## 15.63

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

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

### 15.63

I'm looking at the answer for this question and I understand how to do everything, except for the value '-0.59'
I'm assuming that's the lnA from the Arrhenius Equation, however, how do we calculate that with the values given?

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

### Re: 15.63

The lnA's cancel each other out. It was just an error in the solutions manual. The minus sign is supposed to be an equal sign. Its .59 is equal to -(activation energy)/(gas constant)*(1/t1)-(1/t2)

or

$\frac{-E_{a}}{R}*(\frac{1}{t_{1}}-\frac{1}{t_{2}})$

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

### Re: 15.63

You can calculate this using another form of the Arrhenius equation where:

ln(k2/k1) = (Ea/R)(1/T1-1/T2)

Where k1= 1.5 * 1010 L*mol-1s-1, T1= 298K, T2=310K, and Ea= 38kJ/mol and then solve for k2 from there. You get the .59 from the calculations on the right hand side.