The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 `JK^(-1) mol^(-1)`)

Explanation

To solve the problem of calculating the energy of activation (Ea) for a reaction that increases its rate four times when the temperature changes from 300 K to 320 K, we can use the Arrhenius equation in its logarithmic form. Here are the steps to arrive at the solution:

### Step 1: Understand the relationship between rate constants and temperature
Given that the rate of the reaction increases four times when the temperature changes, we can express this as:
\[ k_2 = 4k_1 \]
where \( k_1 \) is the rate constant at 300 K and \( k_2 \) is the rate constant at 320 K.

### Step 2: Use the Arrhenius equation
The Arrhenius equation in logarithmic form is given by:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
where:
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)),
- \( T_1 \) is the initial temperature (300 K),
- \( T_2 \) is the final temperature (320 K).

### Step 3: Substitute values into the equation
Since we know that \( k_2 = 4k_1 \), we can write:
\[ \ln(4) = \frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{320} \right) \]

### Step 4: Calculate \( \ln(4) \)
We know:
\[ \ln(4) \approx 1.386 \]

### Step 5: Calculate the temperature difference
Now calculate \( \frac{1}{300} - \frac{1}{320} \):
\[ \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800} \]

### Step 6: Substitute into the equation
Now substituting back into the equation:
\[ 1.386 = \frac{E_a}{8.314} \left( \frac{1}{4800} \right) \]

### Step 7: Rearranging to find \( E_a \)
Rearranging gives:
\[ E_a = 1.386 \times 8.314 \times 4800 \]

### Step 8: Calculate \( E_a \)
Now perform the calculation:
\[ E_a = 1.386 \times 8.314 \times 4800 \approx 55,327.2 \text{ J/mol} \]

### Final Answer
Thus, the energy of activation \( E_a \) is approximately:
\[ E_a \approx 55,327.2 \text{ J/mol} \]

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