Method of Initial Rates
Example 1:
Consider the following reaction and data:
Trial
A+2B → 3C
v = k [A]a [B]b
Rate
Initial Concentration Initial Concentration
(mole L-1 sec-1)
of A (mole L-1)
of B (mole L-1)
1
2.73
0.100
0.100
2
6.14
0.150
0.100
3
2.71
0.100
0.200
One can employ the following algebraic technique for determining the exponents.
First, write the ratio of the rate laws for two trials.
r1
k [A]1a [B]1b
=
r2
k [A]2a [B]2b
Next, substitute the numerical values into the equation.
k (0.100 mole L-1)a (0.100 mole L-1)b
2.73 mole L-1 sec-1
=
-1
6.14 mole L sec
k (0.150 mole L-1)a (0.100 mole L-1)b
-1
Notice that the units for each quantity and the rate constant can be removed, and in this case the
exponent b is removed when the concentrations of B divide. The equation simplifies to
0.100a
2.73
=
6.14
0.150a
0.4446 = 0.6667a
To convert a from an exponent into a coefficient, take the logarithm of both sides of the equation.
ln[0.4446] = ln[0.6667a]
-0.8106 = -0.4054a
The value of a may now be readily determined.
-0.8106
a=
= 1.9995
-0.4054
In this case the reaction is second order in A (a = 2).
A similar strategy can be employed to determine the value of b.
 it should be obvious from inspection of trials 1 and 3 that the reaction is zero-order in B
 Once the exponents are known, the rate constant can be calculated.
 Because the data generally suffers from experimental error, it is best to calculate the rate
constant for each trial and use the average value.
Consider a reaction A + B -> Products.
Run #
Initial [A] ([A]0)
Initial [B] ([B]0)
Initial Rate (v0)
1
1.00 M
1.00 M
1.25 x 10-2 M/s
2
1.00 M
2.00 M
2.5 x 10-2 M/s
3
2.00 M
2.00 M
2.5 x 10-2 M/s
What is the rate constant for this reaction?
in runs 1 and 2, the initial concentration of B has been varied. In fact, [B] 0 has doubled from run 1 to run 2
and the reaction rate has also doubled. Therefore the reaction must be First Order in B.
Examination of runs 2 and 3 show that in these experimenal runs the concentration dependence of A has
been isolated. In this case the doubling of the initial concentration of A has No EFFECT on the reaction rate
so the reaction must be Zeroth Order in A.
Thus the rate law for the reaction is: rate = k[B]1[A]0 = k[B]
Now, the rate constant can be determined from any of the experimental runs. Let's use run # 3
rate = 2.5 x 10-2 M/s= k (2.0 M)
k = 1.25 x 10-2 s-1
-------------------------------------------------------------------Consider a reaction A + B + C-> Products
Run
[A]0
[B]0
[C]0
v0
1
0.151 M
0.213 M
0.398 M
0.480 M/s
2
0.251 M
0.105 M
0.325 M
0.356 M/s
3
0.151 M
0.213 M
0.525 M
1.102 M/s
4
0.151 M
0.250 M
0.480 M
0.988 M/s
What is the initial rate of the reaction when all the reactants are at 0.100 M concentrations?
rate = k [A]a [B]b [C]c
The ratio of the initial rates of runs 1 and 3 is then:
Note that we have chosen this ratio so that many terms on the right hand side cancel, i.e. k and the
concentrations of species A and B. The ratio reduces to:
Note that none of the numbers in this equation have any units anymore because we have divided them out in
a dimensionless ratio. Now, how to solve for c? Take the natural log (i.e. ln) of both sides of the equation
or
So the reaction is Third Order in reagent C!
To continue, we pick another pair of runs that have a change in the initial concetration of C and just one
other reactant. Let us choose the runs 1 and 4. The ratio of the initial rates of these runs is:
Note that everything cancels, is known, or is the order b. So:
We can determine b as before by taking the ln of both sides of this equation:
and b = 1.00. The reaction is First Order in B!
Now that we know the order of Reactants A and B, we can use another pair of experimental runs to
determine the remaining unknown order. We must use a pair of runs where the initial concentration of A
changes, so we pick runs 1 and 2. As before, the ratio of initial rates is
The rate constant cancels as before and we can solve for a:
and, as usual, take the ln of both sides:
which results in a = 2.00 and the reaction is Second Order in A.
The rate constant for the reaction may now be evaluated from any of the experimental runs:
or
k = 1.57 x 103 l5 / mol5.s
These units are appropriate for a reaction that is sixth order overall. Use another experimental run to check
your answer. If the rate law is correct, every experimetal run will give the same value of the rate constant.
And now for the finale. The rate of the reaction when all the reactants are 0.100 M is:
The Half-life Method
The half-life method involves measuring the time taken to reduce the concentration to half its initial
value for a variety of values of initial concentration, then plotting a graph of concentration vs halflife - suitably linearised. The table below summarises the appropriate linear plots
Table 2. Determination of kinetic parameters from half-life
Order
Rate eqn
Half-life
Linear plot
slope
t½ =
0
1
t½ vs [A]0
Half life is constant for first
order reactions
2
t½ vs
n
log t½ vs log[A]0
-(n-1)
The method for reactions of order n is a general one. The value of the rate constant, k may be found
from the intercept, which is equal to log
The half life method combines some of the advantages of the differential method; order may be
found from a single plot but, as with the integral method; there is no need to determine tangents. For
reactions of order zero, one and two, the rate constant is best determined as indicated in table 2.
Isolation method
The three methods described so far are really best suited to reactions with one reactant. If more than
one reactant is involved, the integrated rate equation quickly becomes very complex and a different
procedure is involved.
The isolation method involves arranging for all but one of the reactants to be in such large excess,
that their concentrations are essentially constant during the progress of the reaction. In practice this
means an excess of at least 10[A]0 and preferably >100[A]0. ( [A]0 represents the concentration of
the limiting reactant)
As an example, let us consider a second order reaction of the form
A + B  Products
For such a reaction, the second order rate equation would be
If [B] were in large excess, the rate equation would simplify to
The rate equation is thus, simplified to a first order equation. This is called a pseudo-first order
reaction. The pseudo first order rate constant is defined from
The second order rate constant may be found by determining the value of k’ for various values of
[B]0 and plotting a graph of k’ vs [B]0. This method may be extended to reactions of order higher
than two and with more than two reactants.