15.63

[MCracchiolo 1C]

In question 15.63, why do we need to subtract 0.59 and where does this value come from?

[Jonathan Tangonan 1E]

The .59 comes from the Ea/R(1/T1-1/T2) calculation where Ea=38kJ/mol T1=298K and T2=310K while on the other side of the equation ln(k2/k1) where k1=1.5x10^10.

[Daniisaacson2F]

I had the same question, I'm wondering how they were able to derive that from the arrhenius equation

[Jonathan Tangonan 1E]

The derivation for this equation is on page 642 in the textbook but in summary you use the Arrhenius equation at one temperature and at another and subtract the first equation from the second equation getting rid of the lnA.

At T1 : lnk1=lnA-Ea/RT1
At T2: lnk2= lnA-Ea/RT2

The subtraction after further simplification results in ln(k1/k2)=(Ea/RT)(1/T1-1/T2).

[Charles Ang 1E]

Assuming the other posts about this are correct, the solutions manual has an error. It should be LnK=.59, so the .59 is not actually subtracted.

[Wenting Hu 2H]

Wait but i get 0.059 instead of 0.59? It that just me?

[Wenting Hu 2H]

Update: the answer book is wrong. It's 0.008314, not 0.08314. So it is equal to 0.59