times during a reaction and asked to determine whether the reaction is first or second order Plan We can plot lnNO 2 and 1NO 2 against time If one plot or the other ID: 709201
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Solution
Analyze
We are given the concentrations of a reactant at various
times during
a reaction and asked to determine whether the reaction is first or second order.Plan We can plot ln[NO2] and 1/[NO2] against time. If one plot or the other is linear, we will know the reaction is either first or second order.Solve To graph ln[NO2] and 1/[NO2] against time, we first make the following calculations from the data given:
The following data were obtained for the gas-phase
decomposition of
nitrogen dioxide at 300 °C
, Is the reaction first or second order in NO2?
Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate LawSlide6
Continued
As
▶ Figure 14.9
shows, only the plot of
1/[NO
2
] versus time
is linear. Thus,
the reaction obeys a second-order rate law: Rate = k[NO2]2. From the slope of this straight-line graph, we determine that k = 0.543 M –1 s–1
for the disappearanceof NO2.Practice Exercise 1
For a certain reaction A → products, a plot of ln[A] versus time produces a straight
line with a slope of –3.0 × 10–2
s–1.Which of the following statements is or are true?:
(
i
) The
reaction follows
first-order kinetics.
(
ii) The rate constant for the reaction is 3.0 × 10–2 s–1.(iii) The initial concentration of [A] was 1.0 M.(a) Only one of the statements is true.(b) Statements (i) and (ii) are true.(c) Statements (i) and (iii) are true.(d) Statements (ii) and (iii) are true.(e) All three statements are true.
Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate LawSlide7
Continued
Practice
Exercise
2
The decomposition of NO
2
discussed in the Sample Exercise
is second
order in NO2 with k = 0.543 M –1 s–1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the concentration of this reactant after 0.500 h?Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate LawSlide8
14.5 Temperature and RateSlide9
The light gets absorbed by the cold water
more
effectively.
The
cold water reduces the reaction rate. The hot water reduces the reaction rate.Why does the light stick glow with less light in cold water than in hot water?Slide10
Yes
, the reaction will eventually reach equilibrium.
No
, the temperature continues to increase kinetic energy of reactants.
Impossible to tell from the information givenWould you expect this curve to eventually go back down to lower values? Why or why not?Slide11
The Collisions Model
The
collision model
is based on the kinetic molecular theory
. In a chemical reaction, bonds are broken and new bonds are formedMolecules must collide to react.If there are more collisions, more reactions can occur.So, if there are more molecules, the reaction rate is faster.Also, if the temperature is higher, molecules move faster, causing more collisions and a higher rate of reaction.Slide12
Orientation of Molecules
Furthermore, molecules must collide with the correct
orientation
and with
enough energy to cause bond breakage and formationSlide13Slide14
Energy Needed for a Reaction to Take Place (Activation Energy)
The minimum energy needed for a reaction to take place is called
activation energy
, E
a.An energy barrier must be overcome for a reaction to take place, much like the ball must be hit to overcome the barrier in the figure below.Slide15
Yes
No
If the barrier were lower than as shown in the figure, would the golfer have to hit the ball as hard?Slide16
Transition State (Activated Complex)
Reactants gain energy as the reaction proceeds until the particles reach the maximum energy state.
The organization of the atoms at this highest energy state is called the
transition state
(or activated complex).The energy needed to form this state is called the activation energy.Slide17Slide18
Reaction ProgressSlide19Slide20Slide21
Similar in magnitude
Extremely
different in magnitude
About
twice as large as overall change in energyAbout half as large as the overall change in energyHow does the energy needed to overcome the energy barrier compare with the overall change in energy for this reaction? Slide22
The
forward
rate
would
be larger because the activation energy is less.The reverse rate would larger because the activation energy is larger.Suppose you could measure the rates for both the forward and reverse reactions
of the process in Figure 14.17. In which direction would the rate be larger? Why?Slide23
Yes
,
because
B
is an intermediate.Yes, because B can be isolated.No, because B is not stable.No, because B can be isolated and transition states are by definition not stable.
Suppose we have two reactions, A ⟶
B and B ⟶ C. You can isolate B, and it is stable. Is B the transition state for the reaction A
⟶ C?Slide24Slide25Slide26Slide27Slide28Slide29
(1)
(
1) and (2)
(
1), (2), and (3)(1), (2), (3), and (4)What would the curve look like for a temperature higher than that for the red curve in the figure?(1) The new curve would be shifted further to the right.(2) The new curve would have a larger fraction of molecules above Ea.(3) The new curve would heave a lower peak.(4) The new curve would have a larger area underneath.Slide30
Solution
The lower the activation energy, the faster the reaction. The value
of
Δ
E does not affect the rate. Hence, the order from slowest reaction to fastest is 2 < 3 < 1.
