Kinetics

Reaction Rates

Kinetics• The study of reaction rates.• Thermodynamics will tell us whether or not

the reaction will occur• Spontaneous reactions (ΔG = neg) are

reactions that will happen - but we can’t tell how fast.

• Diamond will spontaneously turn to graphite – eventually.

• Kinetics is about how fast

Review- Collision Model of Chemical Reactions

Particles must collide

with enough energy

to break bonds

EFFECTIVE COLLISIONS

Factors that affect the rate of reactionIncrease number of effective collisions

increase rate of reaction

• Temperature

• Concentration

• Surface area

• Pressure

• Agitation

• Catalyst

Reaction Rate

• Rate = Conc. of A at t2 -Conc. of A at t1t2- t1

• Rate =[A]t

• Change in concentration per unit time• Consider the reaction

N2(g) + 3H2(g) 2NH3(g)

• As the reaction progresses the concentration H2 goes down

Concentration

Time

[H[H22]]

N2 + 3H2 → 2NH3

Δ[H2]/time = neg

• As the reaction progresses the concentration N2 goes down 1/3 as fast

Δ[H2]/time = neg

Concentration

Time

[H[H22]]

[N[N22]]

N2 + 3H2 → 2NH3

Δ[N2]/time = neg

Δ[H2]/time = neg

• As the reaction progresses the concentration NH3 goes up 2/3 times

Concentration

Time

[H[H22]]

[N[N22]]

[NH[NH33]]

N2 + 3H2 → 2NH3

Δ[NH3]/time = pos

Reaction Rates are always positive

• Over time – the concentration of products increases– the concentration of reactants decreases

• So Δ[P]/time is positive

• But Δ[R]/time is negative, so the rate would be expressed as - Δ[R]/time

If rate is measured in terms of concentration of Hydrogen

-Δ[H2] - mol H2

time L timeand N2 + 3H2 2NH3

Then using the stoichiometry of the reaction

-Δmol H2 1 mol N2

L time 3 mol H2

Then

-Δmol H2 2 mol NH3

L time 3 mol H2

Calculating Rates

• Average rates are taken over a period of time interval

• Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

Average Rate – Find the slope of the line between the two points

Note: slope =

Concentration

Time

[H[H22]]

tt

Note: slope = Δy / Δx

Δconcentration / Δtime

THE RATE

Instantaneous Rate – Find the slope of the line tangent to that point

Concentration

Time

[H[H22]]

ttd[Hd[H22]]

dtdt

12_291

0.000370s

O2

0.0025

0.005

0.0075

0.0100

0.0006

70s

0.0026

110 s

NO2

NO

50 100 150 200 250 300 350 400

Con

cent

ratio

ns (

mol

/L)

Time (s)

[NO2 ]

t

INSTANTANEOUS RATES

2NO2 2NO + O2

C4H9Cl + H2O C4H9OH + HCl

Rate Laws• Reactions are reversible.

• As products accumulate they can begin to turn back into reactants.

• Early on the rate will depend on only the amount of reactants present.

• We want to measure the reactants as soon as they are mixed.

• In addition, as the reaction progresses the concentration of reactants decrease

• This is called the initial rate method.

• Rate laws show the effect of concentration

on the rate of the reaction

• The concentration of the products do not appear in the rate law because this is an initial rate.

• The order (exponent)

– must be determined experimentally,

– cannot be obtained from the equation

Rate LawsRate Laws

• You will find that the rate will only depend on the concentration of the reactants. (Initially)

• Rate = k[NO2]n

• This is called a rate law expression.

• k is called the rate constant.

• n is the order of the reactant -usually a positive integer.

2 NO2 2 NO + O2

NO2 NO + ½ O2

The data for

the reaction at 300ºC

Plot the data to show concentration as a

function of time.

Time (s)[NO2]

(mol/L)

0.00 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

NO2 NO + ½ O2

To determine the rate law for the reaction, we must find the exponent in the equation.

Rate = k [NO2]x

Experiment [NO2] M Rate, M s-1

1 0.060 0.0360

2 0.030 0.0090

3 0.020 0.0040

NO2 NO + ½ O2

To determine the rate law for the reaction, we must find the exponent in the equation.

Rate = k [NO2]x

Exp [NO2] M Rate, M s-1

1 0.060 0.0360

2 0.030 0.0090

3 0.020 0.0040

2x [NO2] 4x rate

NO2 NO + ½ O2

The exponent for the concentration of NO2

must be 2 to show the effect of changing the concentration of NO2 on the rate.

