AP Chem Topic#9Kinetics Notes

Objectives

► Identify the factors which affect reaction rates.► Calculate the rate of production of a product or consumption of a reactant using mole rations and the given

rate.► Determine the rate law for a reaction given data, overall order, and value of the rate constant, inclusive of

units.► Determine the instantaneous rate of a reaction.► Use integrated rate laws to determine concentrations at a certain time, t, and create graphs to determine the

order of reaction. Also, determine the half-life of a reaction.► Write the rate law from a given mechanism given the 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.

AP Tips

● Questions on kinetics may appear in the multiple-choice section. In the free-response portion, kinetics often appears in questions with other topics such as electrochemistry, thermodynamics, gases, stoichiometry, or lab-based questions.

Table of Contents

1. Reaction Rates 2. Rate Laws: An Introduction 3. Determining the Form of the Rate Law 4. The Integrated Rate Law 5. Rate Laws: A Summary 6. Reaction Mechanisms 7. A Model for Chemical Kinetics 8. Catalysis

Reaction Rates Section#1

● Spontaneityo Characteristic of a reaction regarding an inherent tendency for the process to occur

Spontaneous does not mean fast. Strong inherent tendency for hydrogen to combine with oxygen from a thermodynamic

standpoint 2H2(g) + O2(g) → 2H2O(l)● But, the two gases can coexist indefinitely at 25°C

● It is not enough to understand the thermodynamics and stoichiometry; we must also understand the factors governing the rate of the reaction.

o Called chemical kinetics● Introduction to rate of reaction

o Consider 2NO2(g) → 2NO(g) + O2(g) Flask full of NO2(g) and heated to 300°C

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Concentrations of Reactant and Products as a Function of Time for the Reaction: 2NO2(g) → 2NO(g) + O2(g) (@ 300°C)

Concentration (mol/L)

Time (±1s) NO2 NO O2

0 0.0100 0 050 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018150 0.0055 0.0045 0.0023200 0.0048 0.0052 0.0026250 0.0043 0.0057 0.0029300 0.0038 0.0062 0.0031350 0.0034 0.0066 0.0033400 0.0031 0.0069 0.0035

o Rate = conc A @ time t2 - conc A @ time t1 = Δ[A] = mol/L t2 – t1 Δt s

A is the reactant/product being considered - Δ means a decrease in concentration of A

● Being used up +Δ means an increase in concentration of A

● Being created Calculate the average rate [NO2]

● Δ[NO 2] = [NO2]t = 50 – [NO2]t = 0 = 0.0079 mol/L – 0.0100 mol/L Δt 50.s – 0s 50.s

● = - 4.2 x 10-5 mol/L-so [NO2] decreases with timeo Normally work with positive reaction rates

Rate = - Δ[NO2] = - (- 4.2 x 10-5 mol/L-s) = 4.2 x 10-5 mol/L-s Δt

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Average Rate (in mol/L-s) of Decomposition of Nitrogen Dioxide as a Function of Time

- Δ[NO2]Δt

Time Period (s)

4.2x10-5 0 → 502.8x10-5 50 → 1002.0x10-5 100 → 1501.4x10-5 150 → 2001.0x10-5 200 → 250

● Note that the rate is not a constant but decreases with timeo Above are calculated average rateso Instantaneous rate

Rate at a particular time – slope of line tangent to the curve at time, t

Slope of tangent line for t = 100s is Δ y = Δ[NO2] Δx Δt

Rate = -(slope of tangent line) Rate = - Δ[NO2] = (0.0076-0.0050)mol/L = -0.0026mol/L =

Δt (160-50)s 110s = 2.4x10-5 mol/L-s

Use stoichiometry to calculate the rates for the other reactants/products● - Δ[NO2] = Δ[NO] = (2)(Δ[O2] )

Δt Δt Δt● Rate of consumption of NO2 = rate of production of NO = 2(rate of

production of O2)

Rate Laws Section#2

● Chemical reactions are reversible● We study a reaction at a point where products do not interfere with reaction rate

o For the reaction 2NO2(g) → 2NO(g) + O2(g)o The rate will only depend on the reactant, NO2

Rate law = k[NO2]n ● k is the rate law constant● n is called the order of the reactant● Both must be determined by experiment

(1) The concentration of the reactants do not appear in rate law(2) The value of n must be determined experimentally, not from a balanced

equation● Types of rate laws

o Differential rate law Rate depends on concentration

o Integrated rate law Concentrations depend on time

● Rate Law Summaries(1) Two types of rate laws

(a) Deferential - Simply called the rate law, rate depends on concentration(b) Integrated - Concentration depends on time

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(2) Consider reaction where the reverse is unimportant, rate law only uses the concentrations of the reactants.

