For the reaction A $$\rightarrow$$ B, the rate constant k (in s$$^{-1}$$) is given by
$$\log_{10} k = 20.35 - \frac{2.47 \times 10^3}{T}$$
The energy of activation in kJ mol$$^{-1}$$ is _________. (Nearest integer)
[Given : R = 8.314 J K$$^{-1}$$ mol$$^{-1}$$]

The Arrhenius equation in its natural-logarithm form is stated first: $$\ln k=\ln A-\frac{E_a}{RT},$$ where $$E_a$$ is the energy of activation and $$R$$ is the universal gas constant.

The data given, however, are in common-logarithm (base-10) form: $$\log_{10} k = 20.35-\frac{2.47\times10^3}{T}.$$

To bring this expression to the natural-log form, we recall the relation $$\ln x = 2.303\,\log_{10}x.$$ Substituting, we have

$$\ln k = 2.303\left(20.35-\frac{2.47\times10^3}{T}\right).$$

Expanding the right-hand side gives

$$\ln k = 2.303\times20.35 \;-\; 2.303\left(\frac{2.47\times10^3}{T}\right).$$

This can be rewritten as

$$\ln k = \underbrace{\left(2.303\times20.35\right)}_{\ln A} \;-\; \left[2.303\,(2.47\times10^3)\right]\frac{1}{T}.$$

Comparing with the standard Arrhenius form $$\ln k=\ln A-\frac{E_a}{R}\frac{1}{T},$$ we identify

$$\frac{E_a}{R}=2.303\,(2.47\times10^3).$$

Hence,

$$E_a = 2.303\,(2.47\times10^3)\,R.$$

Substituting the value $$R = 8.314\ \text{J K}^{-1}\text{ mol}^{-1},$$ we perform the multiplication step by step:

First, $$2.303\times2.47 = 5.68841,$$ so

$$2.303\,(2.47\times10^3) = 5.68841\times10^3 = 5688.41.$$

Now, $$E_a = 5688.41 \times 8.314\ \text{J mol}^{-1}.$$

Carrying out the final multiplication,

$$E_a = 47293.44\ \text{J mol}^{-1}.$$

Converting joules to kilojoules (1 kJ = 1000 J):

$$E_a = \frac{47293.44}{1000} = 47.293\ \text{kJ mol}^{-1}.$$

Rounding to the nearest integer gives $$E_a \approx 47\ \text{kJ mol}^{-1}.$$

So, the answer is $$47$$.