14.8

a.


b.




14.9

In general for a reaction aA + bB ® cC + dD
a.


b.




14.10

a.

or
0.055 M/s

b.

The rate at which the
nitrogen concentration is changing must be:
0.027
M/s
The
rate at which nitrogen is reacting is 0.027 M/s.

Think About It:

Will the rate at which ammonia forms always be twice
the rate of reaction of nitrogen, or is this true only at the instant
described in this problem?






14.11

Strategy:

The rate is defined as the change in concentration of
a reactant or product with time.
Each changeinconcentration term is divided by the corresponding
stoichiometric coefficient. Terms
involving reactants are preceded by a minus sign because reactant
concentrations decrease as a reaction progresses—and reaction
rates are always expressed as positive quantities.

Solution:

a.

If the concentration of NO is
changing at the rate of 0.066 M/s, the
rate at which NO_{2} is being formed is
The rate at which NO_{2}
is forming is 0.066 M/s.

b.

The negative sign
indicates that the concentration of molecular oxygen is decreasing as
the reaction progresses. The
rate at which molecular oxygen is reacting is 0.033 M/s.




14.18

Assume the rate law has the form:
rate = k[F_{2}]^{x}[ClO_{2}]^{y}
To determine
the order of the reaction with respect to F_{2}, find two
experiments in which the [ClO_{2}] is held constant. Compare the data from experiments 1 and
3. When the concentration of F_{2}
is doubled, the reaction rate doubles.
Thus, the reaction is firstorder in F_{2}.
To determine
the order with respect to ClO_{2}, compare experiments 1 and
2. When the ClO_{2}
concentration is quadrupled, the reaction rate quadruples. Thus, the reaction is firstorder
in ClO_{2}.
The rate law
is:
rate = k[F_{2}][ClO_{2}]
The value of k can be found using the data from
any of the experiments. If we take
the numbers from the second experiment we have:
Verify that the same value of k can be obtained
from the other sets of data.
Since we now
know the rate law and the value of the rate constant, we can calculate the
rate at any concentration of reactants.
rate = k[F_{2}][ClO_{2}] =
(1.2 M^{1}s^{1})(0.020
M)(0.035 M) = 8.4 ´ 10^{4}
M/s


14.19

rate = k[NH][NO] = (3.0 ´ 10^{4}
/M×s)(0.36
M)(0.075 M) = 8.1 ´ 10^{6}
M/s


14.20

a.

Experiments 2 and 5 show that when we double the
concentration of X at constant concentration of Y, the rate quadruples. Taking the ratio of the rates from these
two experiments
Therefore,
or, x = 2. That is, the reaction is second order
in X. Experiments 2 and 4 indicate
that doubling [Y] at constant [X] doubles the rate. Here we write the ratio as
Therefore,
or, y = 1.
That is, the reaction is first order in Y. Hence, the rate law is given by:
rate =
k[X]^{2}[Y]
The order of the reaction is
(2 + 1) = 3. The reaction is 3rdorder.

b.

The rate constant k can
be calculated using the data from any one of the experiments. Rearranging the rate law and using the
first set of data, we find:
Next, using the known rate
constant and substituting the concentrations of X and Y into the rate
law, we can calculate the initial rate of disappearance of X.
rate
= (10.6 M^{2}s^{1})(0.30
M)^{2}(0.40 M)
= 0.38 M/s



14.21

Strategy:

We are given a set of concentrations and rate data and
asked to determine the order of the reaction and the value of the rate
constant. To determine the order
of the reaction, we need to find the rate law for the reaction. We assume that the rate law takes the
form
rate
= k[A]^{x}[B]^{y}
How do we use the data to determine x and y? Once the orders of the reactants are
known, we can calculate k using the set of rate and concentrations
from any one of the experiments.

Solution:

By comparing the first and second sets of data, we see
that changing [B] does not affect the rate of the reaction. Therefore, the reaction is zero order
in B. By comparing the first and
third sets of data, we see that doubling [A] doubles the rate of the
reaction.
Therefore,
and x
= 1. This shows that the
reaction is first order in A and first order overall.
rate =
k[A]
From the
set of data for the first experiment:
3.20
´
10^{1}
M/s = k(1.50 M)
k =
0.213 s^{1}

Think About It:

What
would be the value of k if you had used the second or third set of
data? Should k be constant?



14.22

a.

For a reaction
firstorder in A,
Rate =
k[A]
1.6
´
10^{2}
M/s = k(0.15 M)
k =
0.11 s^{1}

b.

For a reaction
secondorder in A,
Rate =
k[A]^{2}
1.6
´
10^{2}
M/s = k(0.15 M)^{2}
k =
0.71 /M×s



14.23


14.24

Let P_{0}
be the pressure of ClCO_{2}CCl_{3} at t = 0, and let
x be the decrease in pressure after time t. Note that from the coefficients in the
balanced equation that the loss of 1 atmosphere of ClCO_{2}CCl_{3}
results in the formation of two atmospheres of COCl_{2}. We write:
ClCO_{2}CCl_{3} ® 2COCl_{2}
Time [ClCO_{2}CCl_{3}] [COCl_{2}]
t =
0 P_{0} 0
t = t P_{0}

x 2x
Thus the change (increase) in pressure (DP)
is 2x 
x = x. We have:
0 15.76 0.00 15.76 2.757 0.0635
181 18.88 3.12 12.64 2.537 0.0791
513 22.79 7.03 8.73 2.167 0.115
1164 27.08 11.32 4.44 1.491 0.225
If the
reaction is first order, then a plot of ln vs. t
would be linear. If the reaction is
second order, a plot of 1/ vs. t
would be linear. The two plots are
shown:
From the
graphs we see that the reaction must be firstorder. For a firstorder reaction, the slope is
equal to k. The equation of the line is given on the
graph. The rate constant is: k = 1.08 ´
10^{3}
s^{1}.


14.25

The graph below is a plot of ln P vs. time. Since the plot is linear, the reaction is 1st order.
Slope = k
k
= 1.19 ´
10^{4}
s^{1}


14.30

a.

