Page 1 of 1

### 15.9

Posted: Mon Feb 26, 2018 11:34 pm
Why do we divide by [(mol A) * L^-1]^n when determining units?

### Re: 15.9

Posted: Mon Feb 26, 2018 11:36 pm
On the syllabus, it says to omit this question from the homework.

### Re: 15.9

Posted: Tue Feb 27, 2018 9:34 am
It's because for each one you need to determine what the units for k are. So the equation would be rate = k [(molA)/L]^n and you would have to divide both sides by [(molA)/L]^n to isolate k, then you cancel out the like terms (rate always equals mol/(Ls) ), then depending on the order, the n would be different and cancel out differently.
zero order: mol/(Ls)
First order: 1/s
Second order: L/(mol s)

### Re: 15.9

Posted: Wed Feb 28, 2018 10:29 am
Matthew Lee 3L wrote:On the syllabus, it says to omit this question from the homework.

I think Dr. Lavelle meant to omit Section 15.9 of the reading, not the actual homework question.

### Re: 15.9

Posted: Thu Mar 01, 2018 10:20 am
The reaction rate = ∆concentration of reactants or products/∆time, which would give you units of molarity/time, which is usually mol*L-1seconds-1. If you look at the the general equation for the differential rate law, it is rate = k[R]-1. For example, for a first order reaction, the rate = k[A]1. Since [A]1 is a concentration which has units mol*L-1, we divide rate units (mol*L-1seconds-1) by units of [A] (mol*L-1).

### Re: 15.9

Posted: Thu Mar 01, 2018 11:44 am
to determine the units of k you set up the equations rate=k[A]

Because the units for rate are M/s and the units for concentrations is just M you divide rate by [A] to get k alone and solve for the units of K

you end up doing M/s * 1/M and depending on the n for the given problem your answer will be some variation like 1/M^(n-1)*s