The rate constant the activation energy and the Arrhenius parameter of a chemical reaction at `25^@C` are `3 times 10^-4 sec^-1` 104.4KK/mol and `6 times 10^14 sec^-1` respectively. The value of the rate constant at `T to prop` is :

Options

A. `2 times 10^18 sec^-1`
B. `6 times 10^14 sec^-1`
C. `prop`
D. `3.6 times 10^30 sec^-1`

Answer

To solve the problem, we need to use the Arrhenius equation, which relates the rate constant (k) of a reaction to the temperature (T) and activation energy (EA). The Arrhenius equation is given by:

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

Where:
- \( k \) = rate constant
- \( A \) = Arrhenius parameter (pre-exponential factor)
- \( E_A \) = activation energy
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/mol·K} \))
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify Given Values**:
- Rate constant at \( 25^\circ C \) (or \( 298 \, K \)): \( k = 3 \times 10^{-4} \, \text{s}^{-1} \)
- Activation energy: \( E_A = 104.4 \, \text{kJ/mol} = 104400 \, \text{J/mol} \) (conversion from kJ to J)
- Arrhenius parameter: \( A = 6 \times 10^{14} \, \text{s}^{-1} \)

2. **Convert Temperature**:
- The temperature \( T \) at \( 25^\circ C \) is \( 298 \, K \).

3. **Use the Arrhenius Equation**:
- We need to find the rate constant \( k \) as \( T \) approaches infinity. According to the Arrhenius equation:
\[ k = A \cdot e^{-\frac{E_A}{RT}} \]

4. **Evaluate the Limit as \( T \to \infty \)**:
- As \( T \) approaches infinity, \( \frac{E_A}{RT} \) approaches \( 0 \) because \( R \) is a constant and \( T \) is increasing indefinitely.
- Therefore, \( e^{-\frac{E_A}{RT}} \) approaches \( e^0 = 1 \).

5. **Substitute into the Arrhenius Equation**:
- When \( T \to \infty \):
\[ k = A \cdot 1 = A \]
- Thus, the rate constant \( k \) at \( T \to \infty \) is:
\[ k = 6 \times 10^{14} \, \text{s}^{-1} \]

### Final Answer:
The value of the rate constant at \( T \to \infty \) is \( 6 \times 10^{14} \, \text{s}^{-1} \).