The time required for `10%` completion of a first order reaction at `298K` is equal to that required for its `25%` completion at `308K` . If the value of `A` is `4xx10^(10)s^(-1)`, calculate `k` at `318K` and `E_(a)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between time and rate constant for first-order reactions
For a first-order reaction, the relationship between time and the rate constant (k) can be expressed using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]
Where:
- \( T \) is the time taken for a certain percentage of completion.
- \( A_0 \) is the initial concentration.
- \( A \) is the concentration at time \( T \).

### Step 2: Set up equations for the given conditions
1. For 10% completion at 298 K:
- \( A_0 = 100 \) (initial concentration)
- \( A = 90 \) (after 10% completion)
- Therefore, the equation becomes:
\[ T_{298} = \frac{2.303}{k_{298}} \log \left( \frac{100}{90} \right) \]

2. For 25% completion at 308 K:
- \( A_0 = 100 \)
- \( A = 75 \) (after 25% completion)
- Therefore, the equation becomes:
\[ T_{308} = \frac{2.303}{k_{308}} \log \left( \frac{100}{75} \right) \]

### Step 3: Calculate the logarithmic values
1. Calculate \( \log \left( \frac{100}{90} \right) \):
\[ \log \left( \frac{100}{90} \right) = \log(1.1111) \approx 0.0458 \]

2. Calculate \( \log \left( \frac{100}{75} \right) \):
\[ \log \left( \frac{100}{75} \right) = \log(1.3333) \approx 0.1249 \]

### Step 4: Substitute the logarithmic values into the equations
1. For \( T_{298} \):
\[ T_{298} = \frac{2.303}{k_{298}} \times 0.0458 \]
Rearranging gives:
\[ k_{298} = \frac{2.303 \times 0.0458}{T_{298}} \]

2. For \( T_{308} \):
\[ T_{308} = \frac{2.303}{k_{308}} \times 0.1249 \]
Rearranging gives:
\[ k_{308} = \frac{2.303 \times 0.1249}{T_{308}} \]

### Step 5: Set up the ratio of the rate constants
Since the times for 10% completion at 298 K and 25% completion at 308 K are equal:
\[ T_{298} = T_{308} \]
Thus, we can set up the equation:
\[ \frac{0.0458}{k_{298}} = \frac{0.1249}{k_{308}} \]
From which we can derive:
\[ \frac{k_{308}}{k_{298}} = \frac{0.1249}{0.0458} \approx 2.72897 \]

### Step 6: Use the Arrhenius equation to find \( E_a \)
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 \( k_{298} \) and \( k_{308} \):
\[ \log \left( \frac{k_{308}}{k_{298}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{308} \right) \]

### Step 7: Calculate \( k \) at 318 K
Using the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where \( A = 4 \times 10^{10} \, \text{s}^{-1} \).

### Step 8: Final calculations
1. Calculate \( E_a \) using the values from the previous steps.
2. Substitute \( E_a \) into the Arrhenius equation to find \( k \) at 318 K.