## 15.65 (c)

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

Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### 15.65 (c)

15.65 For the reversible, one-step reaction 2A ⇌ B + C, the forward rate constant for the formation of B is 265 L.mol^-1.min^-1 and the rate constant for the reverse reaction is 392 L.mol^-1.min^-1. The activation energy for the forward reaction is 39.7 kJ.mol^-1 and that of the reverse reaction is 25.4 kJ.mol^-1. (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Can someone please explain to me why the statement "raising the temperature will increase the rate constant of the reaction with the higher activation barrier more than it will the rate constant of the reaction with the lower energy barrier" is true?

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

### Re: 15.65 (c)

In the book it states that the reaction with a higher activation barrier will be more sensitive to changes in temperature therefore raising its k value. For example, an endothermic reaction has a higher activation energy barrier in the forward reaction than in the reverse. This is why adding heat is favorable to an endothermic reaction.