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```									CHEMISTRY 12

Chapter 6

Rates of Chemical Reactions
Solutions for Practice Problems
Student Textbook page 272

1. Problem
Cyclopropane, C3H6 , is used in the synthesis of organic compounds and as a fast-
acting anesthetic. It undergoes rearrangement to form propene. If cyclopropane
disappears at a rate of 0.25 mol/s, at what rate is propene being produced?
What Is Required?
Calculate the rate of propene production.
What Is Given?
The rate of rearrangement of cyclopropane is 0.25 mol/s.
Cyclopropane and propene both have the formula C3H6 . The balanced chemical
equation is
C3H6 (cyclopropane) → C3H6 (propene)
Determine the mole to mole ratio between cyclopropane and propene. Use this to
determine the rate of formation of propene.
The mole to mole ratio between cyclopropane and propene is 1:1. The rate of
formation of propene is the same as the rate of reaction of cyclopropane, or
0.25 mol/s.
The 1:1 stoichiometry of the reaction indicates that the rate of reaction of
cyclopropane is equivalent to the rate of formation of propene.

2. Problem
Ammonia, NH3, reacts with oxygen to produce nitric oxide, NO, and water vapour.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
At a speciﬁc time in the reaction, ammonia is disappearing at a rate of
0.068 mol/(L·s). What is the corresponding rate of production of water?
What Is Required?
Calculate the rate of production of water.
What Is Given?
You have the balanced chemical equation and the rate of disappearance of ammonia.
Determine the mole to mole ratio between NH3 and H2O. Since 6 mol of H2O are
produced for every 4 mol of NH3 that react, the rate of appearance of H2O is 6 or 3
4    2
the rate of reaction of NH3.

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Rate of appearance of H2O = 3 × (rate of disappearance of NH3)
2
= 3 × 0.068 mol/(L·s)
2
= 0.10 mol/(L·s)
The rate of production of water is greater than the rate of disappearance of ammonia,
which makes sense, given the mole ratio.

3. Problem
Hydrogen bromide reacts with oxygen to produce bromine and water vapour.
4HBr(g) + O2(g) → 2Br2(g) + 2H2O(g)
How does the rate of decomposition of HBr (in mol/(L·s)) compare with the rate of
formation of Br2 (also in mol/(L·s))? Express your answer as an equation.
What Is Required?
Determine the rate of decomposition of HBr relative to the rate of formation of Br2 ,
What Is Given?
You are given the balanced chemical equation.
First check that the equation is balanced. Then use the molar coefﬁcients in the
balanced equation to determine the relative rates of decomposition of HBr and
formation of Br2 .
Since 4 mol HBr are consumed for every 2 mol of Br2 produced, the rate of
decomposition of HBr is equal to 2 times the rate of formation of Br2 .
Rate of decomposition of HBr = 2 × (rate of formation of Br2 ) or,
∆[Br2] = −0.5∆[HBr]
∆t          ∆t
From the coefﬁcients in the balanced equation, you can see that the rate of
decomposition of HBr is twice that of the rate of formation of Br2 .

4. Problem
Magnesium metal reacts with hydrochloric acid to produce magnesium chloride and
hydrogen gas.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Over an interval of 1.00 s, the mass of Mg(s) changes by −0.011 g.
(a) What is the corresponding rate of consumption of HCl(aq) (in mol/s)?
(b) Calculate the corresponding rate of production of H2(g) (in L/s) at 20˚C and
101 kPa.
What Is Required?
First, calculate the rate of consumption of HCl(aq) , in mol/s. Then calculate the rate
of production of H2 , in L/s at 20˚C and 101 kPa.
What Is Given?
You have the balanced chemical equation, and you know the rate of consumption of
Mg, in g/s.

