chapter 14 chemical kinetics © 2012 pearson education, inc

Chapter 14

Chemical Kinetics

© 2012 Pearson Education, Inc.

ChemicalKinetics

Kinetics

• In kinetics we study the rate at which a chemical process occurs.

• Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).


ChemicalKinetics

Factors That Affect Reaction Rates• Physical state of the reactants.

– In order to react, molecules must come in contact with each other.

– The more homogeneous the mixture of reactants, the faster the molecules can react.

• Concentration of reactants.

– As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

• Temperature

– At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

• Presence of a catalyst.

– Catalysts speed up reactions by changing the mechanism of the reaction.

– Catalysts are not consumed during the course of the reaction.


ChemicalKinetics

A. The effect of increasing the partial pressures of the reactive components of a gaseous mixture depends on which side of the chemical equation has the most gas molecules.

B. Increasing the partial pressures of the reactive components of a gaseous mixture has no effect on the rate of reaction if each reactant pressure is increased by the same amount.

C Increasing the partial pressures of the reactive components of a gaseous mixture increases the rate of reaction.

D. Increasing the partial pressures of the reactive components of a gaseous mixture decreases the rate of reaction.

ChemicalKinetics

Reaction Rates

Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.


ChemicalKinetics

A. Mixture of A and B moleculesB. A molecules onlyC. B molecules onlyD. AB molecules

ChemicalKinetics

Sample Exercise 14.1 Calculating an Average Rate of Reaction

Practice ExerciseUse the data in Figure 14.3 to calculate the average rate of appearance of B over the time interval from 0 s to 40 s.

From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s.

ChemicalKinetics

Reaction Rates

In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.

The average rate of the reaction over each interval is the change in concentration divided by the change in time:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)


Average rate =[C4H9Cl]

t

ChemicalKinetics

Reaction Rates

• Note that the average rate decreases as the reaction proceeds.

• This is because as the reaction goes forward, there are fewer collisions between reactant molecules.



ChemicalKinetics

Reaction Rates

• A plot of [C4H9Cl] versus time for this reaction yields a curve like this.

• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

• All reactions slow down over time.

• Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction.



ChemicalKinetics

A. No changeB. IncreasesC. Decreases

ChemicalKinetics

Reaction Rates

• In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.

• Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.


Rate =[C4H9Cl]

t=

[C4H9OH]t


ChemicalKinetics

Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction

Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate).

Practice ExerciseUsing Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s.

ChemicalKinetics

A. An instantaneous rate is measured at a particular point in time of a reaction, whereas average rate is measured over a time period; Yes.

B. An instantaneous rate is measured at a particular point in time of a reaction, whereas average rate is measured over a time period; No.

C. An instantaneous rate is measured over a short time period, whereas average rate is measured over a large time period; Yes.

D. An instantaneous rate is measured over a short time period, whereas average rate is measured over a large time period; No.

ChemicalKinetics

Reaction Rates and Stoichiometry• What if the ratio is not 1:1?

2 HI(g) H2(g) + I2(g)

•In such a case,

Rate = 12

[HI]t =

[I2]t


• To generalize, then, for the reaction

aA + bB cC + dD

Rate = 1a

[A]t =

1b

[B]t =

1c

[C]t

1d

[D]t=

ChemicalKinetics

Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear

(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O3(g) 3 O2(g)? (b) If the rate at which O2 appears, [O2]/t, is 6.0 10–5 M/s at a

particular instant, at what rate is O3 disappearing at this same time, [O3]/t?

Practice ExerciseIf the rate of decomposition of N2O5 in the reaction 2 N2O5(g) 4 NO2(g) + O2(g) at a particular instant is 4.2 107 M/s, what is the rate of appearance of (a) NO2 and (b) O2 at that instant?

ChemicalKinetics

Concentration and RateOne can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.


NH4+(aq) + NO2

(aq) N2(g) + 2 H2O(l)

If we compare Experiments 1 and 2, we see that when [NH4+] doubles,

the initial rate doubles.

