The rate constant, the activation energy and the Arrhenius parameter of a chemical reactions at `25^(@)C` are `3.0xx10^(-4)s^(-), "104.4 kJ mol"^(-1)` and `6xx10^(14)s^(-1)` respectively. The value of the rate constant as `T rarr oo` is

Options

A. `2.0xx10^(18)s^(-1)`
B. `6.0xx10^(14)s^(-1)`
C. infinity
D. `3.6xx10^(30)s^(-1)`

Answer

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

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

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

### Step 1: Identify the given values
From the problem statement, we have:
- \( k = 3.0 \times 10^{-4} \, s^{-1} \) (at \( T = 25^\circ C \))
- \( E_a = 104.4 \, kJ \, mol^{-1} \)
- \( A = 6 \times 10^{14} \, s^{-1} \)

### Step 2: Convert activation energy to J/mol
Since the gas constant \( R \) is typically given in J/(mol·K), we need to convert \( E_a \) from kJ/mol to J/mol:

\[ E_a = 104.4 \, kJ/mol \times 1000 \, J/kJ = 104400 \, J/mol \]

### Step 3: Use the Arrhenius equation as \( T \to \infty \)
As \( T \to \infty \), the term \( \frac{E_a}{RT} \) approaches 0 because \( R \) is a positive constant and \( T \) is increasing without bound. Thus, we can simplify the equation:

\[ \lim_{T \to \infty} k = A e^{-\frac{E_a}{RT}} = A e^{0} = A \]

### Step 4: Substitute the value of \( A \)
Now we substitute the value of \( A \):

\[ k = 6 \times 10^{14} \, s^{-1} \]

### Final Answer
Thus, the value of the rate constant as \( T \to \infty \) is:

\[ \boxed{6 \times 10^{14} \, s^{-1}} \]