For following reactions `Aoverset(700K)rarr`Product
`Aoverset(500K)rarr`Product
it was found that the `E_a` is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged , the activation energy for catalysed reaction if (Assume pre exponential factor is same)

Options

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

Answer

To solve the problem, we need to determine the activation energy for the catalyzed reaction given that the activation energy decreases by 30 kJ/mol in the presence of a catalyst.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two reactions:
- Reaction at 700 K with activation energy \( E_{A1} \)
- Reaction at 500 K with activation energy \( E_{A2} \)
The activation energy decreases by 30 kJ/mol when a catalyst is used.

2. **Setting Up the Equation**:
Since the rate remains unchanged, we can use the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature.

3. **Equating the Rates**:
Since the rates are unchanged, we can write:
\[ A e^{-\frac{E_{A1}}{RT_1}} = A e^{-\frac{E_{A2}}{RT_2}} \]
By cancelling \( A \) from both sides, we have:
\[ e^{-\frac{E_{A1}}{RT_1}} = e^{-\frac{E_{A2}}{RT_2}} \]

4. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ -\frac{E_{A1}}{RT_1} = -\frac{E_{A2}}{RT_2} \]
Rearranging gives:
\[ \frac{E_{A1}}{T_1} = \frac{E_{A2}}{T_2} \]

5. **Substituting the Values**:
We know that \( E_{A1} = E_{A2} + 30 \) kJ/mol (since it decreases by 30 kJ/mol). Substituting this into the equation gives:
\[ \frac{E_{A2} + 30}{700} = \frac{E_{A2}}{500} \]

6. **Cross-Multiplying**:
Cross-multiplying to eliminate the fractions:
\[ 500(E_{A2} + 30) = 700E_{A2} \]

7. **Expanding and Rearranging**:
Expanding the left side:
\[ 500E_{A2} + 15000 = 700E_{A2} \]
Rearranging gives:
\[ 15000 = 700E_{A2} - 500E_{A2} \]
\[ 15000 = 200E_{A2} \]

8. **Solving for \( E_{A2} \)**:
Dividing both sides by 200:
\[ E_{A2} = \frac{15000}{200} = 75 \text{ kJ/mol} \]

### Final Answer:
The activation energy for the catalyzed reaction \( E_{A2} \) is **75 kJ/mol**.