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From the balanced chemical equation, you can obtain the relative rates of consump-
tion or production of reactants and products, in mol/s.
(a) You must ﬁrst convert the rate of disappearance of Mg from g/s to mol/s. From
the balanced chemical equation you can see that the rate of reaction of HCl(aq) is
twice the rate of reaction of Mg.
(b) From the balanced chemical equation you can see that the coefﬁcients of Mg and
of H2 are both 1. This means that the rate of production of H2 , in mol/s, is equal
to the rate of disappearance of Mg, in mol/s. For H2 , you then need to convert
mol/s to L/s at 20˚C, 101 kPa. Recall that 1.00 mol of any gas occupies a volume
of 24.0 L at 20˚C, 101 kPa.
0.011 g/s
(a) Rate of disappearance of Mg =
24.3 g/mol
= 4.5 × 10−4 mol/s
Rate of consumption of HCl = 2 × (rate of disappearance of Mg)
= 2 × 4.5 × 10−4 mol/s
= 9.0 × 10−4 mol/s
(b) Rate of appearance of H2 = rate of disappearance of Mg
= (4.5 × 10−4 mol/s) × (24.0 L/mol)
= 0.011 L/s at 20˚C, 101 kPa
You have correctly used the coefﬁcients in the balanced equation. The units are
correct.

Solutions for Practice Problems
Student Textbook page 284

5. Problem
When heated, ethylene oxide decomposes to produce methane and carbon monoxide.
C2H4O(g) → CH4(g) + CO(g)
At 415˚C, the following initial rate data were recorded.
Experiment   [C 2H 4O] 0 (mol/L) Initial rate (mol/(L • s))
1           0.002 85               5.84 × 10−7
2           0.004 28               8.76 × 10−7
3           0.005 70               1.17 × 10−6

Determine the rate law equation and the rate constant at 415˚.
What Is Required?
You need to ﬁnd the value of m and k in the general rate law equation for the
reaction.
Rate = k[C2H4O]m
What Is Given?
You know the initial concentration of the reactant and the initial rates for three
experiments.
Since there is only one reactant, simply compare the initial rates of two experiments
with different initial [C2H4O]. Compare the change in initial rate with the change in
initial concentration.

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Begin with experiments 1 and 2. Set up a ratio of the two rates as shown.
m
Rate2 = k 0.002 85 mol/L = 5.84 × 10−7 mol/(L·s)
Rate1    k 0.004 28 mol/L
m
8.76 × 10−7 mol/(L·s)
Because k is constant, it can be cancelled.
m
Rate2 = k 0.002 85 mol/L = 5.84 × 10−7 mol/(L·s)
Rate1    k 0.004 28 mol/L
m
8.76 × 10−7 mol/(L·s)
0.666m = 0.666
m = 1 (by inspection)
Similarly, you can set up a ratio involving experiments 1 and 3 or experiments 2
and 3 to obtain m = 1. Therefore, the reaction is ﬁrst order with respect to
ethylene oxide.
To determine the value of k, substitute m = 1 into the rate law equation. You can
use data from any of experiments 1, 2 or 3 and solve for k.
Substituting data from experiment 1 into the rate law equation gives:
1
5.84 × 10−7 mol/(L·s) = k 0.002 85 mol/L
−7
k = 5.84 × 10 mol/(L·s)
0.002 85 mol/L
= 2.05 × 10−4 s−1 (at 415˚C)
The rate law equation for the reaction is:
Rate = (2.05 × 10−4 s−1)[C2H4O]1 (at 415˚C)
The value of m = 1 can be checked by inspection. When the concentration of
ethylene oxide doubles, so will the initial rate. The value of k can be checked by
using data from one of the other experiments. The units of k are consistent with a
ﬁrst order reaction. The number of signiﬁcant digits for k are appropriate for the
given data.

6. Problem
Iodine chloride reacts with hydrogen to produce iodine and hydrogen chloride.
2ICl + H2 → I2 + 2HCl
At temperature T, the following initial rate data were recorded.
Experiment        [ICI] 0 (mol/L)   [H 2 ] 0 (mol/L)   Initial rate (mol/(L • s))
1                 0.20             0.050                   0.0015
2                 0.40             0.050                   0.0030
3                 0.20             0.200                   0.0060

Determine the rate law equation and the rate constant at temperature T.
What Is Required?
You need to ﬁnd the values of m, n, and k in the general rate law equation for the
reaction.
Rate = k[ICl]m[H2]n
What Is Given?
You know the initial concentrations of the reactants and the initial rates for three
experiments.