To calculate the rate
constant, k, from the halflife of a firstorder reaction, we use
Equation 14.5 of the text.
For a firstorder reaction, we only need the halflife to calculate
the rate constant. From Equation
14.5

b.

The relationship between the
concentration of a reactant at different times in a firstorder reaction
is given by Equations 14.3 and 14.4 of the text. We are asked to determine the time
required for 95% of the phosphine to decompose. If we initially have 100% of the
compound and 95% has reacted, then what is left must be (100% 
95%), or 5%. Thus, the ratio of
the percentages will be equal to the ratio of the actual concentrations;
that is, [A]_{t}/[A]_{0} = 5%/100%, or 0.05/1.00.
The time required for 95% of
the phosphine to decompose can be found using Equation 14.3 of the text.



14.31

We know that half of the
substance decomposes in a time equal to the halflife, t_{1/2}. This leaves half of the compound. Half of what is left decomposes in a time
equal to another halflife, so that only one quarter of the original
compound remains. We see that 75% of
the original compound has decomposed after two halflives. Thus two halflives equal one hour, or
the halflife of the decay is 30 min.
100%
starting compound 50%
starting compound 25%
starting compound
Using first order kinetics, we
can solve for k using Equation 14.3 of the text, with [A]_{0}
= 100 and [A] = 25,
Then, substituting k into
Equation 14.5 of the text, you arrive at the same answer for t_{1/2}.


14.32

t =
7.4 s


14.33

a.

Since the
reaction is known to be secondorder, the relationship between reactant
concentration and time is given by Equation 14.6 of the text. The problem supplies the rate constant
and the initial (time = 0) concentration of NOBr. The concentration after 22s can be found
easily.
[NOBr] = 0.034 M
Think About
It:

If the
reaction were first order with the same k and initial
concentration, could you calculate the concentration after 22 s? If the reaction were first order and
you were given the t_{1/2}, could you calculate the
concentration after 22 s?


b.

The halflife for a
secondorder reaction is dependent on the initial
concentration. The halflives can
be calculated using Equation 14.7 of the text.
For an initial
concentration of 0.054 M, you should find . Note
that the halflife of a secondorder reaction is inversely proportional
to the initial reactant concentration.



14.34

a.

At t = 0 s, there are 16 red spheres. At t = 10 s, there are 8 red
spheres. Therefore, the halflife
is
10 s.

b.

Red spheres are shown in grey. Blue spheres are shown in white.



14.35

a.

The relative rates containers i, ii, and iii are 4:3:6.

b.

The relative
rates would be unaffected, each absolute rate would decrease by 50%.

c.

Because halflife of a firstorder reaction does not
depend on reactant concentration, the relative halflives are 1:1:1.



14.36

Strategy:

The halflife
is the time required for the reactant concentration to drop to half its original value.

Setup:

At t = 0 s, there are 12 A molecules (blue
spheres). At t = 10 s,
there are 6 blue spheres.
Therefore, the halflife is 10 s.

Solution:

At t = 20 s half of the 6 remaining A molecules
will be converted to B molecules (yellow spheres). This leaves 3 blue spheres and 9 yellow
spheres at t = 20 s.
Figure (d)
represents the numbers of molecules present after 20 s.



14.37

Strategy:

For a zerothorder reaction, we get a straight line if
we plot reactant concentration [A] versus time. For a firstorder reaction, we get a
straight line if we plot the natural log of reactant concentration (ln [A])
versus time. For a secondorder
reaction, we obtain a straight line when we plot the reciprocal of
reactant concentration (1/[A]) against time.

Solution:

a.

Firstorder
reaction.

b.

Secondorder
reaction.

c.

Zerothorder
reaction.




14.42

One form of
the Arrhenius equation (Equation 14.11 of the text) relates the rate
constant at one temperature to the rate constant at another
temperature. The data are: T_{1} = 250°C
= 523 K, T_{2} = 150°C = 423 K, and k_{1}/k_{2}
= 1.50 ´
10^{3}. Substituting into
Equation 14.11,
E_{a} =
1.3 ´ 10^{5} J/mol =
1.3 ´ 10^{2} kJ/mol


14.43

Graphing Equation 14.10 of the text requires plotting ln
k versus 1/T. The
graph is shown below.
The slope of the line is 1.24 ´
10^{4} K, which is E_{a}/R. The activation energy is:
E_{a} =
slope ´
R = (1.24 ´
10^{4} K) ´ (8.314 J/K×mol)
E_{a} =
1.03 ´ 10^{5} J/mol = 103 kJ/mol
Think About It:

Do you need to know the order of the reaction to find
the activation energy? Is it
possible to have a negative activation energy? What would a potential energy versus
reaction coordinate diagram look like in such a case?



14.44

Use a
modified form of the Arrhenius equation to calculate the temperature at
which the rate constant is
8.80 ´
10^{4}
s^{1}. We carry an extra significant figure
throughout this calculation to minimize rounding errors.
19.43T_{2} =
1.251 ´
10^{4} K
T_{2} =
644 K = 371°C


14.45

Strategy:

Equation 14.8 relates the rate constant to the
frequency factor, the activation energy, and the temperature. Remember to make sure the units of R
and E_{a} are consistent.

Solution:

The appropriate value of R
is 8.314 J/K mol, not 0.0821 L×atm/mol×K. You must also use the activation energy
value of 63,000 J/mol (why?). Once
the temperature has been converted to kelvins, the rate constant is:
k =
3.0 ´ 10^{3} s^{1}

Think About It:

Can you tell from the units of k what the order
of the reaction is?



14.46

Since the ratio of rates is equal to the ratio of rate
constants, we can write:
E_{a} =
5.1 ´ 10^{4} J/mol =
51 kJ/mol


14.47

Let k_{1} be the rate constant at 295 K
and 2k_{1} the rate constant at 305 K. Using Equation 14.11, we write:
E_{a} =
5.18 ´ 10^{4} J/mol = 51.8 kJ/mol


14.48

3.35 ´ 10^{3}
K^{1}
=
T_{2} = 298.34 K or 25.34°C
To the appropriate number of significant figures, this is the same as
the original temperature. It takes a
very small temperature change to increase the reaction rate by 20%.


