a diamond is forever. is it so? c diamond c graphite Δg = ∑ Δg products - ∑ Δg reactants Δg...

A diamond is forever.

Is It So?

C diamond C graphite

ΔG = ∑ ΔGproducts - ∑ ΔGreactants

ΔG = ΔGgraphite - ΔGdiamond

ΔG = (0) - (3 kJ/mol)

ΔG = -3 kJ/mol

Consider: Gibb’s Free Energy

?

Look quick, before it turns into graphite.

While it’s true her diamond is spontaneously turning into graphite before her eyes, it’s happening very

slowly. Don’t hold your breath waiting to see any change. It takes billions of

years.

While thermodynamics answers the question as to whether or not a reaction or event is spontaneous, it DOES NOT tell how fast a reaction goes.

This is what kinetics does....describes the rate of the reaction.

Chemical Kinetics are about the only tool a chemist has to probe the actual mechanism of a chemical reaction.

It is the study of reaction rates.

NO2 NO + O2Reaction Rate: The slope of a Concentration vs Time graph.

Reaction Rates

Reaction Rates

Factors affecting reaction rates:

•The nature of the reaction mechanism

•The concentration of the reactants

•The temperature at which the reaction occurs

•The presence of a catalyst

•The surface area of solid or liquid reactants or catalysts

Red Blue

Rate Laws describe reaction rates mathematically

Rate = k[reactant 1]m [reactant 2]n…

Akdt

Ad

To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations.

There are two forms of the rate law…Differential rate laws: Shows how

the rate depends on concentration. Sometimes called just the “rate law”.Integrated rates laws: Shows how the concentration depends on time.

Rate Laws

The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated.

Integrated Differential

The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated.

Integrated Differential

This of course requires the use of my calculus.

Other Stuff to Know: Reaction Order

The differential rate law for most reactions has the general form….

Rate = k[reactant 1]m [reactant 2]n…

The exponents m and n are called reaction orders. Their sum (m + n) is called the overall reaction order.

Other Stuff to Know: Reaction Order

Usually the values of the reaction order must be determined experimentally, they cannot be predicted by looking at the chemical reaction.

In most rate laws, reaction orders are 0, 1, or 2. However, occasionally they are a fraction or even negative.

The rate of a reaction depends on concentration; however the rate constant does not.

The most common method for experimentally determining the differential rate law is the method of initial rates.

In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t = 0 as possible)

Finding The Differential Rate Law

Using Initial Rates to Determine the Form of the Rate Law

Exp # [A] [B] Initial Rate (M/s)

1 .100M .100M 4x10-5

2 .100M .200M 4x10-5

3 .200M .100M 16x10-5

A + B C

From this data, find the form of the rate law..

Rate = k[A]m[B]n

Example:

Exp # [A] [B] Initial Rate (M/s)

1 .100M .100M 4x10-5

2 .100M .200M 4x10-5

3 .200M .100M 16x10-5

n

n

m

m

k

k

Rate

Rate

]100[.

]200[.

]100[.

]100[.

104

104

1

25

5

n

n

]100[.

]200[.1 n21 n = 0

Rate = k[A]m[B]n

[B]0 = 1

Exp # [A] [B] Initial Rate (M/s)

1 .100M .100M 4x10-5

2 .100M .200M 4x10-5

3 .200M .100M 16x10-5

n

n

m

m

k

k

Rate

Rate

]100[.

]100[.

]100[.

]200[.

104

1016

1

35

5

m

m

]100[.

]200[.4 m24 m = 2

Rate = k[A]2[B]0 = k [A]2

Exp # [A] [B] Initial Rate (M/s)

1 .100M .100M 4x10-5

2 .100M .200M 4x10-5

3 .200M .100M 16x10-5

02 ][][ BA

ratek 3

2

5

104]100[.

104

Rate = 4x10-3 [A]2

Now, solve for k…

First Order Reactions

Rate =

Ln [A]t – ln [A]0 = -kt

ktA

At

0

ln

Akdt

Ad

Integrating, etc. leads to…

[A] = [A]o e-kt

Derive the Integrated First Order Rate Law:

][][

Akt

ARate

kxt

x

Given a first order differential rate law…

That is, the rate of change of the concentration of reagent A is proportional to how much A there is.

Let x = [A] (the concentration of A), Then…

The minus sign is because the value of [A], or x, is decreasing.

Transforming to calculus notation…

kxdt

dx

Separate the variables…

dtkx

dx

Integrate both sides…

dtkdxx

1

dtkdxx

1

Cktx ln

Cktx ee ln

ktC

Ckt

eex

eex

Integrating…

Solving for x…

Now use initial conditions to determine C.

At t = 0, x = [A]o [A]o = initial concentration of A

o

Co

CkCo

AC

eA

eeeA

][ln

ln][ln

][ )0(

ktCeex Recall that..

oC Ae ][

ktoeAAx ][][

ktoeAAx ][][

Take ln’s of both sides…

kto

kto

eA

eAA

ln][ln

][ln][ln

So…

ktAA

ktAA

o

o

]ln[][ln

][ln][ln

First Order Reactions

Ln [A]t – ln [A]0 = -kt

ktA

At

0

ln

These equations can be used to determine …

•The concentration of a reactant remaining at any time after the reaction has started.

•The time required for a given fraction of a sample to react.

•The time required for a reactant concentration to reach a certain level.

Akdt

AdRate

Half Life (First Order Process)

Half Life of a ReactionThe half life of a reaction, t½ , is the time

required for the concentration of a reactant to drop to one half of its initial value.

ktA

At

0

ln

2

1

0

021

ln ktA

A

ln (½) = -kt½

k

t 21ln

21 T½ =

0.693/k

Half life for a first order rate law is independent of the initial concentration of reactant. Thus, the half life is the same at any time during the reaction.

