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)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the rate constant (k) at 27°C

For a first-order reaction, the relationship between the rate constant (k) and the time (t) taken for the reaction to reach a certain completion percentage can be expressed using the formula:

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

Where \( t_{1/2} \) is the half-life of the reaction. Since the reaction is 50% complete in 30 minutes at 27°C, we can substitute \( t_{1/2} = 30 \text{ min} \) into the equation.

\[ k_{27°C} = \frac{0.693}{30 \text{ min}} = \frac{0.693}{30 \times 60 \text{ s}} = \frac{0.693}{1800 \text{ s}} \approx 0.000385 \text{ s}^{-1} \]

### Step 2: Calculate the rate constant (k) at 47°C

Similarly, for the reaction that is 50% complete in 10 minutes at 47°C, we can use the same formula:

\[ k_{47°C} = \frac{0.693}{10 \text{ min}} = \frac{0.693}{10 \times 60 \text{ s}} = \frac{0.693}{600 \text{ s}} \approx 0.001155 \text{ s}^{-1} \]

### Step 3: Use the Arrhenius equation to calculate the activation energy (Ea)

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

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

Where:
- \( k_1 = k_{27°C} \)
- \( k_2 = k_{47°C} \)
- \( R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \)
- \( T_1 = 27°C = 300 \text{ K} \)
- \( T_2 = 47°C = 320 \text{ K} \)

Substituting the values:

\[ \ln\left(\frac{0.001155}{0.000385}\right) = -\frac{E_a}{8.314}\left(\frac{1}{320} - \frac{1}{300}\right) \]

Calculating the left side:

\[ \ln(3) \approx 1.0986 \]

Calculating the right side:

\[ \frac{1}{320} - \frac{1}{300} = \frac{300 - 320}{96000} = -\frac{20}{96000} = -\frac{1}{4800} \]

Now substituting back into the equation:

\[ 1.0986 = -\frac{E_a}{8.314} \left(-\frac{1}{4800}\right) \]

Rearranging gives:

\[ E_a = 1.0986 \times 8.314 \times 4800 \]

Calculating \( E_a \):

\[ E_a \approx 1.0986 \times 8.314 \times 4800 \approx 43.2 \text{ kJ mol}^{-1} \]

### Final Answers
- Rate constant at 27°C, \( k_{27°C} \approx 0.000385 \text{ s}^{-1} \)
- Rate constant at 47°C, \( k_{47°C} \approx 0.001155 \text{ s}^{-1} \)
- Activation energy, \( E_a \approx 43.2 \text{ kJ mol}^{-1} \)