Units of k for orders above 2nd


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Mary Becerra 2D
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Joined: Fri Sep 29, 2017 7:06 am

Units of k for orders above 2nd

Postby Mary Becerra 2D » Fri Mar 09, 2018 11:14 am

So I noticed a trend with the units of k, we keep dividing each one by M (molarity) i.e. zero order is k M/s, first order is k 1/s, second order is k 1/M.s. Does this trend continue for 3rd, 4th, and 5th orders?

Deap Bhandal L1 S1J
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Re: Units of k for orders above 2nd

Postby Deap Bhandal L1 S1J » Fri Mar 09, 2018 11:26 am

Yes, for example, the k for a 10th order reaction would be M^(-9) * s^(-1).

JamesAntonios 1E
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Joined: Fri Sep 29, 2017 7:04 am

Re: Units of k for orders above 2nd

Postby JamesAntonios 1E » Fri Mar 09, 2018 12:00 pm

Yes this trend should continue. Just remember that for zero order, it is actually mol/l*s, not inverted.

204918982
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Re: Units of k for orders above 2nd

Postby 204918982 » Sat Mar 10, 2018 2:57 pm

Yes I'm pretty sure this will always work, I've never thought about it like this so thanks for the tip!

Ilan Shavolian 1K
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Joined: Fri Sep 29, 2017 7:03 am

Re: Units of k for orders above 2nd

Postby Ilan Shavolian 1K » Sat Mar 10, 2018 8:41 pm

isnt there supposed to be an L,L2... at the top??

Fatima_Iqbal_2E
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Re: Units of k for orders above 2nd

Postby Fatima_Iqbal_2E » Sat Mar 10, 2018 11:01 pm

A helpful way of finding the units for different orders is to use . The x value would be the order that you are finding your units for. For example, if you are trying to find the units for a 10th order reaction, it would be =. I hope this helps!

Fatima_Iqbal_2E
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Re: Units of k for orders above 2nd

Postby Fatima_Iqbal_2E » Sat Mar 10, 2018 11:02 pm

Ilan Shavolian 1K wrote:isnt there supposed to be an L,L2... at the top??


The M refers to molarity, so the liters would be implied. If m was written, then yes, L would've needed to have been specified.


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