For a chemical reaction `Ato` products, the following equation is found to be followed, `logK=16.398-2800/T`
At `27^(@)C` the rate contant of the reaction is ______

Options

A. `1.16xx10^(5)`
B. `1.16xx10^(6)`
C. `1.321xx10^(7)`
D. `1.16xx10^(7)`

Explanation

To find the rate constant \( k \) for the reaction \( A \to \) products at \( 27^\circ C \), we will use the provided equation:

\[ \log K = 16.398 - \frac{2800}{T} \]

### Step-by-Step Solution:

**Step 1: Convert the temperature to Kelvin.**
- The given temperature is \( 27^\circ C \).
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Thus,
\[ T = 27 + 273.15 = 300.15 \approx 300 \, K \]

**Step 2: Substitute the temperature into the equation.**
- Now, we substitute \( T = 300 \, K \) into the equation:
\[ \log K = 16.398 - \frac{2800}{300} \]

**Step 3: Calculate the value of \( \frac{2800}{300} \).**
- Performing the division:
\[ \frac{2800}{300} = 9.3333 \approx 9.33 \]

**Step 4: Substitute this value back into the equation.**
- Now, we can substitute this back:
\[ \log K = 16.398 - 9.33 \]

**Step 5: Calculate \( \log K \).**
- Performing the subtraction:
\[ \log K = 16.398 - 9.33 = 7.068 \]

**Step 6: Find \( K \) by taking the antilog.**
- To find \( K \), we take the antilogarithm:
\[ K = 10^{7.068} \]

**Step 7: Calculate \( K \).**
- Using the property of logarithms:
\[ K \approx 1.16 \times 10^7 \]

### Final Answer:
The rate constant \( K \) of the reaction at \( 27^\circ C \) is approximately:
\[ K \approx 1.16 \times 10^7 \]