|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
|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 + H2 -----> HNO2
|(balance it if you'd like, but the balanced equation
will not give you the exponential values unless the equation is a one-step
|Rates of reaction between NO and H2 at 800oC
H2 Initial Rate
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[H2]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 H2 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
|The exponential constant 'm' for the [NO] is the mathematical relationship
between these two values. i.e.
|2m = 4 therefore m = 2 because 22 = 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
|3m = 9 so m = 2 because 32 = 9
|The rate law expression can now be updated to: rate = k[NO]2[H2]n
|We will now use experiments where the [NO] concentration is kept constant
and the [H2] changes.
|Look at experiments 4 and 5. The H2 concentration DOUBLES
and the rate DOUBLES.
|2n = 2 therefore n = 1 since 21 = 2
|To confirm this number look at experiments 4 and 6. The H2 concentration
TREBLES from 0.001 to 0.003 The rate also TREBLES from 0.008 to 0.024
|3n = 3 therefore n = 1 since 31 = 3
|So the rate law expression can be rewritten as
|rate = k [NO]2 [H2]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 mol2 /L2) *
|0.002 mol/L sec = k * 0.000 000 009 mol3/L3
k = 0.002 mol/L sec
= 500,000 sec/mol2 L2 or sec mol-2
|Therefore the rate law equation for this reaction is
rate = 500,000 sec mol-2 L-2 [NO mole/L]2 [H2 mol/L]