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Step 1 Find two experiments in which [ICl] remains constant and [H2 ] changes.
Compare the rates and concentrations to solve for n.
Then ﬁnd two experiments in which [H2 ] remains constant and [ICl]
changes. Compare the rates and concentrations to solve for m.
Step 2   Use the data and your calculated values for m and n to solve for k, using the
following equation.
Rate = k[ICl]m[H2]n
Step 1 Begin with experiments 1 and 3, where [ICl] remains constant, to determine
the order with respect to H2 .
Set up a ratio for the two rates as shown. Notice that you can cancel [ICl]
because you chose experiments where [ICl] did not change.
m
Rate3     k 0.020 mol/L [0.200]n
=                  m            = 0.0060 mol/(L·s)
Rate1     k 0.020 mol/L [0.050]n         0.0015 mol/(L·s)
Because k is a constant, you can cancel it out.
m
Rate3     k 0.020 mol/L [0.200]n
=                   m          = 0.0060 mol/(L·s)
Rate1     k 0.020 mol/L [0.050]n         0.0015 mol/(L·s)
(4)n = 4
n = 1 (by inspection)
Therefore, the reaction is ﬁrst order with respect to H2 .
Set up a ratio for experiments 1 and 2, where [H2 ] is constant.
You can cancel [H2 ] because it is the same in both experiments.
m        1
Rate2 = k 0.040 mol/L [0.050] = 0.0030 mol/(L·s)
m
Rate1     k 0.020 mol/L [0.050]1         0.0015 mol/(L·s)
Because k is a constant, you can cancel it out.
m        1
Rate2 = k 0.040 mol/L [0.050] = 0.0030 mol/(L·s)
m
Rate1    k 0.020 mol/L [0.050]1          0.0015 mol/(L·s)
(2) = 2
m

m = 1 (by inspection)
The reaction is ﬁrst order with respect to ICl.
Step 2   Overall, the rate law equation is:
Rate = k[ICl]1[H2]1
To ﬁnd the value of the rate constant, substitute data from any of the three
experiments into the rate law equation. Using the data from experiment 1
gives
1                1
0.0015 mol/(L·s) = k 0.20 mol/L 0.050 mol/L
k=         0.0015 mol/(L·s)
1            1
0.20 mol/L 0.050 mol/L
= 0.15 L/(mol·s) (at temperature T )
The overall rate law equation for the reaction is:
Rate = 0.15 L/(mol·s) [ICl][H2]
Check the values of m and n by inspection. When [ICl] doubles (when [H2 ] is
constant), the rate also doubles. When [H2 ] quadruples (when [ICl] is constant), the
rate also quadruples. To check the value for k, substitute data from experiments 2 or
3 into the equation and solve for k. Check that the units for k are consistent with an
overall second order reaction. Also check the number of signiﬁcant digits for k.

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7. Problem
Sulfuryl chloride (also known as chlorosulfuric acid and thionyl chloride), SO2Cl2 ,
is used in a variety of applications, including the synthesis of pharmaceuticals,
rubber-based plastics, dyestuff, and rayon. At a certain temperature, the rate of
decomposition of sulfuryl chloride was studied.
SO2Cl2(g) → SO2(g) + Cl2(g)
[SO2Cl 2 ] (mol/L)     Initial rate (mol/(L • s))
0.150                   3.3 × 10−6
0.300                   6.6 × 10−6
0.450                   9.9 × 10−6

(a) Write the rate law equation for the decomposition of sulfuryl chloride.
(b) Determine the rate constant, k, for the reaction, with the appropriate units.
What Is Required?
You need to ﬁnd the values of m and k in the general rate law equation for the
reaction.
Rate = k[SO2Cl2]m
What Is Given?
You know the initial concentration of the reactant and the initial rates for three
experiments.
Since there is only one reactant, simply compare the initial rates of two experiments
with different initial [SO2Cl2]. Compare the change in initial rate with the change in
initial concentration.
(a) Begin with the ﬁrst and second experiments. Set up a ratio of the two rates as
shown.                     m
−6
Rate2 = k 0.150 mol/L
m   = 3.3 × 10−6 mol/(L·s)
Rate1   k 0.300 mol/L            6.6 × 10 mol/(L·s)
Because k is constant, it can be cancelled.
m
Rate2 = k 0.150 mol/L = 3.3 × 10−6 mol/(L·s)
m
Rate1    k 0.300 mol/L          6.6 × 10−6 mol/(L·s)
0.5m = 0.5
m = 1(by inspection)
Similarly, you can set up a ratio involving other experiments to obtain m = 1.
Therefore, the reaction is ﬁrst order with respect to sulfuryl chloride. The rate law
equation is:
Rate = k[SO2Cl2]
(b) To determine the value of k, substitute m = 1 into the rate law equation. You can
use data from any of experiments 1, 2 or 3 and solve for k. Substituting data from
the ﬁrst experiment into the rate law equation gives:
3.3 × 10−6 = k[0.15]1
−6
k = 3.3 × 10 mol/(L·s)
0.15 mol/L
= 2.2 × 10−5 s−1
The rate law equation for the reaction is:
Rate = (2.2 × 10−5 s−1)[SO2Cl2]1

