15.63

Arrhenius Equation:

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

15.63

Postby Wenting Hu 2H » Mon Mar 12, 2018 1:32 am

how do you do 15.63?

The rate constant of the reaction between CO2 and
OH in aqueous solution to give the HCO3
ion is 1.5 x10^10 L*mol^1s^1 at 25 C. Determine the rate constant at blood
temperature (37 C), given that the activation energy for the
reaction is 38 kJ/mol.

I don't get how to set up the equation.

torieoishi1A
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

Re: 15.63

Postby torieoishi1A » Mon Mar 12, 2018 3:17 am

lnk(k'/k)=E/R((1/T)-(1/T'))
ln(k'/k)=(38kJ/mol)/(.08314kJ/K*mol)((1/298K)-(1/310K))-.59
k=1.5x10^10 L/mol*s
Solve for k'
k'=2.7 x 10^10 L/mol*s

Janice Xiao 1I
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

Re: 15.63

Postby Janice Xiao 1I » Wed Mar 14, 2018 1:53 am

Where does the -0.59 come from?

mhuang 1E
Posts: 22
Joined: Fri Sep 29, 2017 7:05 am

Re: 15.63

Postby mhuang 1E » Wed Mar 14, 2018 2:43 am

I think there was a typo in the solutions manual in that the setup is:
lnk(k'/k)=E/R((1/T)-(1/T'))
ln(k'/k)=(38kJ/mol)/(.008314kJ/K*mol)((1/298K)-(1/310K)) = .59
(k'/k)= 1.8
k=1.5x10^10 L/mol*s
Solve for k'
k'=2.7 x 10^10 L/mol*s

Ashley Davis 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Re: 15.63

Postby Ashley Davis 1I » Wed Mar 14, 2018 3:24 pm

Thank you! Was very confused by this question but this was helpful.


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