(3) Differential and Integrated rate laws are related in a well defined way, the experimental determination of either of the rate laws is sufficient

(4) Experimental convenience usually dictates which type of rate law is determined experimentally(5) 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

Determining the Form of the Rate Law Section#3

● Method of Initial Rateso Instantaneous rate determined just after the reaction begins.

● Using experimental data, we can find the order for each reactant and thereby the order of reaction and the overall rate law

o Order of reaction is equal to the sum of the order of reactants● Take the reaction, NH4

+(aq) + NO2-(aq) → N2(g) + 2H2O(l)

o Several trials of this experiment were run and the following data was generated.Initial Rates from Three Experiments for the Reaction: NH4

+(aq) + NO2-(aq) → N2(g) + 2H2O(l)

ExperimentInitial

Concentration of NH4

+

Initial Concentration of

NO2-

Initial Rate (mol/L•s)

1 0.100 M 0.0050 M 1.35x10-7

2 0.100 M 0.010 M 2.70x10-7

3 0.200 M 0.010 M 5.40x10-7

o Initial rates - Δ[NH4

+ ] = - Δ[NO2] = Δ[N2] = 2(Δ[H2O] )

Δt Δt Δt Δto Rate Law

Rate = k [NH4+]n[NO2

-]m Determination of n and m

● Exp#1: Rate = 1.35x10-7 = k(0.100 M)n(0.0050 M )m ● Exp#2: Rate = 2.70x10-7 = k(0.100 M)n(0.010 M )m ● Ratio of these rates = Rate 2 = 2.70x10 -7 = 2 = k (0.100 M ) n (0.010 M ) m

Rate 1 1.35x10-7 k(0.100 M)n(0.0050M )m

o = (0.010 M ) m = (2.0)m (0.0050M )m

o Thus Rate 2 = 2.00 = (2.0)m Rate 1

m = 1● order for reactant NO2

- is 1 (m = 1)● Exp#2: Rate = 2.70x10-7 = k(0.100 M)n(0.010 M )m ● Exp#3: Rate = 5.40x10-7 = k(0.200 M)n(0.010 M )m ● Ratio of these rates = Rate 3 = 5.40x10 -7 = 2 = k (0.200 M ) n (0.010 M ) m

Rate 2 2.70x10-7 k(0.100M)n(0.010M )m

o = (0.200 M ) n = (2.0)n (0.100M )n

o Thus Rate 3 = 2.00 = (2.0)n Rate 2

n = 1

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● order for reactant NH4+ is 1 (n = 1)

Substitute in for m and n ● Rate = k [NH4

+][NO2-]

o Overall reaction order = m + n = 1 + 1 = 2 Calculate k

● Substitute experimental values for any of the experimentso Initial Rate from Exp#1 = 1.35x10-7 o Using [NH4

+] = 0.100 M from Exp#1o Using [NO2

-] = 0.0050 M from Exp#1 All values MUST be from the same experiment

● Rate = k [NH4+][NO2

-] ● 1.35x10-7 = k(0.100 M)(0.0050M)

o k = 2.7x10-4 L-mol-1-s-1

o Sample Exercise 12.1 – Determining a Rate LawThe reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation BrO3

-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2(l) + 3H2O(l). The table below gives the results for four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant.

