rate when the temperature is raised from 300 K to 310 K is (log 2 = 0.3)

Options

A. `50.6 kJ mol^(-1)`
B. `75.45 kJ mol^(-1)`
C. `45.5 kJ mol^(-1)`
D. `53.4 kJ mol^(-1)`

Answer

To solve the problem of finding the activation energy (EA) for a reaction that doubles the rate when the temperature is raised from 300 K to 310 K, we can use the Arrhenius equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We need to find the activation energy (EA) given that the rate of reaction doubles when the temperature increases from 300 K to 310 K. We are also given that log 2 = 0.3.

### Step 2: Use the Arrhenius Equation
The Arrhenius equation in logarithmic form is:
\[ \log \frac{k_2}{k_1} = \frac{E_A}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
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)
- \( T_1 = 300 \, \text{K} \)
- \( T_2 = 310 \, \text{K} \)

### Step 3: Set Up the Equation
Since the rate doubles, we have:
\[ k_2 = 2k_1 \implies \frac{k_2}{k_1} = 2 \]
Thus, we can substitute this into the equation:
\[ \log 2 = \frac{E_A}{2.303 R} \left( \frac{1}{300} - \frac{1}{310} \right) \]

### Step 4: Calculate the Temperature Difference
Calculate the difference in the reciprocal of the temperatures:
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

### Step 5: Substitute Values into the Equation
Now substituting the values into the equation:
\[ 0.3 = \frac{E_A}{2.303 \times 8.314} \left( \frac{1}{9300} \right) \]

### Step 6: Solve for Activation Energy (EA)
Rearranging the equation to solve for \( E_A \):
\[ E_A = 0.3 \times 2.303 \times 8.314 \times 9300 \]

### Step 7: Perform the Calculations
Calculating the right-hand side:
1. Calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.2 \]
2. Now multiply by 9300:
\[ 19.2 \times 9300 \approx 178176 \]
3. Finally, multiply by 0.3:
\[ E_A \approx 0.3 \times 178176 \approx 53453 \]
4. Convert from Joules to kilojoules:
\[ E_A \approx 53.4 \, \text{kJ/mol} \]

### Step 8: Conclusion
Thus, the activation energy \( E_A \) is approximately **53.4 kJ/mol**.

### Final Answer
The activation energy for the reaction is **53.4 kJ/mol**.

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