Unit 6 - Kinetics

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Unit 6 – Chemical KineticsZumdahl Chapter 12, pages 555-595Upon completion of this unit you should be able to:• Identify factors which affect reaction rates• Calculate the rate of production of product or consumption of a reactant using mole

ratios and given rate• Determine the rate law for a reaction from given data, overall order and value of the rate

constant, inclusive of units• Determine the instantaneous rate of reaction• Use integrated rate laws to determine concentrations at a certain time and create graphs

to determine the order of a reaction.• Determine half-life of a reaction• Write the rate law from a given mechanism given th e speeds of each elementary step• Write the overall reaction for a mechanism and identify catalysts and intermediates

present• Determine the activation energy for the reaction using the Arrhenius equation• Graphically determine the activation energy using the Arrhenius equation

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Reaction Rates• The reaction rate of a chemical reaction is defined as the

change in concentration of a reactant or product per unit time.

Rate =

=

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Reaction Rates• Consider the decomposition of nitrogen dioxide:

2NO2(g) → 2NO(g) + O2 (g)• We start with a flask of nitrogen dioxide at 300 oC and measure the

concentrations of nitrogen dioxide, nitric oxide and oxygen as the nitrogen dioxide decomposes.

Concentration (mol/L)

Time (±1 sec) NO2 NO O2

0 0.0100 0 0

50 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

200 0.0048 0.0052 0.0026

250 0.0043 0.0057 0.0029

300 0.0038 0.0062 0.0031

350 0.0034 0.0066 0.0033

400 0.0031 0.0069 0.0035

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Rate Laws

Time (s)50 150100 200 250 300 350

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Reaction Rates• Note from these results that the concentration of the reactant

decreases with time and the concentrations of the products increase with time. Chemical kinetics deals with the speed, or rate, at which these changes occur.• The average rate at which the concentration of NO2 changes

over the first 50 seconds of the reaction is

= =

= -4.2 x 10-5 mol/L s∙

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Reaction Rates• Since the concentration of NO2 decreases with time, any rate

expression involving a reactant will include a negative sign.• It is customary to work with positive reaction rates, so we

define the rate of this particular reaction as Rate = -

• The average rate of this reaction from 0 to 50 seconds then is 4.2 x 10-5 mol/L s∙• The rate is not constant but decreases with time. This is

because the concentration changes with time.• The instantaneous rate can be determined by calculating the

slope of a line tangent to the curve at that point.

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Reaction Rates• The reaction rate can also be defined in terms of products.• When looking at reaction rates for products, stoichiometry must be

considered. For our example, the coefficients for NO2 and NO are both 2 so NO is produced at the same rate that NO2 is consumed. O2 has a coefficient of 1 so it is produced half as fast as NO2 is consumed.• Because the reaction rate changes with time and because the rate is

different depending on which reactant or product is being studied, we must be very specific when we describe a rate for a chemical reaction.

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Rate Laws• Chemical reactions are reversible. So far, we have only considered the

forward reaction. However, the reverse reaction can also occur. As the concentration of products builds, it will reach a point where the reverse reaction can start to occur.• For the reaction studied, the change in concentration of NO2 depends on

the difference in the rates of the forward and reverse reactions.• This complication can be avoided if we study the rate of reaction under

conditions where the reverse reaction makes only a negligible contribution. Typically, this means we study the reaction at a point soon after the reactants are mixed, before the product has had time to build to significant levels.• If we choose conditions where the reverse reaction can be neglected, the

reaction rate will depend only on the concentrations of the reactants.

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Rate Laws• For the decomposition of nitrogen dioxide, we can write• Rate = k[NO2]n

• This expression shows how the rate depends on the concentration of the reactants and is called a rate law. The proportionality constant, k, called the rate constant and n, called the order of the reactant, are determined experimentally.• The order of a reactant can be an integer or a fraction. For our relatively

simple reactions, the orders will be positive integers.• We could define the rate in terms of the reactant or either of the two

products, each of which would give different values of the rate constant.• We must be careful to specify which species we are discussing in a given

case.

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Rate Laws• A rate law that expresses how the rate depends on concentration is

technically called the differential rate law, but is often simply called the rate law.• A rate law that expresses how the concentrations depend on time is called

the integrated rate law.

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Rate Laws• There are two types of rate laws.• The differential rate law shows how the rate of a reaction depends on

concentration.• The integrated rate law shows how the rate of a reaction depends on time.

• Our rate laws will only involve concentrations of reactants. The reverse reaction is unimportant.• Because the differential and integrated rate laws for a given reaction are

related in a well defined way, the experimental determination of either of the rate laws is sufficient.• Experimental convenience usually dictates which type of rate law is

determined experimentally.• Knowing the rate law for a reaction is important mainly because we can

usually infer the individual steps involved in the reaction from the specific form of the rate law.

