## k' in Pseudo-First-Order Rate Laws

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Clarisse Wikstrom 1H
Posts: 63
Joined: Fri Sep 29, 2017 7:05 am

### k' in Pseudo-First-Order Rate Laws

Why is k' used in the pseudo-first-order rate law? Doesn't k' mean the rate constant for the reverse reaction? Thanks!

Mitch Walters
Posts: 45
Joined: Fri Sep 29, 2017 7:07 am

### Re: k' in Pseudo-First-Order Rate Laws

k' just means a different rate constant. This is because we are treating a second order reaction as if it's a first order reaction, so it must have a different rate constant.

Daniisaacson2F
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### Re: k' in Pseudo-First-Order Rate Laws

Yes, k prime if the reverse reaction's rate constant. You can use that and the forward reaction's k to figure out the equilibrium constant, K.

Akash_Kapoor_1L
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: k' in Pseudo-First-Order Rate Laws

How do we know when to use a pseudo-first-order method instead of calculating the second order rate law normally?

Christina Bedrosian 1B
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Re: k' in Pseudo-First-Order Rate Laws

k' stands for k*the reactants that you don't use in the rate law