Lyndon #16

Arrhenius Equation:

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davidbakalov_lec2_2L
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Joined: Fri Sep 28, 2018 12:23 am

Lyndon #16

Postby davidbakalov_lec2_2L » Sat Mar 16, 2019 6:28 pm

Can someone explain #16 from Lyndon's worksheet? How did he get that graph?

Charles Gu 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: Lyndon #16

Postby Charles Gu 1D » Sat Mar 16, 2019 8:12 pm

The Arrhenius equation is usually k=Ae^(-Ea/RT). If you ln both sides you would get lnk=lnA- Ea/RT. This allows you to figure out the graph since it is in y=mx+b form. 1/T would be the x axis, lnk would be the y axis, and m would be -Ea/R. Therefore, it would be a straight line with a negative slope.


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