Given: $T_1 = 600 \text{ K}$, $K_1 = 2.7 \times 10^{-4} \text{ L mol}^{-1} \text{ s}^{-1}$ $T_2 = 650 \text{ K}$, $K_2 = 3.5 \times 10^{-3} \text{ L mol}^{-1} \text{ s}^{-1}$ $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ Using the Arrhenius equation: $$ \ln \left( \frac{K_2}{K_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) $$ Substituting the values: $$ \begin{aligned} \ln \left( \frac{3.5 \times 10^{-3}}{2.7 \times 10^{-4}} \right) &= \frac{E_a}{8.314} \left( \frac{1}{600} - \frac{1}{650} \right)\\ \ln(12.963) &= \frac{E_a}{8.314} \left( 0.0016667 - 0.0015385 \right)\\ 2.562 &= \frac{E_a}{8.314} \times 0.0001282\\ E_a &= \frac{2.562 \times 8.314}{0.0001282}\\ E_a &= 166219.7 \text{ J mol}^{-1}\\ E_a &\approx 166.22 \text{ kJ mol}^{-1} \end{aligned} $$ The activation energy $E_a$ is $166.22 \text{ kJ mol}^{-1}$.