14.49

Using Equation 14.11,
Remember to convert temperatures to the Kelvin scale.
E_{a} = = 1.3 ´
10^{2} kJ/mol
For maximum freshness, fish should be frozen immediately
after capture and kept frozen until cooked.


14.50

Strategy:

More finely divided solid reactant has more surface
area where molecular collisions can occur. With greater surface area, reactant
molecules can collide more frequently, giving rise to a greater number of
effective collisions—thereby increasing reaction rate.

Setup:

The smaller the particle size, the greater the surface
area.

Solution:

In order of increasing rate of reaction
zinc shot < zinc powder < zinc dust



14.51

Strategy:

Increasing the surface area of a solid reactant
increases the reaction rate. The
more finely divided a solid reactant is, the more surface area is
exposed, and the more collisions that can take place with the aqueous
reactant molecules.
Increasing temperature also increases the reaction
rate. As the temperature
increases, so does the average kinetic energy of a sample of
molecules. As a result, more
molecules in the sample have sufficient kinetic energy to exceed the
activation energy.

Setup:

The smaller the particle size, the greater the surface
area. The crushed tablets will
have a higher surface area and react faster than the tablets broken into
pieces or the whole tablets.
Tablets place into very warm water will react faster
than tablets placed in lukewarm or cold water.

Solution:

In order of
increasing time required for the effervescence to stop:
e < d < b < a < c



14.60


14.61

a.

The order of the reaction is simply the sum of the
exponents in the rate law (Section 14.2 of the text). The reaction is secondorder

b.

The rate law reveals the
identity of the substances participating in the slow or ratedetermining
step of a reaction mechanism. This
rate law implies that the slow step involves the reaction of a molecule
of NO with a molecule of Cl_{2}.
If this is the case, then the first step is the slower
(ratedetermining) step.



14.62

a.

We are given information as to
how the concentrations of X_{2}, Y, and Z affect the rate of the
reaction and are asked to determine the rate law. We assume that the rate law takes the
form
rate
= k[X_{2}]^{x}[Y]^{y}[Z]^{z}
Because the reaction rate doubles when the X_{2}
concentration is doubled, the reaction is firstorder in X. The reaction rate triples when the concentration
of Y is tripled, so the reaction is also firstorder in Y. The concentration of Z has no effect on
the rate, so the reaction is zeroorder in Z.
The
rate law is:
rate =
k[X_{2}][Y]

b.

If a change in the concentration of Z
has no effect on the rate, the concentration of Z is not a term in the
rate law. This implies that Z does
not participate in the ratedetermining step of the reaction mechanism.

c.

The rate law, determined in
part (a), shows that the slow step involves reaction of a molecule of X_{2}
with a molecule of Y. Since Z is
not present in the rate law, it does not take part in the slow step and
must appear in a fast step at a later time. (If the fast step involving Z happened
before the ratedetermining step, the rate law would involve Z in a more
complex way.)
Solution:

A mechanism that is consistent with
the rate law could be:
X_{2} + Y → XY + X (slow)
X + Z
→ XZ (fast)

Think About
It:

The rate
law only tells us about the slow step.
Other mechanisms with different subsequent fast steps are
possible. Try to invent one.




14.63

The
experimentally determined rate law is first order in H_{2} and
second order in NO. In Mechanism I
the slow step is bimolecular and the rate law would be:
rate = k[H_{2}][NO]
Mechanism I can be discarded.
The
ratedetermining step in Mechanism II involves the simultaneous collision
of two NO molecules with one H_{2} molecule. The rate law would be:
rate = k[H_{2}][NO]^{2}
Mechanism II is a possibility.
In Mechanism III we assume the forward and reverse reactions
in the first fast step are in dynamic equilibrium, so their rates are
equal:
k_{f}[NO]^{2} = k_{r}[N_{2}O_{2}]
The slow step
is bimolecular and involves collision of a hydrogen molecule with a
molecule of N_{2}O_{2}.
The rate would be:
rate = k_{2}[H_{2}][N_{2}O_{2}]
If we solve
the dynamic equilibrium equation of the first step for [N_{2}O_{2}]
and substitute into the above equation, we have the rate law:
Mechanism III is also a possibility.
Think About It:

Can you suggest an experiment that might help to
decide between the two mechanisms?



14.64

The first
step involves forward and reverse reactions that are much faster than the
second step. The rates of the
reaction in the first step are given by:
forward rate = k_{1}[O_{3}]
reverse
rate = k_{1}[O][O_{2}]
It is assumed
that these two processes rapidly reach a state of dynamic equilibrium in
which the rates of the forward and reverse reactions are equal:
k_{1}[O_{3}] = k_{1}[O][O_{2}]
If we solve this equality for [O] we have:
The equation for the rate of the second step is:
rate
= k_{2}[O][O_{3}]
If we substitute the expression for [O] derived from the
first step, we have the experimentally verified rate law.
The above rate law predicts that higher
concentrations of O_{2} will decrease the rate. This is because of the reverse reaction
in the first step of the mechanism.
Notice that if more O_{2} molecules are present, they will
serve to scavenge free O atoms and thus slow the disappearance of O_{3}.


14.72

The
ratedetermining step involves the breakdown of ES to E and P. The rate law for this step is:
rate = k_{2}[ES]
In the first
elementary step, the intermediate ES is in equilibrium with E and S. The equilibrium relationship is:
or
Substitute [ES] into the rate law
expression.


14.73

An enzyme is typically a large protein molecule that
contains one or more active sites where interactions with substrates take
place. At higher temperatures the enzyme becomes denatured, that is, it
loses its activity due to a change in the overall structure. The rate of reaction will increase with
increasing temperature until the enzyme denatures. Once this happens, the
reaction rate drops off abruptly.


14.74

In each case the gas
pressure will either increase or decrease.
The pressure can be related to the progress of the reaction through
the balanced equation. In (d), an
electrical conductance measurement could also be used.


