Chapter 13 Worksheet 4 (ws13.4)
Experimental Determination of Rate Laws
Determining Rate Laws using the Method of Initial Rates
1. Technically, the initial rate is the rate at time zero (immediately after mixing the reactants). In
practice, it is not possible to measure the rate at time zero. The initial rate can be approximated by
measuring the average rate after a very short period time so that very little reactant has been
consumed and very little product has formed. As a rule of thumb, you can assume that you are
measuring the initial rate if less than 2% of the reactants have been consumed.
If the reaction A = B is run for a very short period of time, you can assume that:
Average rate = Initial rate = V0
[A] = [A]0
[B] = 0
rate of the reverse reaction = 0
2. For the generalized reaction, aA + bB = products, the rate law is
rate = V = k[A]x[B]y
Recall that the values of x, y, and k must be determined experimentally. We can do this by
measuring the initial rate starting with various concentrations of reactants. Minimally, how many
different combinations of [A]0 and [B]0 must you use in order to calculate x, y, and k?
Since there are 3 unknowns, you need 3 equations. Thus, you need 3 different combinations of [A]0
and [B]0.
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3. For the reaction:
2 NO (g) + Br2 (g) → 2 NOBr (g) the following initial rate data were obtained:
[NO]0 (M)
0.10
0.25
0.10
a. V = k[NO]x[Br2]y
V0 (d[NOBr]/dt; M-s-1)
24
150
60
Initial [Br2]0 (M)
0.20
0.20
0.50
Calculate x, y, and k
You can write 3 equations with 3 unknowns:
24 M-s-1 = k(0.10 M)x(0.20 M)y
150 M-s-1 = k(0.25 M)x(0.20 M)y
60 M-s-1 = k(0.10 M)x(0.50 M)y
Divide the second equation by the first equation to find x.
6.25 = 2.5x
x = 2 (reaction is second order with respect to NO)
Divide the third equation by the first equation to find y.
2.5 = 2.5y
y = 1 (reaction is first order with respect to Br2)
d[NOBr]/dt = k[NO]2[Br2]
24 M-s-1 = k(0.10 M)2(0.20 M)
k = 1.2 x 104 M-2-s-1 [Notice that the unit is correct for a third order (overall) reaction.]
(The numbers do not represent real data. They were contrived to give the same value of k for each
data set. Normally, you would calculate 3 values of k and average them.)
In general, for the reaction A + B = products, to determine the order with respect to A
(x), measure the rate at two different concentrations of A (hold [B] constant):
 [A] 
rate
2  2
 [A] 
rate
1  1 
x
b. Calculate the rate of formation of NOBr and the rates of consumption of NO and Br2 when [NO]
= 0.50 M and [Br2] = 0.30 M.
d[NOBr]/dt = k[NO]2[Br2] = (1.2 x 104 M-2-s-1)(0.50 M)2(0.30 M) = 9.0 x 102 M-s-1
Notice that this answer makes sense. Compare to second set of data in the table above. [NO] was
doubled resulting in a 4-fold increase in rate (second order) and [Br2] was increased 1.5-fold resulting
in an addition 1.5-fold increase in rate (first order). Overall, the rate increased 6-fold.
-d[NO]/dt = d[NOBr]/dt = 9.0 x 102 M-s-1
-d[Br2]/dt = ½ d[NOBr]/dt = 4.5 x 102 M-s-1
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4. Sometimes the data is simple enough to determine the reaction orders by inspection. Use the data
below to determine the rate law for the generalized reaction: aA + bB = products
V = k[A]x[B]y
Initial [A] (M)
0.010
0.020
0.040
0.20
0.20
0.20
Initial [B] (M)
0.20
0.20
0.20
0.01
0.02
0.04
Initial Rate (-d[A]/dt; M-s-1)
5.4 x 10-7
10.8 x 10-7
21.5 x 10-7
2.7 x 10-8
1.1 x 10-7
4.4 x 10-7
x = 1 (rate doubles when A doubles)
y = 2 (rate quadruples when B doubles)
overall order = 3
5.4 x 10-7 M-s-1 = k(0.010 M)(0.20 M)2
k = 1.35 x 10-4 M-2-s-1
Initial [A] (M)
0.010
0.020
0.040
0.20
0.20
0.20
Initial [B] (M)
0.20
0.20
0.20
0.01
0.02
0.04
Initial Rate (-d[A]/dt; M/s)
5.4 x 10-7
21.5 x 10-7
86.4 x 10-7
2.16 x 10-4
2.16 x 10-4
2.16 x 10-4
x = 2 (rate quadruples when A doubles)
y = 0 (rate does not change when B changes)
overall order = 2
5.4 x 10-7 M-s-1 = k(0.010 M)2
k = 5.4 x 10-3 M-1-s-1
Again, this is not real data so you get the exact same value of k no matter which data set you use. For
real data, you would use each data set to calculate a different k and then average the values.
