The rate constant , the activation energy and the Arhenius parameter of a chemical reaction at `25^(@)C` are `3xx10^(-4) s^(-1) , 104.4 " kJ mol"^(-1) and 6xx10^(14) s^(-1)` respectively . The value of the rate constant as `T to oo` is

Options

A. There can be only three values of molecularity ,that is 1,2 and 3 .
B. There can be only four values of order ,that is 0,1,2 and 3.
C. There can be infinite number of values for order .
D. The order ivolves rate while molecularity does not .

Answer

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

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

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

### Step 1: Identify the known values
From the problem, we have:
- \( k_1 = 3 \times 10^{-4} \, s^{-1} \) at \( T_1 = 25^\circ C = 298 \, K \)
- \( E_a = 104.4 \, kJ/mol = 104400 \, J/mol \) (convert to Joules)
- \( A = 6 \times 10^{14} \, s^{-1} \)
- \( R = 8.314 \, J/(mol \cdot K) \)

### Step 2: Use the Arrhenius equation to find \( k \) as \( T \to \infty \)
As \( T \) approaches infinity, the term \( \frac{E_a}{RT} \) approaches zero. Therefore, the exponential term \( e^{-\frac{E_a}{RT}} \) approaches 1.

Thus, we can write:

\[ k_{\infty} = A e^{-\frac{E_a}{RT}} \to A \quad \text{as } T \to \infty \]

### Step 3: Substitute the value of \( A \)
Now, substituting the value of \( A \):

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

### Conclusion
The value of the rate constant as \( T \to \infty \) is:

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

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