14.75

Strictly, the temperature must be specified
whenever the rate or rate constant of a reaction is quoted


14.76

First, calculate the radius of the 10.0 cm^{3}
sphere.
^{}
r =
1.34 cm
The surface area of the sphere is:
area =
4pr^{2} =
4p(1.34
cm)^{2} = 22.6 cm^{2}
Next, calculate the radius of the 1.25 cm^{3}
sphere.
r =
0.668 cm
The surface area of one sphere is:
area =
4pr^{2} =
4p(0.668
cm)^{2} = 5.61 cm^{2}
The
total area of 8 spheres = 5.61 cm^{2} ´ 8
= 44.9 cm^{2}
Obviously, the surface area of the eight spheres
(44.9 cm^{2}) is greater than that of one larger sphere (22.6 cm^{2}). A greater surface area promotes the
catalyzed reaction more effectively.
It can be dangerous to work in grain
elevators, because the large surface area of the grain dust can result in
an extremely rapid combustion reaction (explosion) if the dust is ignited.


14.77

Using the diagrams, we see that the halflife of the
reaction is 20 s. We rearrange
Equation 14.5 to solve for rate constant:
k =
0.035 s^{1}


14.78

The diagrams
indicate that after 15 minutes, the amount of A is cut by half. After another 15 minutes, it is cut by
half again. A constant halflife
indicates that the reaction is first
order. We rearrange Equation
14.5 to solve for rate constant:
k =
0.046 min^{1}


14.79

Most transition metals have several
stable oxidation states. This allows
the metal atoms to act as either a source or a receptor of electrons in a
broad range of reactions.


14.80

The overall rate law is of the general form: rate
= k[H_{2}]^{x}[NO]^{y}
a.

Comparing Experiment #1 and
Experiment #2, we see that the concentration of NO is constant and the
concentration of H_{2} has decreased by onehalf. The initial rate has also decreased by
onehalf. Therefore, the initial
rate is directly proportional to the concentration of H_{2}; x
= 1.
Comparing Experiment #1 and Experiment #3, we see that
the concentration of H_{2} is constant and the concentration of
NO has decreased by onehalf. The initial
rate has decreased by onefourth.
Therefore, the initial rate is proportional to the squared
concentration of NO; y = 2.
The overall rate law
is: rate = k[H_{2}][NO]^{2},
and the order of the reaction is 1 + 2 = 3.

b.

Using Experiment #1 to calculate the rate constant,
rate =
k[H_{2}][NO]^{2}

c.

Consulting the rate law, we
assume that the slow step in the reaction mechanism will probably involve
one H_{2} molecule and two NO molecules. Additionally the hint tells us that O
atoms are an intermediate.
H_{2} + 2NO → N_{2}
+ H_{2}O + O
slow step
O + H_{2} → H_{2}O fast
step
2H_{2} + 2NO → N_{2}
+ 2H_{2}O



14.81

Since
the methanol contains no oxygen18,
the oxygen atom must come from the phosphate group and not the water. The mechanism must involve a bondbreaking process like:


14.82

If water is also the solvent in this
reaction, it is present in vast excess over the other reactants and
products. Throughout the course of
the reaction, the concentration of the water will not change by a
measurable amount. As a result, the
reaction rate will not appear to depend on the concentration of water.


14.83

Temperature, energy of activation,
concentration of reactants, and a catalyst.


14.84

Since the reaction is first order in both A and B, then we can write
the rate law expression:
rate = k[A][B]
Substituting in the values for
the rate, [A], and [B]:
4.1
´
10^{4}
M/s = k(1.6 ´ 10^{2})(2.4
´
10^{3})
k =
11 /M×s
Think About It:

Knowing that the overall reaction was second order, could you have
predicted the units for k?



14.85

a.

To determine the rate law, we must determine the
exponents in the equation
rate =
k[CH_{3}COCH_{3}]^{x}[Br_{2}]^{y}[H^{+}]^{z}
To determine the order of the reaction with respect to
CH_{3}COCH_{3}, find two experiments in which the [Br_{2}]
and [H^{+}] are held constant.
Compare the data from experiments (1) and (5). When the concentration of CH_{3}COCH_{3}
is increased by a factor of 1.33, the reaction rate increases by a factor
of 1.33. Thus, the reaction is firstorder
in CH_{3}COCH_{3}.
To determine the order with respect to Br_{2},
compare experiments (1) and (2).
When the Br_{2} concentration is doubled, the reaction
rate does not change. Thus, the
reaction is zeroorder in Br_{2}.
To determine the order with respect to H^{+},
compare experiments (1) and (3).
When the H^{+} concentration is doubled, the reaction rate
doubles. Thus, the reaction is firstorder
in H^{+}.
The
rate law is:
rate =
k[CH_{3}COCH_{3}][H^{+}]

b.

Rearrange the rate law
from part (a), solving for k.
Substitute the data from any one of the experiments to calculate
k. Using the data from
Experiment (1),
(The units /M×s
can also be expressed as M^{1}s^{1}.)

c.

Let k_{2}
be the rate constant for the slow step:
Let k_{1}
and k_{1} be the rate
constants for the forward and reverse steps in the fast equilibrium.
Therefore,
Equation (1) becomes
which is the
same as (a), where k = k_{1}k_{2}/k_{1}.



14.86

Recall that
the pressure of a gas is directly proportional to the number of moles of
gas. This comes from the ideal gas
equation.
The balanced equation is:
2N_{2}O(g) 2N_{2}(g)
+ O_{2}(g)
From the stoichiometry of the balanced equation, for every one mole
of N_{2}O that decomposes, one mole of N_{2} and 0.5 moles
of O_{2} will be formed.
Let’s assume that we had 2 moles of N_{2}O at t =
0. After one halflife there will
be one mole of N_{2}O remaining and one mole of N_{2} and
0.5 moles of O_{2} will be formed.
The total number of moles of gas after one halflife will be:
At t = 0, there were 2 mol of gas. Now, at, there are 2.5 mol of gas. Since the pressure of a gas is directly
proportional to the number of moles of gas, we can write:


14.87

Fe^{3+} undergoes a
redox cycle: Fe^{3+} ® Fe^{2+} ® Fe^{3+}
Fe^{3+} oxidizes I^{}: 2Fe^{3+} +
2I^{} ® 2Fe^{2+} + I_{2}
Overall Reaction: 2I^{} + S_{2}O ® I_{2} + 2SO
The uncatalyzed reaction is slow
because both I^{}
and S_{2}O are
negatively charged which makes their mutual approach unfavorable.


