The first-order rate constant `k` is related to temperature as `log k = 15.0 - (10^(6)//T)`. Which of the following pair of value is correct ?

Options

A. `A = 10^(15)` and `E = 1.9 xx 10^(4) kJ`
B. `A = 10^(-15)` and `E = 40 kJ`
C. `A = 10^(15)` and `E = 40 kJ`
D. `A = 10^(-15)` and `E = 1.9 xx 10^(4) kJ`

Explanation

To solve the problem, we start with the given equation relating the first-order rate constant \( k \) to temperature \( T \):

\[ \log k = 15.0 - \frac{10^6}{T} \]

### Step 1: Identify the form of the equation
The equation can be compared to the Arrhenius equation in logarithmic form:

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

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.

### Step 2: Compare coefficients
From the given equation \( \log k = 15.0 - \frac{10^6}{T} \), we can compare it with the Arrhenius form:

1. The constant term \( 15.0 \) corresponds to \( \log A \).
2. The term \( -\frac{10^6}{T} \) corresponds to \( -\frac{E_a}{2.303RT} \).

### Step 3: Solve for \( A \)
From the comparison, we can find \( A \):

\[ \log A = 15.0 \implies A = 10^{15.0} = 1.0 \times 10^{15} \]

### Step 4: Solve for \( E_a \)
Now, we need to find \( E_a \). From the second part of the comparison:

\[ \frac{E_a}{2.303R} = 10^6 \]

Rearranging gives:

\[ E_a = 10^6 \times 2.303R \]

Substituting \( R = 8.314 \, \text{J/mol·K} \):

\[ E_a = 10^6 \times 2.303 \times 8.314 \]

Calculating this:

\[ E_a = 10^6 \times 2.303 \times 8.314 \approx 1.9 \times 10^4 \, \text{kJ/mol} \]

### Step 5: Final results
Thus, we have:

- \( A = 1.0 \times 10^{15} \)
- \( E_a \approx 1.9 \times 10^4 \, \text{kJ/mol} \)

### Conclusion
The correct pair of values is:

- \( A = 1.0 \times 10^{15} \)
- \( E_a \approx 1.9 \times 10^4 \, \text{kJ/mol} \)