For a chemical reaction `Ato` products, the following equation is found to be followed, `logK=16.398-2800/T`
Arrhenius factor for the reaction is ______

Options

A. `2.5xx10^(16)`
B. `5xx10^(16)`
C. `7.5xx10^(-16)`
D. `4xx10^(-16)`

Explanation

To find the Arrhenius factor (A) for the reaction given the equation \( \log K = 16.398 - \frac{2800}{T} \), we can follow these steps:

### Step 1: Identify the form of the Arrhenius equation
The Arrhenius equation is generally expressed as:
\[ \log K = \log A - \frac{E_a}{2.303RT} \]
where:
- \( K \) is the rate constant,
- \( A \) is the Arrhenius factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Compare the given equation with the Arrhenius equation
From the given equation:
\[ \log K = 16.398 - \frac{2800}{T} \]
we can see that it can be compared to the general form of the Arrhenius equation.

Here, we can identify:
- The term \( 16.398 \) corresponds to \( \log A \).
- The term \( -\frac{2800}{T} \) corresponds to \( -\frac{E_a}{2.303RT} \).

### Step 3: Find the value of \( A \)
Since we have identified that:
\[ \log A = 16.398 \]
To find \( A \), we need to take the antilogarithm:
\[ A = 10^{16.398} \]

### Step 4: Simplify the calculation
We can express this as:
\[ A = 10^{16} \times 10^{0.398} \]
Now, we need to calculate \( 10^{0.398} \).

### Step 5: Calculate \( 10^{0.398} \)
Using logarithmic values, we can estimate:
- \( 10^{0.3} \approx 2 \)
- \( 10^{0.48} \approx 3.02 \)

Since \( 0.398 \) is between \( 0.3 \) and \( 0.48 \), we can estimate:
\[ 10^{0.398} \approx 2.5 \]

### Step 6: Final calculation of \( A \)
Now substituting back:
\[ A \approx 10^{16} \times 2.5 = 2.5 \times 10^{16} \]

### Conclusion
Thus, the Arrhenius factor for the reaction is:
\[ \boxed{2.5 \times 10^{16}} \]