Experimental Determination
of Reaction Order 
The rate is usually given in terms of moles/Litre seconds but
this is not always the case. You will have to be very careful about the
units of each participant reactant in order to get the proper units for
the rate constant itself. 
Before we can determine the value of the rate constant 'k', we
need to find the exponential value for each participant. To find the relationship
of one reactant it is necessary to keep the other reactant(s) constant
so look at the following reaction and data table: 
NO + H_{2 }> HNO_{2} 
(balance it if you'd like, but the balanced
equation will not give you the exponential values unless the equation is
a onestep reaction). 
Rates of reaction between NO and H_{2 }at 800^{o}C 
Experiment NO
H_{2 }Initial Rate
of Reaction
Number moles/L
moles/L moles/L sec
1
0.001 0.004
0.002
2
0.002 0.004
0.008
3
0.003 0.004
0.018
4
0.004 0.001
0.008
5
0.004 0.002
0.016
6
0.004 0.003
0.024
 
From the equation we can write a partial rate law as
rate = k[NO]^{m}[H_{2}]^{n}

You have 6 experiments to choose from. Using these 6 experiments
you must determine the values of 'm', 'n' and 'k'. There are only two reactants,
so choose one to start to work with. We'll start with NO. Choose any two
experiments where the concentration of NO changes but the concentration
of H_{2 }stays the same. We want to determine how the NO changes
the rate! We will choose experiments 1 and 2. 
Using experiments 1 and 2 you can see that the concentration jumps
from 0.001 to 0.002 moles/L. IT DOUBLES!! Take a look at the rates for
these same experiments. The rate jumps from 0.002 to 0.008 moles/L seconds.
IT QUADRUPLED!!. 
The exponential constant 'm' for the [NO] is the mathematical
relationship between these two values. i.e. 
2^{m }= 4 therefore m = 2 because 2^{2 }= 4 
To confirm this, compare experiments 1 and 3. The concentration
of the NO gas TRIPLES. The rate jumps by a factor of nine. Therefore 
3^{m }= 9 so m = 2 because 3^{2 }= 9 
The rate law expression can now be updated to: rate = k[NO]^{2}[H_{2}]^{n} 
We will now use experiments where the [NO] concentration is kept
constant and the [H_{2}] changes. 
Look at experiments 4 and 5. The H_{2 }concentration DOUBLES
and the rate DOUBLES. 
2^{n }= 2 therefore n = 1 since 2^{1 }= 2 
To confirm this number look at experiments 4 and 6. The H_{2
}concentration TREBLES from 0.001 to 0.003 The rate also
TREBLES from 0.008 to 0.024 
3^{n }= 3 therefore n = 1 since 3^{1 }= 3 
So the rate law expression can be rewritten as 
rate = k [NO]^{2 }[H_{2}]^{1} 
From the sum of the exponents this is a third order reaction. 
Now to determine the value of 'k'. 'k' is a constant. It's value
should not change (except under temperature changes). Choose any one of
the experiments. It should not matter which one. 
We will use experiment 1. Using the rate law, above fill in the
values from the data table. 
0.002 mol/L sec = k (0.001 mol/L)^{2 }* (0.004 mol/L) 
0.002 mol/L sec = k * (0.000001 mol^{2 }/L^{2})
* (0.004 mol/L) 
0.002 mol/L sec = k * 0.000 000 009 mol^{3}/L^{3}
k = 0.002 mol/L sec
0.000 000 004 mol^{3}/L^{3}
= 500,000 sec/mol^{2 }L^{2 }or sec mol^{2
}L^{2}

Therefore the rate law equation for this reaction is
rate = 500,000 sec mol^{2 }L^{2 }[NO mole/L]^{2
}[H_{2 }mol/L]
