### 15.9

Posted:

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

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=145&t=28609

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Posted: **Mon Feb 26, 2018 11:34 pm**

Why do we divide by [(mol A) * L^-1]^n when determining units?

Posted: **Mon Feb 26, 2018 11:36 pm**

On the syllabus, it says to omit this question from the homework.

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)

zero order: mol/(Ls)

First order: 1/s

Second order: L/(mol s)

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.

Posted: **Thu Mar 01, 2018 10:20 am**

The reaction rate = âˆ†concentration of reactants or products/âˆ†time, which would give you units of molarity/time, which is usually mol*L^{-1}seconds^{-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^{-1}seconds^{-1}) by units of [A] (mol*L^{-1}).

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

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