For following reactions
`A overset( 400K)( rarr) ` Product
`A underset("catalyst")overset("200K")(rarr)` Product
it was found that the `E_(a)` is decreased by 20 kJ` //` mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is ( Assume per exponential factor is same ) `:`

Options

A. 105 kJ/mol
B. 198 kJ/mol
C. 75 kJ/mol
D. 135 kJ/mol

Explanation

To solve the problem, we need to analyze the given reactions and apply the Arrhenius equation, which relates the rate constant of a reaction to its activation energy and temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For the uncatalyzed reaction at 400 K, let the activation energy be \( E_{a1} \).
- For the catalyzed reaction at 200 K, the activation energy is \( E_{a2} \).
- It is given that \( E_{a2} = E_{a1} - 20 \, \text{kJ/mol} \).
- The rate of both reactions remains unchanged.

2. **Write the Arrhenius Equation:**
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = Arrhenius constant (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin

3. **Set Up the Equation for Both Reactions:**
For the uncatalyzed reaction at 400 K:
\[ k_1 = A e^{-\frac{E_{a1}}{R \cdot 400}} \]
For the catalyzed reaction at 200 K:
\[ k_2 = A e^{-\frac{E_{a2}}{R \cdot 200}} \]

4. **Since the Rates are Unchanged:**
Since \( k_1 = k_2 \), we can equate the two equations:
\[ A e^{-\frac{E_{a1}}{R \cdot 400}} = A e^{-\frac{E_{a2}}{R \cdot 200}} \]
Canceling \( A \) from both sides gives:
\[ e^{-\frac{E_{a1}}{R \cdot 400}} = e^{-\frac{E_{a2}}{R \cdot 200}} \]

5. **Taking Natural Logarithm:**
Taking the natural logarithm of both sides:
\[ -\frac{E_{a1}}{R \cdot 400} = -\frac{E_{a2}}{R \cdot 200} \]
This simplifies to:
\[ \frac{E_{a1}}{400} = \frac{E_{a2}}{200} \]

6. **Substituting \( E_{a2} \):**
Substitute \( E_{a2} = E_{a1} - 20 \):
\[ \frac{E_{a1}}{400} = \frac{E_{a1} - 20}{200} \]

7. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 200 E_{a1} = 400 (E_{a1} - 20) \]
Simplifying this:
\[ 200 E_{a1} = 400 E_{a1} - 8000 \]
Rearranging gives:
\[ 200 E_{a1} - 400 E_{a1} = -8000 \]
\[ -200 E_{a1} = -8000 \]
\[ E_{a1} = 40 \, \text{kJ/mol} \]

8. **Finding \( E_{a2} \):**
Now, substitute \( E_{a1} \) back to find \( E_{a2} \):
\[ E_{a2} = E_{a1} - 20 = 40 - 20 = 20 \, \text{kJ/mol} \]

### Final Answer:
The activation energy for the catalyzed reaction is \( E_{a2} = 20 \, \text{kJ/mol} \).