A first order reaction is 50% completed in 20 minutes at `27^(@)C` and in 5 min at `47^(@)C`. The energy of activation of the reaction is

Options

A. 43.85 kJ/mol
B. 55.14 kJ/mol
C. 11.97 kJ/mol
D. 6.65 kJ/mol

Answer

To find the energy of activation (Ea) for the given first-order reaction, we can use the Arrhenius equation and the data provided. Here is a step-by-step solution:

### Step 1: Identify the given data
- The reaction is 50% completed in:
- \( t_1 = 20 \) minutes at \( T_1 = 27^\circ C \)
- \( t_2 = 5 \) minutes at \( T_2 = 47^\circ C \)

### Step 2: Convert temperatures from Celsius to Kelvin
- Convert \( T_1 \) and \( T_2 \) to Kelvin:
- \( T_1 = 27 + 273 = 300 \, K \)
- \( T_2 = 47 + 273 = 320 \, K \)

### Step 3: Relate the rate constants to the times
- For a first-order reaction, the time taken to reach 50% completion is inversely proportional to the rate constant \( k \):
\[ k_1 \propto \frac{1}{t_1} \quad \text{and} \quad k_2 \propto \frac{1}{t_2} \]
- Therefore, we can write:
\[ \frac{k_1}{k_2} = \frac{t_2}{t_1} = \frac{5}{20} = \frac{1}{4} \]

### Step 4: Use the Arrhenius equation
- According to the Arrhenius equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
- We can express \( \frac{k_2}{k_1} \) as:
\[ \frac{k_2}{k_1} = \frac{1}{4} \implies \ln \left( \frac{k_2}{k_1} \right) = \ln(0.25) = -1.386 \]

### Step 5: Substitute values into the equation
- Now, substituting the values into the Arrhenius equation:
\[ -1.386 = -\frac{E_a}{8.314} \left( \frac{1}{320} - \frac{1}{300} \right) \]
- Calculate \( \left( \frac{1}{320} - \frac{1}{300} \right) \):
\[ \frac{1}{320} - \frac{1}{300} = \frac{300 - 320}{96000} = -\frac{20}{96000} = -\frac{1}{4800} \]

### Step 6: Rearranging the equation to find \( E_a \)
- Substitute this back into the equation:
\[ -1.386 = -\frac{E_a}{8.314} \left( -\frac{1}{4800} \right) \]
- Rearranging gives:
\[ E_a = 1.386 \times 8.314 \times 4800 \]

### Step 7: Calculate \( E_a \)
- Performing the calculation:
\[ E_a = 1.386 \times 8.314 \times 4800 \approx 55.146 \, kJ/mol \]

### Final Answer
The energy of activation \( E_a \) for the reaction is approximately **55.146 kJ/mol**.

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