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

Answer

To solve the problem, we need to analyze the given equation for the rate constant \( k \) and relate it to the Arrhenius equation.

### Step-by-Step Solution:

1. **Given Equation**:
The equation provided is:
\[ \log k \, (\text{min}^{-1}) = 5 - \frac{2000}{T(K)} \]

2. **Rearranging the Equation**:
We can rewrite this equation in terms of \( k \):
\[ k = 10^{5 - \frac{2000}{T}} \]

3. **Using the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (approximately \( 8.314 \, \text{J/mol K} \)),
- \( T \) is the temperature in Kelvin.

4. **Taking Logarithm**:
Taking the logarithm of both sides of the Arrhenius equation gives:
\[ \log k = \log A - \frac{E_a}{RT \ln(10)} \]

5. **Comparing the Two Expressions**:
From the given equation, we have:
\[ \log k = 5 - \frac{2000}{T} \]
From the Arrhenius equation, we have:
\[ \log k = \log A - \frac{E_a}{RT \ln(10)} \]

By comparing the two expressions, we can identify:
- The constant term \( 5 \) corresponds to \( \log A \).
- The term \( -\frac{2000}{T} \) corresponds to \( -\frac{E_a}{RT \ln(10)} \).

6. **Finding Pre-exponential Factor \( A \)**:
From \( \log A = 5 \):
\[ A = 10^5 = 100000 \]

7. **Finding Activation Energy \( E_a \)**:
To find \( E_a \), we can rearrange:
\[ \frac{E_a}{R \ln(10)} = 2000 \]
Therefore:
\[ E_a = 2000 \cdot R \cdot \ln(10) \]
Using \( R \approx 8.314 \, \text{J/mol K} \) and \( \ln(10) \approx 2.303 \):
\[ E_a = 2000 \cdot 8.314 \cdot 2.303 \approx 38329.6 \, \text{J/mol} \approx 38.3 \, \text{kJ/mol} \]

8. **Conclusion**:
Based on the calculations:
- The pre-exponential factor \( A \) is \( 10^5 \) or \( 100000 \).
- The activation energy \( E_a \) is approximately \( 38.3 \, \text{kJ/mol} \).

### Final Answer:
- The pre-exponential factor \( A \) is \( 10^5 \) (not listed in the options).
- The activation energy \( E_a \) is approximately \( 38.3 \, \text{kJ/mol} \).