Zadanie 6c

$\dfrac{k_2}{k_1} = 2.5$ «at T₁ = 303.15 K and T₂ = 313.15 K» ✔

«$\ln \dfrac{k_2}{k_1} = \dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$, so $\ln 2.5 = \dfrac{E_\mathrm{a}}{8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}} \times \left(\dfrac{1}{303.15} - \dfrac{1}{313.15}\right)$» ✔

E_a «$= \dfrac{8.31 \times 0.916}{1.053 \times 10^{-4}}$ = 72 300 J mol⁻¹» = 72.3 «kJ mol⁻¹» ✔

Award [3] for a correct final answer; accept 72–73 kJ mol⁻¹ (72.2 if 303 K and 313 K are used). Accept other methods, for example the equation with ln k for each temperature. Award [2] for 0.914 kJ mol⁻¹ (temperatures not converted to kelvin).