## pseudo rxn

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

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 2:01 am

### pseudo rxn

Can someone explain the conceptual meaning and purpose behind a pseudo reaction

Chem_Mod
Posts: 16491
Joined: Thu Aug 04, 2011 12:53 pm
Has upvoted: 349 times

### Re: pseudo rxn

Pseudo-order reactions are used when there are multiple reactants, since integrated rate laws only account for 1 reactant. The goal is to make a reaction that is dependent on the concentrations of more than one species to appear as a first order reaction. So if you have a second order reaction then you want to assume that one of them has a much greater concentration, so when the other reactant runs out, its almost as if there was no change in the first one's concentration. This then changes the reaction to appear as a first order reaction.

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 2:01 am

### Re: pseudo rxn

So, this is a theorhetically application and not really applicable to real life reactions of multiple reactions in terms of integrated rate laws?

Leah Thomas 2E
Posts: 51
Joined: Fri Sep 29, 2017 6:06 am

### Re: pseudo rxn

Is this the same as using the isolated method of having one reactant in great excess?

Sophia Bozone 2G
Posts: 51
Joined: Fri Sep 29, 2017 6:07 am

### Re: pseudo rxn

A pseudo 1st order reaction means that the reaction is actually a second order reaction, but we've set one of the reactants at such a high concentration that the change to it during the reaction is negligible. Thus, because for all intents and purposes only one reactant concentration is changing, it can be treated as a first order reaction. The same idea applies for 2nd or higher orders as well