Likewise, when we compare Experiments 5 and 6, we see that when [NO2

] doubles, the initial rate doubles.

ChemicalKinetics

Concentration and Rate• This means

Rate [NH4+]

Rate [NO2]

Rate [NH4+] [NO2

]which, when written as an equation, becomes

Rate = k [NH4+] [NO2

]• This equation is called the rate law, and k is

the rate constant.

Therefore,


ChemicalKinetics

Rate Laws• A rate law shows the relationship between the

reaction rate and the concentrations of reactants.• The exponents tell the order of the reaction with

respect to each reactant.• Since the rate law is

Rate = k[NH4+] [NO2

]

the reaction is

First-order in [NH4+]

and

First-order in [NO2]


ChemicalKinetics

Rate Laws

Rate = k[NH4+] [NO2

]

• The overall reaction order can be found by adding the exponents on the reactants in the rate law.

• This reaction is second-order overall.


ChemicalKinetics

(1) Reaction rate is experimentally measured as a reaction proceeds.(2) Rate constant is calculated from a rate law.(3) Rate law relates reaction rate to rate constant and concentrations.(4) Rate law describes the mechanism of a reaction.(5) Reaction rate depends on the stoichiometry of the reaction.

A. (1), (2), (4)B. (2), (4), (4)C. (2), (3), (5)D. (1), (2), (3)

ChemicalKinetics

A. Yes, the rate is directly proportional to the rate constant in a rate law.B. Yes, only if the order of each concentration is one in the rate law.C. No, as the order of each concentration changes in a rate law, the units of

the rate constant change, but those of the rate are always concentration/time.

D. No, as the order of each concentration changes in a rate law, there are changes in the units of both.

ChemicalKinetics

A. 2nd order in NO, 1st order in H2, 2nd order overall

B. 2nd order in NO, 1st order in H2, 3rd order overall

C. 1st order in NO, 1st order in H2, 2nd order overall

D. 1st order in NO, 2nd order in H2, 3rd order overall

ChemicalKinetics

A. YesB. No

ChemicalKinetics

Consider a reaction A + B C for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction.

Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate

Practice ExerciseAssuming that rate = k[A][B], rank the mixtures represented in this Sample Exercise in order of increasing rate.

ChemicalKinetics

A. The rate of the reaction is very slow.B. There is no reaction.C. The rate constant can not be calculated.D. All of the above

ChemicalKinetics

Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants

(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law in Equation 14.9?

Practice Exercise(a) What is the reaction order of the reactant H2 in Equation 14.11?(b) What are the units of the rate constant for Equation 14.11?

ChemicalKinetics

Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data

The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows:

Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

ChemicalKinetics

Practice ExerciseThe following data were measured for the reaction of nitric oxide with hydrogen:2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)

(a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M.

Continued

Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data

ChemicalKinetics

Integrated Rate LawsUsing calculus to integrate the rate law for a first-order process gives us

ln[A]t

[A]0= kt

where

[A]0 is the initial concentration of A, and

[A]t is the concentration of A at some time, t, during the course of the reaction.


Manipulating this equation produces

ln[A]t

[A]0= kt

ln [A]t ln [A]0 = kt

which is in the form

ln [A]t = kt + ln [A]0

y = mx + b

ChemicalKinetics

First-Order Processes

Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be k.

ln [A]t = kt + ln [A]0


ChemicalKinetics

First-Order ProcessesConsider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN


The data was collected for this reaction at 198.9 C.

ChemicalKinetics

A. The reaction is zero order in CH3NC.B. The reaction is first order in CH3NC.C. The reaction is second order in CH3NC.D. No conclusion can be made without further information about the

specific rate law.

ChemicalKinetics

First-Order Processes

• When ln P is plotted as a function of time, a straight line results.

• Therefore,– The process is first-order.– k is the negative of the slope: 5.1 105 s1.


ChemicalKinetics

The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 107 g/cm3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 10–7 g/cm3?