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Determining the Form of the Rate Law• The first step in understanding how a reaction occurs is to determine the

form of the rate law.• Consider the decomposition reaction of dinitrogen pentoxide in carbon

tetrachloride solution:2N2O5(soln) → 4NO2(soln) + O2 (g)

• Data for this reaction are listed on the next slide.

• In this reaction, the oxygen gas escapes from the solution and thus does not react with the nitrogen dioxide, so we do not have to be concerned about the effects of the reverse reaction.

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Determining the Form of the Rate Law[N2O5] (mol/L) Time (s)

1.00 0

0.88 200

0.78 400

0.69 600

0.61 800

0.54 1000

0.48 1200

0.43 1400

0.38 1600

0.34 1800

0.30 2000

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Determining the Form of the Rate Law• Evaluation of the reaction rates at concentrations of N2O5 of 0.90 M and 0.45 M

yields the following:

• Note that when [N2O5] is halved, the reaction rate is also halved. This means that the rate of this reaction depends on the concentration of N2O5 to the first power. The rate law for this reaction is:

Rate = k [N2O5] 1 = k [N2O5] • The reaction is first order in N2O5. The order is not the same as the coefficient of

N2O5 in the balanced equation. The order must be obtained by observing how the reaction rate depends on the concentration of that reactant.• In general terms, rate = k [A]• Doubling the concentration of A doubles the reaction rate.

[N2O5] Rate (mol/L s)∙

0.90 M 5.4 x 10 -4

0.45 M 2.7 x 10-4

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Determining the Form of the Rate LawMethod of Initial Rates

• The initial rate of a reaction is the instantaneous rate determined just after the reaction begins (just after t=0).• Several experiments are carried out using different initial concentrations

and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the form of the rate law to be determined.

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• Consider the reaction NH4

+ (aq)+ NO2- (aq) → N2(g) + 2H2O(l)

• With initial rates shown below

• The general form of the rate law isRate = k[NH4

+]n[NO2-]m

Experiment Initial Concentration of NH4+ Initial Concentration of NO2

- Initial Rate (Mol/L s)∙

1 0.100 M 0.0050 M 1.35 x 10-7

2 0.100 M 0.010 M 2.70 x 10 -7

3 0.200 M 0.010 M 5.40 x 10 -7

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• For experiment 1:Rate = 1.35 x 10 -7 mol/L s = k(0.100 mol/L)∙ n (0.0050 mol/L)m

• For experiment 2:Rate = 2.70 x 10 -7 mol/L s = k(0.100 mol/L)∙ n (0.010 mol/L)m

The ratio of these rates is: =

= 2.00 = (2.0) which means m is 1 and the rate law for this reaction is first order in NO2

-

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A similar analysis yields =

= 2.00 = (2.0) n which means m is 1 and the rate law for this reaction is first order in NH4

+

The rate law is Rate = k[NH4+] [NO2

-]The rate law is first order in both NO2

- and NH4+. It is merely a coincidence

that the coefficients of NO2- and NH4

+ are the same.The overall rate order is n + m = 2. The reaction is second order overall.

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The value of the rate constant k can be calculated using the results of any of the three experiments. Rate = k[NH4

+][NO2-]

Experiment 11.35 x 10 -7 mol/L s = k (0.100 mol/L)(0.0050 mol/L)∙k= 2.7 x 10 -4 L/mol s∙



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Determining the Form of the Rate LawSample Exercise 12.1 – page 566

• Consider the reaction BrO3

- (aq) + 5Br- (aq) + 6 H+ (aq) → 3Br2(l) + 3H2O(l)• Using the data from the table below, determine the orders for all three

reactants, the overall reaction order and the value of the rate constant.Experiment Initial Concentration

of BrO3- (Mol/L)

Initial Concentration of Br- (Mol/L)

Initial Concentration of H+ (Mol/L)

Initial Rate (Mol/L s)∙

1 0.10 0.10 0.10 8.0 x 10 -4

2 0.20 0.10 0.10 1.6 x 10 -3

3 0.20 0.20 0.10 3.2 x 10 -3

4 0.10 0.10 0.20 3.2 x 10 -3

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First Order Rate Laws• The differential rate laws consider rate as a function of reactant

concentrations. It is also useful to express the reactant concentrations as a function of time.• For the reaction

