## Units of K

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

shannon_tseng_3L
Posts: 20
Joined: Wed Sep 21, 2016 2:56 pm

### Units of K

Is there an easy way to find the units of different K values for different reaction orders?

Rachel Wile 2D
Posts: 49
Joined: Sat Jul 09, 2016 3:00 am

### Re: Units of K

For a zero order, the units of k are M/s, or mol*L^-1*s^-1. For first order, the units are 1/s or s^-1. For second order, the units are 1/M*s or L*mol^-1*s^-1. You will notice that there is a pattern here. Every time the order is increase by one we divide by M or mols/L. So therefore, for a third order, the units of k would be 1/M^2*s or L^2*mol^-2*s^-1. Hope this helps!

Amy Ko 3C
Posts: 24
Joined: Sat Jul 09, 2016 3:00 am

### Re: Units of K

Memorizing the k units helps. You can also set up the rate equations to find the k units
For example: a 2nd order rate equation is rate = k [R]^2
The unit of rate is mol/Ls and each [R] has the unit of mol/L
Then, mol/Ls = k (mol/L)(mol/L)
k must cancel out 1 L in denominator and 1 mole in numerator and must add s in denominator
Thus, the unit of k must be L/(mol x s)

Ryan Cerny 3I
Posts: 25
Joined: Wed Sep 21, 2016 3:00 pm
Been upvoted: 1 time

### Re: Units of K

K(units) = (L/mol)^(overall n -1) x s^-1

Joslyn_Santana_2B
Posts: 51
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Units of K

Starting from M/s you just then divide my M continuously to get the next order units.