The Results from Four Experiments to Study the Reaction BrO3-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2 (l) + 3H2O(l)

ExperimentInitial Concentration

of BrO3- (mol/L)

Initial Concentration of Br- (mol/L)

Initial Concentration of H+ (mol/L)

Measured Initial Rate (mol/L •s)

1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 1.6x10-3

3 0.20 0.20 0.10 3.2x10-3

4 0.10 0.10 0.20 3.2x10-3

Solution: [BrO3-] doubles, rate doubles, 2n = 2, n = 1

[Br-] doubles, rate doubles, 2m = 2, m=1, [H+] doubles, rate quadruples, 2p = 4, p =2Rate = k[BrO3

-][Br-][H+]2, Overall reaction order is 4 k = Rate/[BrO3

-][Br-][H+]2 = 8.0x10-4 / (0.10 x 0.10 x 0.102) = 8.0 L3-mol-3 / sAnswer: n = 1, m = 1, p = 2, overall = 4, and k = 8.0 L3 •mol-3-s

The Integrated Rate Law Section#4

Zero Order Graphs

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1 st Order Graphs

2 nd Order Graphs

● Expressing reactants concentrations as a function of time● Reaction s involving a single reactant

o aA → products Rate = Δ[A] = k[A]n

Δt● First-Order Rate Laws

o For the reaction, 2N2O5(solution) → 4NO(solution) + O2(g)o Rate = k[N2O5]

Experiments show this to have a reaction order of 1● First-order reaction

Integrating the rate law● ln[N2O5] = -kt + ln[N2O5]0

o [N2O5]0 is the initial concentration of N2O5 when it t = 0o Integrated first-order rate law

ln[A] = -kt + ln[A]0 expresses the concentration of A as a function of time

o Note(1) Equation shows [A] depends on time. If [A]0 and k are known then [A]t can be

calculated(2) ln[A] = -kt + ln[A]0 in line form (y = mx + b)

● y vs. x with slope m and intercept b

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● graph ln[A] vs. time gives a straight line with a slope = -k● Used to test whether the reaction is first order

o Reaction is first order when plot of ln[A] vs. t is a straight line(3) Can be expressed in ration of [A] to [A]0 , ln([A]0/[A]) = kt

o Sample Exercise 12.2 – First-Order Rate Laws 1The decomposition of N2O5 in the gas phase was studied at constant temperature. 2N2O5(g) → 4NO(g) + O2(g). The following results were collected:

[N2O5] (mol/L) Time (s)0.1000 00.0707 500.0500 1000.0250 2000.0125 300

0.0062560 400Using these data, verify that the rate law is first order in [N2O5]. And calculate the value of the rate constant, where rate = -Δ[N2O5]/ Δt.

ln[N2O5] Time (s)0

50100200300400

Ans: k = 6.93x10-3 s-1 o Sample Exercise 12.3 – First-Order Rate Laws II

Using the data given in 12.2, calculate [N2O5] at 150s after the start of the reaction.Ans: [N2O5]t=150 = 0.0353 mol/L

● Half Life of a First-Order Reactiono The time for a reactant to reach half of its original concentration

t1/2 = 0.693/k General equation for the half life for a first order reaction Independent of the initial concentration

o Sample Exercise 12.4 – Half-Life for First-Order ReactionA certain first-order reaction has a half life of 20.0 minutes.(a) Calculate the rate constant for this reaction. (b) How much time is required for this reaction to

be 75% complete? Ans: (a) 3.47x10-2 min-1 (b) 40. minutes

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● Second-Order Rate Lawso aA → productso Rate = k[A]2 o Integrated second-order rate law

1 = kt + 1 n [A] [A]0

o Note(1) 1/[A] vs. t produces a straight line with slope equal to k(2) [A] depends on t and can be used to calculate [A] at any time t, provided k and [A]0 are known

o Half-life for a second-order reaction t1/2 = 1 n

k[A]0

o Sample Exercise 12.5 – Determining Rate LawsButadiene reacts to form its dimer according to the equation 2C4H6(g) → C8H12(g).The following data were collected for this reaction at a given temperature:

[C4H6] (mol/L) Time (±1 s) [C4H6] (mol/L) Time (±1 s)0.01000 0 0.00313 36000.00625 1000 0.00270 44000.00476 1800 0.00241 52000.00370 2800 0.00208 6200

(a) Is this reaction first or second-order?(b) What is the value of the rate constant for the reaction?(c) What is the half-life for the reaction under the conditions of this experiment?

