# Arrhenius equation - 2 | Chemical Kinetics: Mechanisms

## General Chemistry 2 - Arrhenius equation - 2

The rate of an elementary reaction has been studied as a function of temperature.

The following data were obtained for the rate constant:

Exp 1: T = 300K; k = 6.47 mol.L^{-1}.s^{-1}

Exp 2: T = 500K; k = 2.84 x 10^{4} mol.L^{-1}.s^{-1}

Exp 3: T = 700K; k = 1.03 x 10^{6} mol.L^{-1}.s^{-1}

Calculate the activation energy for the reaction.

Arrhenius equation:

k = Ae^{-Ea/RT}

k_{2} / k_{1} = Ae^{-Ea/RT2} / Ae^{-Ea/RT1}

⇒^{ }k_{2} / k_{1} = e^{-Ea/RT2 + Ea/RT1}

⇒^{ }ln (k_{2} / k_{1}) = $\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$ $\left(\frac{1}{{\mathrm{T}}_{1}}-\frac{1}{{\mathrm{T}}_{2}}\right)$

⇒ E_{a} = $\frac{\mathrm{ln}\left({\displaystyle \frac{2.84\times {10}^{4}}{6.47}}\right)\times 8.314}{{\displaystyle \frac{1}{300}}-{\displaystyle \frac{1}{500}}}$ = 5.23 x 10^{4} J = 52.3 kJ