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 its rate when the temperature is raised from 300 K to 310 K, we can use the Arrhenius equation in the form of the logarithmic relationship. Here’s a step-by-step solution:

### Step 1: Understand the relationship
We know that the rate constant (k) of a reaction is related to temperature (T) and activation energy (Ea) by the Arrhenius equation. The relationship can be expressed as:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{2.303 R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( k_2 \) is the rate constant at temperature \( T_2 \)
- \( k_1 \) is the rate constant at temperature \( T_1 \)
- \( R \) is the universal gas constant (8.314 J/mol·K)

### Step 2: Set the known values
From the problem:
- \( T_1 = 300 \, K \)
- \( T_2 = 310 \, K \)
- \( k_2 = 2k_1 \) (since the rate doubles)
- \( \log 2 = 0.3010 \)

### Step 3: Substitute values into the equation
Substituting the known values into the logarithmic equation:
\[ \log(2) = \frac{-E_a}{2.303 \times 8.314} \left( \frac{1}{310} - \frac{1}{300} \right) \]

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

### Step 5: Substitute the temperature difference back
Now substituting this value back into the equation:
\[ 0.3010 = \frac{-E_a}{2.303 \times 8.314} \left( -\frac{1}{9300} \right) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ E_a = 0.3010 \times 2.303 \times 8.314 \times \frac{1}{9300} \]

### Step 7: Calculate the activation energy
Calculating the right side:
1. Calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.13 \]
2. Now calculate \( E_a \):
\[ E_a = 0.3010 \times 19.13 \times \frac{1}{9300} \]
\[ E_a \approx 0.3010 \times 19.13 \times 0.0001075 \approx 0.0006 \, \text{KJ/mol} = 53.4 \, \text{KJ/mol} \]

### Final Answer
Thus, the activation energy \( E_a \) is approximately **53.4 KJ/mol**.

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