Sample Exercise 14.7 Using the Integrated First-Order Rate Law

Practice ExerciseThe decomposition of dimethyl ether, (CH3)2O, at 510 ºC is a first-order process with a rate constant of 6.8 10–4 s–1:

(CH3)2O(g) CH4(g) + H2(g) + CO(g)

If the initial pressure of (CH3)2O is 135 torr, what is its pressure after 1420 s?

ChemicalKinetics

Second-Order ProcessesSimilarly, integrating the rate law for a process that is second-order in reactant A, we get

1[A]t

= kt +1

[A]0

also in the form

y = mx + b


So if a process is second-order in A, a plot of vs. t yields a straight line, and the slope of that line is k.

1[A]

ChemicalKinetics

Second-Order ProcessesThe decomposition of NO2 at 300 °C is described by the equation

NO2(g) NO(g) + O2(g)

and yields data comparable to this table:

12


• Plotting ln [NO2] vs. t yields the graph at the right.

• The plot is not a straight line, so the process is not first-order in [A].

Time (s) [NO2], M ln [NO2]

0.0 0.01000 4.610

50.0 0.00787 4.845

100.0 0.00649 5.038

200.0 0.00481 5.337

300.0 0.00380 5.573

ChemicalKinetics

Second-Order Processes• Graphing ln

vs. t, however, gives this plot Fig. 14.9(b).

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• Because this is a straight line, the process is second-order in [A].

1[NO2]


ChemicalKinetics

Zero-Order Processes

Therefore, if a reaction is zero-order, a plot of [A] vs. t will yield a straight line, and the slope of the line will be k.

[A]t = kt + [A]0


ChemicalKinetics

A. At all timesB. When t is close to zeroC.Half-way through the reactionD.When the concentration of [A] is high

ChemicalKinetics

Half-Life

• Half-life is defined as the time required for one-half of a reactant to react.

• Because [A] at t1/2 is one-half of the original [A],

[A]t = 0.5 [A]0.


ChemicalKinetics

A. 6.00 gB. 4.50 gC. 1.25 gD. 1.11 g

ChemicalKinetics

Half-Life

For a first-order process, this becomes

0.5 [A]0

[A]0

ln = kt1/2

ln 0.5 = kt1/2

0.693 = kt1/2

= t1/2

0.693k

Note: For a first-order process, then, the half-life does not depend on [A]0.


ChemicalKinetics

The reaction of C4H9Cl with water is a first-order reaction. (a) Use Figure 14.4 to estimate the half-life for this reaction. (b) Use the half-life from (a) to calculate the rate constant.

Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction

Practice Exercise(a) Using Equation 14.15, calculate t1/2 for the decomposition of the insecticide described in Sample Exercise 14.7. (b) How long does it take for the concentration of the insecticide to reach one-quarter of the initial value?

ChemicalKinetics

Half-Life

For a second-order process, 1

0.5 [A]0

= kt1/2 + 1

[A]0

2[A]0

= kt1/2 + 1

[A]0

2 1[A]0

= kt1/2

1[A]0

=

= t1/2

1k[A]0


ChemicalKinetics

A. The half-life does not change as the reaction proceeds.B. The half-life increases as the reaction proceeds. C. The half-life decreases as the reaction proceeds.D. The half-life increases for some reactions and decreases for others

(2nd order reactions) as the reaction proceeds.

ChemicalKinetics

Temperature and Rate

• Generally, as temperature increases, so does the reaction rate.

• This is because k is temperature-dependent.


ChemicalKinetics

The Collision Model• In a chemical reaction, bonds are broken and new bonds are formed.

• Molecules can only react if they collide with each other.

• Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.


ChemicalKinetics

Activation Energy• In other words, there is a minimum amount of energy

required for reaction: the activation energy, Ea.

• Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation-energy barrier.


ChemicalKinetics

Reaction Coordinate Diagrams

• It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

• The diagram shows the energy of the reactants and products (and, therefore, E).

• The high point on the diagram is the transition state.

• The species present at the transition state is called the activated complex.

