Chemical Kinetics

Rates of Chemical Reactions

Chemical Kinetics

Kinetics - study of rates of chemical reactions and the mechanisms by which they occur.

Reaction rate - increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.

Reaction mechanism - the series of molecular steps by which a reaction occurs.

Kinetics versus Thermodynamics• Thermodynamics determines if a reaction can occur.• Kinetics determines how quickly a reaction occurs.

– Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.

OUSINSTANTANE

kJ -79=G OHOH+H

SLOW VERY

kJ 396G COO C

o2982

-aq

+aq

o298g2g2diamond

l

The Rate of ReactionConsider the hypothetical reaction,

aA(aq) + bB(aq) cC(aq) + dD(aq)equimolar amounts of reactants, A and B, will be

consumed while products, C and D, will be formed as indicated in this graph:

0

0.2

0.4

0.6

0.8

1

1.2

Time

Con

cent

rati

ons

of

Rea

ctan

ts &

Pro

duct

s

[A] & [B]

[C] & [D]

[A] = concentration of A in M ( mol/L).

Reaction does not go entirely to completion.◦ The [A] and [B]

> 0 plus the [C] and [D] < 1.

The Rate of Reaction

aA(aq) + bB(aq) cC(aq) + dD(aq)

- Mathematically, the rate of a reaction can be written as:

t d

D+

t c

C+or

t b

B-

t a

A-= Rate

In terms of reactants

In terms of products

(aq)(aq)94)l(2(aq)94 HClOHHC OH ClHC

[C4H9Cl] (M) Time, t (s)0.1 0

0.0905 500.082 100

0.0741 1500.0671 2000.0549 3000.0448 4000.0368 500

0.02 8000.005 1500

0.0025 2250

Average rate of reaction – rate over a period of time

timeinitial timefinal

ClHCClHC

t

ClHC Rate Average

timeinitial94 timefinal94

94

[C4H9Cl] (M) Time, t (s) Average rate0.1 00.0905 50 0.000190.082 100 0.000170.0741 150 0.0001580.0671 200 0.000140.0549 300 0.0001220.0448 400 0.0001010.0368 500 0.000080.02 800 0.0000560.005 1500 2.14286E-050.0025 2250 3.33333E-06

Instantaneous rate of reaction – rate at a particular time

Instantaneous rate at 0 s – Initial Rate

Instantaneous rate is actually the slope (negative for reactants, positive for products) of the tangent line passing through the time t. ◦ It is a derivative of the curve at that point.

◦ INITIAL RATE the instantaneous rate at t = 0 Important parameter being measured because it gives a

more accurate measure of rate Changes when the initial concentration of reactant also

changes

t

ClHC x

1

1 - = Rate 94

Rate Law Expressions / Rate Laws• The dependence of reaction rates to the

concentration of species is shown in a RATE LAW expression (or simply rate law)

Rate = k[reactant1]n[reactant2]m...

– m, n reaction orders (rate is nth order with respect to reactant 1, mth order wrt reactant 2)

– m + n overall reaction order– k rate constant – a very important rate

parameter, as it describes how the reaction proceeds

• Consider the following date for the reaction:

2(g)(l)2-3(aq)4(aq) NO2H NO NH

Experiment #Initial [NH4

+ ] M

Initial [NO3- ]

MObserved Initial Rate,

M/s1 0.01 0.2 5.4 x 10-7

2 0.02 0.2 10.8 x 10-7

3 0.2 0.0202 21.6 x 10-7

4 0.2 0.0404 43.3 x 10-9

- as NH4+ doubles, the initial rate

doubles!- as NO3

- doubles, the initial rate also doubles!

]NO[],[NH Rate -34

]NO][[NHk Rate -34

Types of reaction based on Rate law Expressions

ReactionPossible Experimental Rate Law

Order of Reaction Unit of k (when t in secs)

A products Rate = k Zero order M/s

A products Rate = k[A] First order wrt [A], First order overall 1/s

2A products Rate = k[A]2 2nd order wrt [A], 2nd order overall 1/M-s

A + B products Rate = k[A][B]1st order wrt [A], 1st order wrt [B], 2nd order overall

1/M-s

The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the

reaction?2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)

ExperimentInitial [A]

(M)Initial [B]

(M)Initial [C]

(M)

Initial rate of formation of

D (M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3

Differential Rate Law Forms

Reaction Possible Experimental Rate Law Differential Rate law form

A products Rate = k

A products Rate = k[A]

2A products Rate = k[A]2

A + B products Rate = k[A][B]

k

t

A

k[A]

t

A

2k[A]t

A

k[A][B]

t

B

t

A

Integrated Rate Law Equations

• For a first-order reaction, the rate is proportional to the first power of [A].