14.88

The rate expression for a third order reaction is:
The units for the rate law are:
k = M^{2}s^{1}


14.89

For a rate law,
zero order means that the exponent is zero. In other words, the reaction rate is just
equal to a constant; it doesn't change as time passes.
a.

(i): The rate law would be:
rate
= k[A]^{0} =
k
(ii) The integrated zeroorder rate law
is: [A] = kt
+ [A]_{0}. Therefore,
a plot of [A] versus time should be a straight line with a slope equal to
k.

b.

[A] =
[A]_{0}  kt
Substituting into the above equation:

c.

When [A] = 0,
[A]_{0} =
kt
Substituting for k,
This indicates that the
integrated rate law is no longer valid after two
halflives.



14.90

Both
compounds, A and B, decompose by firstorder kinetics. Therefore, we can write a firstorder
rate equation for A and also one for B.
We can calculate each of the rate constants, k_{A}
and k_{B}, from their respective halflives.
The initial concentration of A
and B are equal. [A]_{0} = [B]_{0}. Therefore, from the firstorder rate
equations, we can write:
4 =
ln
4 =
0.0246t
t =
56.4 min


14.91

There are three
gases present and we can measure only the total pressure of the gases. To measure the partial pressure of
azomethane at a particular time, we must withdraw a sample of the mixture,
analyze and determine the mole fractions.
Then,
P_{azomethane} = P_{T}C_{azomethane}
This is a rather tedious process if many measurements are
required. A mass spectrometer will
help (see Section 2.3 of the text).


14.92

a.

Changing the
concentration of a reactant has no effect on k.

b.

If a reaction is run in a solvent other than in the gas phase,
then the reaction mechanism will probably change and will thus change k.

c.

Doubling the
volume simply changes the concentration.
No effect on k, as in (a).

d.

The rate constant k
changes with temperature.

e.

A catalyst changes
the reaction mechanism and therefore changes k.



14.93


14.94

Mathematically, the amount left after ten halflives
is:


14.95

a.

A catalyst works
by changing the reaction mechanism, thus lowering the activation energy.

b.

A catalyst changes
the reaction mechanism.

c.

A catalyst does
not change the enthalpy of reaction.

d.

A catalyst
increases the forward rate of reaction.

e.

A catalyst
increases the reverse rate of reaction.



14.96

The net ionic equation is:
Zn(s) + 2H^{+}(aq)
→ Zn^{2+}(aq) + H_{2}(g)
a.

Changing from the same mass of
granulated zinc to powdered zinc increases the rate because the
surface area of the zinc (and thus its concentration) has increased.

b.

Decreasing the mass of zinc (in
the same granulated form) will decrease the rate because the total
surface area of zinc has decreased.

c.

The concentration of protons
has decreased in changing from the strong acid (hydrochloric) to the weak
acid (acetic); the rate will decrease.

d.

An increase in temperature will increase the rate
constant k; therefore, the rate of reaction increases.



14.97

At very high [H_{2}],
k_{2}[H_{2}] >>
1
At very low [H_{2}],
k_{2}[H_{2}] <<
1
The result from Problem 14.80 agrees with the rate law determined
for low [H_{2}].


14.98

If the
reaction is 35.5% complete, the amount of A remaining is 64.5%. The ratio of [A]_{t}/[A]_{0}
is 64.5%/100% or 0.645/1.00. Using
the firstorder integrated rate law, Equation 14.3 of the text, we have
0.439 = k(4.90
min)
k = 0.0896
min^{1}


14.99

First we plot the data for
the reaction: 2N_{2}O_{5} ® 4NO_{2} + O_{2}
The data is linear, which means that the initial rate is
directly proportional to the concentration of N_{2}O_{5}.
Thus, the rate law is:
Rate = k[N_{2}O_{5}]
The rate
constant k can be determined from the slope of the graph or by using
any set of data.
k =
1.0 ´ 10^{5}
s^{1}
Note that the rate law is not Rate = k[N_{2}O_{5}]^{2},
as we might expect from the balanced equation. In general, the order of a reaction must
be determined by experiment; it cannot be deduced from the coefficients in
the balanced equation.


14.100

The firstorder rate equation can be arranged to take the form of a
straight line.
ln[A] = kt + ln[A]_{ 0}
If a reaction obeys firstorder kinetics, a plot of ln[A] vs. t
will be a straight line with a slope of k.
The slope of a plot of ln[N_{2}O_{5}] vs. t is
6.18
´
10^{4}
min^{1}. Thus,
k
=
6.18 ´
10^{4}
min^{1}
The equation
for the halflife of a firstorder reaction is:


14.101

The red bromine vapor absorbs photons
of blue light and dissociates to form bromine atoms.
Br_{2} ® 2Br×
The bromine atoms collide with methane molecules and abstract
hydrogen atoms.
Br×
+ CH_{4} ® HBr + ×CH_{3}
The methyl radical then
reacts with Br_{2}, giving the observed product and regenerating a
bromine atom to start the process over again:
×CH_{3} + Br_{2
}® CH_{3}Br + Br×
Br×
+ CH_{4} ® HBr + ×CH_{3} and so on...


14.102

a.

In the twostep mechanism the ratedetermining step is the
collision of a hydrogen molecule with two iodine atoms. If visible light increases the
concentration of iodine atoms, then the rate must increase. If the true ratedetermining step were
the collision of a hydrogen molecule with an iodine molecule (the
onestep mechanism), then the visible light would have no effect (it
might even slow the reaction by depleting the number of available iodine
molecules).

b.