Rate = k [NO2]2

Exp [NO2] M Rate, M s-1

1 0.060 0.0360

2 0.030 0.0090

3 0.020 0.0040

3x [NO2] 9x rate

NO2 NO + ½ O2

Rate = k [NO2]x

Pick two experiments to plug into the equation.

Experiment [NO2] M Rate, M s-1

1 0.060 0.0360

2 0.030 0.0090

3 0.020 0.0040

NO2 NO + ½ O2

Use the rate and one of the experiments to calculate the rate constant.

Experiment [NO2] M Rate, M s-1

1 0.060 0.0360

2 0.030 0.0090

3 0.020 0.0040

2ClO2 + 2OH- ClO3- + ClO2

- + H2O

• Determine the rate law for the reaction.• Calculate the rate constant.• What is the overall reaction order?

Experiment [ClO2-] M [OH-] M Rate, M s-1

1 0.060 0.030 0.0248

2 0.020 0.030 0.00276

3 0.020 0.090 0.00828

Types of Rate Laws• Differential Rate Law - describes how rate depends on

concentration. Method to determine- change initial concentration and measure effect on rate

• Integrated Rate Law - describes how concentration depends on time. Method to determine- measure the concentration of reactants as function of time

• For each type of differential rate law there is an integrated rate law and vice versa.

Determining Rate Laws• The first step is to determine the form of

the rate law (especially its order).• Must be determined from experimental

data.• For this reaction

2 N2O5 (aq) 4NO2 (aq) + O2(g)• The reverse reaction won’t play a role

because the gas leaves

[N[N22OO55] (mol/L) ] (mol/L) Time (s) Time (s)

1.001.00 00

0.880.88 200200

0.780.78 400400

0.690.69 600600

0.610.61 800800

0.540.54 10001000

0.480.48 12001200

0.430.43 14001400

0.380.38 16001600

0.340.34 18001800

0.300.30 20002000

Now graph the data

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

• To find rate we have to find the slope at two points• We will use the tangent method.

[N[N22OO55] ]

(mol/L)(mol/L)

Time (s)

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4 (200 - 600) -400

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800

• At 0.8 M rate is 5.0 x 10-4

• At 0.4 M rate is 2.5 x 10-4

• Since the rate is twice as fast when the concentration is twice as big the rate law must be.. First power

• Rate = -[N2O5] = k[N2O5]1 = k[N2O5] t

• We say this reaction is first order in N2O5

• The only way to determine order is to run the experiment.

The method of Initial Rates• This method requires that a reaction be

run several times.

• The initial concentrations of the reactants are varied.

• The reaction rate is measured just after the reactants are mixed.

• Eliminates the effect of the reverse reaction.

An example• For the reaction

BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O

• The general form of the Rate Law is Rate

= k[BrO3-]n[Br-]m[H+]p

• We use experimental data to determine the values of n,m,and p

Initial concentrations (M)

Rate (M/s)

BrOBrO33-- BrBr-- HH++

0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--

44

0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--

33

0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--

33

0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--

33

Calculate the rate law and k

Integrated Rate Law• Expresses the reaction concentration as a

function of time.

• Form of the equation depends on the order of the rate law (differential).

• Changes Rate = [A]n t

• We will only work with n = 0, 1, and 2

First Order• For the reaction 2N2O5 4NO2 + O2

• We found the Rate = k[N2O5]1

• If concentration doubles rate doubles.• If we integrate this equation with respect to time

we get the Integrated Rate Law[N2O5]0 = initial concentration of N2O5

[N2O5]t = concentration of N2O5 after some time (t)

ln = natural log

= - ktln ( )[N2O5]0

[N2O5]t

First Order rate = k[N2O5]• For the reaction 2N2O5 4NO2 + O2

• Rearrange this equation to y = mx + b form– Since- log of quotient equals difference of logs

• SO. ln[N2O5]t = - k t + ln[N2O5]0

- ln[N2O5]t=ln ( )[N2O5]0

[N2O5]t

= ktln ( )[N2O5]0

[N2O5]t

ln[N2O5]t

y mx b+=

First Order• Linear form integrated rate law

•ln[N2O5]t = - k t + ln[N2O5]0

= ktln ( )[N2O5]0

[N2O5]t

y mx b+=

ln[N2O5]t

t

m = slope = -k

• Graphing natural log of concentration of reactant as a function of time

• If you get a straight line, you may determine it is first order

• Integrated First Order

First Order

= - ktln

( ) [R]0

[R]t

Half Life• The time required to reach half the

original concentration.

