The rate constant, activation energy, and Arrphenius parameter of a chemical reaction are `3.0xx10^(-4)s^(-1), 104.4KJ mol^(-1)`, and `6.0xx10^(14)s^(-1)`, respectively. The value of rate constant as `Trarroo` is

Options

A. `2.0xx10^(18)s^(-1)`
B. `6.0xx10^(14)s^(-1)`
C. `3.6xx10^(30)s^(-1)`
D. None of these

Explanation

To find the value of the rate constant \( k \) as the temperature \( T \) approaches infinity, we can use the Arrhenius equation:

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

where:
- \( k \) is the rate constant,
- \( A \) is the Arrhenius parameter (frequency factor),
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant, and
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- Rate constant \( k = 3.0 \times 10^{-4} \, s^{-1} \)
- Activation energy \( E_a = 104.4 \, kJ \, mol^{-1} \)
- Arrhenius parameter \( A = 6.0 \times 10^{14} \, s^{-1} \)

2. **Convert activation energy to Joules**:
Since \( R \) is typically in \( J \, mol^{-1} \, K^{-1} \), we need to convert \( E_a \) from kilojoules to joules:
\[ E_a = 104.4 \, kJ \, mol^{-1} = 104400 \, J \, mol^{-1} \]

3. **Consider the limit as \( T \) approaches infinity**:
As \( T \) approaches infinity, the term \( RT \) also approaches infinity. Therefore, the fraction \( \frac{E_a}{RT} \) approaches zero:
\[ \frac{E_a}{RT} \to 0 \quad \text{as} \quad T \to \infty \]

4. **Evaluate the exponential term**:
The exponential term becomes:
\[ e^{-\frac{E_a}{RT}} \to e^0 = 1 \]

5. **Substitute back into the Arrhenius equation**:
Now substituting this back into the Arrhenius equation:
\[ k = A \cdot 1 = A \]

6. **Final calculation**:
Since \( A = 6.0 \times 10^{14} \, s^{-1} \):
\[ k = 6.0 \times 10^{14} \, s^{-1} \]

Thus, the value of the rate constant \( k \) as \( T \) approaches infinity is:

\[ \boxed{6.0 \times 10^{14} \, s^{-1}} \]