To split hydrogen
molecules into atoms, one needs ultraviolet light of much higher energy.



14.103

Lowering the temperature would slow all chemical reactions, which
would be especially important for those that might damage the brain.


14.104

We are told that the reaction is secondorder. Therefore, the rate law is rate = k [P]^{2}.
Given the rate constant and the concentration of the
protein (P), we can calculate the rate:
rate = 6.2
´
10^{3}/M×s
(2.7 ´
10^{4}
M)^{2} = 4.5 ´ 10^{10}
M/s
Using Equation 14.6, we can calculate the amount of time
necessary for the concentration of P to drop to 2.7 ´ 10^{5}
M.
t = 5.4
´
10^{6} s


14.105

a.

We can write the rate law for
an elementary step directly from the stoichiometry of the balanced
reaction. In this ratedetermining
elementary step three molecules must collide simultaneously (one X and
two Y's). This makes the reaction
termolecular, and consequently the rate law must be third order: first order in X and second order in Y.
The
rate law is:
rate =
k[X][Y]^{2}

b.

The value of the rate
constant can be found by solving algebraically for k.
or 0.019
M^{2}s^{1}
Think About
It:

Could you write the rate law if the reaction shown
were the overall balanced equation and not an elemen
ary step?




14.106

a.

O + O_{3 }→
2O_{2}

b.

Cl is a
catalyst; ClO is an intermediate.

c.

The CF
bond is stronger than the CCl
bond.

d.

Ethane will remove the Cl atoms:
Cl + C_{2}H_{6}
→ HCl + C_{2}H_{5}

e.

The overall reaction is: O + O_{3} ® 2O_{2}.
The reaction is exothermic.



14.107

Reaction is secondorder
because a plot of 1/[ClO] vs. time is a straight line. The slope of the line equals the rate
constant, k.
k = Slope
=
2.4 ´
10^{7} /M×s or 2.4 ´
10^{7} M^{1}s^{1}


14.108

We can calculate the ratio of k_{1}/k_{2}
at 40°C
using the Arrhenius equation.
DE_{a} = 5.4
´
10^{3} J/mol
Having calculated
DE_{a},
we can substitute back into the equation to calculate the ratio k_{1}/k_{2}
at 300°C
(573 K).


14.109

During the first five minutes or so the
engine is relatively cold, so the exhaust gases will not fully react with
the components of the catalytic converter.
Remember, for almost all reactions, the rate of reaction increases
with temperature.


14.110

The actual
appearance depends on the relative magnitudes of the rate constants for the
two steps.


14.111

a.

The dependence of the rate constant of a reaction on
temperature can be expressed by the Arrhenius equation (Equation 14.8), . If the rate, k, changes significantly with a small change in temperature, T, then the activation energy must be high.

b.

If a bimolecular reaction occurs every time an A and a
B molecule collide, the molecules
must have the appropriate orientation and the energy that the colliding
particles possess must be greater than the activation energy.



14.112

A plausible
twostep mechanism is:
NO_{2} + NO_{2} →
NO_{3} + NO (slow)
NO_{3} + NO_{2} →
NO_{3} + NO (fast)


14.113

First, solve for the rate constant, k, from the
halflife of the decay.
Now, we can calculate the time
for the plutonium to decay from 5.0 ´ 10^{2} g to
1.0 ´
10^{2} g using the equation for a firstorder reaction relating
concentration and time.
1.61 = (2.84
´
10^{6}
yr^{1})t
t =
5.7 ´ 10^{5} yr


14.114

At high pressure
of PH_{3}, all the sites on W are occupied, so the rate is
independent of [PH_{3}].


14.115

a.

Catalyst: Mn^{2+}; intermediates: Mn^{3+},
Mn^{4+}
First step is
ratedetermining.

b.

Without the catalyst, the reaction would be a termolecular one involving
3 cations! (Tl^{+} and two
Ce^{4+}). The reaction
would be slow.

c.

The catalyst is a
homogeneous catalyst because it has the same phase (aqueous) as the
reactants.



14.116

a.

Since a plot of ln (sucrose) vs. time is linear, the reaction is 1st order.
Slope =
3.68
´
10^{3}
min^{1} =
k
k =
3.68 ´ 10^{3}
min^{1}

b.

t =
814 min

c.

[H_{2}O]
is roughly unchanged. This is a
pseudofirstorder reaction.



14.117

Initially, the number of moles of
gas in terms of the volume is:
We can calculate the
concentration of dimethyl ether from the following equation.
Since, the
volume is held constant, it will cancel out of the equation. The concentration of dimethyl ether after
8.0 minutes (480 s) is:
[(CH_{3})_{2}O]_{t} =
5.06 ´ 10^{3} M
After 8.0 min, the concentration of (CH_{3})_{2}O
has decreased by (5.90 ´ 10^{3}

5.06 ´
10^{3})M
or 8.4 ´
10^{4}
M. Since three moles of
product form for each mole of dimethyl ether that reacts, the
concentrations of the products are (3)(8.4 ´ 10^{4}
M) = 2.5 ´
10^{3}
M.
The pressure of the system after 8.0 minutes is:
P =
[(5.06 ´
10^{3})
+ (2.5 ´
10^{3})]M
´
(0.0821 L×atm/mol×K)(723
K)
P =
0.45 atm


14.118

This is a unit conversion
problem. Recall that 1000 cm^{3} = 1 L.
k = 4.8 × 10^{6} L/mol×s = 4.8 × 10^{6} /M×s


14.119

a.


b.

If , then, from part (a) of this problem:
k_{1}[A] =
k_{2}[B]



14.120

Drinking too much alcohol too
fast means all the alcohol dehydrogenase (ADH) active sites are tied up and
the excess alcohol will damage the central nervous system.


14.121

a.

The firstorder rate constant can be determined
from the halflife.

b.

See Problem 14.94. Mathematically, the amount left after
ten halflives
is:

c.