• [R]t = [R]0/2 when t = t1/2

• If the reaction is first order

= - ktln

( ) [R]0

[R]t Since [R]t = [R]0/2

[R]t = [R]0 equals 2

ln (2) = kt1/2

Half Life

• t1/2 = 0.693 / k

• The time to reach half the original concentration does not depend on the starting concentration.

• An easy way to find k

• The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?

Second Order

• Rate = -[R]/t = k[R]2

• integrated rate law

• 1/[R]t = kt + 1/[A]0

• y= 1/[A] m = k

• x= t b = 1/[A]0

• A straight line if 1/[A] vs t is graphed

• Knowing k and [A]0 you can calculate [A] at any time t

Second Order Rate = k [R]2

integrated rate law

1/[R]t = k t + 1/[A]0

y = m x + b

t

1/[R] Slope = k

Second Order Half Life• [A] = [A]0 /2 at t = t1/2

1

20

2[ ]A = kt +

1

[A]10

22[ [A]

- 1

A] = kt

0 01

tk[A]1 =

1

02

1

[A] = kt

01 2

Zero Order Rate Law

• Rate = k[A]0 = k

• Rate does not change with concentration.

• Integrated [A] = -kt + [A]0

• When [A] = [A]0 /2 t = t1/2

• t1/2 = [A]0 /2k

• Most often when reaction happens on a surface because the surface area stays constant.

• Also applies to enzyme chemistry.

Zero Order Rate Law

Time

Concentration

Time

Concentration

A]/t

t

k =

A]

Zero order First order Second orderRate law

Graph [A] vs time

Integrated

rate law

Zero order First order Second order

Linear graph with time

Integrated

rate law

Rearrange

to y=mx+b

Summary of Rate Laws

Reaction Mechanisms

• The series of steps that actually occur in a chemical reaction.

• Kinetics can tell us something about the mechanism

• A balanced equation does not tell us how the reactants become products.

• 2NO2 + F2 2NO2F Rate = k[NO2][F2]

• The overall equation is a summary.• In order for this reaction to occur as

written, three molecules must collide simultaneously

• An unlikely event.• So a reaction mechanism is proposed

showing a series of likely steps

Reaction Mechanisms

• 2NO2 + F2 2NO2F

• Rate = k[NO2][F2]• The proposed mechanism is• NO2 + F2 NO2F + F (slow)• F + NO2 NO2F (fast) • F is called an intermediate It is formed

then consumed in the reaction

Reaction Mechanisms

• Each of the two reactions is called an elementary step .

• The rate for a reaction can be written from its molecularity. The reaction rate reflects the stoichimetry.

• Molecularity is the number of pieces that must come together.

• Elementary steps add up to the balanced equation

Reaction Mechanisms

• Unimolecular step involves one molecule - Rate is first order.

• Bimolecular step - requires two molecules - Rate is second order

• Termolecular step- requires three molecules - Rate is third order

• Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

• A products Rate = k[A]

• A+A products Rate= k[A]2

• 2A products Rate= k[A]2

• A+B products Rate= k[A][B]

• A+A+B products Rate= k[A]2[B]

• 2A+B products Rate= k[A]2[B]

• A+B+C products Rate= k[A][B][C]

Molecularity and Rate LawsFor an elementary step –

the rate law reflects the stoichiometry

Reaction Mechanisms

• Proposed series of elementary steps.– Describing each collision

• The rate of the reaction is determined by the rate limiting step.

• The overall experimentally determined rate law must be consistent with the rate law that reflects the rate limiting step.

• The proposed mechanism isNO2 + F2 NO2F + F (slow) rate = k [NO2][F2]

F + NO2 NO2F (fast) rate = k [F][NO2]

Since the first step is the slow step, it is the rate limiting step and thus the rate of the overall reaction depends upon the rate of this step. Note: The rate of this step is also the rate of the overall reaction.

2NO2 + F2 2NO2F Rate = k[NO2][F2]

The steps must add up to the overall reaction.

• 2 NO2Cl → 2 NO2 + Cl2

• Proposed steps

• NO2Cl → NO2 + Cl (slow)

NO2Cl + Cl → NO2 + Cl2 (fast)

• Write the overall rate law that would be consistent with this proposed reaction mechanism.

