The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJ mol$$^{-1}$$ is ______. (Rounded-off to the nearest integer) [$$R = 8.314$$ J K$$^{-1}$$ mol$$^{-1}$$]

Using the Arrhenius equation in logarithmic form: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$.

Given that $$k_2/k_1 = 5$$, $$T_1 = 27 + 273 = 300$$ K, and $$T_2 = 52 + 273 = 325$$ K.

Computing the temperature term: $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{325} = \frac{325 - 300}{300 \times 325} = \frac{25}{97500} = 2.564 \times 10^{-4}$$ K$$^{-1}$$.

Substituting: $$\ln 5 = \frac{E_a}{8.314} \times 2.564 \times 10^{-4}$$.

$$1.6094 = \frac{E_a \times 2.564 \times 10^{-4}}{8.314}$$.

Solving for $$E_a$$: $$E_a = \frac{1.6094 \times 8.314}{2.564 \times 10^{-4}} = \frac{13.381}{2.564 \times 10^{-4}} = 52189$$ J mol$$^{-1}$$ $$= 52.19$$ kJ mol$$^{-1}$$.

Rounding to the nearest integer, the activation energy is $$\boxed{52}$$ kJ mol$$^{-1}$$.