2N2O5(soln) → 4NO2(soln) + O2 (g)• We found that the rate law is

Rate = k [N2O5] = • It is a first order reaction, which means if the concentration of N2O5 is

doubled, the reaction rate is doubled. Integration of this expression yields the form

ln[N2O5] = -kt + ln[N2O5]0

• Note that this equation expresses reactant concentration as a function of time

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Integrated Rate Law – First Order• Consider the reaction aA → products• Where the kinetics are first order in [A], the rate law is

rate = =k[A]

• And the integrated first order rate law isln[A] = -kt + ln[A]0

• The equation shows how the concentration of A depends on time. If the initial concentration of A and the rate constant are known, the concentration of A at any time can be calculated.• The equation is of the form y=mx+b, so if a reaction is first order, a plot of

ln[a] versus time will be linear. If this plot is not a straight line, the reaction will not be first order.• The equation can be rewritten as ln( ) = kt.

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Integrated Rate Law – First OrderSample exercise 12.2 – page 569

The decomposition of N2O5 in the gas phase was studied at constant temperature.

2N2O5(g) → 4NO2(g) + O2 (g)The following results were collected:

Using these data, verify that the rate law is first order in [N2O5], and calculate the value of the rate constant, where the rate = -∆[N2O5]/t.

[N2O5] (mol/L) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

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Integrated Rate Law – First Order• The time required for a reaction to reach half its original concentration is

called the half-life of a reactant and is designated by the symbol t1/2.• The half life for the decomposition reaction in Sample Exercise 12.2 is 100

seconds. It always takes 100 seconds for [N2O5] to be halved.• In general terms of aA → products, for a first order reaction

ln( ) = kt and when t = t1/2, [A] =

For t = t1/2 ln( ) = kt1/2

Substituting for ln(2) and solving for t1/2 gives t1/2 = • This is the general equation for the half-life of a first order reaction. For a

first order reaction, half-life does not depend on concentration.

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Integrated Rate Law – Second Order• Consider the reaction aA → products• Where the kinetics are second order in [A], the rate law is

rate = =k[A]2

• And the integrated second order rate law is = kt +

• A plot of versus time will be linear with a slope equal to k.

• The expression for the half-life of a second order reaction is t1/2 =

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Integrated Rate Law – Zero Order• Consider the reaction aA → products• Where the kinetics are zero order in [A], the rate law is

rate = =k[A]0

• And the integrated zero order rate law is[A] = -kt +[A]0

• A plot of k versus time will be linear with a slope equal to -k.

• The expression for the half-life of a zero order reaction is t1/2 =

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Integrated Rate Law• Consider the reaction

BrO3- (aq) + 5Br- (aq) + 6 H+ (aq) → 3Br2(l) + 3H2O(l)

where [BrO3-]0 = 1.0 x 10 -3 M, [Br-]0 = 1.0 M and [H+]0 = 1.0 M.

As the reaction proceeds, [BrO3-] decreases significantly, but the Br- and H+ ion

concentrations are so large initially, relative little of these two reactants is consumed. Thus Br- and H+ remain relatively constant. We can assume [Br-]0 = [Br-] and [H+ ]0 = [H+ ]. This means the rate law can be written

Rate = k [Br-]0 [H+ ]0 2 [BrO3-] = k’ [BrO3

-].Where k’ = k [Br-]0 [H+ ]0 2

• Since this law was obtained by simplifying an more complicated arrangement, it is called a pseudo-first-order rate law.

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Reaction Mechanisms• Most chemical reactions occur by series of steps called the reaction

mechanism.• Consider the reaction

NO2(g) + CO(g) → NO(g) + CO2 (g)

• The rate law for this reaction is known from experiment to be

Rate = k [NO2]2

• This reaction is more complicated than it appears from the balanced equation. This is quite typical; the balanced equation fro a reaction tells us the reactants, the products and the stoichiometry but gives no direct information about the reaction mechanism.

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Reaction Mechanisms• For the reaction between nitrogen dioxide and carbon monoxide, the

mechanism is thought to involve the following steps:

NO2(g) + NO2(g) → NO3(g) + NO (g)

NO3(g) + CO(g) → NO2(g) + CO2 (g)

• where k1 and k2 are rate constants of the individual reactions. In this mechanism, gaseous NO3 is an intermediate, a species that is neither a reactant or a product but is formed and consumed during the reaction sequence.

k2

k1

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Reaction Mechanisms• Each of these two reactions is called an elementary step, a reaction whose

rate law can be written from its molecularity. • Molecularity is the number of species that must collide to produce the

reaction indicated by that step.• A reaction involving one molecule is called a unimolecular step. Bimolecular

and termolecular reactions involve the collisions of two and three species, respectively.• A reaction mechanism is a series of elementary steps that must

1. Give the overall balanced equation for the reaction2. Agree with the experimentally determined rate law

• Multistep reactions often have one step that is much slower than all the others. Reactants can becomes products only as fast as they can get through the slowest step. This slowest step is called the rate determining step.

k2

k1

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Reaction Mechanisms• For the reaction between nitrogen dioxide and carbon monoxide, the sum

of the two steps results in the overall balanced equation, so the first requirement for a correct mechanism is met.

