Rate constant k varies with temperature by equation, log k `(min+^(-1))=5-(2000)/(T(K))` We can conclud:

Options

A. pre-exponential factor A is 5
B. `E_(a)` is 2000 kcal
C. pre-exponential factor A is `10^(5)`
D. `E_(a)" is "9.212` kcal

Explanation

To solve the problem, we will analyze the given equation for the rate constant \( k \) and derive the necessary values step by step.

### Given Equation:
\[ \log k \, (\text{min}^{-1}) = 5 - \frac{2000}{T(K)} \]

### Step 1: Identify the Pre-exponential Factor (A)
The equation can be compared to the Arrhenius equation in logarithmic form:
\[ \log k = \log A - \frac{E_a}{2.303RT} \]
From the given equation, we can see that:
\[ \log A = 5 \]
To find \( A \), we will convert from logarithmic form to exponential form:
\[ A = 10^5 \]

### Step 2: Calculate Activation Energy (E_a)
From the comparison of the two equations, we can identify that:
\[ \frac{E_a}{2.303R} = \frac{2000}{T} \]
We can rearrange this to find \( E_a \):
\[ E_a = \frac{2000 \cdot 2.303R}{T} \]
Since \( R \) (the gas constant) is typically \( 8.314 \, \text{J/(mol K)} \) or \( 2 \, \text{cal/(mol K)} \) when using calories, we will use \( R = 2 \, \text{cal/(mol K)} \) for our calculation.

### Step 3: Substitute R and Simplify
Substituting \( R \) into the equation:
\[ E_a = \frac{2000 \cdot 2.303 \cdot 2}{T} \]
This simplifies to:
\[ E_a = \frac{2000 \cdot 4.606}{T} = \frac{9212}{T} \, \text{cal/mol} \]

### Step 4: Convert Activation Energy to Kilocalories
To convert \( E_a \) from calories to kilocalories:
\[ E_a = \frac{9212 \, \text{cal/mol}}{1000} = 9.212 \, \text{kcal/mol} \]

### Conclusion
From the calculations, we conclude:
- The pre-exponential factor \( A \) is \( 10^5 \).
- The activation energy \( E_a \) is \( 9.212 \, \text{kcal/mol} \).

Thus, we can conclude that both options regarding the pre-exponential factor and activation energy are correct.