A reaction is carried out at `600 K`. If the same reaction is carried out in the presence of catalyst at the same rate and same frequency factor, the temperature required is `500 K`. What is the activation energy of the reaction, if the catalyst lowers the activation energy barrier by `20 KJ//mol` ?

Options

A. 100 kJ/mol
B. 120 kJ/mol
C. 80
D. None of these

Answer

To solve the problem, we will use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy and temperature. The equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) is the rate constant,
- \( A \) is the frequency factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (approximately \( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.

### Step-by-step Solution:

1. **Write the Arrhenius equation for the reaction without the catalyst:**
\[ k = A e^{-\frac{E_a}{RT_1}} \]
Here, \( T_1 = 600 \, K \).

2. **Write the Arrhenius equation for the reaction with the catalyst:**
\[ k' = A e^{-\frac{E_a'}{RT_2}} \]
Here, \( T_2 = 500 \, K \) and \( E_a' = E_a - 20 \, \text{kJ/mol} \).

3. **Since the rate constants are the same (\( k = k' \)), we can set the two equations equal to each other:**
\[ A e^{-\frac{E_a}{RT_1}} = A e^{-\frac{E_a - 20}{RT_2}} \]

4. **Cancel \( A \) from both sides (since \( A \) is the same):**
\[ e^{-\frac{E_a}{RT_1}} = e^{-\frac{E_a - 20}{RT_2}} \]

5. **Taking the natural logarithm of both sides:**
\[ -\frac{E_a}{RT_1} = -\frac{E_a - 20}{RT_2} \]

6. **Eliminate the negative sign:**
\[ \frac{E_a}{RT_1} = \frac{E_a - 20}{RT_2} \]

7. **Cross-multiply to eliminate the fractions:**
\[ E_a \cdot RT_2 = (E_a - 20) \cdot RT_1 \]

8. **Distributing on the right side:**
\[ E_a \cdot RT_2 = E_a \cdot RT_1 - 20 \cdot RT_1 \]

9. **Rearranging the equation:**
\[ E_a \cdot RT_2 - E_a \cdot RT_1 = -20 \cdot RT_1 \]

10. **Factoring out \( E_a \):**
\[ E_a (RT_2 - RT_1) = -20 \cdot RT_1 \]

11. **Solving for \( E_a \):**
\[ E_a = \frac{-20 \cdot RT_1}{R(T_2 - T_1)} \]

12. **Substituting \( R = 8.314 \, \text{J/mol·K} \) (or \( 0.008314 \, \text{kJ/mol·K} \)) and the temperatures:**
\[ E_a = \frac{-20 \cdot 0.008314 \cdot 600}{0.008314(500 - 600)} \]

13. **Calculating the values:**
\[ E_a = \frac{-20 \cdot 0.008314 \cdot 600}{0.008314 \cdot (-100)} \]
\[ E_a = \frac{-20 \cdot 600}{-100} = 120 \, \text{kJ/mol} \]

### Final Answer:
The activation energy \( E_a \) of the reaction is **120 kJ/mol**.