## Arrhenius equation and activation energies

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

Jorja De Jesus 2C
Posts: 121
Joined: Sat Jul 20, 2019 12:15 am

### Arrhenius equation and activation energies

Does the Arrhenius equation find the activation energy and then could the equation $lnK=(-E_{A}/RT)+lnA$ be used on the activation energy? Also what does exp stand for in the Arrhenius equation?

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### Re: Arrhenius equation and activation energies

Referring to homework 7D.1 as an example, the problem gives you 2 rate constants at 2 different temperatures and asks you to find the activation energy. In this case, you would rearrange the Arrhenius equation as a difference of T1 and T2 to get $ln\frac{k_{2}}{k_{1}} = \frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$. Then once you plug the values in, you could get the value for Ea. In the constants and equations sheet, exp stands for e^.
Last edited by JamieVu_2C on Sat Mar 14, 2020 8:42 pm, edited 1 time in total.

J Medina 2I
Posts: 102
Joined: Wed Sep 25, 2019 12:17 am

### Re: Arrhenius equation and activation energies

Yes, if given all other information, the Arrhenius equation can be used to find the activation energy, Ea. I think exponential could be referring to when both sides are raised to the power of e which cancels the ln on the left and then raises the right side to the power of e.