15.9
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 52
- Joined: Fri Sep 29, 2017 7:04 am
-
- Posts: 51
- Joined: Fri Sep 29, 2017 7:07 am
-
- Posts: 32
- Joined: Fri Sep 29, 2017 7:05 am
Re: 15.9
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)
-
- Posts: 35
- Joined: Sat Jul 22, 2017 3:01 am
Re: 15.9
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.
-
- Posts: 71
- Joined: Fri Sep 29, 2017 7:04 am
Re: 15.9
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).
-
- Posts: 58
- Joined: Thu Jul 27, 2017 3:00 am
Re: 15.9
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
Who is online
Users browsing this forum: No registered users and 8 guests