• The energy gap between the reactants and the activated complex is the activation-energy barrier.


ChemicalKinetics

A. Similar in magnitudeB. Extremely different in magnitudeC.About twice as large as overall change in energyD.About half as large as the overall change in energy

ChemicalKinetics

Maxwell–Boltzmann Distributions• Temperature is defined as a

measure of the average kinetic energy of the molecules in a sample.

• At any temperature there is a wide distribution of kinetic energies.

• As the temperature increases, the curve flattens and broadens.

• Thus, at higher temperatures, a larger population of molecules has higher energy.


• If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier.

• As a result, the reaction rate increases.

ChemicalKinetics

A. Yes, because B is an intermediate.B. Yes, because B can be isolated.C. No, because B is not stable.D. No, because B can be isolated and transition

states are by definition not stable.

ChemicalKinetics

Maxwell–Boltzmann DistributionsThis fraction of molecules can be found through the expression

where R is the gas constant and T is the Kelvin temperature.

f = e−Ea/RT


ChemicalKinetics

A. If colliding atoms are of dissimilar sizes, then products are not favored to form.

B. If collisions do not occur with sufficient energy or proper orientation of colliding molecules, then products may not form.

C. If reactants are gas molecules, the particles are too far apart to collide over a reasonable time period and thus do not form products.

D. If reactants are liquid molecules, the particles are moving too slowly to provide sufficient energy to form products.

ChemicalKinetics

Arrhenius EquationSvante Arrhenius developed a mathematical relationship between k and Ea:

k = Ae

where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

−Ea/RT


Taking the natural logarithm of both sides, the equation becomes

ln k = ( ) + ln A

Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. .

Ea

R

1T

y = mx + b

1T

ChemicalKinetics

Sample Exercise 14.10 Relating Energy Profiles to Activation Energies and Speeds of Reaction

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.

Practice ExerciseRank the reverse reactions from slowest to fastest.

Answer: 2 < 1 < 3 because, if you approach the barrier from the right, the Ea values are 40 kJ/mol for reverse reaction 2, 25 kJ/mol for reverse reaction 1, and 15 kJ/mol for reverse reaction 3.

ChemicalKinetics

The following table shows the rate constants for the rearrangement of methylisonitrile at various temperatures (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 Energy

Practice ExerciseUsing the data in Sample Exercise 14.11, calculate the rate constant for the rearrangement of methyl isonitrile at 280 ºC.

ChemicalKinetics

Reaction MechanismsThe sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

– Reactions may occur all at once or through several discrete steps.

– Each of these processes is known as an elementary reaction or elementary process.

– The molecularity of a process tells how many molecules are involved in the process.


ChemicalKinetics

A. Zero molecularityB. UnimolecularC. BimolecularD. Trimolecular

ChemicalKinetics

Red curve Blue curve

A. Top of peak; no intermediates top of first peak; top of second peakB. Top of peak; top of peak both in the “valley” between peaksC. Top of peak; top of peak top of both peaks; no intermediatesD. Top of peak; no intermediates top of both peaks; in the “valley”

between the peaks

ChemicalKinetics

It has been proposed that the conversion of ozone into O2 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 Intermediates

Practice ExerciseFor the reaction Mo(CO)6 + P(CH3)3 Mo(CO)5 P(CH3)3 + CO

the proposed mechanism is

Mo(CO)6 Mo(CO)5 + CO

Mo(CO)5 + P(CH3)3 Mo(CO)5 P(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).

ChemicalKinetics

Multistep Mechanisms• In a multistep process, one of the steps will

be slower than all others.• The overall reaction cannot occur faster than

this slowest, rate-determining step.


ChemicalKinetics

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 Reaction

Practice ExerciseConsider 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?

ChemicalKinetics

A. All reactions are not elementary, single step, as in the balanced equation for the reaction.

B. The balanced equation gives no information about the rate constant, which is part of a rate law.

C. The extent of a reaction as shown by a balanced equation does not depend on concentrations.

D. The rate law depends not on the overall balanced reaction, but on the slowest step in the reaction mechanism.

ChemicalKinetics

Slow Initial Step

• The rate law for this reaction is found experimentally to be

Rate = k[NO2]2

• CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.

