In the presence of a catalyst, the rate of a reaction grows to the extent of `10^(5)` times at `298 K`. Hence, the catalyst must have lowered `E_(a)` by

Options

A. `25 kJ mol^(-1)`
B. `20 kJ mol^(-1)`
C. `10 kJ mol^(-1)`
D. `28.5 kJ mol^(-1)`

Explanation

To solve the problem, we need to determine how much the activation energy (Ea) has been lowered by a catalyst, given that the rate of reaction increases by a factor of \(10^5\) at a temperature of 298 K.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Rate Constants and Activation Energy**:
The rate constant \(k\) for a reaction can be expressed using the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \(A\) is the pre-exponential factor,
- \(E_a\) is the activation energy,
- \(R\) is the universal gas constant (8.314 J/mol·K),
- \(T\) is the temperature in Kelvin.

2. **Setting Up the Problem**:
We have two rate constants, \(k_1\) and \(k_2\), corresponding to the reaction rates before and after the introduction of the catalyst. Given that the rate increases by a factor of \(10^5\), we can express this as:
\[ \frac{k_2}{k_1} = 10^5 \]

3. **Using the Arrhenius Equation**:
We can express the ratio of the rate constants using the Arrhenius equation:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a2}}{RT}}}{A e^{-\frac{E_{a1}}{RT}}} = e^{-\frac{E_{a2} - E_{a1}}{RT}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = e^{-\frac{\Delta E_a}{RT}} \]
where \(\Delta E_a = E_{a2} - E_{a1}\).

4. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{\Delta E_a}{RT} \]

5. **Substituting Known Values**:
We know:
- \(\frac{k_2}{k_1} = 10^5\)
- \(R = 8.314 \, \text{J/mol·K}\)
- \(T = 298 \, \text{K}\)

Therefore:
\[ \ln(10^5) = 5 \ln(10) \approx 5 \times 2.303 \approx 11.515 \]

6. **Rearranging the Equation**:
Now substituting into the equation:
\[ 11.515 = -\frac{\Delta E_a}{(8.314)(298)} \]

7. **Calculating \(\Delta E_a\)**:
Rearranging gives:
\[ \Delta E_a = -11.515 \times (8.314) \times (298) \]
Calculating this:
\[ \Delta E_a = -11.515 \times 2477.572 \approx -28566.2 \, \text{J/mol} \approx -28.566 \, \text{kJ/mol} \]

8. **Final Answer**:
Rounding to two decimal places, the catalyst must have lowered the activation energy by approximately:
\[ \Delta E_a \approx -28.5 \, \text{kJ/mol} \]

### Conclusion:
The catalyst must have lowered the activation energy by approximately **28.5 kJ/mol**.