Zadanie 6c

temperatures converted to kelvin «303 K and 313 K» AND $\dfrac{k_2}{k_1} = 2.5$ ✔

«$\ln\dfrac{k_2}{k_1} = \dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ so $\ln 2.5 = \dfrac{E_\mathrm{a}}{8.31}\left(\dfrac{1}{303} - \dfrac{1}{313}\right)$» ✔

«$E_\mathrm{a} = \dfrac{8.31 \times 0.916}{1.054 \times 10^{-4}}$ J mol⁻¹ =» 72.2 «kJ mol⁻¹» ✔

Award [3] for a correct final answer; accept 72 to 73 (273.15 K for the conversion gives 72.3). Accept other methods, e.g. the two forms of the Arrhenius equation subtracted. Award [2] for 0.914 «kJ mol⁻¹», obtained without converting to kelvin.