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Integrated Rate Laws and Determining Rate Laws Using the Graphical Method
Consider a reaction with only a single reactant: A = products
Rate = -d[A]/dt = k[A]x
For a reaction that is first order with respect to A:
 d[A]
 k[A]
dt
Integrating gives the first order integrated rate law (see derivation at the end of this worksheet):
ln[A] = -kt + ln[A]0
or
[A]
 e kt
[A] 0
(Notice that the ratio is the fraction of A remaining at time t.)
or
[A] = [A]oe-kt
(The last equation describes the shape of the [A] vs. time curve for a 1st order reaction with only one
reactant. This is an example of “exponential decay”. We will see this equation again when we discuss the
decay of radioactive nuclei!)
If a reaction is first order, a plot of ___ln[A]_______ vs. ____t______ will give a straight line with
slope = -k
y-intercept = ln[A]0
Sketch the graph:
ln[A] is on the y-axis and t is on the x-axis. It is a straight line with a negative slope. The y-intercept
is a positive value (not zero).
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For a reaction that is second order with respect to A:
 d[A]
 k[A]2
dt
Integrating gives the second order integrated rate law (see derivation at the end of this worksheet):
1
1
 kt 
[A]
[A] 0
If a reaction is second order, a plot of _________ vs. ________ will give a straight line with
slope =
y-intercept =
Sketch the graph:
1/[A] is on the y-axis and t is on the x-axis. It is a straight line with a positive slope. The y-intercept is
a positive value (not zero).
For a reaction that is zeroth order with respect to A:
 d[A]
 k[A]0  k
dt
Integrating gives the zeroth order integrated rate law:
[A] = -kt +[A]0
If a reaction is zeroth order, a plot of _________ vs. ________ will give a straight line with
slope =
y-intercept =
Sketch the graph:
[A] is on the y-axis and t is on the x-axis. It is a straight line with a negative slope. The y-intercept is
a positive value.
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1. You have seen two methods for the experimental determination of rate laws. When would you use the
initial rate method and when would you use the graphical method?
Generally, the graphical method is easier but this can only be used if the reverse reaction does not
occur. The reverse reaction will not occur if one or more of the products are removed as they are
made. In addition, there are some reactions for which the reverse reaction is so slow that it can be
ignored. The initial rate method can only be used if the reaction is slow enough (relative to the
measurement method) that you can measure the rate before 2% of the reactants have been consumed.
Half-life
The half-life of a reaction, t½, is the time required for the concentration of a reactant to drop to one half of
its initial value (when [A] = ½ [A]o).
First order half-life equation (see derivation on page 575 of your textbook):
t 1/2 
0.693
k
Second order half-life equation (see derivation on page 578 of your textbook):
t 1/2 
1
k[A] 0
Zero order half-life equation (see derivation on page 580 of your textbook):
t 1/2 
[A] 0
2k
First half-life: Time required to reach 1/2 of the initial concentration.
Second half-life: Time required to go from 1/2 to 1/4 of the initial concentration.
Third half-life: Time required to go from 1/4 to 1/8 of the initial concentration.
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Supplementary Material (Optional)
Derivation of the integrated rate laws (the change of concentration with time)
First order:
 d[A]
 k[A]
dt
Separate variables and integrate both sides of the equation:
d[A]
 kdt
[A]
ln [A] = -kt + C
If t = 0, then [A] = [A]0 and C = ln [A]0
ln [A] = -kt + ln[A]0
Second order:
 d[A]
 k[A]2
dt
Separate variables and integrate both sides of the equation:
d[A]
 kdt
[A] 2
-1
 kt  C
[A]
If t = 0, then [A] = [A]0 so C =
-1
[A] 0
1
1
 kt 
[A]
[A] 0
See if you can derive the zeroth order integrated rate law. (See above.)
Note: The equations for the zero, first, and second order half-lives are derived in your textbook.
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How do you use the integrated rate law method for reactions with more than one reactant?
The trick is to perform the reaction with one of the reactants present in great excess! Then, the
concentration of the excess reactant doesn’t change during the reaction and you can treat the reaction
as though it has only one reactant!
Consider the generalized reaction with two reactants: aA + bB = products
Rate = -d[A]/dt = k[A]x[B]y
If [B] >> [A], then we can assume that the concentration of B at time t is:
[B]t = [B]0 (only a tiny fraction of B is consumed)
and
-d[A]/dt = k’[A]x
where k’ (pseudo rate constant) = k[B]0y
Now the analysis is the same as for a reaction with only one reactant. The slope of the linear graph,
gives you the pseudo rate constant. The actual rate constant is calculated from the equation above.
To determine the order with respect to B, just reverse the procedure and do the reaction with a huge
excess of A.
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