NO2(g) + NO2(g) → NO3(g) + NO (g)

NO3(g) + CO(g) → NO2(g) + CO2 (g)

NO2(g) + CO(g) → NO(g) + CO2 (g)

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Reaction Mechanisms• Assume that the first step is rate determining and the second step is relatively fast.

NO2(g) + NO2(g) → NO3(g) + NO (g) (slow)NO3(g) + CO(g) → NO2(g) + CO2 (g) (fast)

• This assumes that the formation of NO3 occurs much more slowly than is reaction with CO. The rate of CO2 production is then controlled by the rate of formation of NO3 in the first step. Since this is an elementary step, we can write the rate law from the molecularity.

Rate of formation of NO3= k [NO2]2

• Since the overall rate can be no faster than the slowest step,Overall rate = k [NO2]2

• This agrees with the experimentally determined rate law given earlier.• The mechanism we assumed satisfies both requirements and may be the correct

mechanism.• A mechanism can never be proved absolutely.

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Reaction MechanismsSample Exercise 12.6 – page 581

• The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is

NO2(g) + F2 (g) → NO2F (g)• The experimentally determined rate law is

rate = k [NO2] [F2]• A suggested mechanism for this reaction is

NO2(g) + F2 (g) → NO2F (g) + F (g) (slow)F (g) + NO2(g) → NO2F (g) (fast)

• Is this an acceptable mechanism?

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A Model for Chemical Kinetics• The collision model is built around the central idea that molecules must

collide to react.• The kinetic molecular theory of gases predicts that an increase in

temperature raises molecular velocities and increases the frequency of collisions between molecules. This idea agrees with the observation that reaction rates are greater at higher temperatures.• So there is a qualitative agreement between the collision model and

experimental observations.• However, it is found that the rate of reaction is much smaller than the

calculated collision frequency in a collection of gas particles. That means that only a small fraction of the collisions produces a reaction.• Arrhenius proposed that there is a threshold energy, called the activation

energy, that must be overcome to produce a chemical reaction.

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A Model for Chemical Kinetics• Consider the reaction

2BrNO(g) → 2NO (g) + Br2 (g)• In this reaction two Br-N bonds must be broken and one Br-Br bond must

be formed. Breaking a Br-N bond requires considerable energy (243 kJ/mol). • The collision model postulates that the energy comes from the kinetic

energy possessed by the reacting molecules before the collision. This kinetic energy is changed into potential energy as the molecules are distorted during a collision to break bonds and rearrange into product molecules.• The arrangement of atoms found at the top of the potential energy barrier

is called the activated complex, or transition state.• The conversion of BrNO to NO and Br2 is exothermic as indicated by the

fact that the products have lower potential energy than the reactant.

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A Model for Chemical Kinetics

Pote

ntial

Ene

rgy

Reaction progress

(reactant)

(transition state)

2NO + Br2 } ∆E

Ea

ON---BrON---Br

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A Model for Chemical Kinetics• A minimum energy is required for two BrNO molecules to react so that

products can form.• This energy comes from the energy of the collision.• A collision between two BrNO molecules with small kinetic energies will not

have enough energy to get over the barrier. • At a given temperature only a certain fraction of the collisions possess

enough energy to result in product formation.

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A Model for Chemical Kinetics• The fraction of effective collisions increase exponentially with temperature.

Arrhenisu postulated that the number of collisions having an energy greater than or equal to the activation energy is given by the expression

Number of collisions with the activation energy = (total number of collisions) e ∙ –Ea/RT

where Ea is the activation energy, R is the universal gas constant and T is the Kelvin temperature.

• Taking into account some other factors, this can written in the formK = A e –Ea/RT

which is called the Arrhenius equation.

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Catalysis• A catalyst is a substance that speeds up a reaction without being

consumed itself.• A catalysts works by providing a new pathway for the reaction with lower

activation energy.

• A homogeneous catalyst is present in the same phase as the reacting molecules. A heterogeneous catalyst exists in a different phase.

Uncatalyzed pathway

Catalyzed pathway

Reactants

Products

∆E}

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KineticsPractice – pages 599-605Problems 21, 23, 25, 27, 31, 35, 39, 41, 43, 45, 47, 49. 51, 53 and 71

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Unit 6 - Kinetics