Calculate the ratio of the rate constants for a catalyzed and an uncatalyzed reaction at room temperature (25.0°C) if the activation energy is lowered by 10.0 kJ.mol^-1 by the use of a catalyst.

Question

| General Chemistry 2

Accepted Answer

Arrhenius equation:

k = Ae^-Ea/RT

E_a (uncatalyzed reaction) = E_a (catalyzed reaction) + 10.0 kJ.mol^-1

k_cat / k_uncat = Ae^{-Ea (cat)/RT} / Ae^{-Ea (uncat)/RT}

⇒ k_cat / k_uncat = e^{(- 1/RT (E_a(cat) – E_a(uncat))}

⇒ k_cat / k_uncat = $e^{\frac{10 \times 10^{3}}{8.314 \times 298}}$ = 56.6

Catalysis | Chemical Kinetics: Mechanisms

General Chemistry 2 - Catalysis