If 99.0%
has disappeared, then 1.0% remains.
The ratio of [A]_{t}/[A]_{0} is 1.0%/100%
or 0.010/1.00. Substitute into the
firstorder integrated rate law, Equation 14.3 of the text, to determine
the time.
4.61 =
(0.0247
yr^{1})t
t =
187 yr



14.122

1.

Assuming the reactions
have roughly the same frequency factors, the one with the largest
activation energy will be the slowest, and the one with the smallest
activation energy will be the fastest.
The reactions ranked from slowest to fastest are:
(b)
< (c) < (a)

2.

Reaction (a): DH
= 40
kJ/mol
Reaction (b): DH
= 20 kJ/mol
Reaction (c): DH
= 20
kJ/mol
(a) and (c) are exothermic, and (b) is endothermic



14.123

a.

There are
three elementary steps: A ®
B, B ®
C, and C ®
D.

b.

There are
two intermediates: B and C.

c.

The third
step, C ®
D, is rate determining because it has the largest activation energy.

d.

The overall reaction is exothermic.



14.124

The fire should not be doused with water because titanium acts as a
catalyst to decompose steam as
follows:
2H_{2}O(g) ® 2H_{2}(g) + O_{2}(g)
H_{2} gas is flammable and forms an explosive mixture with
O_{2}.


14.125

Let k_{cat}
= k_{uncat}
Then,
Since the
frequency factor is the same, we can write:
Taking the
natural log (ln) of both sides of the equation gives:
or,
Substituting in the given values:
T_{2}
=
1.8 ´ 10^{3} K
This temperature is much too high to be practical.


14.126

First, let's calculate the
number of radium nuclei in 1.0 g.
We can now calculate the rate constant, k, from the
activity and the number of nuclei, and then we can calculate the halflife
from the rate constant.
activity = kN
The halflife is:
Next, let's convert 500 years to seconds. Then we can calculate the number of
nuclei remaining after 500
years.
Use the firstorder
integrated rate law to determine the number of nuclei remaining after 500
years.
N_{t} =
2.2 ´
10^{21} Ra nuclei
Finally, from the number of nuclei remaining after 500 years and
the rate constant, we can calculate the activity.
activity = kN
activity =
(1.4 ´
10^{11}
/s)(2.2 ´
10^{21} nuclei) = 3.1 ´ 10^{10}
nuclear disintegrations/s


14.127

a.

The rate law for the reaction is:
rate =
k[Hb][O_{2}]
We are given the rate
constant and the concentration of Hb and O_{2}, so we can
substitute in these quantities to solve for rate.
rate =
(2.1 ´
10^{6} /M×s)(8.0 ´
10^{6}
M)(1.5 ´
10^{6}
M)
rate =
2.5 ´ 10^{5}
M/s

b.

If HbO_{2} is
being formed at the rate of 2.5 ´
10^{5}
M/s, then O_{2} is being consumed at the same rate,
2.5 ´
10^{5}
M/s. Note the 1:1 mole
ratio between O_{2} and HbO_{2}.

c.

The rate of formation
of HbO_{2} increases, but the concentration of Hb remains the
same. Assuming that temperature is
constant, we can use the same rate constant as in part (a). We substitute rate, [Hb], and the rate
constant into the rate law to solve for O_{2} concentration.
rate =
k[Hb][O_{2}]
1.4 ´
10^{4}
M/s = (2.1 ´
10^{6} /M×s)(8.0 ´
10^{6}
M)[O_{2}]
[O_{2}] =
8.3 ´ 10^{6}
M



14.128

Initially, the rate
increases with increasing pressure (concentration) of NH_{3}. The straightline relationship in the
first half of the plot shows that the rate of reaction is directly
proportional to the concentration of ammonia. Rate = k[NH_{3}]. The more ammonia that is adsorbed on the
tungsten surface, the faster the reaction.
At a certain pressure (concentration), the rate is no longer
dependent on the concentration of ammonia (horizontal portion of
plot). The reaction is now
zeroorder in NH_{3} concentration.
At a certain concentration of NH_{3}, all the reactive sites
on the metal surface are occupied by NH_{3} molecules, and the rate
becomes constant. Increasing the
concentration further has no effect on the rate.


14.129

, where C is a proportionality
constant.
Substituting in for zero,
first, and secondorder reactions gives:
n
= 0,
n
= 1,
n
= 2,
Think About It:

Compare these results with those in Table 14.5 of the
text. What is C in each
case?



14.130

a.

The relationship between
halflife and rate constant is given in Equation 14.5 of the text. Rearranging
to solve for rate
constant gives
k =
0.0350 min^{1}

b.

Following the same
procedure as in part (a), we find the rate constant at 70°C
to be 1.58 ´
10^{3}
min^{1}. We now have two values of rate
constants (k_{1} and k_{2}) at two
temperatures (T_{1} and T_{2}). This information allows us to calculate
the activation energy, E_{a}, using Equation 14.11 of the
text.
E_{a} =
1.1 ´
10^{5} J/mol = 110 kJ/mol

c.

Since all the steps are
elementary steps, we can deduce the rate law simply from the equations
representing the steps. The rate
laws are:
Initiation: rate
= k_{i}[R_{2}]
Propagation: rate =
k_{p}[M][M_{1}]
Termination: rate =
k_{t}[M'][M"]
The reactant molecules are the ethylene monomers, and the product
is polyethylene. Recalling that
intermediates are species that are formed in an early elementary step and
consumed in a later step, we see that they are the radicals M'×,
M"×,
and so on. (The R×
species also qualifies as an intermediate.)

d.

The growth of long polymers would be favored by a high rate of
propagations and a low rate of termination. Since
the rate law of propagation depends on the concentration of monomer, an
increase in the concentration of ethylene would increase the propagation
(growth) rate. From the rate law
for termination we see that a low concentration of the radical fragment
M'×
or M"×
would lead to a slower rate of termination. This can be accomplished by using a low
concentration of the initiator, R_{2}.



14.131

a.

