Question

Based on the three step mechanism below, what is the rate law for the reaction 2A + 2B →E + G?

A + B ⇌ D (Fast equilibrium)

D + B →E + F ( slow)

A + F → G (fast

Answer 1

in the case of multi step mechanism

the rate of the reaction is determined by the rate determining step

and

the rate determing step is the slowest reaction in the mechanism

in the given steps

the slowest one is

D + B ---> E + F

so

the rate is determined by this step

and

the rate law is given by

rate = k1 [D] [B]

but D is not in the final equation

so

now consdier the equilibrium reaction

A + B --->> D

the equilibrium constant is given by

K2 = [D] / [A] [B]

so

[D] = k2 [A] [B]

now

subsitute this in the first equation

rate = k1 [B] [D]

rate = k1 [B] x k2 [A] [B]

rate = (k1 x k2) [A] [B]^2

consider k1 x k2 = k

so

rate = k [A] [B]^2