• In Calculus

-1a

At

k A

-1a

At

k Add

-A

Aa k t

dd

-

AA

a k tA

A tdd

0 0

This equation can be evaluated as:

-ln A a k t or

-ln A A a k t - a k 0

which becomes

-ln A A a k t

t0t

t

t

0

0

0

ln

ln

First-Order integrated rate law

k t aA

Aln

t

0

- similar to the equation introduced for the half-life of reactions in chem 16- majority of simple nuclear decompositions/reactions are first order process

0

0

Alnk t aAln

or

k t aAlnAln

Second-Order Integrated Rate Law

• Only for second-order wrt [A] and 2nd order overall

2Ak t a

A

d

d

tk A a

A2 d

d

t

0

A

A2 tk a

A

A

0

dd

k t aA

1

A

1

0

Second-Order Integrated Rate Law

Zero-Order Integrated Rate Law

k t a

A

d

d

tk aA dd

t

0

A

A

tk aA0

dd

k t a-AA

or

k t -aAA

0

0

Integrated Rate Law Equations and determination of Reaction Orders

Correlation with Linear Equation

If you are determining the order of reaction for a particular reactant and you have a set of concentration-time data (3 data points is enough) wrt the reactant, then you can use correlation with integrated rate laws.

bmy x

0Alnk t aAln

k t a-AA 0

0A

1k t a

A

1

bmy x

• Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant.

700Kat HBrHCBrHC gg42g52

Time

(min) 0 1 2 3 4 5

[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37

k t a-AA 0

Time

(min) 0 1 2 3 4 5

1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7

Time

(min) 0 1 2 3 4 5

ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99

0Alnk t aAln

0A

1k t a

A

1

[C2H5Br] vs. time

0

0.20.4

0.60.8

11.2

0 1 2 3 4 5

Time (min)

[C2

H5

Br]

ln [C2H5Br] vs. time

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5

Time (min)

ln [

C2H

5B

r]

1/[C2H5Br] vs. time

0

1

2

3

0 1 2 3 4 5

Time (min)

1/[C

2H5B

r]

• Half-life of reactions, t1/2 – time when half of the initial concentration of reactant is consumed.– To obtain half-life equation, just substitute [A] = ½ [A]0 when

t = t1/2

0Alnk t aAln ak t-AA 0 0A

1k t a

A

1

021 Aak

1t

2ak

A][t 0

21

ak

ln2t

21

Sucrose, C12H22O11, which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose.

C12 H22 O11(aq) + H2O(l) 2C6H12O6(aq)

At 23oC and in 0.5M HCl, the following data were obtained for the disappearance of sucrose:

(a)Determine the order of reaction and the rate law expression.(b)What is the half-life for the conversion of sucrose into glucose and fructose based on the conditions of analysis?

Time (min) [C12H22O11], M

0 0.316

39 0.274

80 0.238

140 0.190

210 0.146

Determination of Rate Laws

Write the general form of the rate law expression (depends on the reactants present)

Determine the orders of reaction wrt each reactant◦Log-rate ratio technique

Initial rates and initial concentrations of reactants◦ integrated rate law technique

Concentration-time dataDetermine the rate constant k

Experimental Methods of Determining Rate Law

• Initial Rates Method (Method of Initial Rates)– Initial rates of reaction are obtained– Time involved not > 5 minutes from start of

reaction– If time is given instead of rate, the rate can be is

approximated by 1/t

t

1

t

1 Rate

t

]tion[concentra Rate

Consider the gas-phase reaction between nitric oxide and bromine at 273 oC:

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

The following data for the time of appearance of NOBr were obtained:

(a)Determine the rate law(b)Calculate the average value of the rate constant for the appearance of NOBr.(c)How is the rate of appearance of NOBr related to the rate of disappearance of Br2?(d) What is the rate of disappearance on Br2 when [NO] =0.075 M and [Br2] = 0.185 M?

Experiment # [NO], M [Br2], M Time, sec

1 0.11 0.25 41.67

2 0.27 0.25 6.67

3 0.12 0.50 16.71

4 0.36 0.50 1.36

Experimental Methods of Determining Rate Law

• Pseudo-Order Systems (Isolation method)– The concentration of one of the reactants is

kept high and constant, so that the reaction depends only on the variable reactant concentration.