Ans: (a) second-order (b) 6.14x10-2 L-mol-1-s-1 (c) 1.63x103 s● Zero-Order Rate Laws

o Rate = k[A]0 = k Rate is constant, does depend on concentration of A

o Integrated Rate law for a zero-order reaction [A] = -kt + [A]0 Plot of [A] vs. t gives a straight line w/ slope –k

o Half-life for a zero-order reaction t1/2 = [A]0

2k

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● Integrated Rate Laws with More than One reactanto Consider BrO3

-(aq) + 5Br-(aq) 6H+(aq) → 3Br2(l) + 3 H2O(l)o From experimental evidence

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

Run under these condition, [BrO3-]0 = 1.0x10-3, [Br-]0 = 1.0M, and [H+]0 = 1.0M

● As the reaction proceeds, [BrO3-] decreases significantly but due the large initial

concentration of the other two reactants, very little of these reactants are consumed.

o Thus, [Br-] and [ H+] stay relatively constant [Br-]t = [Br-]0 and [H+]t = [H+]0 New Rate = k‘[BrO3

-]● Where k‘ = k[Br-]0[H+]0

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o Pseudo-first-order rate law

Rate Laws: A Summary Section#5

(1) For simplification, only the forward reaction is important(2) Two types of rate laws

(a) Differential rate law (often called the rate law) Shows how rate depends on concentrations

(b) Integrated rate laws Shows how concentration depends on time

(3) The type of data collected determines type of rate law(4) Most common method of experimentally determining the rate law is the method of initial rates. Found

by running experiment several times with varying concentrations of reactants. Order for each reactant can be found.

(5) To determine the integrated rate law, concentrations are measured at different times as reaction proceeds. Determine which plot yields a straight line.

(6) Integrated rate law for a reaction with several reactants can be found (kind of) by choosing conditions where the other reactants are in such high concentrations they do not significantly decrease as the reaction proceeds.

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Summary of the kinetics for reactions of the type aA → products that Are Zero, First, and Second Order in[A]

OrderZero First Second

Rate Law: Rate = k Rate = k[A] Rate = k[A]2

Plot Integrated Rae Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 1/[A] = kt + 1/[A]0

Needed to Give Straight Line: [A] vs t ln[A] vs t 1/[A] vs. t

Relationship of Rate Constant to the Slope of Straight Line: Slope = -k Slope = -k Slope = k

Hal-life: t1/2 = [A]0/2k t1/2 = 0.693/k t1/2 = 1/ k[A]0

Reaction Mechanisms Section#6

● Reaction mechanismo The series of steps a reaction undergoes to produce productso Consider the reaction: NO2(g) + CO(g) → NO(g) + CO2(g)

Rate = k[NO2]2 from experimental data k1

● NO2(g) + NO2(g) →NO3(g) + NO(g) k2

● NO3(g) + CO(g) → NO2(g) + CO2(g)o NO3 is an intermediate

An intermediate is used as a reactant in a subsequent step in the reaction and does not show in the balance overall reaction or in the rate law

● Each of these two steps are called elementary stepso A reaction whose rate law can be written from its molecularity

● Molecularityo The number of species that must collide to produce a reaction indicated by

that stepo Reaction involving one step

Unimolecularo Reaction involving two molecules or three molecules

Bimolecular or termolecular● Termolecular steps are rare

Examples of Elementary StepsElementary Step Molecularity Rate LawA → products unimolecular Rate = k[A]

A + A → products(2A → products)

bimolecular Rate = k[A]2

A + B → products bimolecular Rate = k[A][B]A + A + B → products(2A + B) → products)

termolecular Rate = k[A]2[B]

A + B + C → products termolecular Rate = k[A][B][C]

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o Rate law follows directly from the molecularity of that elementary step Bimolecular step the rate law is always second order, either k[A]

or k[A][B]o We can now define a reaction mechanism more precisely

It is a series of elementary steps that must satisfy two requirements(1) Sum of steps must give the overall balanced equation for the without(2) Mechanism must agree w/ experimentally determined rate law

o Take the above reaction of NO2(g) + NO2(g) →NO3(g) + NO(g) slow NO3(g) + CO(g) → NO2(g) + CO2(g) fast

o Overall reaction: NO2(g) + CO(g) → NO(g) + CO2(g) 1st requirement is met

o For the second part to be met, we need to introduce a new idea: the rate determining step The slowest step in the reaction

● This is the bottleneck and determines how fast the reaction will proceed Above next to the reaction is the speed at which the elementary step proceeds

