## 6th edition 15.9

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Jesse Kuehn 1B
Posts: 40
Joined: Wed Nov 14, 2018 12:23 am

### 6th edition 15.9

when trying to find the units for the rate constants, why is the units for k mol per liter per second when the units in the table for constants are liter per mol per second?

Neil Hsu 2A
Posts: 61
Joined: Fri Sep 28, 2018 12:16 am

### Re: 6th edition 15.9

I'm not sure which question you're talking about (I have 7th edition), but generally, the units for rate constant k depend on the order of the reaction. The units change so that the rate in the rate law will always equal M/s. For example, for first order reactions where rate = k [A], [A] is in molars, so k must have units of s^-1 to make the rate have units of M/s.

Heesu_Kim_1F
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am

### Re: 6th edition 15.9

I agree, the units for rate constant k depend on the order of reaction.
For zero order reactions, the unit for k is (mol x L^-1 x s^-1).
For first order reactions, the unit for k is (s^-1).
For second order reactions, the unit for k is (L x mol^-1 x s^-1).
Hope this helps!

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Re: 6th edition 15.9

In general, the rule of the thumb for finding the units of k is to find what units will result in the rate having the units of mol/(L*s). First, look at the units that the concentrations give you and make the units of k cancel those concentrations leaving you with mol/(L*s).

tierra parker 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: 6th edition 15.9

i personally don't know the exact reason why but if you try it with the first reaction they give you in the chart (btw i have the seventh edition). the rate law = k[H2][I2] and if you're just considering the units for the problem the left side or the rate has units of M/s while the right side is L/mol*sec x mol/L x mol/l and if you cancel out the moles and liters you're left with mol/L*sec which is M/s the same as the left side

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