– Only initial rates are measured

The reaction of peroxydisulfate ion, S2O82-, with iodide ion, I-, is given by

the equation :

S2O82-

(aq) + 3I-(aq) 2SO4

2-(aq) + I3

-(aq)

At a particular temperature, the rate of disappearance of S2O82-

varies with reactant concentrations in the following manner:

(a)Determine the rate law for the reaction.(b)What is the rate of disappearance of I- when [S2O8

2-] =0.075 M and [I-] = 0.050 M?

[S2O82-], M [I-], M Rate, M/s

0.018 1.52 9.77 x 10-5

0.027 1.53 1.48 x 10-4

0.036 1.55 1.99 x 10-4

0.79 0.015 4.23 x 10-5

0.80 0.022 6.29 x 10-5

0.81 0.037 1.07 x 10-4

Factors That Affect Reaction Rates

There are several factors that can influence the rate of a reaction:

1. The nature of the reactants.2. The concentration of the reactants.3. The temperature of the reaction.4. The presence of a catalyst.

Nature of Reactants• Chemical properties of reactant affect rates

burns. and ignites H -reaction rapid andViolent

HNaOH 2OH 2Na 2

2

g2aq2s

reaction. Slow

HOHCaOH 2Ca g2aq22s

eyehuman toeperceptiblNot

reaction No OH Mg 2s

Acid-Base neutralizations, salt formation, and precipitation reactions or ION-EXCHANGE reactions are usually FASTER than reactions involving COVALENT-bond breakage and formation.

Nature of Reactants• The Phase of reactants affect the rate of reaction

eyehuman toeperceptiblNot

reaction No OH Mg 2s

processfast Very

HMgO OH Mg (g)2(s))(2(s) g

Homogenous reactions are faster than heterogenous reactions

Heterogenous reactions with finely divided particles are usually fast

Dissolution of “panutsa” versus plain “table sugar”.

Concentration of reactants

• In the molecular level, increasing the concentration of reactants means increasing the chances of reactant encounters, or COLLISIONS– Basic idea of COLLISION THEORY of reactions

Expression Law Rate from -

[reactant] Rate

Collision Theory of Reaction Rates

• Three basic events must happen for a reaction to occur. The atoms, molecules or ions involved must:

1. Collide.2. Collide with enough energy to break and form

bonds.3. Collide with the proper orientation for a

reaction to occur.

Collision Theory of Reaction Rates

• To increase chances/probability of collisions, Temperature (since KE of particles as T increases)

• Also, T allows initial energy for bond breakage of reactant molecules.

kJ 891 OH CO O CH (g)22(g),

2(g)4(g) matches

reaction) no added,heat (no O CH 2(g)4(g)

Collision Theory of Reaction Rates

• Proper orientation of particles (upon collision) is also necessary for formation of products to occur.

– will product form in the following orientation of molecules?

(g)2)g(22(g) O2H O2H

(g)2)g(22(g) O2H O2H

(g))g(22(g) 2HI IH

• Collision Theory only explains that successful formation of products happens when– KE energy sufficient to form and break the

chemical bonds– correct orientation of molecules is achieved upon

collision

• It gives an incomplete description of the corresponding energy involved in the rearrangement of atoms to form the final compound

Transition State TheoryTransition state theory postulates that

reactants form a high energy intermediate, the transition state, which then falls apart into the products.

For a reaction to occur, the reactants must acquire sufficient energy to form the transition state.◦This energy is called the activation energy or Ea.

(g))g(22(g) 2HI IH

Dependence of Rate on Temperature – Arrhenius Equation

• Svante Arrhenius developed this relationship among – (1) the temperature (T)– (2) the activation energy (Ea), and

– (3) the specific rate constant (k).

RT

E-Aln =kln

or

Ae=k

a

RTE- a

Kin eTemperaturTK-mole

joule 8.314constant gasR

energy ActivationE

factorfrequency A

constant ratek

Ae=k

a

RTE- a

49

The Arrhenius Equation

• If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.

ln k ln A -ERT

and

ln k ln A -E

RT

1a

1

2a

2

50

The Arrhenius EquationSubtract one equation from the other.

ln k k A - ln A -E

RTERT

ln k kERT

-E

RT

2 1a

2

a

1

2 1a

1

a

2

ln ln

ln

12

12a

1

2

21

a

1

2

TT

T-T

R

E

k

kln

or

T

1

T

1

R

E

k

kln

The temperature dependence of the rate constant for the reaction CO(g) + NO2(g) CO2(g) + NO(g)

is tabulated below. Calculate Ea and A.