How to get rid of intermediates• They can’t appear in the rate law.• Slow step determines the rate and the rate

law• Use the reactions that form them• If the reactions are fast and irreversible

the concentration of the intermediate is based on stoichiometry.

• If it is formed by a reversible reaction set the rates equal to each other.

Formed in reversible reactions• 2 NO + O2 2 NO2

• Mechanism

• 2 NO N2O2 (fast)

• N2O2 + O2 2 NO2 (slow)

• rate = k2[N2O2][O2]

• k1[NO]2 = k1[N2O2] (equilibrium)

• Rate = k2 (k1/ k-1)[NO]2[O2]

• Rate =k [NO]2[O2]

Formed in fast reactions• 2 IBr I2+ Br2 • Mechanism• IBr I + Br (fast)• IBr + Br I + Br2 (slow)• I + I I2 (fast)• Rate = k[IBr][Br] but [Br]= [IBr] because

the first step is fast• Rate = k[IBr][IBr] = k[IBr]2

Collision theory• Molecules must collide to react.

• Concentration affects rates because collisions are more likely.

• Must collide hard enough.

• Temperature and rate are related.

• Only a small number of collisions produce reactions.

Potential Energy

Reaction Coordinate

Reactants

Products

Potential Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

Potential Energy

Reaction Coordinate

Reactants

Products

Activated complex

Potential Energy

Reaction Coordinate

Reactants

ProductsE}

Potential Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

Terms• Activation energy - the minimum energy

needed to make a reaction happen.

• Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

Arrhenius• Said that reaction rate should increase

with temperature.

• At high temperature more molecules have the energy required to get over the barrier.

• The number of collisions with the necessary energy increases exponentially.

Arrhenius• Number of collisions with the required

energy = ze-Ea/RT

• z = total collisions

• e is Euler’s number (inverse of ln)

• Ea = activation energy

• R = ideal gas constant (in J/K mol)

• T is temperature in Kelvin

Problem with this• Observed rate is too small

• Due to molecular orientation- they have to be facing the right way

• written into equation as p the steric factor.

ON

Br

ON

Br

O N Br ONBr ONBr

O NBr

O N BrONBr No Reaction

O NBr

O NBr

Arrhenius Equation

• k = zpe-Ea/RT = Ae-Ea/RT

• A is called the frequency factor = zp• k is the rate constant• ln k = -(Ea/R)(1/T) + ln A• Another line !!!!• ln k vs 1/T is a straight line• With slope Ea/R so we can find Ea

• And intercept ln A

Arrhenius Equation- Another line !!!!

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

y = m x + b

ln k

1/T

slope = -(Ea/R)

Activation Energy and Rates

The final saga

Mechanisms and rates • There is an activation energy for each

elementary step.

• Activation energy determines k.

• k = Ae- (Ea/RT)

• k determines rate

• Slowest step (rate determining) must have the highest activation energy.

• This reaction takes place in three steps

Ea

First step is fast

Low activation energy

Second step is slowHigh activation energy

Ea

Ea

Third step is fastLow activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

Catalysts• Speed up a reaction without being used up

in the reaction.

• Enzymes are biological catalysts.

• Homogenous Catalysts are in the same phase as the reactants.

• Heterogeneous Catalysts are in a different phase as the reactants.

How Catalysts Work

• Catalysts allow reactions to proceed by a different mechanism - a new pathway.

• New pathway has a lower activation energy.

• More molecules will have this activation energy.

• Does not change E• Show up as a reactant in one step and a

product in a later step

Pt surface

HH

HH

HH

HH

• Hydrogen bonds to surface of metal.

• Break H-H bonds

Heterogenous Catalysts

Pt surface

HH

HH

Heterogenous Catalysts

C HH C

HH

Pt surface

HH

HH

Heterogenous Catalysts

C HH C

HH

• The double bond breaks and bonds to the catalyst.

Pt surface

HH

HH

Heterogenous Catalysts

C HH C

HH

• The hydrogen atoms bond with the carbon

Pt surface

H

Heterogenous Catalysts

C HH C

HH

H HH

Homogenous Catalysts• Chlorofluorocarbons (CFCs) catalyze the

decomposition of ozone.

• Enzymes regulating the body processes. (Protein catalysts)

Catalysts and rate• Catalysts will speed up a reaction but only

to a certain point.

• Past a certain point adding more reactants won’t change the rate.

• Zero Order

Catalysts and rate.

Concentration of reactants

Rate

• Rate increases until the active sites of catalyst are filled.

• Then rate is independent of concentration