# Enzymatic rate laws

We are given:

k1 [ES]= [E][S] / Km

E+S---->ES

<---- with Km=(k1+K3)/K1

k1'

k2 material balance gives:

E+I------>EI

<------ [Eo] = [E][S]/Km + [E][I]Ki

k2'

where Ki=[EI]/ [E][I] is the equilibrium constant

k3 for step 2 on the left.

ES--------->E+P

Use this to show that the initial reaction rate is given by:

k3[E]o[S]

v=d[P]/dt=----------------------------- approximately=to k3[E]o[S]o/Km+[S]o

Km+[S]+KmKi[I]

https://brainmass.com/chemistry/enzymatic-rate-laws-42200

#### Solution Preview

The first thing to realize is what we are looking for...

d[P]/dt = k3[ES] "from last reaction"

substitute in for [ES]

d[P]/dt = k3*([E][S]/km) "from first relationship [ES]= [E][S] / Km"

Notice that [E] is not in the intital reaction rate wanted. Thus we need to get rid ...

#### Solution Summary

This solution is provided in 335 words. It uses step-by-step problem to simplify the expression.