Units of k


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Frenz Cabison 1B
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Joined: Fri Sep 29, 2017 7:07 am

Units of k

Postby Frenz Cabison 1B » Mon Mar 05, 2018 10:22 am

What are the units of k in zero-, first-, and second- order reactions? How are these units derived?

Sirajbir Sodhi 2K
Posts: 47
Joined: Sat Jul 22, 2017 3:00 am

Re: Units of k

Postby Sirajbir Sodhi 2K » Mon Mar 05, 2018 10:49 am

Zero-order: M/s
First order: s^-1
Second order: L/(mol*s)

These units are derived from the rate law -- we know the rate of reaction must be M/s, so the units of k change to accommodate that.
Last edited by Sirajbir Sodhi 2K on Mon Mar 05, 2018 10:52 am, edited 1 time in total.

Gurkriti Ahluwalia 1K
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Joined: Fri Sep 29, 2017 7:07 am

Re: Units of k

Postby Gurkriti Ahluwalia 1K » Mon Mar 05, 2018 5:32 pm

shouldn't it be mol/l/s for zero order reactions??

Qining Jin 1F
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Joined: Fri Sep 29, 2017 7:07 am

Re: Units of k

Postby Qining Jin 1F » Mon Mar 05, 2018 6:23 pm

The units for zero order is mol/(L*s)

Sarah Rutzick 1L
Posts: 50
Joined: Tue Oct 10, 2017 7:13 am

Re: Units of k

Postby Sarah Rutzick 1L » Tue Mar 06, 2018 10:16 pm

If you are having trouble figuring out the units of k, write out the entire rate law with the units. You know that the rate must be in mol/Ls, so the units of k must cancel out the units of concentration of the reactants until you get mol/Ls.

Peri Bingham 1G
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Joined: Fri Sep 29, 2017 7:03 am

Re: Units of k

Postby Peri Bingham 1G » Tue Mar 06, 2018 10:20 pm

Units of the rate are almost always mol/L*s and units of concentration are almost alway mol/L. Therefore the units of k can be determined based on the reaction order and the already determined units.

Angel R Morales Dis1G
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Joined: Sat Jul 22, 2017 3:01 am

Re: Units of k

Postby Angel R Morales Dis1G » Thu Mar 15, 2018 7:54 pm

The easiest way to determine the units of the rate constant k is to use the following "formula":
M^(1-X)/S
where x is the order of the reaction. Examples:
Zero Order: M(^1-0)/S ----> units: M/S
First Order: M^(1-1)/S ----> units: 1/S and so on

Kyle Alves 3K
Posts: 46
Joined: Thu Jul 27, 2017 3:01 am

Re: Units of k

Postby Kyle Alves 3K » Thu Mar 15, 2018 10:22 pm

You can also solve for the units of k for each order! Since rate always begins with mol/(L.s)
Zero order(mol/L.s) = k[A mol/L]^0
mol/(L.s) = k

First order (mol/L.s) = k[A mol/L]
/s = k[A]

Second order (mol/L.s) = k[A mol/L]^2
(mol/(L.s)) = k[A]^2 (mol^2/L^2)
L/(mol.s) = k[A]^2

Rachel Wang
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

Re: Units of k

Postby Rachel Wang » Fri Mar 16, 2018 7:09 pm

0 order: M/s or mol/L.s

1st order: 1/s

2nd order: 1/M.s or L/mol.s

3rd order: 1/M^2.s or L^2/mol^2.s

etc.

Michael Downs 1L
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Joined: Thu Jul 13, 2017 3:00 am

Re: Units of k

Postby Michael Downs 1L » Fri Mar 16, 2018 8:59 pm

Will we have any questions pertaining to any reaction larger than the 3rd order on the final?

Rachel N 1I
Posts: 48
Joined: Thu Jul 27, 2017 3:00 am

Re: Units of k

Postby Rachel N 1I » Sat Mar 17, 2018 12:55 pm

I highly doubt that we would be asked to find anything higher than the 3rd order. But we may be asked questions where the overall order is greater than 3.

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

Re: Units of k

Postby Destiny Diaz 4D » Tue Mar 05, 2019 7:25 pm

There might be questions that ask to derive 3+ order equations but I doubt we will be asked to do much with them at all.

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: Units of k

Postby Nina Do 4L » Wed Mar 13, 2019 10:20 pm

I would remember that zero order would be M/Sec and then divide M each time you go down an order. Here's what I mean:
Zero order would be M(molarity)/second
so divide by M to get units for first order. Notice how M cancels out.
1st order units: s^-1
Divide M again to get 2nd order units
s^-1*M^-1

And continue dividing Molarity each time you go down an order. Hope that helps!


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