Consider a series of reactions having these energy profiles
:
Rank
the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A.Sample Exercise 14.10 Activation Energies and Speeds of ReactionSlide31
Practice
Exercise 1
Which of the following statements is or are true?
(
i) The activation energies for the forward and reverse directions of a reaction can be different. (ii) Assuming that A is constant, if both Ea and T increase, then k will increase
.(iii) For two different reactions, the one with the smaller value
of Ea will necessarily have the larger value for
k.(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.(c) Statements (i) and (iii) are true.(d) Statements (ii) and (iii) are true.(e) All three statements are true.Practice Exercise 2Rank the reverse reactions from slowest to fastest.
Continued
Sample Exercise 14.10
Activation Energies and Speeds of ReactionSlide32Slide33Slide34Slide35Slide36
Solution
Analyze
We are given rate constants,
k
, measured at several temperatures and asked to determine the activation energy, Ea, and the rate constant, k, at a particular temperature.Plan We can obtain Ea from the slope of a graph of ln k versus 1/T. Once we know Ea, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.Solve(a) We must first convert the temperatures from
degrees Celsius to kelvins. We then take the inverse of each temperature, 1/
T, and the natural log of each rate constant, ln k. This gives us the table
shown at the right:
The following table shows the rate constants for the rearrangement of methyl isonitrile at varioustemperatures (these are the data points in Figure 14.14):(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the
rate constant at 430.0 K?Sample Exercise 14.11
Determining the Activation EnergySlide37
A graph of
ln
k
versus 1/T is a straight line (► Figure 14.19).The slope of the line is obtained by choosing any twowell-separated points and using the coordinates of each:
Because logarithms have no units, the numerator in this equation is
dimensionless. The denominator has the units of 1/
T, namely, K–1
. Thus, the overall units for the slope are K. The slope equals –Ea/R. We use the value for the gas constant R in units of J/mol-K (Table 10.2). We thus obtain
We report the activation energy to only two significant figures because
we are limited by the precision with which we
can read the graph in Figure 14.19.
Continued
Sample Exercise 14.11
Determining the Activation EnergySlide38
(b)
To
determine the rate constant,
k
1, at T1 = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ/mol and one of the rate constants and temperatures from the given data, such as k2
= 2.52 × 10–5
s–1 and T2
= 462.9 K:
Thus,Note that the units of k1 are the same as those of k2.Practice Exercise 1Using the data in Sample Exercise 14.11, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C?
(a) 8.1 × 10–15 s–1 (
b) 2.2 × 10–13 s
–1
(c) 2.7 ×
10
–9
s
–1
(d)
2.3
× 10–1 s–1 (e) 9.2 × 103 s–1
Practice Exercise 2To one significant figure, what is the value for the frequency factor
A
for the data presented in Sample Exercise 14.11
.
Continued
Sample Exercise 14.11
Determining the Activation EnergySlide39
Law vs. Theory
Kinetics gives
what
happens. We call the description the
rate law.Why do we observe that rate law? We explain with a theory called a mechanism.A mechanism is a series of stepwise reactions that show how reactants become products.14.6 Reaction MechanismsSlide40
Reaction Mechanisms
Reactions may occur all at once or through several discrete steps.
Each of these processes is known as an
elementary reaction
or elementary process.Slide41Slide42
Molecularity
The
molecularity
of an elementary reaction tells how many molecules are involved in that step of the mechanism.Slide43
Termolecular?
Termolecular
steps require
three
molecules to simultaneously collide with the proper orientation and the proper energy.These are rare, if they indeed do occur.These must be slower than unimolecular or bimolecular steps.Nearly all mechanisms use only unimolecular or bimolecular reactions.Slide44Slide45Slide46Slide47Slide48Slide49Slide50Slide51
Zero
molecularity
Unimolecular
BimolecularTrimolecularWhat is the molecularity of the elementary reaction?NO(g) + Cl2(g) ⟶ NOCl(g) + Cl(g
)Slide52
Solution
Analyze
We are given a two-step mechanism and asked for
(a)
the molecularities of each of the two elementary reactions, (b) the equation for the overall process, and (c) the intermediate.Plan The molecularity of each elementary reaction depends on the number of reactant molecules in the equation for that reaction. The overall equation is the sum of the equations for the elementary reactions. The intermediate is a substance formed in one step of the mechanism and used in another and therefore not part of the equation for the overall reaction.Solve(a) The first elementary reaction involves a single reactant and is consequently unimolecular. The second reaction, which involves two reactant molecules, is bimolecular.
It has been proposed that the conversion of ozone into O
2
proceeds by a two-step mechanism:
O3(g) → O2(g) +
O(g) O3(g) + O(g) →
2 O2(g)(a)
Describe the molecularity of each elementary reaction in this mechanism.(b)
Write the equation for the overall reaction.(c)
Identify
the intermediate(s).
