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

Options

A. 43.85 KJ/mol
B. 55.32 KJ/mol
C. 11.97 KJ/mol
D. 6.65 KJ/mol

Answer

To find the energy of activation for the first-order reaction, we can follow these steps:

### Step 1: Understand the Given Data
We have two scenarios for a first-order reaction:
1. The reaction is 50% completed in 20 minutes at 27°C.
2. The reaction is 50% completed in 5 minutes at 47°C.

### Step 2: Convert Temperatures to Kelvin
- For 27°C:
\[ T_1 = 27 + 273 = 300 \text{ K} \]
- For 47°C:
\[ T_2 = 47 + 273 = 320 \text{ K} \]

### Step 3: Calculate the Rate Constants (k)
For a first-order reaction, the half-life (\(t_{1/2}\)) is related to the rate constant (\(k\)) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Thus, we can express \(k\) as:
\[ k = \frac{0.693}{t_{1/2}} \]
- For the first case:
\[ k_1 = \frac{0.693}{20 \text{ min}} = \frac{0.693}{1200 \text{ s}} = 5.775 \times 10^{-4} \text{ s}^{-1} \]
- For the second case:
\[ k_2 = \frac{0.693}{5 \text{ min}} = \frac{0.693}{300 \text{ s}} = 2.31 \times 10^{-3} \text{ s}^{-1} \]

### Step 4: Use the Arrhenius Equation
The relationship between the rate constants and temperature can be expressed using the Arrhenius equation:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \(E_a\) = activation energy
- \(R\) = universal gas constant = 8.314 J/(mol·K)

### Step 5: Substitute the Values
Substituting \(k_1\) and \(k_2\) into the equation:
\[ \log \frac{2.31 \times 10^{-3}}{5.775 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{320} \right) \]
Calculating the left-hand side:
\[ \log \frac{2.31}{0.5775} \approx \log 4 = 0.602 \]

Calculating the right-hand side:
\[ \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800} \]

### Step 6: Solve for Activation Energy \(E_a\)
Now substituting back into the equation:
\[ 0.602 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{4800} \]
Rearranging gives:
\[ E_a = 0.602 \times 2.303 \times 8.314 \times 4800 \]
Calculating:
\[ E_a \approx 55327 \text{ J/mol} \approx 55.327 \text{ kJ/mol} \]

### Final Answer
The energy of activation for the reaction is approximately **55.327 kJ/mol**.

---