### Units

Posted:

**Wed Feb 17, 2016 1:26 pm**Does the rate constant k of zero-order reactions always have the units of mole/L*sec?

Created by Dr. Laurence Lavelle

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

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=147&t=12577

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Posted: **Wed Feb 17, 2016 1:26 pm**

Does the rate constant k of zero-order reactions always have the units of mole/L*sec?

Posted: **Wed Feb 17, 2016 2:52 pm**

Yes, other ways of writing the same units would be M (molarity)/s or Ms^-1 or mol(L^-1)(s^-1).

Posted: **Fri Mar 11, 2016 12:11 am**

An easy way to determine the units of k is to understand that the rate needs to have units of M/s so for 0 order rate=k and therefore k=M/s. for 1st order: rate=k[A] where [A] has the units of M and so when multiplied with k, k must have units of s^-1. For second order rate=k[A]^2 where [A]^2 has units of M^2 so in order to get rate = M/s, k must be M^-1 s^-1.

Posted: **Sat Mar 11, 2017 2:36 pm**

Yes, it'll be always M/Sec

Posted: **Sun Mar 12, 2017 6:19 pm**

Divide by Molarity for each order essentially

Posted: **Wed Feb 28, 2018 4:37 pm**

There's also one using kPA/sec. Does anyone know how kPA relates to the rate of reactions?

Posted: **Wed Feb 28, 2018 9:55 pm**

yes because for oth order, the rate law is written as:

rate = k[A]^0

which is equal to rate = k

since the unit for rate = mol/l*s, the unit for k is therefore mol/l*s

rate = k[A]^0

which is equal to rate = k

since the unit for rate = mol/l*s, the unit for k is therefore mol/l*s

Posted: **Wed Feb 28, 2018 10:13 pm**

Yes! M/s