Units of k

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

Frenz Cabison 1B
Posts: 54
Joined: Fri Sep 29, 2017 7:07 am

Units of k

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

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

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

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

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

Re: Units of k

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
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Re: Units of k

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
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: Units of k

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

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

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

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

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

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

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!