A first order reaction is `50%` completed in `30 min` at `27^(@)C` and in `10 min` at `47^(@)C`. Calculate the reaction rate constants at `27^(@)C` and the energy of activation of the reaction in `kJ mol^(-1)`.

Options

A. `k = 0.0231 "min"^(-1), E_(a) = 43.848 kJ mol^(-1)`
B. `k = 0.017 "min"^(-1), E_(a) = 52.54 kJ mol^(-1)`
C. `k = 0.0693 "min"^(-1), E_(a) = 43.848 kJ mol^(-1)`
D. `k = 0.0231 "min"^(-1), E_(a) = 28.92 kJ mol^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the rate constants (k) at both temperatures.

For a first-order reaction, the rate constant \( k \) can be calculated using the formula:

\[ k = \frac{0.693}{t_{1/2}} \]

where \( t_{1/2} \) is the half-life of the reaction.

#### For \( 27^\circ C \) (30 min):
\[ k_1 = \frac{0.693}{30 \text{ min}} = 0.0231 \text{ min}^{-1} \]

#### For \( 47^\circ C \) (10 min):
\[ k_2 = \frac{0.693}{10 \text{ min}} = 0.0693 \text{ min}^{-1} \]

### Step 2: Use the Arrhenius equation to find the activation energy (Ea).

The Arrhenius equation relates the rate constants at two different temperatures to the activation energy:

\[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Where:
- \( R \) is the universal gas constant, \( 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \) or \( 0.008314 \, \text{kJ mol}^{-1} \text{K}^{-1} \)
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin.

Convert the temperatures:
- \( T_1 = 27^\circ C = 300 \, K \)
- \( T_2 = 47^\circ C = 320 \, K \)

Now substituting the values into the equation:

\[ \log \left(\frac{0.0693}{0.0231}\right) = \frac{E_a}{2.303 \times 0.008314} \left(\frac{1}{300} - \frac{1}{320}\right) \]

Calculating the left side:
\[ \log \left(\frac{0.0693}{0.0231}\right) = \log(2.99) \approx 0.476 \]

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

Now substituting back:
\[ 0.476 = \frac{E_a}{2.303 \times 0.008314} \times \frac{1}{4800} \]

Rearranging to solve for \( E_a \):
\[ E_a = 0.476 \times 2.303 \times 0.008314 \times 4800 \]

Calculating \( E_a \):
\[ E_a \approx 43.848 \, \text{kJ mol}^{-1} \]

### Final Answers:
- The reaction rate constant at \( 27^\circ C \) is \( 0.0231 \, \text{min}^{-1} \).
- The activation energy \( E_a \) is \( 43.848 \, \text{kJ mol}^{-1} \).