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Chapter 11
Reaction Pathways
Solutions for Chapter 11 End-of-Chapter Problems
Problem 11.1.
Examples of very slow processes are geological changes that, for example, transform
sedimentary deposits into metamorphic rock or mineralize bones to form fossils. Reactions that
take a year or a few years include those that age wine and, if continued for too long, ruin it. Fast
reactions include almost every reaction in living organisms, because many biochemical
pathways must be sensitive to environmental changes in order to maintain the organism
(homeostasis). Other very fast reactions are those in combustions and explosions.
Problem 11.2.
(a) In Investigate This 11.1, the red phenolphthalein color faded more rapidly in the solution
with higher base concentration, so we would predict that it would fade faster in a 2.0 M sodium
hydroxide solution compared to a 1.0 M solution.
(b) By the same reasoning as in part (a), we would expect that the color would fade slower in a
0.10 M sodium hydroxide solution compared to a 1.0 M solution.
Problem 11.3.
The reaction of hydroxide anions with the colorless form of phenolphthalein is an acid-base
reaction transferring the protons from the two acidic –OH groups bonded to the six-membered
rings (These react like phenol; see Table 6.2.) to hydroxide ions. The dianion rearranges to the
more stable red form with more pi-electron delocalization over the entire structure.
OH
HO
C O
C
+ 2OH–(aq)
O
(aq)
O–
–O
C O
C
+ 2H2O(l)
O
O
C
O
+ 2H2O(l)
C
O
O
(aq)
(aq)
red
colorless
Problem 11.4.
The rate of a chemical reaction refers to the change of concentrations of reactants and products
as a function of time. Rate is a change per time. The measurement of any variable that is
proportional to a reactant or product concentration as a function of time, can be used to
characterize the rate of the reaction.
Problem 11.5.
(a) In the stroboscopic study of the bombardier beetle shown in the chapter opening illustration,
there are about seven frames between the beginning of the first discharge and the beginning of
the second. Since the frames are taken every 0.00025 seconds (0.25 ms), the time between
discharges is about 0.00175 s [= (7 frames)(0.00025 s·frame–1)]. The rate of discharge is the
inverse of time between discharges: rate = (1 discharge)
≈ 570 discharge·s–1.
(0.00175 s)
(b) The rate of discharge of the beetle would produce disturbances in the air around the beetle at
a frequency of 570 s–1, which is in the range of human hearing. Audio recordings of the sound
the beetle makes show that the sound is produced at the frequency of the discharge.
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Problem 11.6.
The reaction that turns the phenolphthalein red is an acid-base reaction (Problem 11.3), transfer
of protons from the acid (phenolphthalein) to the base (hydroxide anion). Proton transfers are
almost always quite rapid. The reaction that produces the colorless form of phenolphthalein
from its red dianion is between a nucleophile (the hydroxide anion) and an electrophilic center
at the middle of the dianion. Nucleophile-electrophile reactions are not always fast and this one
is slowed by the fact that two negative species have to come together for the reaction to occur.
Since like charges repel one another, this slows the reaction.
Problem 11.7.
(a) When NH3(aq) solution is added to AgCl(s) the NH3 begins to react to form the diammine
complex, Ag(NH3)2+(aq):
AgCl(s) + 2NH3(aq) ⇔ Ag(NH3)2+(aq) + Cl−(aq)
The concentration of NH3(aq), [NH3(aq)], in the solution decreases with time, as more and
more AgCl(s) reacts.
(b) The amount of solid silver chloride decreases with time, as it reacts with the NH3(aq) to
form Ag(NH3)2+(aq).
(c) The amount of the diammine complex, Ag(NH3)2+(aq), increases with time as more and
more AgCl(s) and NH3(aq) react.
Problem 11.8.
Molecules move faster at higher temperatures. (Their internal motions (vibrations and rotations)
also gain more energy, so vibrations and rotations are more active.) Molecules have to come
together to react, so they should come together faster if they are moving faster and reactions
should speed up as temperature increases. (As we will find in Section 11.6, this is a relatively
minor effect compared to the activation energy effect, but even reactions with zero activation
energy go faster at higher temperature because of the faster motion.)
Problem 11.9.
To test whether a substance, either a reactant or a catalyst, changes the rate of a reaction, you
have to have a method to measure the rate of the reaction, for example, change of color with
time, rate of evolution of gas with time, or some other observable change that is proportional to
the amount of reaction that has occurred. Then you measure the rate of the reaction with varying
amounts of the substance being tested, while all other reaction conditions are held constant. If
the reaction rate increases as the concentration of the substance tested increases, it is very likely
that the substance is acting either to catalyze the reaction or is a reactant that is involved in part
of the reaction pathway that determines the rate of the reaction. [As usual, there are exceptions
to this generalization. For example, if a catalyst for the reaction is saturated (see Section 11.10
and Problem 11.79), then addition of more reactant will not increase the reaction rate.]
Problem 11.10.
Consider a solution reaction that bubbles as the solution goes from clear and colorless to clear
blue. Since a gas is being produced in this reaction (it bubbles), the volume of gas produced as a
function of time could be measured to find the reaction rate. The volume of gas at constant
pressure and temperature is proportional to the number of moles of gas produced, so the rate of
production of gas volume is proportional to the rate at which gaseous product produced.
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Similarly, the pressure of gas produced at constant volume and temperature is proportional to
the number of moles of gas produced, so the rate of increase of pressure at constant volume and
temperature could also be used as a measure of rate. The color change in the solution, colorless
to blue, must also be associated with whatever change is occurring, so we can follow the change
in absorption of light (more red light is absorbed as the process occurs) as a function of time to
get a measure of the rate of the process.
Problem 11.11.
(a) We are asked to sketch a concentration vs. time graph of the concentrations of N2, H2, and
NH3, beginning with a mixture of N2 and H2 undergoing this reaction:
N2(g) + 3H2(g) → 2NH3(g)
According to the problem statement, we start with a stoichiometric mixture of N2 and H2, that
is, with a concentration of H2 that is three times the concentration of N2. The changes of
concentrations with time are not linear, but, for simplicity, they are drawn as linear in this
graph:
30
concentrations
25
20
N2
H2
15
NH3
10
5
0
0
1
2
3
4 time
5 6
7
8
9
10
Things to note are: the rate of change of the H2 concentration is three times the rate of change of
the N2 concentration, that is, the slope of the H2 line is three times the slope of the N2 line; the
ratio of H2 to N2 is equal to three, throughout the reaction, since three moles of H2 react for
every mole of N2 that reacts; and, the concentration of NH3 is equal to twice the concentration
of N2 that has reacted throughout the reaction.
(b) The expression for the rate of ammonia, NH3, production in terms of gas pressure, PNH3,
change is:
∆PNH 3
rate of ammonia production =
∆t
(c) The expression for the rate of ammonia production in terms of concentration, [NH3(g)],
change is:
∆ [NH 3 ]
rate of ammonia production =
∆t
(d) We see from the graph in part (a) that the rate of production of NH3 is twice as large as the
rate of reaction of N2. If N2 is reacting at a rate of 0.15 M·min–1, then ammonia must be being
formed at double this rate or 0.30 M·min–1.
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(e) The reasoning is similar to that in part (d). H2 reacts three times faster than N2, so if N2 is
reacting at a rate of 0.15 M·min–1, then hydrogen must be be disappearing at triple this rate or
0.45 M·min–1.
(f) Since the datum for the rate of N2 disappearance is given in terms of molarity, the rate of
reaction we are interested in is the one written in part (c). From part (d), we know the rate of
change of ammonia concentration, so the rate of reaction (ammonia production) is 0.30
M·min-1.
Problem 11.12.
(a) Data for a reaction that can be symbolized as: A → 2B are:
[A], mol·L–1
1.000
0.833
0.714
0.625
0.555
time, s
0.0
10.0
20.0
30.0
40.0
This figure is a plot of these data, with approximate tangents to the curve shown at 10.0, 20.0,
and 30.0 seconds. (The tangent lines are displaced a bit from the curve to make them easier to
distinguish from one another and the curve.)
1.00
0.95
0.90
[A], M
0.85
0.80
0.75
0.70
0.65
0.60
0.55
0.50
0
5
10
15
20
25
30
35
40
time, s
The tangent slopes are:
(0.75 M – 0.88 M)
–1
at 10.0 s:
(15.0 s – 5.0 s) = – 0.013 M·s
at 20.0 s:
at 30.0 s:
(0.65 M – 0.75 M)
–1
(25.0 s – 15.0 s) = – 0.010 M·s
(0.575 M – 0.65 M)
–1
(35.0 s – 25.0 s) = – 0.0075 M·s
(b) The rate [slopes of the tangents in part (a)] decreases over time, that is, the change in
concentration of the reactant as a function of time becomes less. If the rate depends on the
amount of A left to react, then it makes sense that the reaction should become slower as [A]
decreases.
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(c) The average rates during 10 second intervals are:
0.0 to 10.0 s: (0.833 M – 1.000 M) (10.0 s – 0.0 s) = – 0.0167 M·s–1
10.0 to 20.0 s: (0.714 M – 0.833 M) (20.0 s – 10.0 s) = – 0.0119 M·s–1
20.0 to 30.0 s: (0.625 M – 0.714 M) (30.0 s – 20.0 s) = – 0.0089 M·s–1
30.0 to 40.0 s: (0.555 M – 0.625 M) (40.0 s – 30.0 s) = – 0.0070 M·s–1
(d) The average rates in part (c) follow the same pattern as the instantaneous rates in part (a)
and are intermediate in numeric value for the intervals from 10.0 to 20.0 and 20.0 to 30.0
seconds, as we would expect. The earlier and later average rate values are faster and slower,
respectively, than the earliest and latest instantaneous rates since the average rates are for earlier
and later parts of the reaction.
Problem 11.13.
[NOTE: These are “cooked” data assuming a reaction that is second order in [H+]. “Data” in
end-of-chapter problems in at least two texts suggest this second order. The data are so good
that they are also probably fake. These data are probably derived from an assumed mechanism
that involves an equilibrium between methanol and hydronium ion to give protonated methanol
which then reacts in a slow second order reaction with chloride ion to displace water. Since the
hydronium and chloride ion concentrations are assumed to be the same (solution is moderately
strong HCl), the rate law that is first order in methanol, hydronium, and chloride reduces to one
that is second order in hydronium when the system is flooded with methanol.]
(a) Data for the reaction, CH3OH + H+ + Cl– → CH3Cl + H2O, in a solution of methanol and
hydrochloric acid are:
time, min
0
90
200
360
720
[H+], M
2.12
1.95
1.74
1.54
1.19
The data for the decrease in concentration of the hydronium ion are plotted as the square blue
symbols on this graph:
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2.5
[H+], M
concentrations, M
2.0
[CH3Cl], M
1.5
1.0
0.5
0.0
0
200
400
600
800
time, min
The average rates of hydronium ion disappearance during each time interval are:
–1
(1.95 M – 2.12 M)
0 to 90 min:
(90 min – 0 min) = – 0.0019 M·min
90 to 200 min:
(1.74 M – 1.95 M)
–1
(200 min – 90 min) = – 0.0019 M·min
200 to 360 min: (1.54 M – 1.74 M) (360 min – 200 min) = – 0.0013 M·min–1
360 to 720 min: (1.19 M – 1.54 M) (7200 min – 360 min) = – 0.00097 M·min–1
(b) The graph and the average rates calculated in part (a) both show that the rate of the reaction
decreases with time, that is, the reaction gets slower as the hydronium ion (and chloride anion)
are used up.
(c) To get the concentration of chloromethane, [CH3Cl(solvent)], formed in the solution, we
assume that the reaction stoichiometry is CH3OH + H+ + Cl– → CH3Cl + H2O, so that one mole
of CH3Cl is formed for every mole of H+ that reacts. The difference between the molarity of H+
at any time and its molarity at zero time is the amount that has reacted, which is the amount of
CH3Cl formed. These differences are plotted as red circles on the graph in part (a).
Problem 11.14.
(a) For each reaction, we express the rate of reaction in terms of pressure changes, using the
rates as defined by equations (11.7) and (11.8).
∆PNO 3
∆P
1 ∆PNO 2
(i) 2NO2(g) → NO(g) + NO3(g)
rate = –⎛⎝ ⎞⎠
= NO =
∆t
∆t
2 ∆t
∆PBr2
1 ∆P
1 ∆P
(ii) 2NOBr(g) → 2NO(g) + Br2(g)
rate = –⎛⎝ ⎞⎠ NOBr = ⎛⎝ ⎞⎠ NO =
∆t
∆t
2
2 ∆t
(iii) 2N2O5(g) → 4NO2(g) + O2(g)
∆PO 2
1 ∆PN 2 O 5
1 ∆PNO 2
rate = –⎛⎝ ⎞⎠
= ⎛⎝ ⎞⎠
=
∆t
∆t
2
4 ∆t
(b) In reaction (i), the reactant disappears twice as fast as either product forms, since two
reactant molecules disappear to form one molecule of each product. This difference is
accounted for to get the reaction rate in part (a) by multiplying the rate of disappearance of
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reactant by one half. In reaction (ii), two reactant molecules disappear to form two molecules of
one product and one molecule of the other. The reactant disappears at the same rate as one
product, NO, is formed, but twice as fast as the other, Br2, is formed. These differences are
accounted for to get the reaction rate in part (a) by multiplying the rate of disappearance of the
reactant and the rate of appearance of NO by one half. In reaction (iii), two reactant molecules
disappear to form four molecules of one product, NO2, and one molecule of the other product,
O2. Thus, the reactant disappears at half the rate one product, NO2, is formed and at twice the
rate the other product, O2, is formed. These differences are accounted for to get the reaction rate
in part (a) by multiplying the rate of disappearance of reactant by one half and the rate of
formation of NO2 by one fourth.
Problem 11.15.
(a) 2-chloro-2-methylpropane, (CH3)3CCl, reacts in a solution of water and acetone to give
2-methyl-2-propanol, (CH3)3COH. When a small amount of hydroxide ion, OH–, is present in
the solution, the overall initial reaction is: (CH3)3CCl + OH– → (CH3)3COH + Cl–. For different
initial concentrations of hydroxide ion, [OH–]0, with all other conditions the same, the times
required for the hydroxide to react completely (monitored by change in color of an acid-base
indicator) were:
expt #
1
2
3
[OH–]0, M
0.0025
0.0030
0.0015
time to color change, s
35
43
22
The rate of disappearance of the hydroxide ion in each experiment is the change in
concentration of hydroxide ion (assumed to be completely reacted when the color change
occurs) divided by the time required for it to react:
rate(1) = (0 M – 0.0025 M)
= –7.1 × 10–5 M·s–1
(35 s)
rate(2) = (0 M – 0.0030 M)
rate(3) = (0 M – 0.0015 M)
(43 s)
(22 s)
= –7.0 × 10–5 M·s–1
= –6.8 × 10–5 M·s–1
The rate of disappearance of the hydroxide ion is the same in each experiment, –7.0 × 10–5 M·s-1
(average).
(b) From the stoichiometry of the reaction, we see that one mole of 2-chloro-2-methylpropane,
(CH3)3CCl, reacts for every mole of hydroxide ion that reacts. Therefore, the rate of
disappearance of (CH3)3CCl is the same as the rate of disappearance of hydroxide ion, which is
–7.0 × 10–5 M·s–1 in these experiments. Since the rate of the reaction is the same for all the
concentrations of hydroxide ion investigated, we can conclude that the reaction rate does not
depend on the concentration of hydroxide ion in the solution.
Problem 11.16.
(a) The rate of formation of oxygen from ozone, 2O3(g) → 3O2(g), is 1.8 × 10–3 M·s–1 at a
certain temperature. The rate of the reaction can be written as:
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1 ∆[O3 ] ⎛ 1⎞ ∆[O2 ]
rate = –⎛⎝ ⎞⎠
=⎝ ⎠
2 ∆t
3 ∆t
Therefore, we can write the rate of ozone disappearance as:
∆[O3 ]
2 ∆[O2 ]
= –⎛⎝ ⎞⎠
= – (2/3)(1.8 × 10–3 M·s–1) = – 1.2 × 10–3 M·s–1
∆t
3 ∆t
(b) As we see in part (a), we can get the rate of the reaction from either the rate of
disappearance of ozone or the rate of appearance of oxygen:
rate = (1/3)(1.8 × 10–3 M·s–1) = 6 × 10–4 M·s–1
Show that the same result is obtained from the rate of disappearance of ozone.
Problem 11.17.