Temperature, K k, M-1 s-1

600 0.028

650 0.22

700 1.3

750 6.0

800 23

Factors That Affect Reaction RatesThe presence of a catalyst.The presence of a catalyst.

◦ May be included in the actual reaction, but is not consumed in the process.

◦ Catalysts change reaction rates by providing an alternative reaction pathway with a lower Ea

Presence of CatalystsHomogeneous catalysts exist in same phase as

the reactants.Heterogeneous catalysts exist in different

phases than the reactants.◦ Catalysts are often solids.

g2g2Pt and NiO

g

g2Pt and NiO

g2g

g2g2Pt and NiO

g2g188

ONNO 2

CO 2O+CO 2

OH 18CO16O 25+HC

Presence of Catalysts

ProcessHaber

NH 2H 3 N g3OFeor Fe

g2g232

npreparatio acid Sulfuric

SO 2OSO 2 g3NiO/Ptor OV

g2g252

Molecularity of Reactions – Reaction Mechanisms

• Reaction mechanism - the series of molecular steps by which a reaction occurs.– The molecular steps proposed should add

back to final balanced equation– It must be CONSISTENT with the RATE LAW

obtained for the actual reaction (remember rate law is derived from experimental results)

Molecularity of Reactions – Reaction Mechanisms

NOOk=Rate :law Rate alExperiment

O+NONO+O

3

g2g2gg3

The mechanism that you propose should be simple and consistent with the overall reaction

223

223

33

O+NONO+Oreaction Overall

O+NONO+O stepFast

O+NONO+O step Slow

-Each step in the proposed mechanism is called an ELEMENTARY STEP or ELEMENTARY reactionELEMENTARY STEP or ELEMENTARY reaction

- For each elementary step, the rate law is based on the reactants present and the order is the same as the actual order is the same as the actual stoichiometric coefficient stoichiometric coefficient as shown in the reaction

223

223

33

O+NONO+O reaction Overall

O+NONO+O step)(fast 2 Step

O+NONO+O step) (slow 1 Step2

1

k

k

][O][NO rate :2 Step

][NO][O rate :1 Step

322

311

k

k

- The elementary step that is assigned as SLOW STEP will dictate the actual rate law for the entire reaction

- SLOW step dictates the rate law because it is the main “bottle neck” of the entire reaction

- The rate constant k1 is the actual rate constant obtained in the experiment

NOO rate :1 Step 311 k

g2g2gg3 O+NONO+O

NOOk=Rate

law Rate alExperiment

3

The proposed mechanism is consistent and The proposed mechanism is consistent and therefore a possible reaction pathwaytherefore a possible reaction pathway

A mechanism that is inconsistent with the rate-law expression is:

Experimentally determined reaction orders indicate the number of molecules involved in:

1. the slow step only or2. the slow step and the equilibrium steps preceding

the slow step.

correct. becannot mechanism thisproveswhich

Ok=Rate is mechanism thisfrom law-rate The

ONONO+Oreaction Overall

NONO+O stepFast

O+OO step Slow

3

223

2

23

The proposed mechanism for the substitution of OH- with Br- in tertbutyl bromide is shown below

aq-

(aq) 33aq-

(aq) 33 Br+COH)CH(OH+CBr)CH(

OHCOH)(CHOHCOH)(CH (3)

COH)(CHOHC)CH( )2(

Br+C)CH(CBr)CH( (1)

233233

2332 33

- 33 33

3

2

1

k

k

k

1. Determine the slow step if the experimental rate law obtained is rate = k[(CH3)CBr]

2. Identify the intermediates in the proposed reaction mechanism

3. Identify the catalysts in the system4. Give the equation for the dependence of Ea with the

reaction constants of the proposed mechanism

2(g)g2(g) NO2O+2NO

(slow) NO2OON (2)

m)equilibriu (fast, ON 2NO (1)

2222

22

2

)('1

)(1

k

k

k

backward

forward

A consistent mechanism proposed for the oxidation of nitrous oxide into nitrogen dioxide is shown below.

1. Identify the intermediates in the proposed reaction mechanism

2. Determine the expected rate law for this mechanism (Hint: Equilibrium systems indicate rates of forward and backward steps to be constant)

3. Give the equation for the dependence of Ea with the reaction constants of the proposed mechanism