Rate constant `k` varies with temperature by equation `log k (min^(-1)) = log 5 - (2000 kcal)/(RT xx 2.303)`. We can conclude that

Options

A. The pre-exponential factor `A` is `10^(5)`.
B. `E_(a)` is `2000 kcal`
C. `E_(a)` is `9.12 kcal`
D. The pre-exponential factor `A` is `5`

Explanation

To solve the problem, we need to analyze the given equation for the rate constant \( k \) and relate it to the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Understand the given equation
The equation provided is:
\[ \log k = \log 5 - \frac{2000 \text{ kcal}}{RT \cdot 2.303} \]
This equation describes how the rate constant \( k \) varies with temperature \( T \).

### Step 2: Compare with the Arrhenius equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
Taking the logarithm of both sides, we get:
\[ \log k = \log A - \frac{E_a}{2.303RT} \]
where:
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy.

### Step 3: Equate the two equations
From the two equations:
1. \(\log k = \log 5 - \frac{2000 \text{ kcal}}{RT \cdot 2.303}\)
2. \(\log k = \log A - \frac{E_a}{2.303RT}\)

We can equate the coefficients:
- From \(\log k\), we can see that:
\[ \log A = \log 5 \implies A = 5 \]
- For the activation energy:
\[ E_a = 2000 \text{ kcal} \]

### Step 4: Conclusion
From the analysis, we conclude that:
- The pre-exponential factor \( A \) is \( 5 \).
- The activation energy \( E_a \) is \( 2000 \text{ kcal} \).

### Final Answer
- Pre-exponential factor \( A = 5 \)
- Activation energy \( E_a = 2000 \text{ kcal} \)