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### 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?

### Re: Orders and units

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.

### Re: Orders and units

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.

### Re: Orders and units

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)