Chapter 6 Rates of Chemical Reactions • MHR   89
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The value m = 1 can be checked by inspection. When the concentration of sulfuryl
chloride doubles, so will the initial rate. The value of k can be checked by using data
from one of the other experiments. The units of k are consistent with a ﬁrst order
reaction. The number of signiﬁcant digits for k are appropriate for the given data.

8. Problem
Consider the following reaction.
2A + 3B + C → products
This reaction was found to obey the following rate law equation.
Rate = k[A]2[B][C]
Copy the following table into your notebook. Then use the given information to
predict the blank values. Do not write in this textbook.
Initial [A]    Initial [B]      Initial [C]    Initial rate
Experiment         (mol/L)        (mol/L)          (mol/L)      (mol/(L • s))
1             0.10            0.20           0.050            0.40
2             0.10              (a)           0.10            0.40
3             0.20          0.050               (b)           0.20
4               (c)         0.025            0.040            0.45
5             0.10          0.010             0.15              (d)

What Is Required?
You need to calculate the missing values in the table:
(a) = Initial [B] for experiment 2
(b) = Initial [C] for experiment 3
(c) = Initial [A] for experiment 4
(d) = Initial rate for experiment 5
What Is Given?
You know the rate law equation, and you are given some incomplete rate data.
You know that the reaction is ﬁrst order in B and C. This means that when the con-
centration of B is doubled, if the concentration of other reactants are kept constant,
the rate will also double. The rate also doubles when C is doubled, if all else is kept
constant.
The reaction is second order in A. When the concentration of A is doubled, if [B]
and [C] are kept constant, the rate will increase by a factor of 22 , or 4 times. If [A] is
tripled, the rate will increase by a factor of 32 , or 9 times.
You can use the complete set of data, given in experiment 1, along with the rate law
equation, to solve for k, the rate constant. This will allow you to determine any
unknown concentrations in experiments 2 to 5.
Substituting data from experiment 1 into the rate law equation gives:
0.40 mol/(L·s) = k[0.10 mol/L]2[0.20 mol/L]1[0.050 mol/L]1

k=                  0.40 mol/(L·s)
[0.10 mol/L]2[0.20 mol/(L·s)]1[0.050 mol/L]1
= 4.0 × 103 L3/(s·mol3)

Chapter 6 Rates of Chemical Reactions • MHR   90
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(a) To solve for (a) in experiment 2, use the value of k and the given data. Substitute
into the rate equation.
0.40 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × [0.10 mol/L]2 × (a) × [0.10 mol/L]1
(a) =                   0.40 mol/(L·s)
4.0 × 103 L3/(s·mol3)[0.10 mol/L]2 × [0.10 mol/L]1
= 0.10 mol/L
(b) To solve for (b) in experiment 3, use the value of k and the given data. Substitute
into the rate equation.
0.20 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × [0.20 mol/L]2[0.050]1 × (b)
(b) =                    0.20 mol/(L·s)
4.0 × 103 L3/(s·mol3) × [0.20 mol/L]2[0.050 mol/L]1
= 0.025 mol/L
(c) To solve for (c) in experiment 4, use the value of k and the given data. Substitute
into the rate equation.
0.45 mol/(L·s) = 4.0 × 103 L3/(s·mol3) × (c)2 × [0.025]1[0.040]1
(c)2 =                   0.45 mol/(L·s)
4.0 × 103 L3/(s·mol3)[0.025 mol/L]1[0.040 mol/L]1
= 0.11 mol2/L2
(c) = 0.34 mol/L
(d) To solve for (d) in experiment 5, use the value of k and the given data. Substitute
into the rate equation.
(d) = 4.0 × 103 L3/(s·mol3) × [0.10 mol/L]2[0.010]1[0.15]1
(d) = 0.060 mol/(L·s)

Check your calculations. The units and signiﬁcant digits are correct. You can also
check by inspection, using the fact that the reaction is second order in A and ﬁrst
order in B and C.

Solutions for Practice Problems
Student Textbook page 287

9. Problem
Cyclopropane, C3H6 , has a three-membered hydrocarbon ring structure. It undergoes
rearrangement to propene. At 1000˚C, the ﬁrst-order rate constant for the
decomposition of cyclopropane is 9.2 s−1 .
(a) Determine the half-life of the reaction.
(b) What percent of the original concentration of cyclopropane will remain after
4 half-lives?
What Is Required?
First, you need to ﬁnd the half-life of the reaction. Next, you need to ﬁnd the
percentage of the original concentration of cyclopropane remaining after 4 half-lives.
What Is Given?
You know that the reaction is ﬁrst order in cyclopropane, with k = 9.2 s−1 .
(a) To ﬁnd t1/2 , use the following equation.

t1/2 = 0.693
k

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1
(b) Each half-life reduces the initial amount of cyclopropane by , or 50%.
2
Let the original concentration of cyclopropane be A.
1
After 1 t1/2 , [cyclopropane] =      2
A
1      1         1
After 2   t1/2 , [cyclopropane] = 2 ×       2
A   =   4
A
After 3   t1/2 , [cyclopropane] = 1 ×2
1
4
A   =   1
8
A
After 4   t1/2 , [cyclopropane] = 1 ×2
1
8
A   =   1
16
A
4
After 4   half-lives, there is 16 , or 1
1
2
of the original amount of cyclopropane
remaining.
(a) Solve for t1/2 .

t1/2 = 0.693
9.2 s−1
= 7.5 × 10−2 s
The half-life of the reaction is 7.5 × 10−2 s.
1 4
1
(b) After 4 half-lives, there will be    = 16 of the original amount present. If we
2
take the original concentration of cyclopropane to be 100, this corresponds to
1
16
× 100 = 6.2
Therefore, 6.2% of the original amount of cyclopropane will remain after
4 half-lives.
The units and number of signiﬁcant digits for the half-life are correct. The relatively
large rate constant corresponds to a short half-life. 4 half-lives corresponds to
1
2
1
× 1 × 1 × 1 = 16 of the original amount, or about 6.2% remaining.
2    2    2

10. Problem
Peroxyacetyl nitrate (PAN), H3CCO2ONO2 , is a constituent of photochemical smog.
It undergoes a ﬁrst-order decomposition reaction with t1/2 = 32 min.
(a) Calculate the rate constant in s−1 for the ﬁrst-order decomposition of PAN.
(b) 128 min after a sample of PAN began to decompose, the concentration of PAN in
the air is 3.1 × 1013 molecules/L . What was the concentration of PAN when the
decomposition began?
What Is Required?
First, you must ﬁnd the ﬁrst order rate constant, in s−1 , for the decomposition
of PAN. Next, ﬁnd the concentration of PAN originally present if [PAN] is
3.1 × 1013 molecules/L after decomposing for 128 minutes.
What Is Given?
(a) The half-life of the reaction is 32 minutes.
(b) [PAN] = 3.1 × 1013 molecules/L at t = 128 minutes.
(a) Rearrange the following equation and solve for k.
t1/2 = 0.693
k

Chapter 6 Rates of Chemical Reactions • MHR   92
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(b) Begin by determining the number of half-lives in 128 minutes. Use this to
determine the original [PAN].
We know that
1 n
2
× (original concentration) = (concentration after n half-lives)
or,
(original concentration) = (concentration after n half-lives)
1 n
2

(a) 32 min = 32 min × 60 s/min = 1.9 × 103 s
k = 0.693
t1/2
=     0.693
1.9 × 103 s
= 3.6 × 10−4 s−1

(b) Number of half-lives =
128 min
32 min/half-life
= 4 half-lives
After 4 half-lives, [PAN] = 3.1 × 1013 molecules/L.
Substituting into the equation gives
(original concentration) = concentration after n half-lives
1 n
2

= 3.1 × 10 1molecules/L
13
4
2
= 5.0 × 10 molecules/L
14

The original [PAN] is 5.0 × 1014 molecules/L .
The units and number of signiﬁcant digits for the half-life and original concentration
of PAN are correct. Dividing the original [PAN] in half four times gives
3.1 × 1013 molecules/L .

11. Problem
In general, a reaction is essentially over after 10 half-lives. Prove that this generaliza-
tion is reasonable.
What Is Required?
Show that only a negligible percentage of the original amount of a substance remains
after 10 half-lives.
What Is Given?
You are given the time of 10 half-lives.
Use the following equation:
n
Amount remaining after n half-lives = (original amount) × 1    2
For simplicity, take the original mass of the substance to be 100 g.
10
Amount remaining after 10 half-lives = (100 g) × 1  2
Amount remaining after 10 half-lives = 0.098 g
After 10 half-lives, a 100 g sample would be reduced to 0.098 g, or 0.098% of the
original amount. This is a negligible amount.

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Take 100 and divide it by 2 ten times to get the same answer. Less than one-tenth of
one percent of a substance remains after 10 half-lives.

12. Problem
The half-life of a certain ﬁrst-order reaction is 120 s. How long do you estimate that
it will take for 90% of the original sample to react?
What Is Required?
Estimate how long it will take for 90% of the sample to react.
What Is Given?
You know that the half-life of the reaction is 120 s.
Assume a 100 g sample size. When 90% of the sample has reacted, 10 g will remain.
To estimate how long this will take, divide 100 g by 2 several times, until you reach
approximately 10 g.
Alternatively, use the equation below and solve for n, the number of half-lives. This
will give a more accurate solution, but will require the use of logarithms.
n
Amount remaining after n half-lives = (original amount) × 1     2
Dividing 100 g by 2 several times gives
100 = 50 (1 t )
1/2
2
50 = 25 (2 t )
1/2
2
25 = 12.5 (3 t )
1/2
2
12.5 = 6.25 (4 t )
1/2
2
It takes between 3 and 4 half-lives for 90% of the sample to react. If the half-life is
120 s, this will take about 400 s.
Alternate Solution (using logarithms)
n
Amount remaining after n half-lives = Original amount × 1  2
or,
1 n     Amount remaining after n half-lives
=
2               Original amount
=   10
100
= 0.10
Taking the logarithm of both sides gives:
1 n
log    2
= log(0.10)
1
n × log    2
= log(0.10)
log(0.10)
n=
log 12
= 3.3
It will take 3.3 half-lives, or 3.9 × 102 s, for 90% of the sample to react.
The estimated answer agrees with the calculated value.

Chapter 6 Rates of Chemical Reactions • MHR   94
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Solutions for Practice Problems
Student Textbook page 294

13. Problem
The following reaction is exothermic.
2ClO(g) → Cl2(g) + O2(g)
Draw and label a potential energy diagram for the reaction. Propose a reasonable
activated complex.
Solution
Since the reaction is exothermic, the products should be lower than the reactants on
the potential energy diagram. The activated complex exists between the reactants and
products, and is of higher energy than the reactants.
possible activated complex
Cl . . . Cl

...

...
O ... O
Potential Energy (kJ)

2ClO(g)

∆H

Cl2(g) + O2(g)

Reaction Progress

A reasonable activated complex might be a species where the bonds between Cl and
O in the reactants are breaking, while new bonds between two Cl atoms and between
two O atoms are forming.

14. Problem
Consider the following reaction.
AB + C → AC + B ∆H = +65 kJ, Ea(rev) = 34 kJ
Draw and label a potential energy diagram for this reaction. Calculate and label
Ea(fwd). Include a possible structure for the activated complex.
Solution
The reaction is endothermic, since ∆H > 0. The products will be of higher energy
than the reactants.

Chapter 6 Rates of Chemical Reactions • MHR   95
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possible activated complex
A ... C

...
B

Potential Energy (kJ)
AC + B

Ea(fwd) = 99 kJ

∆H = +65 kJ
AB + C

Reaction Progress
For an endothermic reaction,
Ea(fwd) = ∆H + Ea(rev)
= 65 kJ + 34 kJ
= 99 kJ
The activation energy for the forward reaction is 99 kJ.
A possible structure for the activated complex could involve simultaneous bond
breaking in the reactants and the formation of new bonds in the products.

15. Problem
Consider the reaction below.
C + D → CD ∆H = −132 kJ, Ea(fwd) = 61 kJ
Draw and label a potential energy diagram for this reaction. Calculate and label
Ea(rev) . Include a possible structure for the activated complex.
Solution
Since this reaction is exothermic, the potential energy diagram will be similar to
Figure 6.12.
possible activated complex

C ... D
Potential Energy (kJ)

Ea(rev) = 193 kJ
C+D

∆H = −132 kJ

CD

Reaction Progress

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For an exothermic reaction,
Ea(rev) = ∆H + Ea(fwd)
= 132 kJ + 61 kJ
= 193 kJ
(Note: Use the positive value of ∆H.)
A possible structure of the activated complex might involve partial bond formation
between C and D.

16. Problem
In the upper atmosphere, oxygen exists in other forms other than O2(g) . For example,
it exists as ozone, O3(g) , and as single oxygen atoms, O(g) . Ozone and atomic oxygen
react to form two molecules of oxygen. For this reaction, the enthalpy change is
−392 kJ and the activation energy is 19 kJ. Draw and label a potential energy
diagram. Include a value for Ea(rev) . Propose a structure for the activated complex.
Solution
The reaction is
O3(g) + O(g) → 2O2(g) ∆H = −392 kJ; Ea(fwd) = 19 kJ
Since this reaction is exothermic, the potential energy diagram will be similar to
Figure 6.12.
Potential Energy (kJ)

Ea(rev) = 411 kJ
O3 + O

∆H = −392 kJ

2O2

Reaction Progress
For an exothermic reaction,
Ea(rev) = ∆H + Ea(fwd)
= 392 kJ + 19 kJ
= 411 kJ
A possible structure of the activated complex might involve partial bond breakage in
the O3 molecule coupled with partial bond formation to produce two O2 molecules.

Solutions for Practice Problems
Student Textbook page 301

17. Problem
NO2(g) and F2(g) react to form NO2F(g) . The experimentally determined rate law for
the reaction is written as follows:
Rate = k[NO2][F2]
A chemist proposes the following mechanism. Determine whether the mechanism is
reasonable.
Step 1 NO2(g) + F2(g) → NO2F(g) + F(g) (slow)
Step 2 NO2(g) + F(g) → NO2F(g)               (fast)

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What Is Required?
You need to determine whether the proposed mechanism is reasonable. To do this,
you need to answer the following questions:
- Do the steps add up to give the overall reaction? You need to write the balanced
chemical equation since it was not given.
- Are the steps reasonable in terms of their molecularity?
- Is the mechanism consistent with the experimentally determined rate law?
What Is Given?
You know the proposed mechanism and the rate law for the overall reaction.
First, write the balanced chemical equation. Next, add the two reactions and cancel
out reaction intermediates. Check the molecularity of the steps. Determine the rate
law equation for the rate determining step, and compare it to the overall rate law
equation.
The balanced chemical equation is
2NO2(g) + F2(g) → 2NO2F(g)
Step 1          NO2(g) + F2(g) → NO2F(g) + F(g)
Step 2          NO2(g) + F(g) → NO2F(g)
2NO2(g) + F2(g) + F(g) → 2NO2F(g) + F(g)
or,
2NO2(g) + F2(g) → 2NO2F(g)
The two steps add up to give the overall reaction. Both steps are bimolecular, which
is chemically reasonable. The ﬁrst step of the mechanism is rate-determining, and its
rate law equation is
Rate1 = k1[NO2][F2] .
This rate law equation matches the overall reaction. Based on the steps of the
proposed mechanism, the molecularity of these steps, and the rate law equation of
the rate-determining step, the proposed mechanism seems reasonable.
The reaction intermediate in the proposed mechanism is atomic ﬂuorine, F. When
the steps were added, you were able to cancel F.

