Rate constant `k = 1.2 xx 10^(3) mol^(-1) L s^(-1)` and `E_(a) = 2.0 xx 10^(2) kJ mol^(-1)`. When `T rarr oo`:

Options

A. `A = 2.0 xx 10^(2) kJ mol^(-1)`
B. `A = 1.2 xx 10^(3) mol^(-1) L s^(-1)`
C. `A = 1.2 xx 10^(3) mol L^(-1) s^(-1)`
D. `A = 2.4 xx 10^(3) kJ mol^(-1) s^(-1)`

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant \( k \) to the activation energy \( E_a \) and the temperature \( T \). The equation is given by:

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

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

Given:
- \( k = 1.2 \times 10^{3} \, \text{mol}^{-1} \, \text{L} \, \text{s}^{-1} \)
- \( E_a = 2.0 \times 10^{2} \, \text{kJ} \, \text{mol}^{-1} \)

### Step 1: Convert Activation Energy to Joules
First, we need to convert the activation energy from kilojoules to joules:

\[ E_a = 2.0 \times 10^{2} \, \text{kJ} \times 1000 \, \text{J/kJ} = 2.0 \times 10^{5} \, \text{J/mol} \]

### Step 2: Substitute Values into the Arrhenius Equation
We will rearrange the Arrhenius equation to solve for \( A \):

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

### Step 3: Consider the Limit as \( T \to \infty \)
As \( T \) approaches infinity, the term \( \frac{E_a}{RT} \) approaches zero:

\[ e^{\frac{E_a}{RT}} \to e^{0} = 1 \]

Thus, the equation simplifies to:

\[ A = k \cdot 1 = k \]

### Step 4: Calculate \( A \)
Now we can substitute the value of \( k \):

\[ A = 1.2 \times 10^{3} \, \text{mol}^{-1} \, \text{L} \, \text{s}^{-1} \]

### Conclusion
The frequency factor \( A \) when \( T \to \infty \) is:

\[ A = 1.2 \times 10^{3} \, \text{mol}^{-1} \, \text{L} \, \text{s}^{-1} \]