Pseudo rate laws

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

Ashley Kenney 1E
Posts: 32
Joined: Wed Nov 14, 2018 12:23 am

Pseudo rate laws

Can someone explain what a pseudo rate law is, and in what context we would use a pseudo rate law? Also, practically speaking, why is it important for us to use and understand pseudo rate laws? Thank you in advance!

eden tefera 2B
Posts: 39
Joined: Fri Sep 28, 2018 12:21 am

Re: Pseudo rate laws

When you have a reaction that is not first order and you want to determine one of the reactant's concentration, you can use a pseudo-first-order reaction in which one of the reactants is in great excess, essentially remaining constant as the reaction progresses. This allows you to determine concentration using a first order reaction.

Nathan Tran 4K
Posts: 92
Joined: Fri Sep 28, 2018 12:16 am

Re: Pseudo rate laws

Essentially, because the concentration of one reactant is so high, we can neglect the excess reactant to make the rate law easier to work with. So for a second order reaction involving two concentrations, if one of the concentrations is high, we can ignore it, and say the reaction is a pseudo first order reaction where the reaction only depends on one reactant.