Determination of Rate Exponents using Initial rates of reactions.
Chemists can examine the change in the initial rates of the reaction when concentrations of reactants
are changed to determine the order of reactants (for a specific equation). This method can be used to
save time, since scientists do not have to follow reactions to their end. This can be important in
studying very slow reactions. The chart below indicates the general change in rate observed for each
order of reaction when the concentration of a specific reactant is doubled.
Order of
Reactant
First Order
Change in Initial Rate when
concentration is doubled
Rate doubles
Second Order
Rate quadruples
Zero Order
Rate remains unchanged
Initial Rates Method
By examining charts of how the initial rate changes when reactant concentrations are changed
chemists can accurately determine the order of the reaction for that reactant. The examples given
below illustrate how this is done.
a. Single reactant
Examine the chart below;
Table 1: Rate experiments on [A] (concentration of A ) vs initial rate, for the following equation
Equation
A -----> B + C
Experiment
Initial [A]
(maul/l)
Initial Rate ( mol/(l*s)
)
1
0.01
4.8 * 10 -6
2
0.02
9.6 * 10 -6
3
0.03
1.4 * 10 -5
The [A] is doubled between experiment # 1 and # 2, the initial rate also doubles indicating a direct
relationship between [A] and rate. This is a First Order reaction. Examine the first and third
experiment.



What happens to the [A]?
What happens to the initial rate?
Does this also indicate a direct relationship or first order
reaction?
b. Two Reactants
When there is more than one reactant, we examine experiments in which only one of the two reactant
concentrations change while the other remains constant (is controlled).
The data table below is the one presented in the flash presentation.
Reaction
A
+
B
Products
Experiment
Initial [A]
Initial [B]
Initial Rate
1
0.01
0.03
2.4 * 10 -4
2
0.03
0.03
7.2 * 10 -4
3
0.01
0.06
2.4 * 10 -4
The following is a worked out example similar to the one presented in the above flash
presentation

Determine the reaction order for the reactants in the reaction below given the
table of data on initial rate.
Write the rate law for the reaction as was done in the flash presentation above.
Determine the value of the rate constant k.


Reaction
2ICl
+
H2
I2
+
2HCl
Experiment
[ICl] (mol/L)
[H2] ( mol/L)
Initial Rate
(mol/(L*s))
1
0.10
0.01
0.002
2
0.20
0.01
0.004
3
0.10
0.04
0.008
Use the following questions and answers to help understand how this works.
1. Examine the [Icl] concentration between experiment 1 and 2. What happens? (it
doubles.)
2. What happens to the [H2] between experiment 1 and 2 ? (it remains constant)
3. What happens to the initial rate? (it doubles) Therefore Icl is first order.
4. What happens to the [H2] between experiments 1 and 3? (it quadruples , four times
greater)
5. What happens to the [ICl] between experiments 1 and 3? (it remains constant)
6. What happens to the initial rate of the reaction? (it quadruples, four times greater)
Therefore H2 is first order. (rate quadruples when concentration quadruples)
7. The rate law then is Rate = k [ICl]1 [H2] 1
8. To determine the value of k we plug in the information from any experiment into our rate
law, for example experiment #2 above
 [Icl] = 0.10 mol/l
 [H2] = 0.01 mol/l
 Initial Rate = 0.002 mol/(L*s), plugging in to the above rate law this equals
 0.002 mol/(L*s) = k (0.10 mol/l) 1 (0.01 mol/l) 1
 k=2
Consider the following set of data:
1) 2 O3
3 O2
Experiment # Initial [O3] (M) Initial Rate (M/sec)
1
2.00
0.500
2
4.00
1.000
3
6.00
1.500
First, set up a generic rate law:
rate = k[A]X
Next, create a ratio with the rate laws for 2 experiments:
Simplify and solve for x:
The reaction is first order with respect to [O3] because the exponent was calculated to be 1.
Now we can write the rate law and solve for k by plugging in values for the rate and the [O3] from a
single experiment:
2) N2O4
2 NO2
Experiment. # Initial [N2O4 ] (M) Initial Rate (M/sec)
1
0.50
0.050
2
1.00
0.200
3
1.50
0.450
3) Xe + 3 F2
XeF6
Experiment. # Initial [Xe] (M) Initial [F2] (M) Initial Rate (M/sec)
1
0.50
0.25
0.00156
2
1.50
1.00
0.05625
3
0.75
0.25
0.0032
4
1.50
0.25
0.01406
5
0.50
1.00
0.00625
The rate law includes the concentrations of all the reactants:
rate = k[Xe]X[F2]Y
Therefore, we must solve for both X and Y separately.
First choose two experiments in which the initial [F2] are the same and set up a ratio with these
experiments:
The [F2] cancel each other out ,and the equation simplifies to find that X = 2. The reaction is second
order with respect to [Xe].
Next, we do the same to solve for Y. Choose two experiments in which the initial [Xe] are the same
and set up a ratio.
The [Xe] cancel each other out, and the equation simplifies to find that Y = 1. The reaction is first
order with respect to [F2].
Now we can write the rate law:
* The reaction is first order with respect to F2 and second order with respect to Xe.
Now we have to solve for k: