We are given:
k1 [ES]= [E][S] / Km
<---- with Km=(k1+K3)/K1
k2 material balance gives:
<------ [Eo] = [E][S]/Km + [E][I]Ki
where Ki=[EI]/ [E][I] is the equilibrium constant
k3 for step 2 on the left.
Use this to show that the initial reaction rate is given by:
v=d[P]/dt=----------------------------- approximately=to k3[E]o[S]o/Km+[S]o
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 ...
This solution is provided in 335 words. It uses step-by-step problem to simplify the expression.