(a) The coupled reactions in the iodide-persulfate-thiosulfate system animated in the Web
Companion, Chapter 11, Section 11.2, page 6, can be represented like this:
oxidized
reduced
S2O82–
2I–
S4O62–
2SO42–
I2
2S2O32–
oxidized
reduced
reduced
oxidized
(This representation is patterned after Figure 10.12, that shows the coupling of reactions
responsible for the observations on the Blue-Bottle reaction, Investigate This 10.61, Chapter 10,
Section 10.8, which remains blue for a time and then suddenly changes to colorless. In the
Companion reaction, the solution remains clear and colorless until all the thiosulfate has reacted
and then suddenly changes color to a deep blue, if starch indicator is present.) The reaction on
the left here is the reaction at the top of the Companion page and the reaction on the right is the
reaction in the middle of the page.
(b) The coupling species in the iodide-persulfate-thiosulfate system are the iodide and iodine
that cycle between the reduced iodide form and the oxidized iodine. The reduction of persulfate
dianion to sulfate is coupled to the oxidation of thiosulfate dianion.
(c) The solution remains clear and colorless until the thiosulfate has all reacted, which means
that no appreciable amount of iodine is present in the solution until the thiosulfate has all
reacted. Thus, in order to prevent the build-up of iodine (and, hence, color) in the solution, the
reduction of iodine to iodide by thiosulfate must be fast compared to the oxidation of iodide to
iodine by persulfate.
(d) Since iodide is being rapidly regenerated throughout the colorless phase of the reaction, its
concentration does not change and remains equal to its initial concentration.
(e) The ∆[I2] shown in the Web Companion plot of [I2] as a function of t, is the change in
concentration of iodine that would have occurred in the time period ∆t, if the reaction mixture
didn’t contain thiosulfate. The amount of iodine that would have formed is equal to one half the
number of moles of thiosulfate added to the solution.
(f) The Web Companion plot is a little misleading, because it shows the ∆ [I2] as though it had
actually occurred during the measured time, ∆t. This may be the best one can do to represent a
coupled reaction system like this one. Difficulties with representation, do not translate to
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difficulties interpreting the data, which you have an opportunity to do in Check This 11.22 and
Problem 11.26.
Problem 11.18.
(a) The reaction, 2CO(g) → CO2(g) + C(s), was studied by adding CO(g) to a hot reaction
vessel and following the decrease in total pressure, Ptotal, of the gases in the constant volume
reactor:
time, s
Ptotal, kPa
0
400
1000
1800
33.2
31.6
29.8
27.9
For every two moles of gaseous CO that react, one mole of gaseous CO2 is formed. Therefore, if
n2 moles of CO2 have been formed, 2n2 moles of CO have reacted. If the initial number of
moles of CO in the reactor is n0 and the number of moles of CO left at a later time is n1
(= n0 - 2n2), then the total number of moles of gas, nt, at the later time is:
nt = n1 + n2 = (n0 – 2n2) + n2 = n0 – n2
Thus:
n2 = n0 – nt
n1 = n0 – 2n2 = n0 – 2(n0 – nt) = 2nt – n0
Since the total pressure in the reactor is directly proportional to the number of moles of gas in
the reactor, we can rewrite these two equations in terms of the measured total pressures to give
the pressures of each gas in the reactor:
PCO2 = (Ptotal)0 – (Ptotal)t
PCO = 2(Ptotal)t – (Ptotal)0
Using the second equation, we get PCO at 0, 400, 1000, and 1800 s as 33.2, 30.0, 26.4, and 22.6
kPa, respectively. Thus, the average rates of CO disappearance are:
rate (0 to 400 s) = (30.0 kPa – 33.2 kPa) (400 s – 0 s) = 8.0 × 10–3 kPa·s–1
rate (400 to 1000 s) = (26.4 kPa – 30.0 kPa) (1000 s – 400 s) = 6.0 × 10–3 kPa·s–1
rate (1000 to 1800 s) = (22.6 kPa – 26.4 kPa) (1800 s – 1000 s) = 4.8 × 10–3 kPa·s–1
(b) We could use a procedure similar to that we used in part (a) to get the average rates of
formation of CO2 in this system, but it is much easier to use the relationship between the rate of
disappearance of CO and rate of formation of CO2. From the stoichiometry, the rate of
formation of CO2 is one half the rate of disappearance of CO. Thus, in the three time intervals,
the rates of formation of CO2 are 4.0 × 10–3 kPa·s–1, 3.0 × 10–3 kPa·s–1, and 2.4 × 10–3 kPa·s–1,
respectively.Problem 11.19.
Show how the rate of reaction will change if the concentrations of A and B are both halved in a
reaction that is second order in the concentration of reactant A and first order in the
concentration of reactant B. The rate law for this reaction is:
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rate = k[A]2[B]
When [A] = ao and [B] = bo, we have:
rateo = kao2bo
If [A] = ao/2 and [B] = bo/2, then:
2
2
rate1/2 = k(ao/2)2(bo/2) = k(ao /4)(bo/2) = kao bo/8 = rateo/8
The overall rate of reaction with both reactants halved in concentration is one-eighth the
original rate.
Problem 11.20.
(a) For the reaction, CH3Br(aq) + OH–(aq) → CH3OH(aq) + Br–(aq), the initial rate of reaction
decreased by a factor of two when the concentration of hydroxide ion, [OH–(aq)], was halved
and increased by a factor of 1.4 when the concentration of bromomethane, [CH3Br(aq)], was
increased by a factor of 1.4. The reaction rate goes down by a factor of two when the hydroxide
ion concentration, [OH–(aq)], is decreased by a factor of two, so we can conclude that the
reaction is first order with respect to [OH–(aq)].
(b) The reaction rate goes up by a factor of 1.4 when the bromomethane concentration,
[CH3Br(aq)] is increased by a factor of 1.4, so we can conclude that the reaction is first order
with respect to [CH3Br(aq)].
(c) The overall order of the reaction (with respect to the concentrations we have investigated) is
two and the rate law is:
rate = k[CH3Br(aq)][OH–(aq)]
Problem 11.21.
(a) For the gas-phase reaction, 2NO(g) + Cl2(g) → 2NOCl(g), the initial rate of disappearance
of NO(g) is eight times faster when the concentrations of both reactants are doubled and twice
as fast when the concentration of chlorine alone is doubled. The rate of the reaction doubles
when the chlorine concentration, [Cl2(g)], doubles, so the reaction must be first order with
respect to [Cl2(g)].
(b) The rate of the reaction increases by a factor of eight when both the chlorine concentration,
[Cl2(g)], and the nitric oxide concentration, [NO(g)], are doubled. From part (a), we know that
the effect of the [Cl2(g)] is to double the reaction rate, so the effect of [NO(g)] doubling is to
increase the rate further by a factor of four. Since doubling [NO(g)] results in a quadrupling of
the reaction rate, the reaction must be second order with respect to [NO(g)].
(c) The overall order of this reaction (with respect to the concentrations we have investigated) is
three, that is, it is third order overall, and the rate law is:
rate = k[NO(g)]2[Cl2(g)]
Problem 11.22.
(a) The rate of a reaction is proportional to the concentration of a reactant and the concentration
of a catalyst. The measured initial rates of reaction at a constant catalyst concentration give a
rate constant, 8.9 × 10–5 s–1, for the first order disappearance of the reactant. The overall secondorder rate law for this reaction and the first-order rate law with the catalyst concentration,
[catalyst], constant, are:
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rate = k[catalyst][reactant] = k′[reactant]
The rate constant, k′, is 8.9 × 10–5 s–1. If [catalyst] = 5.0 × 10–3 M, we can calculate k as:
(
)
(
8.9 × 10 –5 s
k
′
k = [catalyst] =
= 1.8 × 10–3 M–1·s–1
–3
5.0 × 10 M
)
Note that the units of k are correct for a second order rate constant.
(b) If [reactant] = 0.12 M and [catalyst] = 3.5 × 10–3 M, the initial rate of reaction is:
rate = (1.8 × 10–3 M–1·s–1)(0.12 M)(3.5 × 10–3 M) = 7.5 × 10–6 M·s–1
Problem 11.23.
The initial rate of a reaction that is first order in each of two reactants is 7.65 × 10–4 M·s–1, when
the initial concentration of each reactant is 0.050 M. Write the rate law for this reaction,
substitute the data given, and solve to get the rate constant for the reaction:
rate = 7.65 × 10–4 M·s–1 = k[reactant 1][reactant 2] = k(0.50 M)(0.50 M)
k=
7.65 × 10 –4 M ⋅ s –1
= 3.1 × 10–3 M–1·s–1
(0.50 M)(0.50 M)
Problem 11.24.
The reaction, 2NO(g) + O2(g) → 2NO2(g), was studied by following the initial rate of
disappearance of nitric oxide, – ∆NO(g) ∆t , for different initial reactant concentrations.
0
(
expt # [NO(g)]0, M
1
2
3
0.0125
0.0250
0.0125
)
(
)
– ∆NO(g) ∆t , M·s–1
0
[O2(g]0, M
0.0183
0.0183
0.0370
0.0202
0.0803
0.0409
(a) The rate of disappearance of NO(g) quadruples when the [NO(g)]0 is doubled (with [O2(g)]0
held constant – runs 1 and 2), so the reaction is second order with respect to [NO(g)].
(b) The rate of disappearance of NO(g) doubles when the [O2(g)]0 is doubled (with [NO(g)]0
held constant – runs 1 and 3), so the reaction is first order with respect to [O2(g)].
( )
(
)
⎡
⎤
(c) From the stoichiometry, the reaction rate = 1 2 ⎢ – ∆NO(g) ∆t ⎥ , so the reaction rate for
⎣
0⎦
each experiment is one-half the value in the last column of the table.
(d) The rate law is:
rate = k[NO(g)]2[O2(g)]
To get k, we substitute the concentrations for any of the experiments into this rate law (with the
rate from part (c)) and solve for k. From experiment #1 we have:
0.0101 M ⋅s –1
= 3.53 × 103 M–2·s–1
k=
2
(0.0125 M) (0.0183 M)
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The units of k, have to be such that, when multiplied by the units of the concentrations (raised to
their appropriate powers), the right hand side of the rate law comes out to have units of M·s–1,
the units of the rate. The units of the product of the concentrations in this rate law are M3, so the
rate constant has the appropriate units to cancel M2 and leave M·s–1.
(e) If the reaction were carried out with [NO(g)]0 = [O2(g]0 = 0.0200 M;
rate = (3.53 × 103 M–2·s–1)(0.0200 M)2(0.0200 M) = 0.0282 M·s–1
The rate of disappearance of nitric oxide under these conditions is 0.0564 M·s–1, since it is twice
the reaction rate [see part (c)].
Problem 11.25.
(a) Data for the reaction, (CH3)3CCl + OH– → (CH3)3COH + Cl–, see Problem 11.15, with the
initial concentration of hydroxide ion, [OH–]0 = 0.0025 M, are:
expt #
[(CH3)3CCl]0, M
time to color change, s
1
4
0.015
0.030
35
18
We get the rate of disappearance of hydroxide ion just as we did in Problem 11.15:
= –7.1 × 10–5 M·s–1
rate(1) = (0 M – 0.0025 M)
(35 s)
rate(4) = (0 M – 0.0025 M) (18 s) = –14 × 10–5 M·s–1
(b) The stoichiometry of the reaction shows that the rate of disappearance of (CH3)3CCl in each
experiment is the same as the rate of disappearance of the hydroxide ion calculated in part (a)
for that experiment. When the concentration of (CH3)3CCl was doubled (experiment 4
compared to 1), the rate of reaction doubled, so the reaction is first order with respect to
[(CH3)3CCl].
(c) In Problem 11.15, we found that the rate of the reaction is independent of the concentration
of hydroxide ion, that is, the reaction is zero order with respect to [OH–]. In part (b) here, we
find that the reaction is first order with respect to [(CH3)3CCl] so the rate law is:
rate = – rate of disappearance of (CH3)3CCl = k[(CH3)3CCl]
(d) We can determine the numeric value of the rate constant from either experiment 1 or 4 by
substituting the appropriate value for [(CH3)3CCl] and the rate into the rate law and solving for
k:
(
)
14 × 10 –5 M ⋅ s –1
–3 –1
k = rate [(CH ) CCl] =
(0.030 M ) = 4.7 × 10 s
3 3
Problem 11.26.
(a) In Check This 11.22, the rate law you found for the reaction in the iodide-persulfatethiosulfate system, S2O82–(aq) + 2I–(aq) → I2(aq) + 2SO42–(aq), from the Web Companion,
Chapter 11, Section 11.3, pages 3-6, is:
rate = k[S2O82–(aq)][I–(aq)]
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Data are given in the Companion for the concentrations of these reactants, so, to calculate the
rate constant, we need the rate of the reaction. Unfortunately, all we have are relative rates of
reaction because the results given are the times required for the same unknown ∆[I2(aq)]. Thus,
we cannot determine the value of k, from these data.
(b) With the added datum, the concentration of thiosulfate anion, [S2O32–(aq)] = 0.0015 M, in
each experiment, we can calculate ∆[I2(aq)] and thence the rate of reaction in each case. From
the stoichiometry of the thiosulfate anion-iodine reaction, Web Companion, Chapter 11, Section
11.2, page 6, we see that two S2O32– react for every one I2 that reacts. Thus, ∆[I2(aq)] =
0.00075 M in each experiment. Any one of the experiments will can give us the value for k:
∆[I2 (aq)]
(0.00075 M)
∆t
(21 s)
rate
=
=
k=
–
–
2–
2–
(0.077 M)(0.077 M)
[S2 O 8 (aq)][I (aq)] [S2 O 8 (aq)][I (aq)]
(
) (
)
k = 6.0 × 10–3 M–1·s–1
Here, we used the data from experiment 1. Check to see that another gives the same result.
Problem 11.27.
(a) The reaction, MnO4–(aq) + Cr3+(aq) → Mn4+(aq) + CrO42–(aq), is a redox reaction since
oxidation numbers change, Cr(III) to Cr(VI) and Mn(VII) to Mn(IV).
(b) The rate of this reaction was studied by measuring the time required for the concentration of
CrO42–(aq) to increase from zero to 0.020 M.
expt #
1
2
3
relative [MnO4–(aq)]0 relative [Cr3+(aq)]0
1
2
1
1
1
0.5
time to [CrO42–(aq)] = 0.020 M
23 min
11 min
45 min
The rate of the reaction is inversely proportional to time required to produce 0.020 M CrO42–
(aq) — the longer it takes for a reaction to get to a certain point, the slower the reaction.
(c) The reaction is first order in both reactants. Comparing experiments 1 and 2, we see that
when permanganate is doubled, keeping [Cr3+(aq)]0 constant, the time is cut in half, which
means the rate is doubled. If the rate doubles when the concentration is doubled, the reaction is
first order in that component. Comparing experiments 3 and 1, we see that when [Cr3+(aq)]0 is
doubled, keeping [MnO4–(aq)]0 constant, the time is cut approximately in half, which means the
rate also doubles.
(d) The rate law for the reaction is:
rate = k[MnO4–(aq)][Cr3+(aq)]
(e) Compared to reaction 1, if both reactant concentrations are cut in half, the rate will be
approximately one fourth the rate in reaction 1. This will mean that it would take approximately
4 times as long to reach the 0.2 M product—that is, approximately 92 min.
Problem 11.28.
(a) The reaction, 2NH3(g) → N2(g) + 3H2(g), occurs on a hot tungsten filament and can be
followed by measuring the pressure increase:
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time, s
0 100 200 400 600
800 1000
∆P, kPa
0 1.46 2.94 5.85 8.81 11.68 14.61
A plot of these data gives a straight line, as shown on this graph.
16
14
12
∆P, kPa
10
8
6
4
2
0
0
200
400
600
800
1000
time, s
The stoichiometry of the reaction shows that every mole of ammonia that reacts produces two
moles of gaseous products. If the ammonia reacts so that the pressure of ammonia, PNH3,
decreases by 1 kPa, the pressure in the reaction system will increase by 1 kPa. Thus, we can
write the reaction rate as:
∆P
∆P
rate = – NH3 = total
∆t
∆t
The straight line plot shows that the rate of reaction, ∆Ptotal/∆t, remains constant as the ammonia
reacts, that is, the reaction rate does not depend on the ammonia pressure (concentration). The
reaction is zero order with respect to the ammonia pressure (concentration).
(b) The slope of the line on the graph is 1.46 × 10–2 kPa·s–1, which is the rate of the reaction.
Since the reaction is zeroth order in ammonia, the pressure of ammonia has no affect on the
rate, so the initial reaction rate at any initial pressure of ammonia is the same, 1.46 × 10–2 kPa·s–
1
. The rate is equal to a constant, the rate constant, which is also equal to 1.46 × 10–2 kPa·s–1.
Problem 11.29.
(a) The decomposition of hydrogen peroxide, 2H2O2(aq) → 2H2O(l) + O2(g), was studied by
determining the concentration of reactant left after 30 minutes of reaction, [H2O2(aq)]30.
expt #
H2O2(aq)]0, M
[H2O2(aq)]30, M
1
2
0.874
0.356
0.812
0.331
3
0.589
0.549
To find the average rate of disappearance of hydrogen peroxide we divide the difference
between the final and initial concentrations by the time of reaction (30 minutes in each case):
disappearance rate1 = (0.812 M – 0.874 M) (30 min) = – 0.0021 M·min–1
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disappearance rate2 = (0.331 M – 0.356 M) (30 min) = – 0.00083 M·min–1
disappearance rate3 = (0.549 M – 0.589 M) (30 min) = – 0.0013 M·min–1
(b) The average rate of disappearance of hydrogen peroxide increases as the initial
concentration of hydrogen peroxide increases. Since less than 10% of the hydrogen peroxide
reacts in each case, we can take these rates of disappearance as an approximation to the initial
rate of disappearance in each case. Plot the average rates as a function of the initial
concentration for the first 30 minutes for each experiment to see whether the rate is directly
proportional to concentration.
average reaction rate, m/min
0.0000
-0.0005
-0.0010
-0.0015
-0.0020
-0.0025
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
average [H2O2], M
We see that the points lie pretty well on a straight line, so we can conclude that the reaction is
probably first order with respect to [H2O2(aq)].
(c) From the result in part (b), we can write the rate law for the reaction as:
rate = k[H2O2(aq)]
The stoichiometry of the reaction, shows that two molecules of H2O2 disappear for every mole
of O2 that is formed. We write the rate of the reaction as:
⎛ 1 ⎞ ∆[H 2O 2 (aq )]
rate = – ⎜ ⎟
⎝ 2⎠
∆t
Thus, using the results from part (a), the average rates of reaction in the three experiments are:
rate1 = 0.0011 M·min–1
rate2 = 0.0004 M·min–1
rate3 = 0.0007 M·min–1
The rate constant for the reaction can be estimated from any of the experiments by dividing the
average rate by the initial [H2O2(aq)]:
k = (0.0011 M·min–1)/(0.874 M) = 0.0013 min–1
(d) Starting with 0.234 M hydrogen peroxide, predict the concentration of unreacted peroxide
after 30 minutes of reaction. Under this condition we have:
rate = k[H2O2(aq)] = (0.0013 min–1)(0.234 M) = 0.00030 M·min–1
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The rate of disappearance of hydrogen peroxide is twice as great and has a negative value,
- 0.00060 M·min–1, so we get the amount of hydrogen peroxide reacted as:
∆[H2O2(aq)] = (– 0.00060 M·min–1)(30 min) = – 0.018 M
The amount of hydrogen peroxide remaining is 0.216 M (= 0.234 M – 0.018 M).
Problem 11.30.
(a) In aqueous acidic solution, hydrogen peroxide oxidizes iodide ions to elemental iodine and
is, itself, reduced to water. The balanced equation for this redox reaction is:
H2O2(aq) + 2I–(aq) +2H+(aq) → 2H2O(aq) + I2(aq)
(b) In separate experiments, with all other conditions the same, doubling the initial hydrogen
peroxide concentration or doubling the iodide ion concentration caused the initial rate of
formation of iodine to double. The rate of formation of iodine was four times higher at pH 1.4
compared to pH 2.0, with all other conditions the same. Since the rate of the reaction doubles
when either [H2O2(aq)] or [I–(aq)] is doubled, the reaction is first order with respect to both
[H2O2(aq)] and [I–(aq)]. At pH 1.4, [H+(aq)] = 10-1.4 M = 0.04 M and at pH 2.0, [H+(aq)] = 10–
2.0
M = 0.01 M. An increase of [H+(aq)] by a factor of four from pH 2.0 to 1.4, increases the rate
by a factor of four, so the reaction is also first order with respect to [H+(aq)]. The rate law is:
rate = k[H2O2(aq)][I–(aq)][H+(aq)]
(c) The relationships of the rate and concentration changes with time are:
⎛ 1 ⎞ ∆[I– (aq )]
∆[I2 (aq )]
∆[H 2 O 2 (aq )]
rate =
= –
= –⎜ ⎟
⎝2⎠
∆t
∆t
∆t
If the rate of formation of iodine is 2.5 × 10–3 M·s–1, the rate of disappearance of hydrogen
peroxide is –2.5 × 10–3 M·s–1, and the rate of disappearance of iodide is –5.0 × 10–3 M·s–1.
(d) Changing the concentration of a reactant in the rate law has no affect on the reaction rate
constant. The rate constant is a temperature-dependent proportionality factor between the rate
and the concentrations and independent of the concentrations.
(e) The rate constant is unaffected by dilution (which changes the reactant concentrations), as
explained in part (d). Since the dilution halves the concentrations of all three reactants, the rate
of reaction will be decreased by a factor of (1/2)3, or one-eighth. The reaction rate will decrease
by a factor of eight.
Problem 11.31.
(a) The decomposition reaction for azomethane is: CH3NNCH3(g) → CH3CH3(g) + N2(g). If the
initial pressure of azomethane in a reaction vessel is 10.0 kPa, the azomethane pressure will be
reduced by 2.5 kPa to 7.5 kPa after one-quarter of it has decomposed. Since one mole of ethane
and one mole of nitrogen are formed for every one mole of azomethane that decomposes, when
azomethane pressure decreases by 2.5 kPa, ethane pressure and nitrogen will each increase by
2.5 kPa. The net change in pressure is + 2.5 kPa, (= –2.5 kPa from the azomethane + 2.5 kPa
from the ethane + 2.5 kPa from the nitrogen), so that the final total pressure = 12.5 kPa.
(b) As we saw in part (a), when the azomethane pressure decreases by x, the total pressure in the
flask increases by x. Thus, if, beginning with azomethane at 20 kPa (and 300 °C), the total
pressure increases by 1.0 kPa, the azomethane pressure must have decreased from 20 to 19 kPa,
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Reaction Pathways
while both ethane and nitrogen pressures increased from 0 to 1 kPa. Therefore, 1/20 of the
original azomethane has decomposed and the rate of decomposition is:
(–1 kPa)
∆P
= 8.3 × 10-3 kPa-s-1
rate = – azomethane = –
120 s
∆t
(c) In another experiment at 300 °C, the initial pressure of azomethane was 5.00 kPa and the
total pressure in the vessel reached 5.25 kPa after 2.00 minutes. The initial pressure of
azomethane here in part (c) (5.00 kPa) is one fourth the initial pressure in part (b) (20.0 kPa).
The average initial rate in part (c) = – (–0.25 kPa)
, which is one-fourth the average initial
(120 s)
rate in part (b). Since the rate is reduced by one fourth when the initial pressure of azomethane
is reduced by one fourth, we conclude that the reaction is first order with respect to azomethane.
rate = kPazomethane
k=
rate
Pazomethane
initial rate
8.3 ×10 –3 kPa ⋅ s –1
=
=
= 4.2 × 10-4 s-1
initial Pazomethane
20 kPa
Problem 11.32.
We wish to show that “the sum of the series of reactions (11.25), (11.26), and (11.27) is the
stoichiometric reaction (11.16).” An easy approach to this sort of problem is to sum all the
reactants on the left and all the products on the right of the series of equations and then cross out
the ones that appear on both sides to give the sum or net reaction produced by the series of
reactions. In this case, we get:
OCl–(aq) + H2O(l) + HOCl(aq) + I–(aq) HOI(aq) + OH–(aq) →
HOCl(aq) + OH–(aq) + HOI(aq) + Cl–(aq) + OI–(aq) + H2O(l)
The resulting net reaction is stoichiometric reaction (11.16):
OCl–(aq) + I–(aq) → Cl–(aq) + OI–(aq)
Problem 11.33.
Possible pathways for the reaction of nitrogen dioxide with carbon monoxide are:
(i) NO2(g) + CO(g) → NO(g) + CO2(g)
k(i)
(ii) NO2(g) + NO2(g) ⇔ NO(g) + NO3(g)
fast equilibrium, K(ii)
NO3(g) + CO(g) → NO2(g) + CO2(g)
slow, k(ii)
(iii) NO2(g) + NO2(g) → NO(g) + NO3(g)
slow, k(iii)
NO3(g) + CO(g) → NO2(g) + CO2(g)
fast
(a) The net reaction is the same for each pathway as it must be, if the pathway is to be a possible
one.
For pathway (i), the reaction written is the net reaction:
net reaction (i): NO2(g) + CO(g) → NO(g) + CO2(g)
For pathway (ii), we sum the reactants and products in the two steps and eliminate those that
appear on both sides to get the net reaction:
NO2(g) + NO2(g) + NO3(g) + CO(g) → NO(g) + NO3(g) + NO2(g) + CO2(g)
net reaction (ii): NO2(g) + CO(g) → NO(g) + CO2(g)
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Pathway (iii) is the same as pathway (ii), with a different slow step, so the net reaction is the
same:
net reaction (iii): NO2(g) + CO(g) → NO(g) + CO2(g)
(b) For reaction pathway (i), the single-step reaction written is the elementary reaction with:
rate (i) = k(i)[NO2(g)][CO(g)]
For reaction pathway (ii), the equilibrium is rapid, the species remain in equilibrium as the
reaction proceeds and we can write (after rearranging the equilibrium constant expression):
⎛ [NO 2 (g)]2 ⎞
[NO3(g)] = K(ii) ⎜
⎟
⎝ [NO(g)] ⎠
The rate law for reaction pathway (ii) is given by the rate for the slow step:
⎛ [NO 2 (g)]2 ⎞
rate (ii) = k(ii)[NO3(g)][CO(g)] = k(ii)K(ii) ⎜
⎟ [CO(g)]
⎝ [NO(g)] ⎠
For reaction pathway (iii), NO3(g) reacts as fast as it is formed and the rate law for the overall
reaction will be the rate law for the slow reaction:
rate (iii) = k(iii)[NO2(g)]2
(c) To distinguish among these pathways, we could vary the initial [NO2(g)] while holding
[CO(g)] constant and determine the initial reaction rates to find out the order of reaction with
respect to [NO2(g)]. If the reaction is found to be first order, the result would support reaction
pathway (i). If the reaction is found to be second order with respect to [NO2(g)], then the
evidence favors pathways (ii) and (iii). A second series of experiments varying [CO(g)] while
holding [NO2(g)] constant would allow us to determine the order of reaction with respect to
[CO(g)]. If there is no dependence on [CO(g)], reaction pathway (iii) is supported (if the
reaction is also second order with respect to [NO2(g)]). If the reaction is first order with respect
to [CO(g)] and also with respect to [NO2(g)], reaction pathway (i) is supported. If the reaction is
first order with respect to [CO(g)] and second order with respect to [NO2(g)], reaction pathway
(ii) is supported. If there is evidence to support reaction pathway (ii), then we could investigate
the affect of adding [NO(g)] to the initial reaction mixture. If the reaction is slowed by this
addition, as the rate law for pathway (ii) predicts, this is further evidence supporting pathway
(ii). Remember that we can never prove that a pathway is correct; we can only show which ones
are not correct, that is, are not consistent with the experimental evidence.
Problem 11.34.
(a) The experimentally-determined rate law for the reaction, 2NO(g) + Cl2(g) → 2NOCl(g), is
rate = k[NO(g)]2[Cl2(g)]. The first step of a pathway proposed for this reaction is:
NO(g) + Cl2(g) ⇔ NOCl2(g)
fast equilibrium, K
A second, rate-limiting step that is consistent with the overall stoichiometry of the reaction is:
NOCl2(g) + NO(g) → 2NOCl(g) slow, k
(b) The overall rate law predicted for this two-step pathway begins with the rate law for the
slow reaction step and then substitution into this equation for any intermediate concentrations in
terms of measurable reactant and product concentrations.
rate = k[NOCl2(g)][NO(g)]
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K=
Reaction Pathways
(NO(g))(Cl2 (g))
(NOCl2 (g))
∴[NOCl2(g)]= K[Cl2(g)][NO(g)]
rate = kK[Cl2(g)][NO(g)]2 = k'[Cl2(g)][NO(g)]2
(c) To distinguish experimentally between the pathway in part (a) and the one-step termolecular
pathway initially suggested in the problem (which gives the same rate law), we might look for
evidence of the intermediate NOCl2. Such studies are not easy, but modern rapid spectroscopic
methods could be useful.
Problem 11.35.
(a) For the proposed pathway for the reaction, (CH3)3CCl + OH– → (CH3)3COH + Cl–, sum the
reactants and products that appear in each step and eliminate those that appear on both sides to
get the net reaction:
(CH3)3CCl + (CH3)3C+ + H2O + (CH3)3COH2+ + OH– →
(CH3)3C+ + Cl– + (CH3)3COH2+ + (CH3)3COH + H2O
net reaction: (CH3)3CCl + OH– → Cl– + (CH3)3COH
(b) The initial slow equilibrium step, (CH3)3CCl ⇔ (CH3)3C+ + Cl–, in the proposed pathway is
rate limiting. The cation, (CH3)3C+, reacts with water almost as soon as it is formed, so the
initial rate of the reaction is the rate of formation of this cation. Let kf be the rate constant for
the forward reaction of (CH3)3CCl:
(CH3)3CCl → (CH3)3C+ + Cl–
The rate law for the overall reaction, which is limited by the rate of this reaction, is:
rate = kf[(CH3)3CCl]
(c) The rate law we wrote for this reaction in Problem 11.25 was first order in [(CH3)3CCl] and
zero order in [OH–], so the rate law derived from this reaction pathway is consistent with what
we have found experimentally. Note that we have shown the rate-limiting step in this pathway
as an equilibrium reaction, but have neglected the reverse reaction reforming (CH3)3CCl from
(CH3)3C+ and OH–. This is not a serious problem when we investigate only initial rates in
solutions that initially contain no Cl–. What happens if we add Cl– to the solutions is discussed
in Problem 11.38.
Problem 11.36.
(a) A possible pathway for the reaction, CH3OH + H+ + Cl– → CH3Cl + H2O, is:
CH3OH + H+ ⇔ CH3OH2+
fast equilibrium, K
–
+
CH3OH2 + Cl → CH3Cl + H2O slow, k
For this reaction pathway, the equilibrium is rapid, so the species remain in equilibrium as the
reaction proceeds and we can write (after rearranging the equilibrium constant expression):
[CH3OH2+] = K[CH3OH][H+]
The reaction rate law for the overall reaction is the rate law for the slow, rate-limiting step:
rate = k[CH3OH2+][Cl–] = kK[CH3OH][H+][Cl–]
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(b) In this reaction system, when the hydrochloric acid, HCl, concentration was doubled, the
rate of reaction quadrupled. The source of both H+ and Cl– in this system is the hydrochloric
acid in the mixture. Thus, [H+] = [Cl–], and we could write the rate law as:
rate = kK[CH3OH][H+]2
The rate of the reaction would quadruple as [H+] doubles, if the only source of H+ and Cl– in the
system is hydrochloric acid.
Problem 11.37.
(a) We are asked to find the net reaction represented by each of these pathways for the
decomposition of N2O5(g).
(i)
N2O5(g) ⇔ NO2(g) + NO3(g)
fast equilibrium, K(i)
NO2(g) + NO3(g) → NO(g) + NO2(g) + O2(g) slow, k(i)
NO(g) + NO3(g) → 2NO2(g)
fast
Note that two NO3(g) are required after the initial step (one in each subsequent reaction). Thus
we need to double the first reaction before adding the reactants and products and eliminating
redundant species:
N2O5(g) + N2O5(g) + NO2(g) + NO3(g) + NO(g) + NO3(g) →
NO2(g) + NO3(g) + NO2(g) + NO3(g) + NO(g) + NO2(g) + O2(g) +2NO2(g)
net reaction: N2O5(g) → 4NO2(g) + O2(g)
N2O5(g) ⇔ NO2(g) + NO3(g)
fast equilibrium, K(ii)
NO3(g) + N2O5(g) → N2O4(g) + NO2(g) + O2(g) slow, k(ii)
N2O4(g) → 2NO2(g)
fast
None of the intermediate species is used more than once in subsequent reactions, so we simply
add reactants and products and eliminate redundant species:
N2O5(g) + NO3(g) + N2O5(g) + N2O4(g) →
NO2(g) + NO3(g) + N2O4(g) + NO2(g) + O2(g) + 2NO2(g)
net reaction: 2N2O5(g) → 4NO2(g) + O2(g)
(ii)
N2O5(g) → NO2(g) + NO3(g)
slow, k(iii)
NO3(g) + N2O5(g) → 3NO2(g) + O2(g) fast
None of the intermediate species is used more than once in subsequent reactions, so we simply
add reactants and products and eliminate redundant species:
N2O5(g) + NO3(g) + N2O5(g) →
NO2(g) + NO3(g) + N2O4(g) + 3NO2(g) + O2(g)
net reaction: 2N2O5(g) → 4NO2(g) + O2(g)
(iii)
The net reaction for each of the pathways is the same.
(b) The rate law for each of the pathways in part (a) is found here. For reaction pathway (i), the
equilibrium is rapid, so the species remain in equilibrium as the reaction proceeds and we can
write (after rearranging the equilibrium constant expression):
⎛ [N O ( g)] ⎞
[NO3(g)] = K(i) ⎜ 2 5
⎟
⎝ [NO 2 ( g) ] ⎠
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The reaction rate law for the overall reaction is the rate law for the slow, rate-limiting step:
⎛
⎞
rate (i) = k(i)[NO3(g)][NO2(g)] = k(i)K(i) ⎜ [N 2 O 5 (g )] ⎟ [NO2(g)] = k(i)K(i)[N2O5(g)]
⎝ [NO 2 ( g )] ⎠
The first step in reaction pathway (ii) is the same as in (i) – with a different equilibrium constant
– and the reaction rate law for the overall reaction is the rate law for the slow, rate limiting step:
⎛ [N O ( g)] ⎞
⎛ [N2 O5 ( g)]2 ⎞
[N
O
(g)]
=
k
K
rate (ii) = k(ii)[NO3(g)][N2O5(g)] = k(ii)K(ii) ⎜ 2 5
2
5
(ii)
(ii)
⎟
⎜
⎟
⎝ [NO 2 ( g) ] ⎠
⎝ [NO 2 (g )] ⎠
The first step in reaction pathway (iii) is the slow rate-limiting step, so the overall reaction rate
law is:
rate (iii) = k(iii)[N2O5(g)]
(c) The initial rate of disappearance of N2O5(g) was found to be a linear function of the initial
concentration of N2O5(g). If the rate of a reaction increases in direct proportion to the increase
in a reactant concentration (linearly with respect to the concentration), the reaction is first order
with respect to the reactant concentration. In the present case, the rate law suggested by the
experimental result is:
rate = k[N2O5(g)]
(d) Reaction pathway (ii), which predicts a rate law that is second order with respect to
[N2O5(g)], is ruled out by the result in part (c).
(e) To help distinguish between the proposed pathways (i) and (iii), we might consider adding
NO2(g) to the initial reaction mixture. This addition should have no affect on the rate of reaction
by pathway (iii), since NO2(g) is not involved as a reactant in any step. The simple assumption
we make about reaction pathway (i) is that the initial equilibrium is maintained. If this is not
quite true, then the addition of NO2(g) might have an affect on the reaction rate. Without a more
detailed analysis, we cannot predict the direction of the affect, but either an increase or decrease
in the rate would suggest that NO2(g) is involved as more than simply a product in the reaction
pathway and this is consistent with pathway (i).
Problem 11.38.
In a continuing study of the system discussed in Problems 11.15, 11.25, and 11.35, a series of
experiments was carried out in which salt, sodium chloride, was dissolved in reaction mixtures
which were otherwise identical. The initial concentration of hydroxide ion, [OH–]0, was 0.0025
M in all these experiments.
expt #
1
5
6
7
8
[NaCl], M
time to color change, s
0
0.85
1.71
2.56
3.42
35
35
43
95
268
(a) We get the rate of disappearance of hydroxide ion just as we did in Problems 11.15 and
11.25:
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rate(1) = (0 M – 0.0025 M)
rate(5) = (0 M – 0.0025 M)
rate(6) = (0 M – 0.0025 M)
rate(7) = (0 M – 0.0025 M)
rate(8) = (0 M – 0.0025 M)
(35 s)
(35 s)
(43 s)
(95 s)
= – 7.1 × 10–5 M·s–1
= – 7.1 × 10–5 M·s–1
= – 5.8 × 10–5 M·s–1
= – 2.6 × 10–5 M·s–1
(268 s)
= – 0.93 × 10–5 M·s–1
(b) The rate of disappearance of (CH3)3CCl is the same as the rate of disappearance of hydroxide
ion, as we have reasoned in Problem 11.15. As more sodium chloride is dissolved in the solution,
the rate of disappearance of (CH3)3CCl decreases. The presence of sodium cations, Na+, and/or
chloride anions, Cl–, in the solution slows the reaction of (CH3)3CCl.
(c) The reaction pathway proposed in Problem 11.35 for the reaction of (CH3)3CCl with water,
involves a slow equilibrium reaction forming the (CH3)3C+ cation and chloride anion, Cl–.
Adding chloride ion to the reaction mixture should decrease the equilibrium concentration of the
(CH3)3C+ cation:
⎛ [(CH3 )3 CCl]⎞
[(CH3)3C+] = K ⎜
⎟
⎠
⎝
[Cl – ]
If less reactant is formed, the reaction will slow down. Thus, it seems reasonable to conclude that
the added chloride anion from the sodium chloride is responsible for the observed decrease in
reaction rate in this system. The affect of added chloride strengthens the case for the reaction
pathway proposed in Problem 11.35.
Problem 11.39.
(a) In acidic solutions, the reaction between hydrogen peroxide and iodide and its rate law are:
H2O2(aq) + 2I–(aq) + 2H+(aq) → I2(aq) + 2H2O(aq)
rate = k[H2O2(aq)][I–(aq)][H+(aq)]
Since H2O2(aq) is a weak Brønsted-Lowry base, hydronium ions can transfer protons to it in
acidic solution:
H2O2(aq) + H3O+(aq) ⇔ H3O2+(aq) + H2O(aq)
(b) The rate limiting step for reaction in acidic solution, assuming it is analogous to the rate
limiting step in pH 7 solution, reaction (11.31), is:
H3O2+(aq) + I–(aq) → HOI(aq) + H2O(aq)
(c) For the reaction HOI(aq) + H2O2(aq) → O2(g) + I–(aq) + H+(aq) + H2O(aq), an increase in
hydronium ion concentration (lower pH) will make the reaction less favorable. We are
disturbing the system by adding a product and Le Chatelier’s principle says that the system will
respond by trying to use up the product, that is, by proceeding in reverse, thus making the
reaction as written less favorable. Use the Nernst equation for this reaction at 298 K, with Eo′ =
0.5 V, to get Eo, the potential at pH 0:
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Eo′ = Eo –
Reaction Pathways
(0.05916 V) log ⎛ (O2 (aq ))(I– (aq))(H + (aq ))(H 2O(aq )) ⎞
⎜
⎝
n
o
(0.5 V) = E –
(0.05916 V)
2
(HOI(aq ))(H 2 O 2 (aq ))
⎟
⎠
⎛ (1)(1)(10–7 )(1) ⎞
o
log ⎜
⎟ = E + (0.2 V)
⎠
⎝
(1)(1)
o
E = (0.5 V) – (0.2 V) = 0.3 V
The driving force for the reaction has decreased in acidic solution as we reasoned from Le
Chatelier’s principle as well.
(d) Find the cell potential for this reaction at pH 0:
HOI(aq) + I–(aq) + H+(aq) → I2(aq) + H2O(l)
The appropriate half reactions are:
2HOI(aq) + 2H+(aq) + 2e– ⇔ 2H2O(aq) + I2(aq) Eo = 1.430 V
I2(aq) + 2e– ⇔ 2I–(aq)
Eo = 0.5355 V
The standard cell potential, Eo, for the reaction is 0.894 V [= (1.430 V) – (0.5355 V)].
(e) The driving force for the reaction producing iodine, part (d), instead of oxygen, part (c), as
the oxidized product of the hydrogen peroxide-iodide reaction at pH 0, is almost 0.6 V greater.
Thermodynamically, the production of iodine is greatly favored in the acidic solution. This
helps to explain the difference in reaction at neutral and acidic pH.
(f) A reaction pathway that is consistent with the stoichiometry of the reaction and the rate law
is:
H2O2(aq) + H3O+(aq) ⇔ H3O2+(aq) + H2O(aq) rapid acid-base equilibrium
H3O2+(aq) + I–(aq) → HOI(aq) + H2O(aq)
slow rate limiting reaction
+
+
HOI(aq) + H3O (aq) ⇔ H2OI (aq) + H2O(aq)
rapid acid-base equilibrium
+
–
H2OI (aq) + I (aq) → I2(aq) + H2O(aq)
rapid nucleophile-electrophile reaction
Problem 11.40.
(a) The oxidized iodine atom in HOI(aq) and the reduced atom in I–(aq) are the pair that couple
the reactions (11.21) and (11.22). HOI is formed from I–(aq) [and H2O2(aq)] in the first reaction
and reacts in the second reaction to reform I–(aq).
(b) The coupling diagram (patterned after Figures 10.12, 1014, and 10.15) is:
oxidized
reduced
H2 O2
I–
OH–
HOI
reduced
oxidized
oxidized
H2O + O2 + H+
H2 O2
reduced
Note that hydrogen peroxide can act as both an oxidizing agent (when the oxidation numbers of
its oxygen atoms are reduced from –1 to a –2) and as a reducing agent (when the oxidation
numbers of its oxygen atoms are increased from –1 to 0). In this reaction, iodide catalyzes the
self-reduction-oxidation (disproportionation) of hydrogen peroxide molecules.
(c) The net reaction for this system contains neither I–(aq) nor HOI(aq). Thus, what we said in
Chapter 10 is true for this case as well: coupling species do not appear in the net reaction for the
coupled series of reactions. Note that the rate law does contain the concentration of iodide
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Chapter 11
anion, [I–(aq)], since it is involved in the rate limiting part of the reaction pathway. This is one
more example of the fact that the rate law and stoichiometry for a reaction are not simply
related.
Problem 11.41.
(a) Two possible pathways for the gas phase reaction of iodine and hydrogen are:
(i)
H2(g) + I2(g) ⇔ 2HI(g) k(i)
(ii)
I2(g) ⇔ I(g) + I(g)
fast equilibrium, K(ii)
H2(g) +I(g) + I(g) → 2HI(g) slow, k(ii)
For reaction pathway (i), the reaction written is an elementary reaction and we can immediately
write its rate law as:
rate (i) = k(i)[H2(g)][I2(g)]
For reaction pathway (ii), the equilibrium is rapid, so the species remain in equilibrium as the
reaction proceeds and we can write (after rearranging the equilibrium constant expression and
solving for [I(g)]):
[I(g)] = K (ii) [I2 (g)]
The reaction rate law is derived from the rate law for the rate limiting reaction:
rate (ii) = k(ii)[H2(g)][I(g)][I(g)] = k(ii)[H2(g)]( K (ii) [I2 (g)] )( K (ii) [I2 (g)] )
rate (ii) = k(ii)K(ii)[H2(g)][I2(g)]
Thus, we see that both pathways lead to the same form of the rate law.
(b) Molecules in the gas phase are far apart and collisions between two of them are relatively
rare. It is highly improbable that during the very short time two molecules are in contact during
a collision a third molecule would collide with the pair. Thus three-body collisions among gas
phase molecules are quite rare. A reaction that depends on such events will be very slow.
Alternatively, if two molecules (or a molecule and an atom) collide and stick together for an
appreciable time before separating, there is a much greater likelihood that a third particle will
collide with the pair before they separate. Iodine atoms are large and their electron cloud is
relatively easily deformed (they are polarizable), so they are pretty “sticky” in dispersion force
interactions with other species. It would not be surprising that an iodine atom and a hydrogen
molecule would collide and stick together longer than if they collided like hard particles. Thus,
we might write pathway (ii) as:
fast equilibrium, K(ii)
I2(g) ⇔ I(g) + I(g)
H2(g) +I(g) ⇔ H2I(g) fast equilibrium, Kcollision
H2I(g) +I(g) → 2HI(g) slow, k(ii)
Show that this reaction pathway gives the same form of the rate law as the other two pathways.
Problem 11.42.
(a) The net reaction of atmospheric ozone with nitric oxide and its rate law are:
O3(g) + NO(g) → NO2(g) + O2(g)
∆ [O 3 (g) ]
–
= k[O3(g)][NO(g)]
∆t
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Two pathways proposed for this reaction are:
(i) O3(g) + NO(g) → O(g) + NO3(g)
O(g) + O3(g) → 2O2(g)
NO3(g) + NO(g) → 2NO2(g)
slow, k(i)
fast
fast
(ii)
O3(g) ⇔ O(g) + O2(g)
fast equilibrium, K(ii)
NO(g) + O(g) → NO2(g)
slow, k(ii)
The net reaction represented by pathway (i) is obtained by adding all the reactants and products
and canceling those that appear on both sides of the resulting reaction:
O3(g) + NO(g) + O(g) + O3(g) + NO3(g) + NO(g)
→ O(g) + NO3(g) + 2O2(g) + 2NO2(g)
2O3(g) +2NO(g) → 2O2(g) + 2NO2(g)
O3(g) + NO(g) → O2(g) + NO2(g)
The stoichiometry of pathway (i) agrees with the overall stoichiometry of the reaction.
The net reaction represented by pathway (ii) is obtained by adding all the reactants and products
and canceling those that appear on both sides of the resulting reaction:
O3(g) + NO(g) + O(g) → O(g) + O2(g) + NO2(g)
O3(g) + NO(g) → O2(g) + NO2(g)
The stoichiometry of pathway (ii) agrees with the overall stoichiometry of the reaction.
(b) The rate law for pathway (i) is the rate law for the slow step:
rate = k(i)[O3(g)][NO(g)]
For reaction pathway (ii), the equilibrium is rapid, so the species remain in equilibrium as the
reaction proceeds and we can write (after rearranging the equilibrium constant expression and
solving for [O(g)]):
⎞
⎛ O (g)
[O(g)] = K(ii) ⎜ [ 3 ]
O
(g)
⎝
[ 2 ]⎟⎠
⎛ O (g)
⎞
rate = k(ii)[NO(g)][O(g)] = k(ii)[NO(g)]K(ii) ⎜ [ 3 ]
O
(g)
⎝
[ 2 ]⎟⎠
⎛ O (g)
⎞
= k'[NO(g] ⎜ [ 3 ]
O
(g)
⎝
[ 2 ]⎟⎠
The rate law predicted for pathway (i) agrees with the observed rate, but that for pathway (ii)
does not. This means that pathway (i) could be the correct pathway but pathway (ii) is excluded.
(c) Another pathway that gives the correct net reaction and rate law is a single-step bimolecular
reaction between the two reactants:
O3(g) + NO(g) → O2(g) + NO2(g)
Show that this pathway does give the correct net reaction and rate law.
Problem 11.43.
The horizontal lines shown on the graph in Figure 11.6(b) are spaced 0.7([0.693 = ln2) units
apart. The spacing between lines is the spacing between values that represent half-lives for the
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Chapter 11
decay. If the counts per minute (cpm) at one time is N, then the counts per minute one half-life
later will be N/2. On the ln(cpm) scale, this difference is:
⎛ N ⎞
difference = lnN – ln(N/2) = ln ⎜
⎟ = ln2
⎝ N / 2⎠
Problem 11.44.
We can use data from Figure 11.6 to find out how long we would have to keep a collection of
32
P waste to assure that its radioactivity is under 1% of the starting value. We can write the
equation for the experimental line in Figure 11.6(b) as:
lnN = (slope)t + lnN0
Rearrange to:
ln(N/N0) = (slope)t
The data in the figure caption tell us that (slope) = –4.85 × 10–2 day–1. Substitute this value into
the equation, together with N/N0 = 0.01 (decay to 1% of the starting activity), to get:
t=
ln (N / N 0 )
slope
=
ln(0.01)
= 95 day
(–4.85 × 10 –2 day –1 )
Ninety-five days is a little less than 14 weeks [= (95 day)/(7 day·week–1], so 14 weeks or a little
over three months is how long you will have to store the 32P waste before it can be discarded as
trash.
Problem 11.45.
We can use data from Check This 11.37 to find the age of an archaeological specimen of bone
that has a carbon-14 radioactivity of 2.93 counts·min–1. Use equation (11.41), kt1/2 = 0.693, and
substitute the carbon-14 half-life of 5730 yr to determine:
0.693
0.693
k=
=
= 1.21 × 10–4 yr–1
t1/2
5730 yr
For a first order radioactive decay, we know that the number of counts, N (proportional to the
number of radioactive nuclei), at any time is related to the number of counts at the initial time,
N0:
ln(N/N0) = –kt
For this sample of bone (presumed to begin with an activity of 12.4 cpm) we get an age of:
t=
ln (N / N 0 )
–k
=
ln [(2.93 cpm)/(12.4 cpm)]
–1.21 × 10
–4
yr
–1
= 11.9 × 103 yr
Problem 11.46.
An antibiotic, A, is metabolized in the body by a first order reaction:
∆ (concentration of A)
= k(concentration of A)
–
∆t
The rate constant, k, depends upon temperature and body mass. At 37 °C, k = 3.0 × 10–5 s–1 for a
70-kg person.
(a) In order to calculate how frequently a 70-kg person must take 400-mg pills of the antibiotic,
in order to be sure to keep the concentration of antibiotic from falling below 140 mg per 70 kg
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Reaction Pathways
body weight, we neglect the time between ingestion of the first and second pill, because the
patient begins with none of the antibiotic in his/her system. Assume that the second pill is taken
when the concentration of the antibiotic has fallen to 140 mg per 70 kg body mass (as will be
the case for all subsequent ingestions). The ingestion of the pill immediately raises the
concentration to 540 mg per 70 kg body mass and we want to know how long it will take for the
concentration to fall to 140 mg per 70 kg body mass, which is when the next pill should be
taken. The time between taking pills should be (where C is the concentration of the antibiotic in
mg per 70 kg):
( C )= ln ((140 mg per 70 kg) (540 mg per 70 kg))= 4.5 × 10 s ≈ 12 hr
ln C
t=
4
0
–k
–3.0 × 10
–5
s
–1
(b) The rate constant for metabolizing the antibiotic is larger (the reaction is faster) at the higher
temperature, 39 °C, so we need to substitute the larger rate constant to find the time between
pills for a person with a fever:
ln (140 mg per 70 kg)
(540 mg per 70 kg)
t=
= 3.4 × 104 s ≈ 9 hr
–5 –1
–4.0 × 10 s
The time comes out about 9.4 hr, but, to be on the safe side, we round down to be sure there is
always enough antibiotic in the patient’s system.
(
)
Problem 11.47.
We can test whether a reaction is first order by plotting the logarithm of the concentration of the
reactant as a function of time. If the resulting plot is linear, the reaction is first order. Because
we are taking logarithms, the concentrations should be expressed as dimensionless ratios, but
the standard state value chosen for the ratio will not affect the slope or linearity of the plot (it
will affect the intercept of the plot). Thus, for simplicity, we often simply take the logarithms of
the numerical concentration values and neglect their units. In the present case, for the
decomposition of gaseous dimethyl ether, (CH3)2O(g) → CH4(g) + H2(g) + CO(g), at 504 °C,
we have:
time, sec
0
390
777
1195
3175
35.1
29.8
24.9
10.4
pressure (CH3)2O(g), kPa 41.5
3.56
3.39
3.21
2.34
ln(pressure (CH3)2O(g)) 3.73
The plot of these data is:
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Chapter 11
4.0
ln(P) = -0.000438t+ 3.73
ln(P)
3.5
3.0
2.5
2.0
0
500 1000 1500 2000 2500 3000 3500
time, s
The plot is linear, so the reaction appears to be first order, with a rate constant,
k = 4.38 × 10-4 s 1, from the slope of the line.
Problem 11.48.
Cyclopropane, C3H6(g), isomerizes to propene, CH3CH=CH2(g), in a first order reaction with a
rate constant of 6.0 × 10–4 s–1 at 773 K. To find the pressure of propene in a reaction vessel after
15 minutes, beginning with a cyclopropane pressure of 56.7 kPa, let P0 be the pressure of the
cyclopropane at time zero and Pt the cyclopropane pressure at some later time, t. Since the
reaction is first order, we can write:
ln P1
= –kt
P0
( )
After 15.0 minutes (900. s) of reaction, starting with 56.7 kPa of cyclopropane, we have:
ln Pt
= –(6.0 × 10–4 s–1)(900. s) = –0.54
(56.7 kPa)
(
)
(P (56.7 kPa))= 0.58
t
Pt = 33 kPa
Therefore, about 24 kPa [= P0 – Pt = (56.7 kPa) – (33 kPa)] of cyclopropane has reacted and,
since one molecule of propene is formed for every molecule of cyclopropane that reacts, this
will produce 24 kPa of propene.
Problem 11.49.
Bacteria in a nutrient medium cause it to be cloudy: the more bacteria, the cloudier the medium.
The number (concentration) of bacteria in a nutrient medium is sometimes determined by
measuring the absorbance of the medium, which is directly proportional to the number of
bacteria present. These are data for a bacterial growth experiment.
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time, min
0
15
30
45
60
absorbance
0.053
0.095
0.167
0.301
0.533
During the part of their growth called the “log phase,” the rate law for bacterial growth is:
∆ (number of bacteria)
= k(number of bacteria)
∆t
(a) The equation for bacterial growth has a positive sign because the change in number of
bacteria is positive; as time increases, the number increases, ∆(number of bacteria) > 0.
(b) The derivation of the equation for the number of bacteria as a function of time is identical to
that in the text for the first order disappearance of a reactant with the negative sign on f replaced
by a positive and growth rather than decay described at each step. Here, let’s use calculus to get
the progress of the growth with time. Let N be the number of bacteria at any time. We rewrite
the rate equation in terms of infinitesimal changes and integrate to get the time dependence:
N t dN
t
∫N0 N = ∫0 kdt
or
ln N t
= kt
lnNt – lnN0 = kt
N0
( )
(c) The equation we have found in part (b) suggests plotting the logarithm of the number of
bacteria as a function of time to test if the growth follows the “log phase” rate law. Since the
absorbance in the bacterial solutions is directly proportional to the number of bacteria, that is,
N = (constant)(absorbance), we can write:
⎤
⎡
ln N t
= ln ⎢(constant)(absorbance)t
⎥
N0
(constant)(absorbance)
⎣
0⎦
( )
⎤
⎡
ln ⎢(absorbance)t
= kt
(absorbance)0 ⎦⎥
⎣
ln(absorbance)t = kt + ln(absorbance)0
Plots of the (absorbance)t and ln(absorbance)t as a function of time are:
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Chapter 11
0.6
-0.5
ln(absorbance) = 0.0385t – 2.94
-1.0
ln(absorbance)
absorbance
0.5
0.4
0.3
0.2
-1.5
-2.0
-2.5
0.1
0
-3.0
0
10
20
30
40
50
60
time, min
0
10
20
30
40
50
60
time, min
The equation we derived shows that the rate constant for the growth is the slope of the line on
the ln(absorbance) vs. time graph: k = 0.0385 min–1. This growth phase is called the “log phase”
because the logarithm of the number of cells increases linearly during this time.
(d) From part (b), we can write:
ln N t
= kt
N0
( )
If the number of bacteria double during this time interval, then Nt = 2N0. If we let the time
required for the doubling be t2, then we get:
ln 2N 0
= ln2 = 0.693 = kt2
N0
(
)
For the growth in this problem, the doubling time is:
0.693
t2 =
= 18 min
(0.0385 min –1 )
(e) Since the doubling time in the second nutrient medium is shorter than in the first, we can
conclude that the bacteria grow faster in the second medium. The second medium is more
favorable for the growth of the bacteria. It must contain nutrients that are more suitable for these
bacteria.
Problem 11.50.
From the solution for Problem 11.49(d), we know that the doubling time for log phase bacterial
growth is t2 = 0.693 . If we assume that the bacteria in freshly pasteurized milk are in their log
k
phase growth, we get;
0.693 0.693
=
= 1.7 × 10–2 hr–1
k=
t2
40 hr
Use k to find the number of bacteria at 10 days (240 hr), beginning with 2.0 × 104 bacteria·mL–1
at time zero:
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( N)
ln N t
0
⎛
⎞
–2
–1
= ln ⎜⎜ N t
4
–1 ⎟
⎟ = kt = (1.7 × 10 hr )(240 hr) = 4.08
2.0
×
10
bacteria
⋅
mL
)⎠
⎝ (
⎛N
⎞
⎜⎜ t
4
–1 ⎟
⎟ = 59.1
⎝ (2.0 × 10 bacteria ⋅ mL )⎠
Nt = 1.2 × 106 bacteria·mL–1
Problem 11.51.
(a) The data for this problem are from Eyring and Daniels, JACS 1930, 52, 1472, as reported in
E. L. King, How Chemical Reactions Occur (Benjamin, New York, 1964). Using the
stoichiometry of the reaction, 2N2O5(CCl4) → 4NO2(CCl4) + O2(g), King recalculated the data
from the literature to convert from volume of oxygen gas evolved as a function of time to
amount of the reactant decomposed as a function of time.
time, min
0
184 319 526 867 1198 1877
[N2O5(CCl4)], M 2.33 2.08 1.91 1.67 1.35 1.11 0.72
⎤
⎡
The experiment took 31.28 hours ⎢= (1877 min)
–1 ⎥ , which is about 1.3 days. (The
⎢⎣
(60 min ⋅ hr )⎥⎦
final point was probably taken when the reaction had been left to run overnight, about 11 hours,
after the penultimate point was taken.)
(b) A concentration vs. time graph and a ln(concentration) vs. time graph (assuming first order
dependence on [N2O5(CCl4)]) are shown here:
2.5
1.0
ln[N2O5] = -6.25x10^-4*t + 0.846
0.8
2.0
1.5
ln[N2O5]
[N2O5], M
0.6
1.0
0.4
0.2
0.0
0.5
-0.2
0.0
-0.4
0
400
800
1200
time, min
1600
2000
0
400
800
1200
time, min
1600
2000
The curve through the points on the [N2O5] vs. time plot is an exponential curve through these
experimental points. The plot of ln[N2O5] vs. time is used to test whether the reaction is first
order and the best linear fit to the data is shown. These data fit a first order decay very well with
a rate constant k = 6.25 × 10–4 min–1 from the slope of the line.
(c) The first order dependence found in part (b) is also the conclusion for the gas phase reaction
from Problem 11.37.
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(d) The reaction stoichiometry shows that each mole of N2O5 that decomposes produces 0.5
mole of O2(g). The number of moles of N2O5 that decompose in the first 184 minutes in 25.0
mL of solution and the number of moles of O2(g) formed are:
mole N2O5 decomposed = [(2.33 M) – (2.08 M)](0.0250 L) = 6.25 × 10–3 mol
mole O2(g) formed = 0.5(6.25 × 10–3 mol) = 3.13 × 10–3 mol
Use the ideal gas equation, PV = nRT, to find the volume occupied by this many moles of
oxygen gas at 318 K (45 °C) and 96.5 kPa pressure:
nRT
(3.13 × 10 –3 mol)(8.314 L ⋅ kPa ⋅ K –1mol–1 )(318 K)
V=
=
= 85.8 × 10–3 L
P
96.5 kPa
V = 85.8 × 10–3 L = 85.8 mL
The total number of moles of N2O5 that decompose during the reaction is 4.03 × 10–3 mol,
which produce 551 mL of O2(g).
Problem 11.52.
Compare the expressions for 99% to 90% completion for a first order reaction:
ln ⎛⎝ N99 N ⎞⎠
− kt99
t
ln(0.01)
0
=
=2=
= 99
N
− kt90
t90
ln(0.1)
ln ⎛⎝ 90 N ⎞⎠
0
The time required for 99% completion is twice that required for 90% completion. It takes the
same length of time to go from 90% to 99% completion (that is, to react 90% of the reactant
remaining after 90% of the reaction) as to go from the beginning of the reaction to 90%
completion. This makes sense, since 90% of the starting amount of reactant is reacting in each
case. You can think of this time as the “90% life” by analogy with the half-life of a first order
reaction.
Problem 11.53.
(a) Use the Arrhenius equation to compare reaction rates (and hence the rate constants, since the
concentration factors will cancel out in the ratio) at different temperatures and to determine the
activation energy for the reaction. In this case, the temperatures are 328 K (55 °C) and 308 K
(35 °C):
–
Ea
R⋅328 K
⎛ E ⎞⎛
1
1
⎞
⎟
–⎜ a ⎟ ⎜
–
rate 328
k328
Ae
⎠
⎠⎝
= 6.7 =
=
= e ⎝ R 328 K 308 K
Ea
–
rate 308
k308
Ae R⋅328 K
Ea
⎞ ⎛ 308 K – 328 K ⎞
ln(6.7) = – ⎛⎝
⎟
–1
–1 ⎠ ⎜
8.314 J ⋅ K ⋅ mol ⎝ (308 K)(328 K) ⎠
Ea = 8.0 × 104 J·mol–1 = 80. kJ·mol–1
(b) Let the temperature for doubling the rate of reaction at 308 K be T2, so we can write:
⎛ 8.0 × 104 J ⋅ mol –1 ⎞ ⎛ 1
1 ⎞
ln(2) = – ⎜
–
–1
–1 ⎟ ⎜
⎝ 8.314 J ⋅ K ⋅ mol ⎠ ⎝ T2 308 K ⎟⎠
T2 = 315 K
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Problem 11.54.
(a) The rate constants have units of s–1, which are the units for first order rate constants, so we
can infer that the reaction, 2NOCl(g) → 2NO(g) + Cl2(g), was found to be first order.
(b) Use the Arrhenius equation to compare the reaction rate constants at different temperatures
and to determine the activation energy for the reaction:
⎛ k400 ⎞
⎛ 6.9 × 10 –4 s–1 ⎞
Ea
⎛
⎞ ⎛ 350 K – 400 K ⎞
ln ⎜
⎟
⎟ = ln ⎜⎝
–6 –1 ⎟ = 4.3 = – ⎝
–1
–1 ⎠ ⎜
⎝ k350 ⎠
9.4 × 10 s ⎠
8.314 J ⋅ K ⋅ mol ⎝ (350 K)(400 K) ⎠
Ea = 1.0 × 105 J·mol–1 = 1.0 × 102 kJ·mol–1
(c) The Arrhenius frequency factor is obtained by substituting Ea into either one of the rate
constant expressions:
k400 = 6.9 × 10–4 s–1 = A e
A = 7.9 × 109 s–1
–
1.0 ×105 J ⋅ mol –1
(8.314 J ⋅ K –1 ⋅ mol –1 )(400 K)
Problem 11.55.
Rate constants for the reaction, 2N2O5 → 4NO2 + O2, at two temperatures in the gas phase and
in carbon tetrachloride solution are:
25 °C
gas phase
CCl4 solution
45 °C
3.38 × 10 min
4.69 × 10–5 min–1
–5
–1
43.0 × 10–5 min–1
Problem 11.51(b)
(a) Use the Arrhenius equation to compare the reaction rate constants at different temperatures
and to determine the activation energy and then the Arrhenius frequency factor for the reaction:
⎛k ⎞
⎛ 43.0 × 10–5 min–1 ⎞
Ea
⎛
⎞ ⎛ 298 K – 318 K⎞
ln ⎜ 318 ⎟ = ln ⎜
⎟
–5
–1 ⎟ = 2.54 = – ⎝
–1
–1 ⎠ ⎜
⎝ 3.38 × 10 min ⎠
⎝ k298 ⎠
8.314 J ⋅ K ⋅ mol ⎝ (298 K)(318 K) ⎠
Ea = 1.00 × 105 J·mol–1 = 100. kJ·mol–1
k318 = 43.0 × 10–5 min–1 = A e
A = 1.15 × 1013 min–1
–
1.00 ×10 5 J ⋅ mol –1
(8.314 J ⋅ K –1 ⋅ mol –1 )(318 K)
(b) For the reaction in solution, we have:
⎛k ⎞
⎛ 62.5 × 10 –5 min –1 ⎞
Ea
⎛
⎞ ⎛ 298 K – 318 K⎞
ln ⎜ 318 ⎟ = ln ⎜
⎟
–5
–1 ⎟ = 2.59 = – ⎝
–1
–1 ⎠ ⎜
⎝ 4.69 × 10 min ⎠
⎝ k298 ⎠
8.314 J ⋅ K ⋅ mol ⎝ (298 K)(318 K) ⎠
Ea = 1.02 × 105 J·mol–1 = 102 kJ·mol–1
k318 = 62.5 × 10–5 min–1 = A e
A = 3.58 × 1013 min–1
–
1.02 ×10 5 J ⋅ mol –1
(8.314 J ⋅ K –1 ⋅ mol –1 )(318 K)
(c) The rate parameters for this reaction in the gas phase and in solution are very similar. Within
the experimental uncertainty, the activation energy is the same under both conditions. The
frequency factor is about three times higher in the solution, which is not a very large effect.
