## Orders and units

Cyianna 2F
Posts: 34
Joined: Wed Nov 16, 2016 3:04 am

### Orders and units

Why is that each different order type (zero, first, second) has different units? Can someone explain it please?

manasa933
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am

### Re: Orders and units

First order is:
k=rate/[A]
Second order reaction:
k=rate/[A]^2
Third order reaction:
k=rate/[A]^3

We measure concentration as mol/L, hence, for reactions of a different order it is (mol/L)^order.

mhuang 1E
Posts: 22
Joined: Fri Sep 29, 2017 7:05 am

### Re: Orders and units

Each different order type has different units because as Harrison replied on my previous post:
"Zero - rate of reaction is equal to rate constant k
1st order - depends on only ONE reactant concentration
2nd order - depends on two different reactants or one reactant squared"
So, that would mean that:
Zero:Rate=k
One:Rate=k[A]
Two:Rate=k[A]^2.
From #9, you then set the above to the given rate and solve for k, the rate constant.

Sandhya Rajkumar 1C
Posts: 50
Joined: Fri Jun 23, 2017 11:40 am

### Re: Orders and units

The rate of the reaction has units mol*(L^-1)*(s^-1)
For a zero order reaction, the rate = k*[A]^0= k, where k is the rate constant. In order for the rate to have units mol*(L^-1)*(s^-1) the units for k have to be
mol*(L^-1)*(s^-1)
For a first order reaction, the rate = k *[A]^1 and the units for [A] are mol*(L^-1) so in order for the rate to have units mol*(L^-1)*(s^-1) k must have units s^-1
For a second order reaction, the rate = k*[A]^2 and the units for [A]^2 are (mol^2)*(L^-2) so in order for the rate to have units mol*(L^-1)*(s^-1) the units for k must be L*(mol^-1)*(s^-1)