## 15.63

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

MCracchiolo 1C
Posts: 55
Joined: Sat Jul 22, 2017 3:00 am

### 15.63

In question 15.63, why do we need to subtract 0.59 and where does this value come from?

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: 15.63

The .59 comes from the Ea/R(1/T1-1/T2) calculation where Ea=38kJ/mol T1=298K and T2=310K while on the other side of the equation ln(k2/k1) where k1=1.5x10^10.

Daniisaacson2F
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### Re: 15.63

I had the same question, I'm wondering how they were able to derive that from the arrhenius equation

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: 15.63

The derivation for this equation is on page 642 in the textbook but in summary you use the Arrhenius equation at one temperature and at another and subtract the first equation from the second equation getting rid of the lnA.

At T1 : lnk1=lnA-Ea/RT1
At T2: lnk2= lnA-Ea/RT2

The subtraction after further simplification results in ln(k1/k2)=(Ea/RT)(1/T1-1/T2).

Charles Ang 1E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: 15.63

Assuming the other posts about this are correct, the solutions manual has an error. It should be LnK=.59, so the .59 is not actually subtracted.

Wenting Hu 2H
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.63

Wait but i get 0.059 instead of 0.59? It that just me?

Wenting Hu 2H
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.63

Update: the answer book is wrong. It's 0.008314, not 0.08314. So it is equal to 0.59