Units  [ENDORSED]


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Lisa Lim 3G
Posts: 12
Joined: Fri Sep 25, 2015 3:00 am

Units

Postby Lisa Lim 3G » Wed Feb 17, 2016 1:26 pm

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

EPerez1B
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

Re: Units

Postby EPerez1B » 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).

Carolyn Stephens 3H
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Re: Units  [ENDORSED]

Postby Carolyn Stephens 3H » 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.

Michelle Kam 1F
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Joined: Wed Sep 21, 2016 2:58 pm

Re: Units

Postby Michelle Kam 1F » Sat Mar 11, 2017 2:36 pm

Yes, it'll be always M/Sec

Samuel_Vydro_1I
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

Re: Units

Postby Samuel_Vydro_1I » Sun Mar 12, 2017 6:19 pm

Divide by Molarity for each order essentially

Chew 2H
Posts: 34
Joined: Fri Sep 29, 2017 7:05 am

Re: Units

Postby Chew 2H » 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?

Dang Lam
Posts: 55
Joined: Thu Jul 27, 2017 3:01 am

Re: Units

Postby Dang Lam » 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

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

Re: Units

Postby Jennie Fox 1D » Wed Feb 28, 2018 10:13 pm

Yes! M/s


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