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Problem 11.56.
This image is from “Sssss is for danger” New Scientist, 11 December 1999 (#2216), p. 31.
Probably from the research of Don Owings, University of California, Davis and/or Matthew
Rowe, Appalachian State University, North Carolina.
Although we do not know the reaction pathway responsible for the snake’s rattling, we can
write a rate equation:
rate = rattle frequency = f = k(snake)
In the equation, (snake) is some function of the concentrations of the molecules that enable the
snake to rattle and is a constant for a given snake species. If we take the logarithm of this
equation, we get:
E
E
lnf = lnk +ln(snake) = lnA – a + ln(snake) = – a + constant
RT
RT
A plot of lnf as a function of 1/T should be a straight line with a slope –Ea/R. (We really should
express the variables f and (snake) as dimensionless quantities, in order to be mathematically
correct in out use of logarithms, but we don’t know what the standard state is for this system.
This does not matter for determining whether the data give a linear plot and obtaining the slope
to get Ea, because the appropriate standard state factors will end up in the constant.) Counting
the number of rattles per 0.20 s in the figure and multiplying by five gives these data:
temperature (T)
°C
10
18
27
35
96
frequency (f)
rattles·s–1
T
K
1/T
103 K–1
lnf
50
80
135
180
283
291
300
308
3.53
3.44
3.33
3.25
3.91
4.38
4.91
5.19
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Reaction Pathways
The plot of lnf vs. 1/T is:
5.5
B
ln(f) = -(4.61x103 K)(1/T) + 20.2
ln(frequency)
5.0
B
4.5
B
4.0
B
3.5
3.2
3.3
3.4
3.5
3.6
1000/T, K–1
The slope of this plot gives Ea = 38 ± 2 kJ·mol–1 [= (4.61 × 103 K)(8.314 J·mol–1·K–1)]. (The
uncertainty represents the uncertainty in our ability to read the data.)
Problem 11.57.
The origin of this problem is a part of the “Newscripts” column in Chemical & Engineering
News, 07 February 2000. p. 88. The study was carried out by John McGregor at Louisiana State
University and sponsored by Dairy Management, Inc., Rosemont, IL, an organization dedicated
to getting people to use more dairy products. The result reported was that “milk should be
stored at no more than 40 °F. Every 5 °F increase in storage temperature cuts shelf life by half.”
Thus, the rate of decay doubles for every 5 °F increase in temperature. The only factor that
varies is the rate constant, so we can take the ratio of rates as the ratio of rate constants for
whatever temperatures we choose. The “data” suggest that the rate of spoilage at 45 °F (7.2 °C
= 280.2 K) is twice that at 40 °F (4.4 °C = 277.4 K):
−
Ea
R⋅280.2
−
Ea
R⋅280.2
k
rate at 280.2 K
= 2 = 280.2 = Ae Ea = e Ea
−
−
k 277.4
rate at 277.4 K
Ae R⋅277.4
e R⋅277.4
Taking logarithms gives:
Ea ⎞ ⎛
Ea ⎞
E 277.4 – 280.2 –1⎞
ln 2 = 0.69 = ⎛⎝ –
= – a⎛
K
− ⎝–
⎠
⎠
⎠
R ⋅ 280.2
R ⋅ 277.4
R ⎝ 280.2 ⋅ 277.4
⎛ 280.2 ⋅ 277.4 K⎞ = 160 kJ·mol–1
Ea = –0.69 (8.314 J·K–1·mol–1) ⎝
277.4 – 280.2 ⎠
Problem 11.58.
(a) Raw (unpasteurized) milk sours in about 9 hours at 20 °C (room temperature), but it takes
about 48 hours to sour in a refrigerator at °5 C. The reasoning here to find the activation energy
for souring of raw milk is essentially the same as in Problem 11.57. In this case the rates of
souring are directly proportional to the inverse of the time so:
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1
rate at 293 K
(9 hr)
=
= 5.33
1
rate at 278 K
(48 hr)
E 278 – 293 –1⎞
K
ln(5.33) = 1.67 = – a ⎛⎝
⎠
R 293 ⋅ 278
⎛ 293 ⋅ 278 K⎞ = 76 kJ·mol–1
Ea = –1.67 (8.314 J·K–1·mol–1) ⎝
278 – 293 ⎠
(b) The activation energy for souring raw milk is about half that required to spoil (sour)
pasteurized milk (from Problem 11.57). The souring is a result of bacterial growth and action in
the milk. Perhaps, pasteurization kills most of the bacteria that are most active in the souring
process, so only ones that act by a different, higher activation energy pathway are left to sour
the pasteurized milk.
Problem 11.59.
(a) We get the rate constant for cyclopentadiene dimer dissociation by substituting the known
(given) values for the frequency factor, activation energy, and desired temperature (200 °C =
473 K) in the Arrhenius equation:
k = (1.3 × 10 s ) e
13 –1
–
146 ×103 J ⋅ mol –1
(8.314 J ⋅ K –1 ⋅ mol –1 )(473 K )
= 9.8 × 10–4 s–1
(b) The units for A (and, hence, k) tell us that this is a first order reaction, so we can write the
rate as:
∆P
– dimer = kPdimer
∆t
The rate at 473 K (200 °C) is 9.8 × 10–4 kPa·s–1
(c) We wish to know how long it takes for 10% of the reactant to react in this first order
reaction. Stated another way, we want to know how long it will take for the initial pressure of
the reactant, P0, to decrease to 90% of the initial value, 0.9P0. We can write this alternative
problem (for a first order reaction) as:
⎛ 0.9P0 ⎞
–4 –1
ln ⎜
⎟ = ln(0.9) = –0.105 = –kt = –(9.8 × 10 s )t
⎝ P0 ⎠
t = 108 s
Problem 11.60.
(a) Data for the temperature dependence of the rate constant for the acid-catalyzed hydrolysis of
sucrose are:
18
25
32
40
45
temperature, °C
–1 –1
0.0022
0.0054
0.013
0.032
0.055
k, M ·s
T, K
291
298
305
313
318
–1
0.00344 0.00336 0.00328 0.00319 0.00314
1/T, K
lnk
–6.12
–5.22
–4.34
3.44
–2.90
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A ln(k) vs. 1/T plot of the data is shown here and gives an activation energy, Ea = 91.5 kJ·mol–1
[= (8.314 J·mol–1·K–1)(1.10 × 104)]. Substitution into the Arrhenius equation for k at 298 K
gives a frequency factor of 5.9 × 1013 M–1·s–1.
-2
-3
ln(k) = -1.10x10^4*(1/T) + 31.8
Ea = 91.5 kJ/mol
ln(k)
-4
-5
-6
-7
0.0031
0.0032
0.0033
1/T, K^-1
0.0034
0.0035
(b) The rate law for this reaction is probably first order in sucrose concentration and first order
in hydronium ion concentration. We know it is second order overall, because the rate constants
are given with the units of second order rate constants and it makes sense that the reactant and
catalyst should enter the reaction pathway together.
(c) At 37 °C we have:
–
91.5 ×10 3 J ⋅ mol–1
(8.314 J ⋅ K –1mol –1 )(310 K)
= 0.023 M–1·s–1
k = (5.9 × 10 M ·s ) e
In a solution with [H+(aq)] = 1.0 M, k[H+(aq)] = 0.023 s–1. This is a pseudo first order reaction
and rate constant (with [H+(aq)] constant during the reaction, because it is a catalyst and is not
consumed). Thus, we get the half-life of sucrose from the first order equation:
0.693 = (0.023 s–1)t1/2
t1/2 = 30. s
13
–1 –1
Problem 11.61.
The amount of development at each temperature is the same, so the rate of the developer
reaction is proportional to the reciprocal of the amount of time required. Since the only factor
that varies with temperature is the rate constant, the rate is proportional to the rate constant and
we can use the rates as a stand-in for the rate constants to find the activation energy. The data
are:
18
20
21
22
24
temperature, °C
t = development time, min
10
9
8
7
6
–1
0.00344 0.00341 0.00340 0.00339 0.00337
1/T, K
–2.30
–2.20
–2.09
–1.95
–1.79
(constant)ln(1/t) ∝ lnk
A plot of these data is shown on this graph:
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-1.7
-1.8
ln(1/t) = -7700*(1/T) + 24
Ea = 64 kJ·mol–1
ln(1/t)
-1.9
-2.0
-2.1
-2.2
-2.3
-2.4
0.00336
0.00338
0.00340
1/T, K–1
0.00342
0.00344
The slope gives an activation energy of 64 kJ·mol–1 [= (8.314 J·mol–1·K–1)(7.7 × 103)]for the
developer reaction.
Problem 11.62.
(a) Because hydrogen peroxide is packaged in an opaque container and there is a label warning
to keep it out of the light, we can infer that light will probably speed up its decomposition
reaction.
(b) Similarly, since we are warned to keep it away from high temperatures, we can infer that
high temperatures will probably speed up the hydrogen peroxide decomposition reaction.
(c) Although care is required in handling the hydrogen peroxide you obtain at the drugstore (a
3% solution), the actions listed in this problem are just those for which the product is intended.
Hydrogen peroxide is used as an antibacterial agent for minor cuts and scrapes on the skin either
applied by pouring it on the wound or with a cotton swab wet with the solution and for
bleaching hair. Thus, with reasonable care, the solution can be used safely and effectively for
such purposes.
Problem 11.63.
Some molecules, M, after absorbing a photon that boosts them to an excited state, M* (Chapter
4, Section 4.5), emit a photon to return to the ground state. Usually the emission is at a longer
wavelength (lower energy) than the absorbed radiation. Excited molecules may also return to
the ground state by transferring their energy to other molecules, Q:
(i) M + hν1 → M*
rate = (constant)I[M]
(ii) M* → M + hν2
rate = k1[M*]
(iii) M* + Q → M + Q* rate = k2[M*][Q]
In some cases, Q* also loses its energy by emission of light, but in many cases it dissipates the
energy in other ways. If a system containing M and Q molecules is irradiated with a constant
intensity, I, of light with a frequency ν1, a steady state is reached for which the intensity of the
emitted radiation, Ie, at frequency ν2, is a constant.
(a) The light emitted, ν2, is at a lower energy than the light absorbed, ν1. Thus, the frequency of
the emitted light must be lower, so that hν2 < hν1.
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(b) If the intensity of emitted radiation, Ie, is constant at the steady state and Ie is proportional to
[M*], then it must be the case that [M*] is also a constant at the steady state. Indeed, that is
what the steady state means – the concentrations of all species in the system are constant
(although undergoing continuous reaction).
NOTE: The steady state and equilibrium conditions are similar, in that the concentrations of all
species in the system are constant while the reactions continue to occur. The fundamental
difference is that the equilibrium state can be maintained in an isolated system (no exchange of
energy with the surroundings), but the steady state requires a continuous exchange of energy
with the surroundings. If this exchange is stopped, the steady state is disrupted and the system
changes to a new state in which the reactions dependent on the energy exchange no longer
occur.
(c) Since no concentrations are changing at the steady state, this means that the rates of
formation and loss of each species must be equal. For [M*], there is one pathway for formation
and two for loss. If we equate the rate of formation to the sum of the rates for loss, we get:
rate formation = (constant)I[M] = rate loss = k1[M*] + k2[M*][Q]
We can rearrange this equation to give:
[M*]/ M] = (constant)I
[
k1 + k 2[Q]
(d) Since [Q] appears in the denominator of the expression for [M*]/[M], this ratio decreases as
[Q] increases. If all other conditions are held constant, [M*] will decrease as [Q] increases and
the emission will decrease as well. Thus, the emission is “quenched” (reduced) by the presence
of Q, which is the origin of the designation “quencher” for molecules to which M* can transfer
energy.
Problem 11.64.
Spontaneous processes (processes that are thermodynamically favored) can be quite rapid, as in
acid-base reactions, or quite slow, as in the reaction of nitrogen and oxygen gases at room
temperature. The rates of all processes depend upon the pathways for reaction and these depend
upon the interactions of individual molecules, not on the overall thermodynamics of the
reaction. There may, for example, be a substantial activation energy barrier for a highly
exothermic reaction, so the reaction proceeds slowly (or undetectably) at low temperature, even
though net entropy change for such a process is likely to be large and positive.
Problem 11.65.
Hydrogen-oxygen mixtures are highly explosive and dangerous. Yet mixtures of hydrogen and
oxygen can be kept for long periods of time (perhaps millennia) without reacting. At room
temperature, the pathway for the hydrogen-oxygen reaction must involve an extremely slow
rate-limiting step, probably one with a high activation energy barrier. We know that raising the
temperature of such a mixture causes it to explode, that is, react so rapidly that the energy
released feeds the continuing reaction. This is consistent with our hypothesis that the ratelimiting step has a high activation energy. This step is likely to involve breaking strong bonds to
form atoms and/or radicals that then continue the reaction.
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Problem 11.66.
(a) For this problem we put together the information from Problem 11.59 and this one for the
forward and reverse directions of cyclopentadiene dimerization (shown here) and dimer
dissociation (shown in Problem 11.59):
2
The energy vs. progress of reaction diagram looks like this:
(b) The difference between the activation energies in the forward and reverse directions is
76 kJ·mol–1, as noted on the diagram in part (a). This is a reasonable estimate of the standard
enthalpy change for the reaction, so, for the dimerization represented in part (a), we have
∆Ho = –76 kJ·mol–1.
(c) Going from two monomer molecules to the dimer involves no C–H bond changes, as we go
from reactants with four C=C bonds and six C–C bonds (from the two monomer molecules) to
the product with two C=C bonds and ten C–C bonds. The net change is breaking two C=C
bonds and making four C–C bonds. Breaking the C=C bonds requires 1240 kJ·mol–1
(= 2·620 kJ·mol–1) and making the four C–C bonds releases 1388 kJ·mol–1 (= 2·347 kJ·mol–1).
Thus, the bond enthalpies predict ∆Ho = –148 kJ·mol–1 [= (1240 kJ·mol–1) – (1388 kJ·mol–1)].
Bond enthalpies give a result that is about twice as exothermic as the activation energy data
yield. It is possible that the dimer structure puts more strain on the C–C bonds than is taken into
account by average bond enthalpies. This effect would decrease the absolute value of the C–C
bond enthalpy and, hence, decrease the exothermicity of the reaction. More data are necessary
to determine whether this is a reasonable explanation for the difference between ∆Ho
determined by these two methods.
(d) We determine the molar concentration of cyclopentadiene monomer liquid, by dividing the
mass of C5H6 in one liter of liquid, 800 g·L–1 [= (0.8 kg·L–1)(1000 g·kg–1)] by its molar mass,
66 g·mol–1, to get 12 mol·L–1. The initial rate of the second order dimerization is:
rate = k[C5H6] = (1.3 × 10 M ·s )e
rate = 1.0 × 10–4 M·s–1
2
102
6
–1 –1
–
70,000 J⋅ mol –1
(8.314 J ⋅K –1 mol –1 )(298 K)
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Reaction Pathways
(e) To estimate the time required for 10% of the cyclopentadiene to dimerize, let us average the
initial rate and the rate when 90% of the monomer remains. Substituting 10.8 M [= 0.9·(12 M)]
into the above equation gives a rate = 0.8 × 10–4 M·s–1, so the average rate is 0.9 × 10–4 M·s–1.
For this case, ∆[C5H6] = –1.2 M, and we can write:
∆ [C5H6 ]
–1.2 M
rate = –
= 0.9 × 10–4 M·s–1 = –
∆t
∆t
4
∆t = 1.3 × 10 s = 3.7 hr
(f) We have found that 10% of a sample of cyclopentadiene monomer will dimerize in a little
less than four hours. Thus, a chemical supplier would not be able to send the pure liquid to a
chemist, since much of it would have dimerized before it arrived.
(g) In Problem 11.59, we found that it takes only seconds for 10% of cyclopentadiene dimer to
react to form monomer at 200 °C. At the high temperature of the distillation, 170 °C, the dimer
will begin to decompose quickly to the volatile monomer, boiling point 40 °C, which leaves the
distillation vessel and is condensed as the liquid monomer. It takes the monomer hours to
dimerize, so, if it is used soon after it is collected, most of it will still be monomer (especially if
it is kept cold, in order to make the reaction even further). Although thermodynamics favors the
dimer, chemists can take advantage of the kinetics to obtain and use the less stable monomer.
Problem 11.67.
