Question

Consider the following mechanism.

2A <-----> B+C Equillibrium

B+D-----> C Slow

----------------------------------

2A+D-----------> C+E

Determine the rate law

Answer 1

rate will depend on the slow step

rate = k [B] [D] --------1

but here B is intermediate

At equilibrium: Rate of forward reaction = rate of reverse reaction which we can write as

kf [A]² = kr [B] [C ] -------2

where Kf = farward equilibrium constant Kr = reverseequilibrium constant

from the 2 equation we can get the concentration [B]

[B] = Kf [A]² / Kr [C]

put this B value in above equation

Rate = k [D] kf [A]² / kr [C]

Rate K' [D][A]² / [C] where K. = k x Kf / Kr