● Notice the first step is slowo This is the rate determining step and the rate law must take into account

the reactantso Since this is an elementary step we can write the rate law from its

molecularity So, rate = k[NO2]2 for this elementary step

● Because the reaction rate can be no faster than the slowest step (rate determining step) the overall rate = k[NO2]2

Note, this rate law is the same as the experimentally determined rate law

o Sample Exercise 12.6 – Reaction MechanismsThe balanced equation for the reaction of the gases nitrogen dioxide and fluorine is 2NO2(g) + F2(g) → 2NO2F(g). The experimentally determined rate law is Rate = k[NO2][F2]. A suggested mechanism for this reaction is NO2 + F2 → NO2F + F slow

F + NO2 → NO2F fastIs this an acceptable mechanism? That is, does it satisfy the two requirements? Ans: Yes

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A Model for Chemical Kinetics Section#7

● How do reactions occur?o Observations have told us reactions speed up when the T is increased

Virtually all rate constants show an exponential increase with absolute To Model

Collision Model(1) Molecules must collide with

enough energy to react (or the collision energy must equal or exceed the activation energyo Observations show only

a small percentage of all collision result in a without

o Thus question first addressed by Svante Arrhenius

Proposed a threshold of E must be reached

● Activation Energy, Ea

o At the moment the activation is met, an activated complex is produced

Also called transition stateo Reaction rate depends on the activation

energy, Ea ln(k) = (-Ea/R)(1/T) + ln(A)

● slope, m, is -Ea/R Observations show the without

rate is considerable smaller than the rate of collisions with enough energy to surmount the barrier

● This means many collisions have enough energy to react but do not react

o Leads to second requirement(2) Molecules must collide in the correct orientation to react

o Molecular orientation The relative orientation of the reactants must allow formation of

any bonds necessary to produce productso Sample Exercise 12.7 – Determining Activation Energy I

The reaction 2N2O5(g) → 4NO2(g) + O2(g) was studied at several T, and the following values of k were obtained:

k(s-1) T (°C)2.0x10-5 207.3x10-5 302.7x10-4 409.1x10-4 502.9x10-3 60

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Calculate the value of Ea for each reaction.To obtain the value of Ea, we need to construct a plot of ln(k) vs. 1/T. First, we must calculate values of ln(k) and 1/T, as shown below:

T (°C) T (K) 1/T (°C) k(s-1) ln(k)20 293 3.41x10-3 2.0x10-5 -10.8230 303 3.30x10-3 7.3x10-5 -9.5340 313 3.19x10-3 2.7x10-4 -8.2250 323 3.10x10-3 9.1x10-4 -7.0060 333 3.00x10-3 2.9x10-3 -5.84

The plot of ln(k) vs. 1/T shown in Fig 12.14, where the slope Δln(k)/ Δ(1/T) is found to be -1.2x104 K. The value of Ea can be determined by solving the following equation:

Slope = -Ea/REa = -R(slope) = -(8.3145 J/K-mol)(-1.2x104K) = 1.0x105 J/mol

Ans: Ea = 1.0x105 J/mol The most common procedure for finding Ea is measuring k at several T and plotting ln(k)

vs. 1/T. However, Ea can be calculated by using the equation:

● ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)

o Sample Exercise 12.8 – Determining Activation Energy IIThe gas phase reaction between methane and diatomic sulfur is given by the equation: CH4(g) + 2S2(g) → CS2(g) + 2H2S(g). At 550°C the rate constant for the reaction is 1.1 L/mol-s, and at 625°C the rate constant is 6.4 L/mol-s. Using these values, calculate Ea for this reaction.

Ans: 1.4x105 J/mol

Catalysis Section#8

● Enzymeso Lower Ea for processes in organic systems

Increases rate of reaction without an increase in T● Catalyst

o A substance that speeds up a reaction without being consumed itself

o Homogeneous catalyst Present in the same phase as the reacting

moleculeso Heterogeneous catalyst

Exists in a different phase than reactants● Usually a solid

Adsorption● Collection of one substance on the surface of another substance

Absorption● The penetration of one substance into another

o Water and sponge Involves (4) steps

(1) Adsorption and activation of the reactants(2) Migration of the adsorbed reactants on the surface(3) Reaction of the adsorbed substances

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(4) Escape, or desorption, of the product

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