When the temperature is raise from 300 K to 400 K the rate of a reaction is increased by `10^(8)` times. What is the energy of activation?

Options

A. 44.1 Kcal
B. 22.5 Kcal
C. 28.2 Kcal
D. 18.6 Kcal

Explanation

To find the energy of activation (Ea) when the temperature is raised from 300 K to 400 K and the rate of the reaction increases by \(10^8\) times, we can use the Arrhenius equation in the form of the logarithmic relation:

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

### Step 1: Identify the values
- Given:
- \(T_1 = 300 \, K\)
- \(T_2 = 400 \, K\)
- Rate increase: \(\frac{k_2}{k_1} = 10^8\)

### Step 2: Substitute values into the equation
- Since \(k_1\) can be assumed as 1, we have:
\[ \log k_2 = \log(10^8) = 8 \]
- Thus, we can rewrite the equation as:
\[ 8 = \frac{E_a}{2.303R} \left( \frac{1}{300} - \frac{1}{400} \right) \]

### Step 3: Calculate \(\left( \frac{1}{T_1} - \frac{1}{T_2} \right)\)
- Calculate:
\[ \frac{1}{300} - \frac{1}{400} = \frac{4 - 3}{1200} = \frac{1}{1200} \]

### Step 4: Substitute and solve for \(E_a\)
- Now substituting back into the equation:
\[ 8 = \frac{E_a}{2.303R} \cdot \frac{1}{1200} \]
- Rearranging gives:
\[ E_a = 8 \cdot 2.303R \cdot 1200 \]

### Step 5: Use the value of the universal gas constant \(R\)
- The value of \(R\) in appropriate units (calories) is \(R = 1.987 \, \text{cal/(mol K)}\).
- Therefore:
\[ E_a = 8 \cdot 2.303 \cdot 1.987 \cdot 1200 \]

### Step 6: Calculate \(E_a\)
- Performing the calculation:
\[ E_a = 8 \cdot 2.303 \cdot 1.987 \cdot 1200 \approx 44,207.6 \, \text{calories} \]

### Step 7: Convert to kilocalories
- Since we need the answer in kilocalories:
\[ E_a = \frac{44,207.6}{1000} \approx 44.21 \, \text{kcal} \]

### Final Answer
- The energy of activation \(E_a\) is approximately \(44.21 \, \text{kcal}\).