## Units of k for orders above 2nd

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

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

### Units of k for orders above 2nd

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

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

JamesAntonios 1E
Posts: 70
Joined: Fri Sep 29, 2017 7:04 am

### Re: Units of k for orders above 2nd

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

204918982
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Joined: Fri Sep 29, 2017 7:04 am

### Re: Units of k for orders above 2nd

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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### Re: Units of k for orders above 2nd

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

A helpful way of finding the units for different orders is to use $M^{1-x}/s$. 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 $M^{1-10}/s$=$M^{-9}/s$. I hope this helps!

Fatima_Iqbal_2E
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### Re: Units of k for orders above 2nd

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.