Arrhenius Equation:

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Emily Duggan 1F
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am


Postby Emily Duggan 1F » Sun Mar 11, 2018 11:58 am

Why does ln (k1/k2)=0.59? On my calculator I keep getting -0.59 because when I use the Arrhenius equation isn't the activation energy variable negative? Thanks!

Lisa Tang 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 2 times

Re: 15.63

Postby Lisa Tang 1C » Sun Mar 11, 2018 12:43 pm

For the problem, you would just plug in the values given into the equation ln(k2/k1)=Ea/R((1/T1)-(1-T2)). The activation energy isn't negative, so it would just be ln(k2/(1.5x1010)=38/.008314((1/298)-(1/310)).

Sandhya Rajkumar 1C
Posts: 50
Joined: Fri Jun 23, 2017 11:40 am

Re: 15.63

Postby Sandhya Rajkumar 1C » Sun Mar 11, 2018 1:38 pm

Also, once you've solved for k2, if you plug it in to ln(k1/k2) you get -0.59. But if you plug k1 and k2 into ln(k2/k1) you get 0.59. So to answer your question, ln(k1/k2) doesn't equal 0.59, it equals -0.59.

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