7E.3 in 7th edition


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aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

7E.3 in 7th edition

Postby aisteles1G » Sun Mar 10, 2019 9:02 am

the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!

armintaheri
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

Re: 7E.3 in 7th edition

Postby armintaheri » Sun Mar 10, 2019 5:45 pm

125-175=50
0.4(125)=50
The two ways of writing it are equivalent. You must be making a mistake elsewhere.

Ruiting Jia 4D
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

Re: 7E.3 in 7th edition

Postby Ruiting Jia 4D » Sat Mar 16, 2019 12:38 am

aisteles1G wrote:the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!


Hi! Is this equation given to us on the sheet?

aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

Re: 7E.3 in 7th edition

Postby aisteles1G » Sat Mar 16, 2019 3:17 pm

Ruiting Jia 4D wrote:
aisteles1G wrote:the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!


Hi! Is this equation given to us on the sheet?


Not exactly, this equation from the problem is just using the K=Ae^(-Ea/RT) equation for each K, which is on the equation sheet and you have to divide the Kr2/Kr1 and your A cancels out and you get the exponent with the two activation energies, so just remember to put the K2 one on top when dividing!


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