A catalyst decreases `E_(a)` form `100 KJ mol^(-1)` to `80 KJ mol^(-1)`. At what temperature the rate of reaction in the absence of catalyst at `500 K` will be equal to rate reaction in the presence of catalyst ?

Options

A. `400 K`
B. `200 K`
C. `625 K`
D. None of these

Explanation

To solve the problem, we need to find the temperature at which the rate of reaction in the presence of a catalyst (with an activation energy of 80 kJ/mol) is equal to the rate of reaction in the absence of a catalyst (with an activation energy of 100 kJ/mol) at a temperature of 500 K.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Activation energy without catalyst, \( E_{a1} = 100 \, \text{kJ/mol} \)
- Activation energy with catalyst, \( E_{a2} = 80 \, \text{kJ/mol} \)
- Temperature without catalyst, \( T_1 = 500 \, \text{K} \)
- Temperature with catalyst, \( T_2 = ? \)

2. **Use the Arrhenius Equation**:
The Arrhenius equation relates the rate constants of reactions to their activation energies and temperatures:
\[ k = A e^{-\frac{E_a}{RT}} \]
When the rates are equal, we can set up the ratio of the activation energies and temperatures:
\[ \frac{E_{a1}}{T_1} = \frac{E_{a2}}{T_2} \]

3. **Substitute the Known Values**:
Plugging in the known values into the equation:
\[ \frac{100 \, \text{kJ/mol}}{500 \, \text{K}} = \frac{80 \, \text{kJ/mol}}{T_2} \]

4. **Cross Multiply**:
Cross multiplying gives us:
\[ 100 \, \text{kJ/mol} \cdot T_2 = 80 \, \text{kJ/mol} \cdot 500 \, \text{K} \]

5. **Calculate the Right Side**:
Calculate \( 80 \times 500 \):
\[ 80 \times 500 = 40000 \, \text{kJ K/mol} \]

6. **Solve for \( T_2 \)**:
Now, we can solve for \( T_2 \):
\[ T_2 = \frac{40000 \, \text{kJ K/mol}}{100 \, \text{kJ/mol}} = 400 \, \text{K} \]

### Final Answer:
The temperature at which the rate of reaction in the absence of a catalyst at 500 K will equal the rate of reaction in the presence of a catalyst is \( T_2 = 400 \, \text{K} \).

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