The time required for `10%` completion of a first order reaction at `298 K` is equal to that required for its `25%` completion at `308 K`. If the pre-exponential factor for the reaction is `3.56 xx 10^(9) s^(-1)`, calculate its rate constant at `318 K` and also the energy of activation.

Explanation

To solve the problem step-by-step, we will follow the process outlined in the video transcript while providing clear explanations and calculations.

### Step 1: Understand the relationship between time and completion for first-order reactions
For a first-order reaction, the time required for a certain percentage completion can be expressed using the formula:
\[ t = \frac{2.303}{k} \log\left(\frac{100}{100 - X}\right) \]
where \( t \) is the time, \( k \) is the rate constant, and \( X \) is the percentage completion.

### Step 2: Set up equations for the two given conditions
1. For 10% completion at \( 298 \, K \):
\[ t_1 = \frac{2.303}{k_1} \log\left(\frac{100}{90}\right) = \frac{2.303}{k_1} \log(10/9) \]
2. For 25% completion at \( 308 \, K \):
\[ t_2 = \frac{2.303}{k_2} \log\left(\frac{100}{75}\right) = \frac{2.303}{k_2} \log(4/3) \]

Since \( t_1 = t_2 \), we can equate the two expressions:
\[ \frac{2.303}{k_1} \log(10/9) = \frac{2.303}{k_2} \log(4/3) \]

### Step 3: Simplify the equation
Cancel \( 2.303 \) from both sides:
\[ \frac{\log(10/9)}{k_1} = \frac{\log(4/3)}{k_2} \]
This can be rearranged to find the ratio of the rate constants:
\[ \frac{k_2}{k_1} = \frac{\log(4/3)}{\log(10/9)} \]

### Step 4: Calculate the logarithmic values
Using a calculator or logarithm tables:
- \( \log(10/9) \approx 0.045757 \)
- \( \log(4/3) \approx 0.124939 \)

Now substituting these values:
\[ \frac{k_2}{k_1} = \frac{0.124939}{0.045757} \approx 2.73 \]

### Step 5: Use the Arrhenius equation to find the activation energy
The Arrhenius equation relates the rate constants at two different temperatures:
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substituting \( R = 8.314 \, J/(mol \cdot K) \), \( T_1 = 298 \, K \), and \( T_2 = 308 \, K \):
\[ \log(2.73) = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{298} - \frac{1}{308}\right) \]

Calculating \( \log(2.73) \approx 0.436 \) and the temperature difference:
\[ \frac{1}{298} - \frac{1}{308} \approx 0.000335 \]

Now substituting these values:
\[ 0.436 = \frac{E_a}{2.303 \times 8.314} \times 0.000335 \]

### Step 6: Solve for \( E_a \)
Rearranging gives:
\[ E_a = \frac{0.436 \times 2.303 \times 8.314}{0.000335} \]
Calculating gives:
\[ E_a \approx 76.6 \, kJ/mol \]

### Step 7: Calculate the rate constant \( k \) at \( 318 \, K \)
Using the Arrhenius equation again:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where \( A = 3.56 \times 10^9 \, s^{-1} \) and \( T = 318 \, K \):
\[ k = 3.56 \times 10^9 e^{-\frac{76600}{8.314 \times 318}} \]
Calculating the exponent:
\[ -\frac{76600}{8.314 \times 318} \approx -29.1 \]
Thus:
\[ k \approx 3.56 \times 10^9 \times e^{-29.1} \approx 3.56 \times 10^9 \times 1.0 \times 10^{-13} \approx 9.3 \times 10^{-4} \, s^{-1} \]

### Final Answers
- Rate constant \( k \) at \( 318 \, K \approx 9.3 \times 10^{-4} \, s^{-1} \)
- Activation energy \( E_a \approx 76.6 \, kJ/mol \)