$K = \frac{k_{forward}}{k_{reverse}}$

Angel Gomez 1K
Posts: 36
Joined: Fri Sep 29, 2017 7:04 am

When Lavelle started lecturing on reaction mechanisms, he gave the equation A+B+C–>P
He said that if the initial concentrations of both B and C were a lot greater than the initial concentration of A, then the rate=k'$[A]^{n}$ since B and C remain essentially constant.

I'm confused as to why we are using k'. Can someone explain what k' is exactly and why we use that in our rate law?

Vincent Kim 2I
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Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

k' is the reverse rate constant of the reaction

Kyle Alves 3K
Posts: 46
Joined: Thu Jul 27, 2017 3:01 am

From Lavelle's Lecture of 3/5
we need to be able to differentiate the rate constants between a forward and reverse rate
for example, a second order reaction would be
A + B => C + D the forward rate= k[A][B]
C + D => A + B the reverse rate = k'[C][D]

at equilibrium(forward = reverse rate)
k/k' = K

Kyle Alves 3K
Posts: 46
Joined: Thu Jul 27, 2017 3:01 am

We then set for your question the forward and revers rates equal! This is to study individual orders.
if want to study the forward rate, we would put B and C in large excess and input exponents (N,M,L)
k[A]^N [B]^M [C]^L = k'[A]^N
the A concentration will cancel out and leave you with:
k = k'/([B]^M [C]^L)
wherein we know both the concentrations of B and C