## 15.69 6th Edition

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

Karyn Nguyen 1K
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### 15.69 6th Edition

A reaction rate increases by a factor of 1000. in the presence of a catalyst at 25 'C. The activation energy of the original pathway is 98 kJ!mol!1. What is the activation energy of the new pathway, all other factors being equal? In practice, the new pathway also has a different pre-exponential factor.

The solution manual solves the question in a confusing manner and I'm not sure what formula to use?

Phan Tran 1K
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Joined: Fri Sep 28, 2018 12:19 am

### Re: 15.69 6th Edition

Essentially, the question is telling you that there's a k before the catalyst was added, let's call it k1, and a k after, k2. 1000k1=k2. From here, you can plug in the Arrhenius equation and the givens in order to find the original activation energy. There's a confusing part where you have to take e^ln1000 in order to bring 1000 into the exponent and solve for Ea, but other than that it should be rather straightforward.

Karyn Nguyen 1K
Posts: 72
Joined: Fri Apr 06, 2018 11:04 am

### Re: 15.69 6th Edition

Sorry I'm still a little confused. How did you get rid of k1?

Karyn Nguyen 1K
Posts: 72
Joined: Fri Apr 06, 2018 11:04 am

### Re: 15.69 6th Edition

Nevermind I understand now, thanks!