KINETICS

KINETICSThe speed of a reaction

Kinetics• The study of reaction rates.

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

• Diamond will spontaneously turn to graphite – eventually.

• Reaction mechanism- the steps by which a reaction takes place.

Reaction Rate

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

• Rate =[A]t

• Change in concentration per unit time

• For this reaction

• N2 + 3H2 2NH3

• As the reaction progresses the concentration of H2 goes down

Concentration

Time

[H[H22]]

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

Concentration

Time

[H[H22]]

[N[N22]]

• As the reaction progresses the concentration of NH3 goes up.

Concentration

Time

[H[H22]]

[N[N22]]

[NH[NH33]]

Calculating Rates• Average rates are taken over long

intervals

• 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

• Derivative.

• Average slope method

Concentration

Time

[H[H22]]

tt

• Instantaneous slope method.

Concentration

Time

[H[H22]]

tt

Defining Rate

• We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.

• In our example: N2 + 3H2 2NH3

• -[N2] = -3 [H2] = 2 [NH3] t t t

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.

• This is called the Initial rate method and time = 0.

• Two key points

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

• The order must be determined experimentally because it can’t be obtained from the equation

Rate LawsRate Laws

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

• 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

• The rate of appearance of O2 can be said to be.

• Rate' = [O2] = k'[NO2]2

t• Because there are 2 NO2 for each O2

• Rate = 2 x Rate'

• So k[NO2]n = 2 x k'[NO2]

n

• So k = 2 x k'

2 NO2 2 NO + O2

Types of Rate Laws

• Differential Rate law - describes how rate depends on concentration.

• Integrated Rate Law - Describes how concentration depends on time.

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

• Rate laws can help us better understand reaction mechanisms.

Determining Rate Laws

• The first step is to determine the form of the rate law (especially its order).

• Must be determined from experimental data which is usually given.

• For this reaction 2 N2O5 (aq) 4NO2 (aq) + O2(g)

• The reverse reaction won’t play a role

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

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .90 M the rate is (.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -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 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800

• Since the rate at twice the concentration is twice as fast the rate law must be..

• 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.

• So the Rate Law is Rate = k[N2O5]

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.

• In our problems, this isn’t a concern.

An Example of a Rate Law• 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

Now we have to see how the rate changes with concentration

Remember our Formula ! - Δ Concentration order = Δ Rate

In words, the change in concentration raised to the order of the exponent = the change in the rate.

- Fortunately, most of the time, these can be solved on inspection. Eg. If the concentration is tripled and the rate is tripled, then 3x = 3 so x = 1.

HOWEVER…

Sometimes more drastic measures are needed.

• Namely the law of logarithms.

• Δ Concentration order = Δ Rate becomes

Order x (Log (conc.)) = Log (rate)

• It is simple and works every time.

• 23 = 8 3 log 2 = log 8 3 x .301 = .903

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

• ln[N2O5] = - kt + ln[N2O5]0

• ln is the natural log

• [N2O5]0 is the initial concentration.

• General form Rate = [A] / t = k[A]

• ln[A] = - kt + ln[A]0

• In the form y = mx + b

• y = ln[A] m = - k

• x = t b = ln[A]0

• A graph of ln[A] vs time is a straight line.

First Order

• By getting the straight line you can prove it is first order.

• Often expressed in a ratio.

First Order Rates

• By getting the straight line you can prove it is first order.

• Often expressed in a ratio

First Order

lnA

A = kt0

Half Life• The time required to reach half the

original concentration.

• If the reaction is first order

• [A] = [A]0/2 when t = t1/2

Half Life• The time required to reach half the

original concentration.

• If the reaction is first order

• [A] = [A]0/2 when t = t1/2

ln

A

A = kt0

01 2

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.

Second Order

• Rate = -[A] / t = k[A]2

• integrated rate law

• 1/[A] = 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 time.

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.

• The material is said to be at the zeroeth order.

Zero Order Rate Law

Time

Concentration

Time

Concentration

A]/t

t

k =

A]

More Complicated Reactions

• BrOBrO33-- + 5 Br + 5 Br -- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO

• For this reaction we found the rate law to For this reaction we found the rate law to be:be:

• Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

• To investigate this reaction rate we need To investigate this reaction rate we need to control the conditions. to control the conditions.

Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

• We set up the experiment so that two of the reactants are in large excess.

• [BrO3-]0= 1.0 x 10-3 M

• [Br-]0 = 1.0 M

• [H+]0 = 1.0 M

• As the reaction proceeds [BrO3-]

changes noticably • [Br-] and [H+] don’t

• This rate law can be rewritten

• Rate = k[BrORate = k[BrO33--][Br][Br--]]00[H[H

++]]0022

• Rate = k[BrRate = k[Br--]]00[H[H++]]00

22[BrO[BrO33--]]

• Rate = k’[BrORate = k’[BrO33--]]

• This is called a pseudo first order rate This is called a pseudo first order rate law.law.

• k =k = k’ k’

[Br[Br--]]00[H[H++]]00

22

Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

Summary of Rate Laws• Know your integrated rate laws.

• Know how to find the order for the reaction

• Know how to find k after you determine the orders.

• As always, read the question carefully to make sure you are using all the given information.

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.

• We will always want to find the rate determining step, which will usually be the slowest step in the mechanism.

• 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, so it isn’t present in the given equation.

Reaction Mechanisms

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

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

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

Reaction Mechanisms

• Unimolecular step requires one molecule so the rate is first order.

• Bimolecular step - requires two molecules so the rate is second order.

• Termolecular step - requires three molecules so the 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.

Molecularity

• A Products Rate = k[A]

• A + A Products Rate= k[A]2

• 2A Products Rate= k[A]2

• A + B ProductsRate= 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]

How to get rid of intermediates

• 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.

• A picture is worth 1,000 words here, so let’s check out the next slide.

Formed in reversible reactions• 2 NO + O2 2 NO2

• Mechanism

• 2 NO N2O2 (fast) Step 1

• N2O2 + O2 2 NO2 (slow) Step 2

• rate = k2[N2O2][O2]

• k1[NO]2 = k-1[N2O2]

• rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]

• NOTICE: Step 1 is in equilibrium so the rate of formation of the N2O2 is equal to the rate of formation of the 2 NO.

Formed in fast reactions• 2 IBr I2+ Br2

• Mechanism• IBr I + Br (fast)

• IBr + Br I + Br2 (slow)

• I + I I2 (fast)

• What is the rate determining step ?

• Rate = k[IBr][Br] but [Br]= [IBr]

• 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 the at 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 (opposite of ln)

• Ea = activation energy

• R = ideal gas constant

• T is temperature in Kelvin

Problems• Observed rate is less than the number of

collisions that have the minimum energy.

• Due to Molecular orientation

• written into equation as p the steric factor.

ON

Br

ON

Br

ON

Br

ON

Br

O N Br ONBr ONBr

O NBr

O N BrONBr No Reaction

Arrhenius Equation

• k = zpe-Ea/RT = Ae-Ea/RT

• A is called the frequency factor = zp

• ln k = -(Ea/R)(1/T) + ln A

• Another line !!!!

• ln k vs t is a straight line

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.

• Do not change E

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 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