## 15.63

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

Emilie Hoffman 1E
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

### 15.63

I tried to solve this problem the slope way, with the activation energy over R being equal to (lnk1 - lnk2)/(1/T1 - 1/T2). Is this the correct way to set up this relationship, because I'm not getting the answer listed in the solutions manual.

Here is the problem for reference:
The rate constant of the reaction between CO2 and OH- in aqueous solution to give the HCO3- ion is 1.5 x 10^10 L/molxs at 25 C. Determine the rate constant at blood temperature (37 C), given that the activation energy for the reaction is 38 kJ/mol.
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Matthew Lee 3L
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.63

I believe if you set it up like that it should be ln(k1/k2) = (Ea/R)(1/T2 - 1/T1)

allyz1F
Posts: 58
Joined: Sat Jul 22, 2017 3:00 am

### Re: 15.63

Can someone please explain where the -.59 came from in the solution manual? I'm very confused by this

Clarisse Wikstrom 1H
Posts: 63
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.63

FYI If you derive the Arrhenius equation yourself, you may also get ln(k2/k1) = (Ea/R)(1/T1- 1/T2), and it works as well. As long as whatever k you have on the top, you have the corresponding T second.