## Method of initial rate vs more than one reactant

Posts: 58
Joined: Fri Sep 29, 2017 7:04 am

### Method of initial rate vs more than one reactant

How do we find the reaction rate/law using method of initial rates vs for more than one reactant? What is the difference?

Silvino Jimenez 1A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Re: Method of initial rate vs more than one reactant

if reaction rate ia to be found in a reaction with more than one reactant, than we must pay attention to what reactant the problem wants us to find the rate is respect of. If Initial rate of reactant given is not the reactant of interest than we must convert it in terms of the consumption of the other reactant.

Justin Chang 2K
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: Method of initial rate vs more than one reactant

If you have, say, A+B-->C, the rate would be in the form of k[A]^n[B]^m.

In order to determine n, you need to see how [A] changes over time. This can be done by making the concentration of [B] extremely high, so it can be regarded as a constant. If it is regarded as a constant, then that means the rate that [B] will go down does not change with time, a "0-order" reaction.

This is called a pseudo rate law: rate=k'[A]^n, where k' is the rate constant under these conditions (where [B] is super high that the concentration of B will not change over time). Now you can use the method of initial rates to figure out the order of A.

Then, do the same thing you did except make the concentration of A super high it can be regarded as a constant. Therefore, the pseudo rate law is k'[B]^m, and you can figure out the order of B from experimental data.

Finally, remember that the rates will be equal. k[A]^n[B]^m=k'[A]^n, and therefore, k=k'/[B]^m.

This is how I understand it. Hope this helps!