18. Problem
A researcher is investigating the following overall reaction.
2C + D → E
The researcher claims that the rate of law equation for the reaction is written
as follows:
Rate = k[C][D]
(a) Is the rate law equation possible for the given reaction?
(b) If so, suggest a mechanism that would match the rate of law. If not, explain
why not.
What Is Required?
Parts (a) and (b) can be answered together. You need to determine if the given rate
law equation is possible, and provide a possible mechanism.
What Is Given?
You know the chemical equation and the rate law equation.

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You need to write a possible mechanism that is consistent with the given information.
The rate law equation is ﬁrst order in both C and D. Step 1 of the mechanism can
involve a rate-determining bimolecular reaction between C and D to form an
intermediate, B. In Step 2, B can react with another molecule of C to form the
product, E.
(a) and (b) The following proposed mechanism satisﬁes the criteria.
Step 1       C+D→B               (slow)
Step 2        B+C→E              (fast)
C+B+D→B+E
or,
C+D→E
This is a reasonable mechanism, since both steps are bimolecular. Step 1 is rate-
determining, which is consistent with the rate law equation.
The proposed reaction intermediate, B, cancels when the two steps are added.

19. Problem
A chemist proposes the following reaction mechanism for a certain reaction.
Step 1 A + B → C             (slow)
Step 2 C + B → E + F (fast)
(a) Write the equation for the chemical reaction that is described by this mechanism.
(b) Write a rate law equation that is consistent with the proposed mechanism.
What Is Required?
First, you need to write the overall chemical equation that is described by the
proposed mechanism. Next, you need to write a rate law equation that is consistent
with the proposed mechanism.
What Is Given?
You know the proposed mechanism, and the rate-determining step.
(a) Add the two steps and cancel any reaction intermediates.
(b) Since Step 1 is rate-determining, this step can be used to write the rate law
equation.
(a) Adding the steps of the mechanism gives
Step 1        A+B→C                    (slow)
Step 2        C+B→E+F                  (fast)
A + 2B + C → C + E + F
or,
A + 2B → E + F
(b) Step 1 is rate-determining, so the rate law equation for the overall reaction is
simply the rate law for Step 1 of the mechanism.
Rate = Rate1 = k[A][B]
The reaction intermediate, C, was cancelled to obtain the overall equation. The
bimolecular rate law equation for the reaction is consistent with having Step 1 as the
rate-determining step.

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20. Problem
Consider the reaction between 2-bromo-2-methylpropane and water.
(CH3)3CBr(aq) + H2O( ) → (CH3)3COH(aq) + H+(aq) + Br−(aq)
Rate experiments show that the reaction is ﬁrst order in (CH3)3CBr, but zero order
in water. Demonstrate that the accepted mechanism, shown below, is reasonable.
Step 1           (CH3)3CBr(aq) → (CH3)3C+(aq) + Br−(aq)       (slow)
+                                +
Step 2 (CH3)3C (aq) + H2O( ) → (CH3)3COH2 (aq)                (fast)
+          +
Step 3       (CH3)3COH2 (aq) → H (aq) + (CH3)3COH(aq) (fast)
What Is Required?
Show that the proposed mechanism is reasonable and that the reaction is ﬁrst order
in (CH3)3CBr and zero order in water.
What Is Given?
You know the proposed reaction mechanism. You know that the reaction is ﬁrst order
in (CH3)3CBr and zero order in water.
From the given information, write the rate law equation. Add the steps of the
mechanism to determine if they combine to give the overall equation. Ascertain that
the proposed mechanism supports the rate law equation.
The rate law equation is
Rate = k[(CH3)3CBr]1[H2O]0 = k[(CH3)3CBr]
Add the steps of the proposed mechanism. Notice that (CH3)3C+(aq) and
(CH3)3COH2+(aq) are both reaction intermediates and can be cancelled.
Step 1             (CH3)3CBr(aq) → (CH3)3C+(aq) + Br−(aq)             (slow)
+                             +
Step 2    (CH3)3C (aq) + H2O( ) → (CH3)3COH2 (aq)                     (fast)
Step 3         (CH3)3COH2+(aq) → H+(aq) + (CH3)3COH(aq)               (fast)
(CH3)3CBr(aq) + H2O( ) → (CH3)3COH(aq) + H+(aq) + Br−(aq)
Since Step 1 is rate-determining, the rate law equation will be
Rate = Rate1 = k[(CH3)3CBr]