The study of reaction rates and steps. 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.

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

t2- t1 Rate = [A]

Dt Change in concentration per unit time For this reaction N2 + 3H2 2NH3

As the reaction progresses the concentration H2 goes down

Concentration

Time

[H2]

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

Concentration

Time

[H2]

[N2]

As the reaction progresses the concentration NH3 goes up.

Concentration

Time

[H2]

[N2]

[NH3]

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

D[H2]

Dt

Instantaneous slope method.

Concentration

Time

[H2]

D t

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 exampleN2 + 3H2 2NH3

-[N2] = -3[H2] = 2[NH3] t t t

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.

Two key points

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

2. The order must be determined experimentally, and cannot 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.

The rate of appearance of O2 can be said to beRate' = D[O2] = k'[NO2]

Dt 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'

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.

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

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

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

rate at those two concentration values.

1.00

0.88

0.78

0.69

0.61

0.54

0.48

0.43

0.380.34

0.30

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 500 1000 1500 2000 2500

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

Rate = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt

We say this reaction is first order in N2O5 The only way to determine order is to run the

experiment.

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.

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)

BrO3- Br- H+

0.10 0.10 0.10 8.0 x 10-

4

0.20 0.10 0.10 1.6 x 10-

3

0.20 0.20 0.10 3.2 x 10-

3

0.10 0.10 0.20 3.2 x 10-

3

Now we have to see how the rate changes with concentration

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

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

[N2O5]0 is the initial concentration.

General form Rate = [A] / t = k[A]

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

In the form y = mx + by = ln[A] m = -kx = t b = ln[A]0

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

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

Often expressed in a ratio

lnA

A = kt0

ln(2) = kt1/2

• 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

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

Rate = -[A] / t = k[A]2

integrated rate law 1/[A] = kt + 1/[A]0

y= 1/[A] m = kx= 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

[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

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.

Time

Concentration

Time

Concentration

A]/t

t

k =

A]

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

For this reaction we found the rate law to be

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

To investigate this reaction rate we need to control the conditions

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

[BrO[BrO33--]]00= 1.0 x 10= 1.0 x 10-3-3 M M

[Br[Br--]]0 0 = 1.0 M= 1.0 M

[H[H++]]0 0 = 1.0 M= 1.0 M As the reaction proceeds [BrOAs the reaction proceeds [BrO33

--] changes ] changes

noticeably noticeably [Br[Br--] and [H] and [H++] don’t] don’t

This rate law can be rewrittenThis rate law can be rewritten

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

22

Rate = k[BrRate = k[Br--]]00[H[H++]]0022[BrO[BrO33

--]]

Rate = k’[BrORate = k’[BrO33--]]

This is called a pseudo first order rate law.This is called a pseudo first order rate law. k =k = k’ k’

[Br[Br--]]00[H[H++]]0022

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 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

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.

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 four 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]

Only for the elementary steps

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.

2 NO + O2 2 NO2 Mechanism

2 NO N2O2 (fast)N2O2 + O2 2 NO2 (slow)

Based on slow step: rate = k2[N2O2][O2]

Substitute for intermed: k1[NO]2 = k-1[N2O2]rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]

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] since fast reversible reaction occurs, conc are related stoichiometrically.

Rate = k[IBr][IBr] = k[IBr]2

Molecules must collide to react.

Concentration affects rates because collisions are more likely if more molecules are present.

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

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.

Said that the 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.

Number of collisions with the required

energy = z e-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

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

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

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 fastLow 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

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 than the reactants.

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

Pt surface

HH

HH

C HH C

HH

Pt surface

HH

HH

The double bond breaks and bonds to the catalyst.

C HH C

HH

Pt surface

HH

HH

The hydrogen atoms bond with the carbon

C HH C

HH

Pt surface

H

C HH C

HH

H HH

Chlorofluorocarbons catalyze the decomposition of ozone.

Enzymes regulating the body processes. (Protein catalysts)

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

Concentration of reactants

Rate Rate increases until the active sites of

catalyst are filled.

Then rate is independent of concentration