15.63

Arrhenius Equation:

Moderators: Chem_Mod, Chem_Admin

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

15.63

Postby Andrea Grigsby 1I » Fri Mar 16, 2018 9:17 pm

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

Postby Curtis Wong 2D » Fri Mar 16, 2018 9:26 pm

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


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

Re: 15.63

Postby Jonathan Tangonan 1E » Fri Mar 16, 2018 9:33 pm

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.


Return to “Arrhenius Equation, Activation Energies, Catalysts”

Who is online

Users browsing this forum: No registered users and 1 guest