For the reaction of H2 and I2, the rate constant is 2.5 10-4\,dm3\,mol-1\,s-1 at 327°C and 1.0\,dm3\,mol-1\,s-1 at…

$T_1 = 600\,\mathrm{K}$, $k_1 = 2.5 \times 10^{-4}$; $T_2 = 800\,\mathrm{K}$, $k_2 = 1.0$

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$\ln\frac{1.0}{2.5 \times 10^{-4}} = \ln 4000 = \ln(4 \times 10^3) = \ln 4 + 3\ln 10 = 1.386 + 6.909 = 8.295$

$8.295 = \frac{E_a}{8.314}\left(\frac{1}{600} - \frac{1}{800}\right) = \frac{E_a}{8.314} \times \frac{800-600}{600 \times 800}$

$= \frac{E_a}{8.314} \times \frac{200}{480000} = \frac{E_a}{8.314} \times 4.167 \times 10^{-4}$

$E_a = \frac{8.295 \times 8.314}{4.167 \times 10^{-4}} = \frac{68.97}{4.167 \times 10^{-4}} = 165500\,\mathrm{J/mol} \approx 166\,\mathrm{kJ/mol}$

Answer: Option (1) — 166 kJ/mol