## 15.67

$V_{0} = \frac{V_{max}[S]}{K_{M} + [S]}; K_{M} = \frac{[E][S]}{[ES]}$

MCracchiolo 1C
Posts: 55
Joined: Sat Jul 22, 2017 3:00 am

### 15.67

Would someone be able to explain why 15.67 uses the ratio of activation energies? i.e. How would I have known to substitute Ea(cat) without looking at the solution manual?

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

### Re: 15.67

They're using the ratios to see how the rate changes with respect to activation energy. I believe you could have plugged in the actual values given in the problem but simply keeping the Ea as a variable makes the calculation easier.

Ahmed Mahmood 4D
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### Re: 15.67

Think about it like this: You have two reaction rates, and you want to compare how the factor is affected by the addition of a catalyst. You know that adding a catalyst changes the activation energy; you know that changing the activation energy changes the reaction rate; you know the formula for the reaction rate. Therefore, in order to calculate the factor by which adding a catalyst changes the reaction rate, you must calculate the ratio of the catalyzed/uncatalyzed reaction rates (Aexp(-Ea[cat]/RT)/Aexp(-Ea[uncat]/RT)). A cancels out.