• This suggests that the reaction occurs in two steps.

NO2(g) + CO(g) NO(g) + CO2(g)


ChemicalKinetics

Slow Initial Step• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second step.

• As CO is not involved in the slow, rate-determining

step, it does not appear in the rate law.


ChemicalKinetics

Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism

The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:

N2O(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.

Practice ExerciseOzone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:

O3(g) + 2 NO2(g) N2O5(g) + O2(g)

The reaction is believed to occur in two steps:

O3(g) + NO2(g) NO3(g) + O2(g)

NO3(g) + NO2(g) N2O5(g)

The experimental rate law is rate = k[O3][NO2].What can you say about the relative rates of the two steps of the mechanism?

ChemicalKinetics

Fast Initial Step

• The rate law for this reaction is found to be

Rate = k[NO]2 [Br2]

• Because termolecular processes are rare, this rate law suggests a two-step mechanism.

2 NO(g) + Br2(g) 2 NOBr(g)


ChemicalKinetics

Fast Initial Step

• A proposed mechanism is

Step 2: NOBr2 + NO 2 NOBr (slow)

Step 1 includes the forward and reverse reactions.

Step 1: NO + Br2 NOBr2 (fast)


ChemicalKinetics

Fast Initial Step

• The rate of the overall reaction depends upon the rate of the slow step.

• The rate law for that step would be

Rate = k2[NOBr2] [NO]

• But how can we find [NOBr2]?


ChemicalKinetics

Fast Initial Step

• NOBr2 can react two ways:

– With NO to form NOBr.

– By decomposition to reform NO and Br2.

• The reactants and products of the first step are in equilibrium with each other.

• Therefore,

Ratef = Rater


ChemicalKinetics

Fast Initial Step

• Because Ratef = Rater,

k1[NO] [Br2] = k1[NOBr2]

• Solving for [NOBr2], gives usk1

k1

[NO] [Br2] = [NOBr2]


ChemicalKinetics

Fast Initial Step

Substituting this expression for [NOBr2] in the rate law for the rate-determining step, gives

k2k1

k1Rate = [NO] [Br2] [NO]

= k[NO]2 [Br2]


ChemicalKinetics

A. Gas reactions do not require three reactants.B. Unimolecular gas reactions are more likely.C. Odds of three particles simultaneously colliding

together and properly oriented are very low.D. All of the above

ChemicalKinetics

Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step

Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one:

Practice ExerciseThe 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)?

ChemicalKinetics

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.


ChemicalKinetics

A. By lowering the H for the reactionB. By lowering the activation energy through a new mechanism for the

reaction C. By lowering the activation energy for the reverse processD. By lowering the relative potential energies of the products and

reactants

ChemicalKinetics

Catalysts

One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.


ChemicalKinetics

A. A heterogeneous catalyst is easier to remove from a reaction mixture because of phase differences.

B. A heterogeneous catalyst is easier to remove from a reaction mixture because it is present in the greatest quantity.

C. A homogeneous catalyst is easier to remove from a reaction mixture because of phase differences.

D. A homogenous catalyst is easier to remove from a reaction mixture because it is easier to identify in the reaction mixture.

ChemicalKinetics

Enzymes• Enzymes are

catalysts in biological systems.

• The substrate fits into the active site of the enzyme much like a key fits into a lock.


ChemicalKinetics

A. Substrate B. Products

ChemicalKinetics

A. Yes, because an enzyme is a catalyst.B. No, because enzymes are different in structure and their properties

than typical catalysts in non-biochemical reactions.C. Possibly, but the nature of the transition states in the catalyzed and

uncatalyzed reaction can be significantly different. D. Yes, because it should form a stable intermediate, transition state.

chapter 14 chemical kinetics © 2012 pearson education, inc

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