The units of the rate constant show the reaction to be
secondorder, meaning the rate law is most likely:
Rate =
k[H_{2}][I_{2}]
We can use the ideal gas equation to solve for the concentrations
of H_{2 }and I_{2}.
We can then solve for the initial rate in terms of H_{2}
and I_{2} and then convert to the initial rate of formation of
HI. We carry an extra significant
figure throughout this calculation to minimize rounding errors.
Since the total pressure is 1658 mmHg and there are equimolar
amounts of H_{2} and I_{2} in the vessel, the partial
pressure of each gas is 829 mmHg.
Let’s convert the units of the rate constant to /M×min, and then we
can substitute into the rate law to solve for rate.
Rate =
k[H_{2}][I_{2}]
We know that,
or

b.

We can use the secondorder integrated rate
law to calculate the concentration of H_{2} after 10.0
minutes. We can then substitute
this concentration back into the rate law to solve for rate.
[H_{2}]_{t} =
0.01534 M
We can now substitute this concentration back into the rate law
to solve for rate. The
concentration of I_{2} after 10.0 minutes will also equal 0.01534
M.
Rate =
k[H_{2}][I_{2}]
We know that,
or
The concentration
of HI after 10.0 minutes is:
[HI]_{t}
= ([H_{2}]_{0}  [H_{2}]_{t})
× 2
[HI]_{t} =
(0.01974 M  0.01534 M)
× 2 = 8.8 × 10^{}^{3} M



14.132

First, we write an overall balanced equation.
P →
P^{*}
P^{*}
→ P_{2}
^{________________________}
P →
P_{2}
The average molar mass is given by:
(1)
where M is the molar mass of P and [P]_{t}
is the concentration of P at a later time in the reaction. Note that in the numerator [P_{2}]
is multiplied by 2 because the molar mass of P_{2} is double that
of P. Also note that the units work
out to give units of molar mass, g/mol.
Based on the stoichiometry of the reaction,
the concentration of [P_{2}] is:
Substituting back into Equation (1) gives:
(2)
In the proposed mechanism, the denaturation
step is ratedetermining. Thus,
Rate = k[P]
Because we are looking at change in
concentration over time, we need the firstorder integrated rate law,
Equation 14.3 of the text.
[P]_{t} =
[P]_{0}e^{kt}
Substituting into Equation (2) gives:
or
The
rate constant, k, can be determined by plotting versus t. The plot will give a straight line with a
slope of k.


14.133

The halflife
is related to the initial concentration of A by
According to
the data given, the halflife doubled when [A]_{0} was halved. This is only possible if the halflife is
inversely proportional to [A]_{0}.
Substituting n = 2 into the above equation gives:
Looking at
this equation, it is clear that if [A]_{0} is halved, the
halflife would double. The reaction
is secondorder.
We use Equation 14.7 of the text to calculate
the rate constant.


14.134

a.

The halflife of a reaction and the initial concentration
are related by
where C
is a constant. Taking the common
logarithm of both sides of the equation,
Because pressure is proportional to concentration at
constant temperature, the above equation can also be written as
A plot
of vs. logP
gives a slope of (n 
1). The data used for the plot
are:
logP


2.422

2.659

2.114

2.358

1.77

2.009

1.20

1.78

There are clearly two
types of behavior exhibited in the graph.
At pressures above 50 mmHg, the
graph appears to be a
straight line. Fitting these three
points results in a best fit line with an equation
of y = 1.00x + 0.25. The slope of the line is 1.00; therefore,
1.00 = (n

1), or n = 0, and the reaction
is zeroorder. Although the data are limited, it is
clear that there is a change in slope below 50 mmHg, indicating a change
in reaction order. It does appear
that the limiting slope as pressure approaches
zero is itself zero. Thus, 0 = (n

1), or n = 1, and the limiting behavior is that of a firstorder
reaction.

b.

As discovered in part (a), the reaction is firstorder at low pressures and zeroorder at
pressures above 50 mmHg.

c.

The mechanism is actually the same at all pressures
considered. At low pressures, the
fraction of the tungsten surface covered is proportional to the pressure
of NH_{3}, so the rate of decomposition will have a
firstorder dependence on ammonia pressure. At increased pressures, all the
catalytic sites are occupied by NH_{3} molecules, and the rate
becomes independent of the ammonia pressure and hence zeroorder in NH_{3}.



14.135

From Equation
14.11 of the text,
The rate constant at 606 K is 1.6 times greater than that at 600
K. This is a 60% increase in the rate constant for
a 1% increase in temperature! The result shows the profound effect of
an exponential dependence.


14.136

l_{1}
(the absorbance of A) decreases with time.
This would happen for all the mechanisms shown. Note that l_{2} (the absorbance of B)
increases with time and then decreases.
Therefore, B cannot be a product as shown in mechanisms (a) or
(b). If B were a product its
absorbance would increase with time and level off, but it would not
decrease. Since the concentration of
B increases and then after some time begins to decrease, it must mean that
it is produced and then it reacts to produce product as in mechanisms (c)
and (d). In mechanism (c), two
products are C and D, so we would expect to see an increase in absorbance
for two species. Since we see an
increase in absorbance for only one species, then the mechanism that is
consistent with the data must be (d). l_{3} is the absorbance of C.


14.137

Strategy:

Increasing the surface area of a solid reactant
increases the reaction rate. The
more finely divided a s
lid reactant is, the more surface area is exposed, and
the more collisions that can take place with
the aqueous reactant molecules.

Setup:

The form of solid magnesium with the largest surface
area (smallest particle size) will react fastest with the acid.

Solution:

Form (b)



14.138

Strategy:

For a gaseous reaction, the halflife is the time required for the partial pressure of
the reactant to drop to half
its original value.

Setup:

Using either a graphing calculator or an Excel
spreadsheet, we plot the partial pressure of the reactant versus time.
The initial partial pressure at t = 0 is 284 mmHg. The
halflife is the time at which the partial pressure is 142 mmHg.

Solution:

Graphing the data with
Excel gives the following plot:



14.139

Diagram (a).
At a higher temperature, the reaction would proceed at a higher
rate, resulting in the
formation
of more product.