(a) The equilibrium constant for a reaction is the ratio of the reverse to the forward rate at
equilibrium. This statement is incorrect for two reasons. 1) The equilibrium constant for an
elementary reaction is the ratio of rate constants, not the rates for the forward and reverse
reactions. 2) The equilibrium constant is the ratio of the forward to the reverse rate constants.
(b) For a reaction with a very small equilibrium constant, the rate constant for the forward
reaction is much smaller than the rate constant for the reverse reaction. This statement is
correct. The forward-to-reverse rate constant ratio is the equilibrium constant, so a small
equilibrium constant means kf < kr.
(c) For an endothermic reaction at equilibrium, an increase in temperature decreases both the
forward and reverse reaction rates, but they are equal when equilibrium is reattained. This
statement is correct. An increase in temperature increases essentially all reaction rates, but, at
equilibrium, the forward and reverse rates are equal. Remember, however, that the equal
forward and reverse rates at the higher temperature are different yhan the equal forward and
reverse rates at the lower temperature.
(d) For an exothermic reaction at equilibrium, an increase in temperature increases both the
forward and reverse rate constants by the same proportion. This statement is incorrect. It does
not matter whether the reaction is exothermic or endothermic, the reaction rate constants do not
change by the same proportion as the temperature is increased, because their activation energies
are different. (In the case of an exothermic reaction, the activation energy for the forward
reaction is less than the activation energy for the reverse reaction, as in Figure 11.8 or the figure
in the solution to Problem 11.66, but that is not relevant to the answer here.) The ratio of the
forward to the reverse rate constants is:
E a ( forward)
–
RT
⎛ A ⎞ – (E a ( forward)– Ea (reverse))
Ae
kf
RT
= f – E a (reverse)
= ⎜ f⎟e
RT
kr
⎝ Ar ⎠
Ar e
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In general, the forward and reverse activation energies are different, so the exponential term in
this expression varies with temperature. Thus, the rate constant ratio varies with temperature;
the rate constants do not remain in the same proportion as temperature changes. If, in a
particular case, the forward and reverse activation energies are the same, then the exponential
term is unity, e0 = 1, and the rate constant ratio does not change with temperature. In this case
the equilibrium constant (the ratio of the rate constants) does not change with temperature, since
the overall enthalpy change for the reaction is zero.
Problem 11.68.
The reversible elementary reaction, A → B + C, has been studied going in the reverse direction.
The reaction, as written, is endothermic with ∆Ho = 57 kJ·mol–1. When equal concentrations of
B and C, both 0.046 M, were reacted at 25 °C, the initial rate of reaction was 1.12 × 10–5 M·s–1.
When equilibrium was attained, the concentration of C in the solution was 0.027 M. A series of
temperature studies on the rate of the reverse reaction gave an activation energy of 43 kJ·mol–1.
(a) The activation energy diagram for this reaction is:
Start building the diagram with ∆Ho = 57 kJ·mol–1, to get the relative energy levels of the
reactants and products shown on the diagram. The reverse reaction is found to have an
activation energy of 43 kJ·mol–1, which enables us to add the activation energy curve to the
diagram with the reverse activation energy labeled. Finally, in order for the energies to be
consistent, we get the forward activation energy, 100 kJ·mol–1, as shown.
(b) In the experiment described, we begin with [B] = [C] = 0.046 M, and find that [C] =
0.027 M, when equilibrium is reached. From the stoichiometry of the reaction, we also know
that, at equilibrium, [B] = 0.027 M and [A] = 0.019 M (= 0.046 M – 0.027 M). Thus, the
equilibrium constant is:
(B)(C) (0.027)2
K=
=
= 0.038
(0.019)
(A)
(c) We substitute the known values for the initial rate and concentrations in the rate law for the
reverse reaction to get kr:
rate = 1.12 × 10–5 M·s–1 = kr[B][C] = kr(0.046 M)(0.046 M)
kr = 5.3 × 10–3 M–1·s–1
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(d) We can calculate the rate constant for the forward reaction, kf, from the equilibrium constant
and the rate constant for the reverse reaction. However, to get the correct units in the result, we
have to assign units to the equilibrium constant as though we had used molarities (rather than
dimensionless molarity ratios) to calculate K. In this case, the units of K are M, so we get:
kf = Kkr = (0.038 M)(5.3 × 10–3 M–1·s–1) = 2.0 × 10–4 s–1
(e) Since we know the rate constants and the activation energies for the forward and reverse
reactions at 298 K (25 °C), we can use the Arrhenius equation to find the frequency factors:
kf
2.0 × 10 –4 s–1
Af = – Ea ( forward
= (100000 J ⋅mol –1 )
= 6.8 × 1013 s–1
)
–
RT
(8.314 J ⋅ K –1 mol–1 )(298 K)
e
e
kr
5.3 × 10 –3 M –1 ⋅ s–1
Ar = – Ea ( reverse
=
= 1.8 × 105 M–1·s–1
–1
)
– (43000 J⋅ mol )
RT
–1
–1
(8.314 J⋅ K mol )(298 K)
e
e
(f) This is an endothermic reaction, so an energy input is required for the reaction. An increase
in temperature increases the energy of the system and disturbs the equilibrium. Le Chatelier’s
principle predicts that the system will react to decrease the disturbance by using up some of the
added energy, thus also forming more product – the equilibrium constant will increase (see part
(g)). We can also reason that the activation energy for the forward reaction is larger than that for
the reverse reaction and we know that the reaction with the higher activation energy is more
sensitive to temperature changes; it gets larger faster as the temperature increases. Thus, kf/kr =
K gets larger as temperature increases.
(g) We can use the relationship written in the solution to Problem 11.67(d) to find the
equilibrium constant (with units of M) at 313 K (40 °C):
E a ( forward )
–
RT
⎛ Af ⎞ – ( E a ( forward )– Ea ( reverse ))
Af e
kf
RT
K=
=
=
⎜ ⎟e
E a ( reverse )
–
RT
kr
A
⎝
⎠
r
Ar e
⎛ 6.8 × 1013 s–1 ⎞ – [(100000 J ⋅mol
K= ⎜
⎟e
⎝ 1.8 × 10 5 M–1 ⋅ s–1 ⎠
–1
)– (43000 J ⋅ mol –1 )]
(8.314 J ⋅ K –1mol –1 )(313 K)
= 0.12 M
The equilibrium constant is higher (about three-fold higher) at the higher temperature, as we
predicted in part (f). We could also use thermodynamic reasoning to find K at the higher
temperature. The quantitative increase in K for a given temperature increase is given in terms of
∆Horeaction by equation (9.59) in Chapter 9, Section 9.8:
⎞⎛
⎛ K ⎞ ⎛ ∆H o
⎞
reaction ⎟ T2 – T1
⎜
⎟
ln ⎜ 2 ⎟ = ⎜⎜
⎟
⎝ K1 ⎠ ⎝
R ⎠ ⎝ T2 ⋅ T1 ⎠
⎛ K ⎞ ⎛ 57000 J ⋅ mol –1 ⎞ ⎛ 313 K – 298 K ⎞
ln ⎜ 313 ⎟ = ⎜
⎟⎜
⎟ = 1.10
⎝ K 298 ⎠ ⎝ 8.314 J ⋅ K –1 ⋅ mol–1 ⎠ ⎝ (313 K )(298 K) ⎠
⎛ K 313 ⎞ ⎛ K313 ⎞
= 3.01;
⎜
⎟ =
⎝ K 298 ⎠ ⎝ 0.038 ⎠
∴ K313 = 0.11
The result is the same (within the round-off uncertainties of the calculations) by both methods.
If you look back at the derivation of equation (9.59), you will also be able to see how the
frequency factors are related to the entropy change for the reaction system. Since the reaction
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produces two molecules from one, the entropy change should be positive. Perhaps you can
calculate it to see whether this is the case.
Problem 11.69.
(a) A first order reaction that proceeds for a time and then reaches an unchanging concentration
of the reactant has probably reached equilibrium. The reaction is still proceeding, but the
reverse reaction is also going on and the forward and reverse rates are equal.
(b) One experiment to do to test our explanation in part (a) is to follow the concentration of the
reactant as a function of time and then plot ln[reactant] vs. time. As we know, a first order
reaction with no complications would yield a straight-line plot. If the reverse reaction is going
on to reform some reactant as the reaction proceeds, then the plot will not be a straight line
(except, perhaps, for a short time near the start of the reaction when the concentration of
products is so low that the reverse reaction is negligible). Since the reactant concentration will
be larger than it would be in the absence of the reverse reaction, the ln[reactant] vs. time plot
will curve upward and finally reach the constant equilibrium value for ln[reactant].
Problem 11.70.
With phenolphthalein and high concentrations of hydroxide ion, [OH–(aq)], the color of the
phenolphthalein dianion, P2–(aq), disappears from the solution – it all reacts by reaction (11.2)
going in the forward direction: P2–(aq) + OH–(aq) → POH3–(aq). At lower [OH–(aq)], say 0.10
M, observations on this system are different. Initially, the rate of disappearance of the color is
first order, rate = kexpt[P2–(aq)], where the experimental first order rate constant is, kexpt = kf[OH–
(aq)], just as at the higher [OH–(aq)]. However, as the reaction proceeds, kexpt gets smaller (even
though [OH-(aq)] is not changing appreciably) and the solution never completely decolorizes
but comes to an unchanging light pink color. Compare these observations with the behavior
described in Problem 11.69.
(a) The reaction is probably at equilibrium when it reaches the final pink color. Reactant and
product concentrations are unchanging. The rate of the forward reaction, P2–(aq) + OH–(aq) →
POH3–(aq) equals the rate of the reverse reaction, POH3–(aq) → P2–(aq) + OH–(aq). If these are
elementary reactions and the rate constants are kf and kr, respectively, the equilibrium is
characterized by these relationships:
POH 3– (aq ))
(
kf
K=
= 2–
kr
(P (aq))(OH – (aq))
(b) The rate law for the reverse reaction is rate = kr[POH3–(aq)].
(c) The second relationship shown in part (a) can be rearranged to:
kr =
(P
2–
(POH
(aq ))
3–
(aq ))
kf(OH–(aq))
Comparing this equation to kr ≈ (1/10)kf[OH–(aq)] (from the problem statement) shows that, at
equilibrium, with [OH–(aq)] = 0.10 M, the [P2–(aq)]/[POH3–(aq)] ratio is about 1 to 10.
Approximately 10% of the phenolphthalein is present as the red form, P2–(aq), and rest as the
colorless form, POH3–(aq).
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(d) Since kr ≈ (1/10)kf[OH–(aq)], when [OH–(aq)] = 0.10 M, we have kf/kr ≈ 100 M–1. When
[OH–(aq)] = 1.0 M, substitution into the equilibrium constant expression in part (a) gives the
[P2–(aq)]/[POH3–(aq)] ratio as about 1 to 100. Under these conditions, approximately 1% of the
phenolphthalein is present as the red form, P2–(aq), and rest as the colorless form, POH3–(aq), at
equilibrium. With so little of the red form present, the solution appears colorless.
(e) The reaction is a bond-forming process, so we would predict that energy would be released
and the process would be exothermic. An activation energy diagram for the reaction would look
like this:
(f) Since the reaction is exothermic, the equilibrium constant will decrease with increasing
temperature (Le Chatelier’s principle and the larger increase of the higher-activation-energy
reverse reaction rate constant). A smaller equilibrium constant will mean a lower [POH3–
(aq)]/[P2–(aq)] ratio or a higher [P2–(aq)]/[POH3–(aq)] ratio. The solution should get darker as
the temperature increases and more P2–(aq) is formed.
Problem 11.71.
These data are initial rates (106 µmol = 1 mol) of an enzyme-catalyzed reaction measured with
the same amount of enzyme and varying initial concentrations of substrate, S.
[S]0, M
2.0 × 10–2
2.0 × 10–3
2.0 × 10–4
1.5 × 10–4
1.3 × 10–5
V0, µmol·min–1
60
60
48
45
12
(a) Vmax for this reaction is 60 µmol·min–1. The table shows that the reaction reaches this
maximum initial rate at an initial substrate concentration, [S]0, of 2.0 × 10–3 M, and does not
increase at higher substrate concentration, so this must be Vmax.
(b) At [S]0, = 2.0 × 10–3 M, essentially all of the enzyme must be tied up as the enzymesubstrate complex, E-S. Thus, the reaction is going as fast as it can go at this concentration of
enzyme and is limited by the number of enzyme molecules present to form the complex.
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Increasing the substrate concentration cannot produce more of the complex, so the reaction has
reached its maximum rate.
(c) Essentially all the enzyme is tied up as the enzyme-substrate complex, E-S, at substrate
concentrations of 2.0 × 10–3 M or higher, so there is approximately zero concentration of the
free enzyme under these conditions.
(d) We can use the alternative Michaelis-Menten equation (11.77) and rearrange it to get 1/K
and hence K, by substituting known values from the table:
V [S]
V0 = 1 max 0
( K )+ [S]0
1
K =
Vmax [S]0 – V0 [S]0
=
V0
(60 µmol ⋅ min –1)(1.5 × 10 –4 M) – (45 µmol ⋅ min–1 )(1.5 × 10–4 M)
–1
45 µmol ⋅ min
1
–5
∴ K = 2.0 × 104 M–1
K = 5.0 × 10 M;
The definition of K, equation (11.70), shows that it should have units of M–1, which we have
obtained here.
Problem 11.72.
Experiments on an enzyme-catalyzed reaction that followed Michaelis-Menten kinetics showed
that K = 1.2 × 105 M–1 and that, with the same amount of enzyme, the initial rate of the reaction,
45 µmol·min–1 (106 µmol = 1 mol), was the same for substrate concentrations of 0.10 M and
0.010 M.
(a) If K = 1.2 × 105 M–1, then 1/K = 8.3 × 10–6 M for this enzyme. Substrate concentrations,
[S]0, of either 0.10 M or 0.010 M are much larger than 1/K, so the sum (1/K) + [S]0 ≈ [S]0, for
both concentrations. Thus, for the Michaelis-Menten equation (11.73), we can write:
⎛ [E] [S] ⎞
⎛ [E] [S] ⎞
V0 = k ⎜ 1 tot 0 ⎟ ≈ k ⎜ tot 0 ⎟ = k[E]tot
⎝ [S]0 ⎠
⎝ ( K )+ [S]0 ⎠
The amount of enzyme is the same in both experiments, so the initial rate is the same for both
the reactions and V0 = Vmax = 45 µmol·min–1 at these enzyme concentrations.
(b) In part (a) we found that Vmax = 45 µmol·min–1 for the same enzyme concentration as used
in this experiment, 2.0 × 10–5 M. Substitute in equation (11.77) to find V0:
(45 µmol ⋅ min –1 )(2.0 × 10 –5 M)
Vmax [S]0
V0 = 1
=
= 32 µmol·min–1
–6
–5
( K )+ [S]0 (8.3 × 10 M) + (2.0 × 10 M)
Problem 11.73.
Suppose that a mutant enzyme in an organism binds the substrate 10 times more tightly than the
native (normal) enzyme, that is, Kmutant = 10Knative.
(a) Consider the Michaelis-Menten equation in this form:
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V [S]
V0 = 1 max 0
( K )+ [S]0
We know that for [S]0 >> 1/K, high substrate concentration, the initial rate becomes Vmax. If
Kmutant is larger than Knative, the maximum rate will be reached more quickly (because 1/Kmutant <
1/Knative), but it will be the same maximum rate, if nothing else about the enzymatic mechanism
has changed. The limiting rate for both enzymes is the same.
(b) At a low substrate concentration that is not saturating for either enzyme, we approach the
case that 1/K >> [S]0 and the initial velocity is:
V0 = KVmax[S]0
When the reaction is catalyzed by the mutant enzyme with the ten-fold larger K, it will go ten
times faster than if it had been catalyzed by the native enzyme. At low substrate concentrations,
the reaction rate will differ by a factor of about ten favoring the mutant enzyme.
(c) A plot of V0 vs. [S]0 is shown here for a case that Vmax = 100 nM·min–1, 1/Knative = 3000 µM,
and 1/Kmutant = 300 µM:
V0, nM min–1
100
mutant
native
0
0
5000
[S]0, µM
Problem 11.74.
