## k' reverse reaction constant

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

Ashley Davis 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

### k' reverse reaction constant

Could someone please clarify why the reaction constant for a reverse reaction is k' (k prime)?

Cooper1C
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: k' reverse reaction constant

It's just the notation - it distinguishes k from k'. I don't think there's anything special about it.

Ashley Davis 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

### Re: k' reverse reaction constant

Oh, okay, I was reading it as "k prime." Thank you.

Angel R Morales Dis1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: k' reverse reaction constant

Just a heads up, k' can also be written as k-1, but yeah, its just a different notation to distinguish the forward from the reverse reaction rate.

Dang Lam
Posts: 55
Joined: Thu Jul 27, 2017 3:01 am

### Re: k' reverse reaction constant

can someone explain to me why K=k/k'?

Seth_Evasco1L
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: k' reverse reaction constant

Say for example A + B --> C + D

The forward rate can be represented as k[A][B] and the reverse rate can be represented as k'[C][D].

At equilibrium you can set these rates equal to each other to get:

k[A][B] = k'[C][D]

This can be rewritten as:

k/k' = [C][D]/[A][B]

K, the equilibrium constant = [product]/[reactant] or in this case [C][D]/[A][B]

Therefore, K is also equal to k/k'

Ashley Macabasco 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: k' reverse reaction constant

In K=k/k', K refers to the equilibrium constant which can also be found by [products]/[reactants]. For the reaction A+B-->C+D k = rate/[A][B] and k' = rate/[C][D]. k/k' would then equal (rate/[A][B])([C][D]/rate) where the rates would cancel leaving [C][D]/[A][B] or [products]/[reactants].