The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy `(E_(a))` of the reaction ? (R = gas constant)

Options

A. 1842.4 R
B. 921.2R
C. 460.6 R
D. 230.3 R

Explanation

To solve the problem, we will use the Arrhenius equation and the relationship between the rate constants at two different temperatures.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let \( T_1 = 200 \, K \)
- Let \( T_2 = 400 \, K \)
- The rate constant at \( T_1 \) is 10 times less than that at \( T_2 \). Therefore, we can express this as:
\[ k_1 = \frac{k_2}{10} \quad \text{or} \quad k_2 = 10 k_1 \]

2. **Use the Arrhenius Equation:**
The Arrhenius equation relates the rate constants at two different temperatures:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

3. **Substitute the Known Values:**
Since \( k_2 = 10 k_1 \), we can substitute this into the equation:
\[ \log \frac{10 k_1}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{200} - \frac{1}{400} \right) \]
This simplifies to:
\[ \log 10 = \frac{E_a}{2.303 R} \left( \frac{1}{200} - \frac{1}{400} \right) \]

4. **Calculate the Logarithm:**
We know that:
\[ \log 10 = 1 \]

5. **Calculate the Temperature Difference:**
Now, calculate \( \frac{1}{200} - \frac{1}{400} \):
\[ \frac{1}{200} - \frac{1}{400} = \frac{2 - 1}{400} = \frac{1}{400} \]

6. **Substitute Back into the Equation:**
Substitute the values into the equation:
\[ 1 = \frac{E_a}{2.303 R} \left( \frac{1}{400} \right) \]

7. **Rearranging to Solve for Activation Energy \( E_a \):**
Rearranging gives:
\[ E_a = 2.303 R \cdot 400 \]

8. **Calculate \( E_a \):**
\[ E_a = 921.2 R \]

### Final Answer:
The activation energy \( E_a \) of the reaction is:
\[ E_a = 921.2 R \]