### Orders and units

Posted:

**Tue Feb 27, 2018 7:09 pm**Why is that each different order type (zero, first, second) has different units? Can someone explain it please?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=145&t=28629

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Posted: **Tue Feb 27, 2018 7:09 pm**

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

Posted: **Tue Feb 27, 2018 7:14 pm**

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.

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.

Posted: **Tue Feb 27, 2018 9:15 pm**

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.

"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.

Posted: **Tue Feb 27, 2018 9:57 pm**

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)

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)