## 15.79

$K = \frac{k_{forward}}{k_{reverse}}$

Lucia H 2L
Posts: 43
Joined: Mon Nov 14, 2016 3:00 am
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### 15.79

Hi, can someone explain how to begin solving 15.79?

In the reaction of HBr with the reactive intermediate CH3CH==CHCH2 , at low temperatures the predominant product is CH3CHBrCH==CH2, but at high temperatures, the predominant product is CH3CH==CHCH2Br. (where == is a double bond)
(a) Which product is formed by the pathway with the larger activation energy?
(b) Does kinetic control predominate at low or high temperatures?

I understand that the most thermodynamically stable solution will be the thermodynamic product and the solution with the lowest activation energy will be the kinetic product. How do you know which is which from the given molecules?

Sean Monji 2B
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.79

From the equation where ln k = Ea/ RT + ln A, we can see that activation energy and temperature are directly related provided all other things constant. Therefore, the product that forms more at higher temperatures has a higher activation energy.
a) If higher temp = higher activation energy, the second product at higher temperatures must have a higher activation energy.
b )Since kinetics controls reactions at lower activation energies, the first product is controlled by kinetics.
Hope that helps