For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, are

Options

A. `1.0 xx 10^(6) s^(-1)` and `9.2 kJ mol^(-1)`
B. `6.0 s^(-1)` and `16.6 kJ mol^(-1)`
C. `1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`
D. `1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)`

Explanation

To solve the problem, we need to extract the pre-exponential factor \( A \) and the activation energy \( E_a \) from the given equation for the temperature-dependent rate constant \( k \):

\[ \log k = -2000 \left(\frac{1}{T}\right) + 6.0 \]

### Step 1: Identify the Arrhenius Equation
The Arrhenius equation is given by:

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

Taking the logarithm of both sides gives:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

### Step 2: Rearranging the Given Equation
We can rearrange the given equation to match the form of the Arrhenius equation:

\[ \log k = -2000 \left(\frac{1}{T}\right) + 6.0 \]

This can be compared to:

\[ \log k = \log A - \frac{E_a}{2.303R} \cdot \frac{1}{T} \]

### Step 3: Identify \( A \)
From the comparison, we can see that:

\[ \log A = 6.0 \]

To find \( A \), we take the antilogarithm:

\[ A = 10^{6.0} = 1 \times 10^6 \]

### Step 4: Identify \( E_a \)
Now, we compare the coefficients of \( \frac{1}{T} \):

\[ -\frac{E_a}{2.303R} = -2000 \]

This simplifies to:

\[ \frac{E_a}{2.303R} = 2000 \]

Now, we can solve for \( E_a \):

\[ E_a = 2000 \times 2.303 \times R \]

Where \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \). Plugging in the value of \( R \):

\[ E_a = 2000 \times 2.303 \times 8.314 \]

Calculating this gives:

\[ E_a = 2000 \times 2.303 \times 8.314 \approx 38300 \, \text{J mol}^{-1} \]

Converting to kilojoules:

\[ E_a \approx 38.3 \, \text{kJ mol}^{-1} \]

### Final Answer
Thus, the pre-exponential factor \( A \) and the activation energy \( E_a \) are:

\[ A = 1 \times 10^6 \quad \text{and} \quad E_a = 38.3 \, \text{kJ mol}^{-1} \]