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

Options

A. 43.58 KJ
B. 55.14 KJ
C. 11.97 KJ
D. 6.65 KJ

Explanation

To solve the problem of finding the activation energy (Ea) for a first-order reaction that is 50% completed in 20 minutes at 27°C and then 5% completed in 5 minutes at 47°C, we can follow these steps:

### Step 1: Convert Temperatures to Kelvin
To convert the temperatures from Celsius to Kelvin:
- \( T_1 = 27°C + 273 = 300 K \)
- \( T_2 = 47°C + 273 = 320 K \)

### Step 2: Use the Half-Life Formula for First-Order Reactions
The half-life (t½) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Where \( k \) is the rate constant.

Given that the reaction is 50% completed in 20 minutes (t½ = 20 minutes) at 300 K:
\[ t_{1/2} = 20 \text{ minutes} = 1200 \text{ seconds} \]
Thus, we can calculate \( k_1 \):
\[ k_1 = \frac{0.693}{1200} \]

### Step 3: Calculate the Rate Constant at 300 K
Calculating \( k_1 \):
\[ k_1 = \frac{0.693}{1200} \approx 0.0005775 \text{ s}^{-1} \]

### Step 4: Determine the Rate Constant for the Second Condition
For the second condition, where the reaction is 5% completed in 5 minutes (t = 5 minutes):
Since 5% completion corresponds to a time that can be approximated using the first-order kinetics, we can find \( k_2 \) using:
\[ k_2 = \frac{0.693}{t_{1/2}} \]
However, we need to find the time for 5% completion. For small percentages, we can use the approximation:
\[ k_2 \approx \frac{0.693}{300} \text{ seconds} \]

### Step 5: Calculate \( k_2 \)
Calculating \( k_2 \):
\[ k_2 = \frac{0.693}{300} \approx 0.00231 \text{ s}^{-1} \]

### Step 6: Use the Arrhenius Equation
The Arrhenius equation relates the rate constants at two different temperatures:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 7: Substitute Values into the Arrhenius Equation
Substituting the values we have:
- \( k_1 \approx 0.0005775 \text{ s}^{-1} \)
- \( k_2 \approx 0.00231 \text{ s}^{-1} \)
- \( R = 8.314 \text{ J/mol K} \)

Calculating:
\[ \ln\left(\frac{0.00231}{0.0005775}\right) = -\frac{E_a}{8.314} \left(\frac{1}{320} - \frac{1}{300}\right) \]

### Step 8: Solve for \( E_a \)
Calculating the left side:
\[ \ln(4) \approx 1.386 \]

Now, calculating the right side:
\[ \frac{1}{320} - \frac{1}{300} = \frac{300 - 320}{96000} = -\frac{20}{96000} = -\frac{1}{4800} \]

Now substituting back:
\[ 1.386 = -\frac{E_a}{8.314} \left(-\frac{1}{4800}\right) \]

Rearranging gives:
\[ E_a = 1.386 \times 8.314 \times 4800 \]

### Step 9: Final Calculation
Calculating \( E_a \):
\[ E_a \approx 55.3 \text{ kJ/mol} \]

### Conclusion
The activation energy \( E_a \) of the reaction is approximately **55.3 kJ/mol**.