For the reaction `A rarr B`, the rate constant k (in `s^(-1)`) is given by
`log_(10) k = 20.35 - ((2.47 xx 10^(3)))/(T)`
The energy of activation in kJ `mol^(-1)` is ________ . (Nearest integer)
[Given : `R = 8.314 J K^(-1) mol^(-1)`]

Answer

To find the energy of activation (Ea) for the reaction \( A \rightarrow B \) given the equation for the rate constant \( k \):

\[ \log_{10} k = 20.35 - \frac{2.47 \times 10^{3}}{T} \]

we can relate this to the Arrhenius equation, which is:

\[ \log_{10} k = \log_{10} A - \frac{E_a}{2.303RT} \]

where:
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)),
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the terms in the equation**:
We can rewrite the given equation in the form of the Arrhenius equation:
\[ \log_{10} k = 20.35 - \frac{2.47 \times 10^{3}}{T} \]

2. **Compare with the Arrhenius equation**:
From the Arrhenius equation, we can see that:
\[ -\frac{E_a}{2.303RT} = -\frac{2.47 \times 10^{3}}{T} \]
This implies:
\[ \frac{E_a}{2.303R} = \frac{2.47 \times 10^{3}}{T} \]

3. **Rearranging the equation**:
Multiply both sides by \( 2.303R \):
\[ E_a = 2.47 \times 10^{3} \times 2.303R \]

4. **Substituting the value of R**:
Substitute \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \):
\[ E_a = 2.47 \times 10^{3} \times 2.303 \times 8.314 \]

5. **Calculating the value**:
First, calculate \( 2.47 \times 10^{3} \times 2.303 \):
\[ 2.47 \times 10^{3} \times 2.303 \approx 5.68841 \times 10^{3} \]
Now multiply by \( 8.314 \):
\[ E_a \approx 5.68841 \times 10^{3} \times 8.314 \approx 47229.84 \, \text{J mol}^{-1} \]

6. **Convert to kJ mol\(^{-1}\)**:
To convert from J to kJ, divide by 1000:
\[ E_a \approx 47.22984 \, \text{kJ mol}^{-1} \]

7. **Round to the nearest integer**:
The nearest integer value of \( E_a \) is:
\[ E_a \approx 47 \, \text{kJ mol}^{-1} \]

### Final Answer:
The energy of activation \( E_a \) is **47 kJ mol\(^{-1}\)**.