Consider $$A \xrightarrow{k_1} B$$ and $$ C \xrightarrow{k_2} D$$ are two reactions. If the rate constant ($$k_{1}$$) of the $$A \rightarrow B$$ reaction can be expressed by the followmg equation $$\log_{10}K = 14.34- \frac{1.5 \times 10^{4}}{T/K}$$ and activation energy $$C\rightarrow D$$ reaction ($$Ea_{2}$$) is $$\frac{1}{5}th$$ of the $$A\rightarrow B$$ reaction ($$Ea_{1}$$), then the value of ($$Ea_{2}$$) is _____________kJ $$mol^{-1}$$. (Nearest Integer)

We need to find the activation energy $$Ea_2$$ for the reaction $$C \rightarrow D$$, given information about the rate constant of $$A \rightarrow B$$.

The Arrhenius equation in logarithmic form is $$\log_{10} k = \log_{10} A - \frac{E_a}{2.303 \, RT}$$.

The expression $$\log_{10} k_1 = 14.34 - \frac{1.5 \times 10^4}{T/K}$$ can be compared to the standard form $$\log_{10} k = \log_{10} A - \frac{E_a}{2.303 \, R} \cdot \frac{1}{T}$$.

By matching the coefficient of $$\frac{1}{T}$$, one obtains $$\frac{E_{a_1}}{2.303 \, R} = 1.5 \times 10^4$$.

Therefore, $$E_{a_1} = 1.5 \times 10^4 \times 2.303 \times R$$. Substituting $$R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$$ gives $$E_{a_1} = 1.5 \times 10^4 \times 2.303 \times 8.314$$ which simplifies to $$1.5 \times 10^4 \times 19.147$$ and results in $$2.872 \times 10^5 \, \text{J mol}^{-1} = 287.2 \, \text{kJ mol}^{-1}$$.

Since $$E_{a_2} = \frac{1}{5} \times E_{a_1}$$, it follows that $$E_{a_2} = \frac{287.2}{5} = 57.44 \, \text{kJ mol}^{-1}$$. Rounding to the nearest integer gives $$E_{a_2} \approx 57 \, \text{kJ mol}^{-1}$$.

The answer is 57.