Sample Exercise 14.12
Determining Molecularity and Identifying IntermediatesSlide53
Continued
(b)
Adding the two elementary reactions gives
2 O
3
(
g
)
+ O(g) → 3 O2(g) + O(g) Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net equation for the chemical process: 2 O3(
g) → 3 O2(g)(c) The intermediate is O(g). It is neither an original reactant nor
a final product but is formed in the first step of the mechanism and consumed in the second.
Practice Exercise 1
Consider the following two-step reaction mechanism:
A(
g
)
+
B(
g
) → X(g) + Y(g) X(g) + C(g) → Y(g) + Z(g)
Sample Exercise 14.12 Determining Molecularity and Identifying IntermediatesSlide54
Continued
Which of the following statements about this mechanism is or are true?
(i) Both of the steps in this mechanism are bimolecular.
(ii) The overall reaction is
A(
g
)
+
B(g) + C(g) → Y(g) + Z(g).(iii) The substance X(g) is an intermediate in this mechanism.(a) Only one of the statements is true.(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.(d) Statements (ii) and (iii) are true.(e)
All three statements are true.Practice Exercise 2
For the reaction
Mo(CO)6 + P(CH3
)
3
→
Mo(CO)
5
P(CH3)3 + COthe proposed mechanism is Mo(CO)6 → Mo(CO)5 + CO Mo(CO)5 + P(CH3)3 → Mo(CO)5P(CH3)3(a) Is the proposed mechanism consistent with the equation for the overall reaction?
(b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s).Sample Exercise 14.12
Determining Molecularity and Identifying IntermediatesSlide55
Solution
Analyze
We are given the equation and asked for its rate law,
assuming that
it is an elementary process.Plan Because we are assuming that the reaction occurs as a single elementary reaction, we are able to write the rate law using the coefficients for the reactants in the equation as the reaction orders.Solve The reaction is bimolecular, involving one molecule of H2 and one molecule of Br2. Thus, the rate law is first order in each reactant and second order overall:Rate = k[H2][Br2]Comment Experimental studies of this reaction show that the reaction actually has a very different rate law:Rate =
k[H2][Br2]1/2
Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism cannot occur by a single elementary step. It must, therefore, involve
two or more elementary steps.
If the following reaction occurs in a single elementary reaction, predict its rate law:
H2(g) + Br2(g
) → 2 HBr(g)
Sample Exercise 14.13
Predicting the Rate Law for an Elementary ReactionSlide56
Continued
Practice Exercise 1
Consider the following reaction: 2 A +
B
→
X
+ 2 Y. You
are told
that the first step in the mechanism of this reaction has the following rate law: Rate = k[A][B]. Which of the following could be the first step in the reaction mechanism (note that substance Z is an intermediate)?(a) A + A → Y + Z(b) A → X + Z(c) A + A + B → X + Y + Y(d) B → X + Y(e) A + B → X +
ZPractice Exercise 2Consider the following reaction: 2
NO(g) + Br2(
g) → 2 NOBr(g
). (a) Write the rate law for the reaction, assuming it involves a
single elementary
reaction.
(b)
Is a single-step mechanism likely for
this reaction
?
Sample Exercise 14.13 Predicting the Rate Law for an Elementary ReactionSlide57
All
reactions
are not
elementary
, single step, as in the balanced equation for the reaction.The balanced equation gives no information about the rate constant, which is part of a rate law.The extent of a reaction as shown by a balanced equation does not depend on concentrations.The rate law depends not on the overall balanced reaction, but on the slowest step in the reaction mechanism.
Why can’t the rate law for a reaction generally be deduced from the
balanced equation
for the reaction?Slide58
Solution
Analyze
Given a multistep mechanism with the relative speeds of
the steps
, we are asked to write the overall reaction and the rate law for that overall reaction.Plan (a) Find the overall reaction by adding the elementary steps and eliminating the intermediates. (b) The rate law for the overall reaction will be that of the slow, rate-determining step.Solve(a) Adding the two elementary reactions gives 2 N2O(g) + O(g) → 2 N2(g) + 2 O
2(g) + O(g)Omitting the intermediate, O(
g), which occurs on both sides of the equation, gives the overall reaction:2 N2O(g
) → 2 N2(g) + O2(g)
(b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order:Rate = k[N2O]
The decomposition of nitrous oxide, N
2
O, is believed to occur by a two-step mechanism:
N
2
O(
g
)
→
N2(g) + O(g) (slow) N2O(g) + O(g) → N2(g) + O2(g) (fast)
(a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction.
