Rate of reaction in absence of catalyst at 700 K is same as in presence of catalyst at 500 K. If catalyst decreases activation energy barrier by 30 kJ/mole, determine activation energy in presence of catalyst. (Assume 'A' factor to be same in both cases)

Options

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

Answer

To solve the problem, we need to establish the relationship between the rates of reaction in the absence and presence of a catalyst, and how the activation energy is affected by the catalyst.

### Step-by-Step Solution:

1. **Understanding the Rate Equations**:
- The rate of reaction in the absence of a catalyst at 700 K can be expressed as:
\[ R_1 = A e^{-\frac{E_a}{RT_1}} \]
where \( T_1 = 700 \, K \) and \( E_a \) is the activation energy in absence of catalyst.

- The rate of reaction in the presence of a catalyst at 500 K can be expressed as:
\[ R_2 = A e^{-\frac{E_a - 30}{RT_2}} \]
where \( T_2 = 500 \, K \) and the activation energy in the presence of the catalyst is \( E_a - 30 \, kJ/mol \).

2. **Setting the Rates Equal**:
- According to the problem, the rates are the same:
\[ R_1 = R_2 \]
- Therefore, we can set the two equations equal to each other:
\[ A e^{-\frac{E_a}{R \cdot 700}} = A e^{-\frac{E_a - 30}{R \cdot 500}} \]
- Since \( A \) is the same in both cases, we can cancel it out:
\[ e^{-\frac{E_a}{R \cdot 700}} = e^{-\frac{E_a - 30}{R \cdot 500}} \]

3. **Taking the Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ -\frac{E_a}{R \cdot 700} = -\frac{E_a - 30}{R \cdot 500} \]

4. **Cross Multiplying**:
- Cross-multiplying to eliminate the fractions:
\[ -E_a \cdot 500 = -(E_a - 30) \cdot 700 \]
- Simplifying this gives:
\[ -500 E_a = -700 E_a + 21000 \]

5. **Rearranging the Equation**:
- Rearranging the equation to isolate \( E_a \):
\[ 700 E_a - 500 E_a = 21000 \]
\[ 200 E_a = 21000 \]
\[ E_a = \frac{21000}{200} = 105 \, kJ/mol \]

6. **Finding Activation Energy in Presence of Catalyst**:
- Since the catalyst decreases the activation energy by 30 kJ/mol:
\[ E_a \text{ (in presence of catalyst)} = E_a - 30 = 105 - 30 = 75 \, kJ/mol \]

### Final Answer:
The activation energy in the presence of the catalyst is **75 kJ/mol**.