For a chemical reaction `Ato` products, the following equation is found to be followed, `logK=16.398-2800/T`
Activation energy of the reaction is _____________K.Cal

Options

A. 128.13
B. 12.767
C. 12813
D. 1.2813

Explanation

To find the activation energy (EA) of the reaction given the equation \( \log K = 16.398 - \frac{2800}{T} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Form of the Equation**:
The given equation resembles the Arrhenius equation in logarithmic form:
\[ \log K = \log A - \frac{E_A}{2.303RT} \]
Here, \( K \) is the rate constant, \( A \) is the pre-exponential factor, \( E_A \) is the activation energy, and \( R \) is the universal gas constant.

2. **Compare the Two Equations**:
From the given equation:
\[ \log K = 16.398 - \frac{2800}{T} \]
We can see that:
\[ -\frac{E_A}{2.303R} = -\frac{2800}{T} \]
This implies:
\[ \frac{E_A}{2.303R} = \frac{2800}{T} \]

3. **Eliminate Temperature (T)**:
Since the equation holds for all temperatures, we can eliminate \( T \) from both sides:
\[ E_A = 2800 \cdot 2.303R \]

4. **Substitute the Value of R**:
The value of the universal gas constant \( R \) can be taken as \( 2 \, \text{cal/mol K} \) (note that \( 1 \, \text{cal} = 0.001 \, \text{kcal} \)):
\[ E_A = 2800 \cdot 2.303 \cdot 2 \, \text{cal} \]

5. **Calculate E_A**:
\[ E_A = 2800 \cdot 2.303 \cdot 2 = 12876.8 \, \text{cal} \]

6. **Convert to Kilo Calories**:
To convert calories to kilocalories:
\[ E_A = \frac{12876.8}{1000} = 12.8768 \, \text{kcal} \]

7. **Round the Answer**:
Rounding to three significant figures gives:
\[ E_A \approx 12.88 \, \text{kcal} \]

### Final Answer:
The activation energy of the reaction is approximately **12.88 K.Cal**.

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