## 7E.3

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

TanveerDhaliwal3G
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am

### 7E.3

Can someone explain the reasoning behind how the solutions manual does 7E.3 and why it is completely different from example 7E.1 even though it is the same type of problem.

Niharika 1H
Posts: 50
Joined: Thu Jul 25, 2019 12:16 am

### Re: 7E.3

For 7E.3, we have to use a ratio of the Arrhenius equation, one at which the activation energy is 125 kJ/mol, and one at which it is 75kJ/mol. 7E.1 takes a different approach, as it says that the reverse and forward reactions rates increase using a catalyst.