Sample Exercise 14.14
Determining the Rate Law for a Multistep MechanismSlide59
Continued
Practice Exercise 1
Let’s consider a hypothetical reaction similar to that in
Practice Exercise
1 of Sample Exercise 14.13: 2 C +
D
→
J
+ 2 K. You are told that the rate of this reaction is second order overall and second order in [C]. Could any of the following be a rate-determining first step in a reaction mechanism that is consistent with the observed rate law for the reaction (note that substance Z is an intermediate)?(a) C + C → K + Z (b) C → J + Z (c) C + D → J + Z (d) D → J + K (e) None of these are consistent with the observed rate law.Practice
Exercise 2Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:
O3(g) + 2 NO2
(g) → N2
O5(g)
+
O
2
(g)
The reaction is believed to occur in two steps:
O
3(g) + NO2(g) → NO3(g) + O2(g)
NO3(g) + NO2(g)
→
N
2
O
5
(
g
)
The experimental rate law is rate =
k
[O
3
][NO
2
]. What can
you say about the relative rates of the two steps of the
mechanism?
Sample Exercise 14.14
Determining the Rate Law for a Multistep MechanismSlide60
Gas
reactions
do not
require three reactants.Unimolecular gas reactions are more likely.Odds of three particles simultaneously colliding together and properly oriented are very low.All of the above
Why are
termolecular elementary steps rare in gas-phase reactions?Slide61
Solution
Analyze
We are given a mechanism with a fast initial step and
asked to
write the rate law for the overall reaction.Plan The rate law of the slow elementary step in a mechanism determines the rate law for the overall reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case, the slow step involves the intermediate N2O2 as a reactant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of starting substances. Thus, we must relate the concentration of N2O2 to the concentration of NO by assuming that an equilibrium is established in the first step.Solve
The second step is rate determining, so the overall rate isRate = k2[N
2O2][Br2]We solve for the concentration of the intermediate N
2O2 by assuming that an equilibrium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal:
K1[NO]2 = k–1[N2O2]
Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the
experimentally observed one:
Sample Exercise 14.15
Deriving the Rate Law for a Mechanism with a Fast Initial StepSlide62
Continued
Solving for the concentration of the intermediate, N
2
O
2
, gives
Substituting
this expression into the rate expression gives
Thus, this mechanism also yields a rate law consistent with the experimental one. Remember: There may be more than one mechanism that leads to an observed experimental rate law!Practice Exercise 1Consider the following hypothetical reaction:2 P + Q → 2 R + S. The following mechanism is proposed for this reaction:
Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial StepSlide63
Continued
Substances T and U are unstable intermediates. What rate law is predicted by this mechanism?
(a)
Rate =
k
[P]
2
(b) Rate = k[P][Q] (c) Rate = k[P]2[Q](d) Rate = k[P][Q]2 (e) Rate = k[U]Practice Exercise
2The first step of a mechanism involving the reaction of bromine is
What is the expression relating the concentration of Br(g) to that of Br2(g
)?
Sample Exercise 14.15
Deriving the Rate Law for a Mechanism with a Fast Initial StepSlide64
14.7 Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.Slide65Slide66Slide67Slide68
Top of peak;
no
intermediates
Top of peak; top
of peakTop of peak; topof peakTop of peak; nointermediatesWhere are the intermediates and transition states in this diagram? Red curve Blue curve
top
of first peak; top
ofsecond peakboth in the “valley”between peaks
top of both peaks; nointermediatestop of both peaks; in the“valley” between the peaksSlide69
Types of Catalysts
Homogeneous catalysts
Heterogeneous catalysts
EnzymesSlide70
Homogeneous Catalysts
The reactants and catalyst are in the same phase.
Many times, reactants and catalyst are dissolved in the same solvent, as seen below.Slide71
Br
–
colors the solution.
Br2 colors the solutionH2O2 colors the solution.O2 colors the solution.Why does the solution in the middle cylinder have a brownish color?Slide72
Heterogeneous Catalysts
The catalyst is in a different phase than the reactants.
Often, gases are passed over a solid catalyst.
The adsorption of the reactants is often the rate-determining step.Slide73Slide74
One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.Slide75
A
heterogeneous
catalyst
is easier to remove from a reaction mixture because of phase differences.A heterogeneous catalyst is easier to remove from a reaction mixture because it is present in the greatest quantity.A homogeneous catalyst is easier to remove from a reaction mixture
because of phase differences.A
homogenous catalyst is easier to remove
from a reaction mixture because it
is easier to identify in the reaction mixture.How does a homogeneous catalyst compare with a heterogeneous one
regarding the ease of recovery of the catalyst from the reaction mixture?Slide76
Enzymes
Enzymes
are biological catalysts.
They have a region where the reactants attach. That region
is called the active site. The reactants are referred to as substrates.