Chemistry - Chemical Kinetics Question with Solution | TestHub

Using the Arrhenius equation:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Substituting the given values:

$\log \left(\frac{3.0\times 10^7}{1.5\times 10^7}\right)=\frac{E_a}{19.15}\left(\frac{330-300}{300\times 330}\right)$

$\log (2)=\frac{E_a}{19.15}\left(\frac{30}{99000}\right)$

$0.301=\frac{E_a}{19.15}\left(\frac{1}{3300}\right)$

$E_a=0.301\times 19.15\times 3300$

$E_a=19021.695{\text{J mol}}^{-1}$

$E_a\approx 19.02{\text{kJ mol}}^{-1}$

Answer: $19.02{\text{kJ mol}}^{-1}$