The enzyme urease catalyzes the hydrolysis of urea to ammonium and carbonate ions (or
bicarbonate under acidic conditions), reaction equation (11.80):
⎯⎯→ 2NH4+(aq) + CO32–(aq)
NH2CONH2(aq) + 2H2O ⎯urease
Initial rate data as a function of substrate, NH2CONH2(aq), concentration are:
[NH2CONH2(aq)]0,
M
0.00065
0.00129
0.00327
0.00830
0.0167
0.0333
V0, M·sec–1
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0.226
0.362
0.600
0.846
0.975
1.03
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(a) Assuming that the reaction obeys Michaelis-Menten kinetics, we can estimate Vmax and K
for the reaction from this plot:
1.2
1.0
V0, M s–1
0.8
0.6
0.4
0.2
0.0
0
0.01
0.02
0.03
0.04
[H2NCONH2]0, M
The horizontal red line on the graph is an estimate for Vmax, about 1.1 M·s–1. Equation (11.79)
indicates that 1/K is equal to the substrate concentration when V0 is half of Vmax. The vertical
red line on the graph is shows about where V0 would be 0.55 M·s–1 and gives 1/K (equal the
substrate concentration at this point) of about 0.0025 M = 2.5 × 10–3 M, so K = 4.0 × 102 M–1.
(b) We can invert both sides of the Michaelis-Menten equation as follows:
V [S]
V0 = 1 max 0 ,
( K )+ [S]0
( )
( )
( )
1 + [S]
1
⎡ 1 ⎤⎛ 1 ⎞
1
1
0
[S]0
K
= K
=
+
= ⎢ K ⎥⎜
⎟ +
⎢ Vmax ⎥⎝ [S]0 ⎠ Vmax
Vmax [S]0
Vmax [S]0 Vmax [S]0
V0
⎦
⎣
This double-reciprocal equation can be plotted (Lineweaver-Burk plot) with the data in this
problem to give:
5
1/V0, s M–1
4
3
2
1/V0 = 2.30x10–3(1/[S] ) + 0.920
0
1
0
0
400
800
1200
1600
, M–1
1/[S]0 = 1/[H2NCONH2]0
The intercept of the line, 0.920 s·M–1, is 1/Vmax, so Vmax = 1.087 M·s–1. The slope of the line,
2.30 × 10–3 s, divided by the intercept gives 1/K = 2.5 × 10–3 M, so K = 4.0 × 102 M–1.
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(c) The values for Vmax and 1/K from parts (a) and (b) are the same. Usually, more reliable
values are obtained from the double-reciprocal plot, if the data cover a wide enough range of
appropriate substrate concentrations (lower than the saturation limit).
(d) For V0 = 0.99Vmax, we get:
V [S]
V0 = 0.99Vmax = 1 max 0
( K )+ [S]0
Vmax cancels from both sides and rearrangement gives:
0.99 1 K + 0.99[S]0 = [S]0
[S]0 (for V0 = 0.99Vmax) = 0.99 0.01 1 K = 99(2.5 × 10–3 M) = 0.25 M
Note that the [S]0 required to get V0 equal to X% of Vmax will always be X 1 K for an
enzymatic reaction that follows the simple pathway described in this chapter.
( )
(
)( )
( )
Problem 11.75.
We know that 1 unit = 10 µmol·(15 min)–1 and that the enzyme has an activity of
2750 units·mg–1, so the amount of pyrophosphate our purified enzyme can hydrolyze is:
amount of pyrophosphate = (2750 units·mg–1)[10 µmol·(15 min)–1·unit–1]
amount of pyrophosphate = 1.83 × 103 µmol·mg–1·min–1
In one second, the amount is:
⎛ 1 min ⎞
= 30.5 µmol·mg–1·s–
amount of pyrophosphate = (1.83 × 103 µmol·mg–1·min–1) ⎜
⎟
⎝ 60 s ⎠
1
Problem 11.76.
Enzymatic reactions are affected by many factors, including molecules that bind to the enzyme
at the site(s) where their substrate(s) must bind to undergo reaction. These molecules are often
called “inhibitors,” because their usual effect is to slow the enzymatic reaction. The initial rate
data here are for an enzymatic reaction in the absence and presence of a constant concentration
of an inhibitor.
[S]0, M
1.0 × 10–4
1.5 × 10–4
2.0 × 10–4
5.0 × 10–4
7.5 × 10–4
V0, µmol·min–1
inhibitor absent
V0, µmol·min–1
inhibitor present
28
36
43
63
74
18
24
30
51
63
(a) The series of enzymatic reactions, including reversible binding of an inhibitor, I, at the
active site of the enzyme is:
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E + S ⇔ E-S
rapid equilibrium, K
E + I ⇔ E-I rapid equilibrium, KI
E-S → E + P slow, k
A pictorial representation of these reactions [modeled after equation (11.69)] is:
The effect of the inhibitor is to tie up some of the enzyme, so there is less free enzyme available
to react to form the enzyme-substrate complex. One way to interpret the decrease in reaction
rate is to say that the system acts as though a reduced amount of enzyme is present when the
inhibitor is added. This situation would lead to a lower value of Vmax for the reaction with
inhibitor added. Another way to interpret the rate decrease is to think of the competition
between the inhibitor and the substrate for the enzyme as decreasing the binding ability of the
enzyme for the substrate (a smaller effective K), which would slow the reaction. A quantitative
analysis of the results may help us see which interpretation is correct (or perhaps suggest
another).
(b) The V0 vs. [S]0 plots of the data look like this:
80
70
V0, µmol min–1
60
50
40
30
no inhibitor
20
inhibitor
10
0
0
0.0002
0.0004
[S]0, M
0.0006
0.0008
The plots are a bit difficult to interpret because they do not seem to be close to saturation by the
substrate, so the Vmax for each curve is hard to estimate. One possibility is that both curves go to
the same Vmax, with the inhibitor reaction requiring a higher substrate concentration to reach
saturation. If we estimate Vmax = 90 mol·min–1, then one-half Vmax is shown by the horizontal
line on the graph. The values for 1/K read from the curve are about 2.5 × 10–4 M and 4.0 × 10–
4
M, respectively, for the reactions without inhibitor and with inhibitor. This result supports the
second interpretation suggested in part (a) where the presence of inhibitor results in a decrease
in the effective K (larger 1/K) for the inhibited reaction. If the first interpretation is more
appropriate, the Vmax for the two plots should be different, with that for the inhibited reaction
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lower. It’s hard to tell from these data whether the Vmax is different for the two cases. However,
if they are different and we estimate that Vmax = 90 mol·min–1 for the reaction without
inhibitor (as we did for the previous analysis) and Vmax = 80 mol·min–1 for the reaction with
inhibitor, then the values for 1/K read from the curve are about 2.5 × 10–4 M and 3.5 × 10–4 M,
respectively, for the reactions without inhibitor and with inhibitor. The implication of this result
is that both the Vmax and K values are affected by the presence of the inhibitor. Perhaps a
double-reciprocal plot of the data will help to clarify the results.
(c) The double-reciprocal plots of the data are:
0.06
1/V0 = 4.6x10-6(1/[S]0) + 0.0103
0.05
1/V0, min µmol–1
inhibitor
0.04
no inhibitor
0.03
0.02
0.01
1/V0 = 2.5x10-6(1/[S]0) + 0.0105
0
0
2000
4000
6000
8000
10000
1/[s]0, M–1
Within the uncertainty of these data, we can conclude that the intercept is the same, 0.0104
min·mmol–1, for both plots. Thus, Vmax [= 1/(intercept) = 96 mmol·min–1] is the same in the
absence or presence of inhibitor. This is consistent with the first analysis we made in part (b)
and our estimate for Vmax was reasonable. The slopes divided by the intercepts give 1/K for each
case and we get 2.4 × 10–4 M and 4.4 × 10–4 M, respectively, for the reactions without inhibitor
and with inhibitor. These results are the same as and support the first analysis in part (b) and the
second interpretation in part (a). The effective value of K is decreased in the presence of
inhibitor, because the competition between the substrate and inhibitor for the enzyme active site
effectively lowers the affinity of the enzyme for the substrate. At high enough substrate
concentration, the competition is essentially always won by the substrate and the maximum rate
is the same as in the absence of the inhibitor.
Problem 11.77.
The normal substrate for the enzyme aspartate transcarbamylase is aspartate. Succinate is an
inhibitor (see Problem 11.76) of the normal reaction. The structures of the substrate and
inhibitor are
O
H2
C
O
C
O
O
NH3
C
H
H2
C
O
C
O
aspartate
C
C
C
H2
O
succinate
O
Based on their structures, this inhibition makes sense. The carboxylate, –C(O)O–, groups at the
ends of both the aspartate and succinate molecules (ions) and the similar molecular shapes are
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obvious similarities between the two structures that could cause them both to be able to bind to
the active site of the enzyme. They could be bound in the active site by ionic interactions with
appropriately placed positive groups, –NH3+, from amino acid side chains such as lysine or the
positively charged form of the nitrogen-containing ring in the histidine side chain.
Problem 11.78.
The rate of almost all reactions, including enzyme-catalyzed reactions, increases with increasing
temperature. For enzymatic reactions, however, the rate decreases rapidly beyond a certain
temperature (which is dependent on the enzyme under study). Enzymes are protein molecules
and we found in Chapter 1, Section 1.9, that proteins denature, that is, lose their active threedimensional structures when raised to too high a temperature, as when we boil or fry an egg.
Thus, for every enzyme, there is probably a temperature above which it begins to denature and
lose the structure responsible for its catalytic activity. There is a wide range of temperature
stability among enzymes. Some are denatured at relatively low temperatures, whereas those in
organisms that live at high temperatures, such as bacteria in hot thermal springs where the
temperatures can be above 100 °C, are stable at these elevated temperatures.
Problem 11.79.
(a) In Problem 11.28, you found that the decomposition of ammonia on a hot tungsten wire goes
at the same rate even as NH3 decomposes and its pressure (concentration) decreases. The
reaction rate is independent of the amount of ammonia left unreacted (at least for the data we
have). This decomposition reaction occurs on the surface of the tungsten metal, which has a
limited area and, therefore, limited sites available for ammonia molecules to bind and react. At
the pressures in the experiment for which we have data, it is possible that the ammonia
concentration is high enough to keep all the active sites occupied (as products are formed and
leave the surface another ammonia molecule immediately binds). If this is a saturation effect,
then we should study the reaction at lower pressures to see if we find pressures where the
decomposition does depend on the ammonia pressure. Also, if the reaction rate depends on the
number of sites on the tungsten, increasing the surface area of the tungsten (longer or more
wires, for example) should increase the rate.
(b) Our results from Problem 11.28 show that the rate of reaction is constant, 1.46 × 10–2 kPa·s–
1
, over the range from 26 kPa (the initial pressure) to at least 11.4 kPa (the lowest NH3 pressure
in the experiment). Assume a very simple Michaelis-Menten-type pathway for the ammonia
reaction:
rapid equilibrium, K
NH3(g) + metal surface sites ⇔ NH3(bound to metal surface)
NH3(bound to metal surface) → products slow, k
The analysis of this pathway is identical to that for the Michaelis-Menten enzyme pathway and
leads to linear dependence of initial reaction rates on ammonia pressure (concentration) at low
pressure and saturation and a maximum rate at high pressure. A sketch of an initial rate, V0, vs.
initial pressure of ammonia, P(NH3), is shown on this graph for the system whose results are
given in Problem 11.28, on the assumption that the reaction will not be saturated at ammonia
pressures below about 10 kPa:
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Chapter 11
12
100
80
9
V0, kPa s–1
60
6
40
0
3
20
0
00
2000
5
4000
6000
10
15
PNH3, kPa
8000
20
10000
25
Problem 11.80.
Rewrite the butyl phosphate hydrolysis reaction to show where the labeled oxygen atoms, O*,
are in the reactants and products:
C4H9OP(O)(OH)2 + H2O* → C4H9OH + (HO*)3PO
In order to end up bonded to the phosphorus atom in the product, it is likely that the reaction
pathway involves the initial formation of a bond between the oxygen atom in water and the
phosphorus atom in the ester. This is the reaction of a nucleophile, one of the nonbonding
electron pairs on the negative end of the polar water molecule, with an electrophile, the
relatively positively polarized phosphorus atom bonded to four electronegative oxygen atoms.
Electron and atomic rearrangements after the initial bond formation lead to the final products. A
possible pathway is shown here (with R– representing the C4H9– group):
O
RO P OH
OH
O
O
OH
RO P
OH
H O
H
OH2
RO H
HO P OH
OH
Problem 11.81.
Reaction pathway (i) in Problem 11.37 proposes that dinitrogen pentoxide, O2NONO2, reacts by
decomposing in a rapid reversible reaction to NO3 and NO2, which then can undergo this rate
limiting reaction:
NO3 + NO2 → NO2 + O2 + NO
(a) Lewis structures and formal charges (excluding zero) for the dinitrogen pentoxide, nitrogen
trioxide, and nitrogen dioxide molecules are:
O
dinitrogen pentoxide
O
O
O
O
N O N
N O N
O
O
4 structures
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O
4 structures
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Reaction Pathways
Chapter 11
O
nitrogen trioxide
N O
O
6 structures
O
O
N
nitrogen dioxide
N
O
O
2 structures
2 structures
(b) We see that all three molecules have one or more Lewis structures for which several
equivalent forms can be written (with the electrons on different atoms), so all are stabilized by
electron delocalization. We also observe that the positive formal charges on the structures for
dinitrogen pentoxide accumulate in the region of the bond that will be broken in the first step of
the proposed mechanism. The ends of the molecule repel one another, which could account for
its dissociation to yield molecules that are also stabilized and do not have such repulsions. For
both dinitrogen pentoxide and nitrogen dioxide, we expect the structures on the left to
contribute more to the electron distribution in the molecule, because they have the fewer formal
charges. Also, in the case of dinitrogen pentoxide, the second structure places a positive formal
charge on oxygen, an atom with a high electronegativity, which is unfavorable.
(c) The reaction between nitrogen trioxide and nitrogen dioxide has to produce an oxygen
molecule, in order to satisfy the stoichiometry of the overall reaction. It seems reasonable to
suppose that a bond between two oxygen atoms should exist in the activated complex leading
from reactants to products. A structure that fits the bill is:
O
O N
N O
O
O
2 structures
This structure is somewhat stabilized by electron delocalization, but the reactants that come
together to form it are even more stabilized. We might anticipate a relatively high activation
energy barrier, since this activated complex is less stabilized by electron delocalization than the
reactants. If the activation energy is high, the slow reaction is explained. [Other Lewis
structures are possible for this species, but all of them place positive formal charges on an
oxygen atom (or two oxygen atoms) and an adjacent atom, which is highly unfavorable.]
Problem 11.82.
In Problem 11.35, a pathway was proposed for the net reaction:
(CH3)3CCl + OH– → (CH3)3COH + Cl–.
This pathway is consistent with the rate law you derived, based on the information in Problems
11.15 and 11.25. Experiments with (CH3)3CBr and (CH3)3CI show that they undergo this same
reaction with the same rate law, but at different rates.
(a) The rate-limiting step for the proposed reaction pathway is the dissociation of the starting
compound to yield ions:
(CH3)3CX → (CH3)3C+ + X–
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In this reaction, X represents Cl, Br, or I. The rate of this reaction is likely to depend on the
strength of the C–X bond in the starting compounds. It is reasonable to hypothesize that the
stronger the bond the higher the activation energy for the reaction and the slower the reaction
(b) Table 7.3 in Chapter 6, Section 7.7, gives the C–X bond enthalpies for X = Cl, Br, and I as
338, 276 and 238 kJ·mol–1, respectively. By our reasoning in part (a), we would expect the
chloro compound to react most slowly and the iodo compound to react most rapidly.
(c) We know that the rates of reactions with higher activation energies are more affected by
changes in the temperature at which the reactions are carried out. From our reasoning in parts
(a) and (b), we would expect the reaction rate for the chloro compound to be most affected by
temperature change.
(d) Assuming that these are elementary reactions, the initial rates of formation of the alkene (by
transfer of a proton to a solvent molecule) and alcohol (by reaction with hydroxide ion in the
solvent) from the cation intermediate, (CH3)3C+, are:
∆ [alkene]
rate (alkene) =
= ke[(CH3)3C+][solvent]
∆t
∆ [alcohol]
= ka[(CH3)3C+][OH–]
rate (alcohol) =
∆t
The ratio of these rates of formation after the same amount of time, ∆t, will be the ratio of
product yields:
k e[solvent]
∆ [alkene]
∆[alcohol] =
ka [OH – ]
If the initial solution conditions are the same for the reaction of each of the halo compounds, the
ratio on the right-hand side of this equation is the same in each reaction and, hence, the ratio of
alkene to alcohol will be the same, no matter which halo compound is the reactant. Thus, the
mechanism is consistent with the observation that the alkene-to-alcohol ratio is the same for all
three reactants.
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