Half Life for a First Order Reaction

(Garden Variety Half Life – Radioactivity)

First order reaction example

The first order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 g/mLyr. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/ml water. Assume that the effective temperature of the lake is 12oC.

a. What is the concentration of the insecticide on June 1 of the following year?b. How long will it take for the concentration to drop to 3.0x10-7 g/ml?

a. What is the concentration of the insecticide on June 1 of the following year?

Ln [A]t = -kt + ln [A0]t=0

ln [insecticide] t=1 year = - (1.45g/mLyr)(1 yr) + ln (5.0x10-7 g/ml)

ln [insecticide] t=1 year = - (1.45g/mL) + (-14.51g/mL)ln [insecticide] t=1 year = -15.96 g/mL

[insecticide] t=1 year = e-15.96

[insecticide] t=1 year = 1.2x10-7 g/mL

Ln [A]t – ln [A]0 = -kt

b. How long will it take for the concentration of the insecticide to drop to 3.0x10-7 g/ml?

ln [A]t = -kt + ln [A0]t=0

ln (3.0x10-7 g/ml) = - (1.45 yr-1)(t) + ln (5.0x10-7 g/ml)t = -[ln (3.0x10-7g/ml) – ln (5.0x10-7g/ml)]/1.45yr-1

t = - [-15.02 + 14.51]/1.45yr-1

t = 0.35 yr

Second Order Reactions

A second order reaction is one whose rate depends on the reactant concentration raised to the second power, or on the concentrations of two different reactants, each raised to the first power. For a reaction that is second order in just one reactant, A, the rate law is given by…

Rate = k[A]2

Second Order Reactions

Rate = k[A]2

2Akt

A

Integrating and stuff leads to the Integrated Rate Law…

0

11

Akt

A t

0

11

Akt

A t

Setting [A] = ½[A]o and solving for time gives…

0

121

Akt

Unlike the half life of first order reactions, the half life of a second order reaction is dependent on the initial concentration of the reactant.

Second Order Half Life’s

Half Life for a Second Order Reaction

Zero Order Reaction

A zero order reaction is one whose rate depends on the reactant concentration raised to the zero power, or in other words, it does not depend on the concentration of the reactant as long as some reactant is present. For the reaction that is zero order in just one reactant, A, the rate law is given by… Rate = k[A]0

Integrating… [A]t = -kt + [A]0

Solving for t½ T½ = [A]0/2k

Half Life for a Zero Order Reaction:

(Rate doesn’t depend on concentration)

Rate Laws: A Summary

To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations.

There are two types of rate laws…

Differential rate laws: Shows how the rate depends on concentration.

Integrated rate laws: Shows how concentration depends on time.

The most common method for experimentally determining the differential rate law is the method of initial rates. In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t=0 as possible).

The point is to evaluate the rate before the concentrations change significantly from the initial values. From a comparison of the initial rates and the initial concentrations the dependence of the rate on the concentrations of various reactants can be obtained – that is the order of each reactant can be determined.

Rate Laws: A Summary

Rate Laws: Summary

To experimentally determine the integrated rate law for a given reaction, concentrations are measured at various values of time as the reaction proceeds. Then the job is to see which integrated rate law correctly fits the data. Typically this is done visually, by ascertaining which type of plot gives a straight line.

Once the correct straight line plot is found, the correct integrated rate law can be chosen and the value of the rate constant, k, can be obtained from the slope of the plot. Also, the differential rate law for the reaction can be calculated.

Reaction Order

Differential Rate Law

Integrated Rate Law

Plot for Straight

Line

Slope of Plot

Zeroth -k

First -k

Second k

oAktA

A

1 tvs

A

1

k

dt

Ad

Akdt

Ad

2Akdt

Ad

ktAA o

ktoeAA tvsAln

tvsA

All the Good Stuff

Reaction Order

Differential Rate Law

Integrated Rate Law

Plot for Straight

Line

Slope of Plot

Zeorth -k

First -k

Second k

oAktA

A

1 tvs

A

1

k

dt

Ad

Akdt

Ad

2Akdt

Ad

ktAA o

ktoeAA tvsAln

tvsA

We can use this information to experimentally find the rate law for a reaction.

NO2 NO + O2

The gas phase decomposition of NO2 is studied at 383oC giving the following data.

Time (sec)Conc. NO2

Molar

500.1000

1000.0170

1500.0090

2000.0062

250 0.0047

ln Concentration vs Time

-6.0000

-5.0000

-4.0000

-3.0000

-2.0000

-1.0000

0.0000

0 5 10 15 20 25

Time (sec)

ln c

once

ntra

tion

TimeMolar Conc.

ln Conc.

500.1000 -2.3026

1000.0170 -4.0745

1500.0090 -4.7105

2000.0062 -5.0832

250 0.0047 -5.3602

Original Data

Graph

Graph the ln (concentration) vs Time. It the plot is a straight line, the reaction is first order.

concentration-1 vs Time

y = 10.16x + 9.1984

0.0000

50.0000

100.0000

150.0000

200.0000

250.0000

0 5 10 15 20 25

Time

Conc

-1

TimeMolar Conc.

Conc.-1

500.1000 10.0000

1000.0170 58.8235

1500.0090 111.1111

2000.0062 161.2903

250 0.0047 212.7660

Original Data

Straight line Second Order Reaction

Graph the 1/(concentration) vs Time. It the plot is a straight line, the reaction is second order.

Reaction constant k = 10.16

Remember that the rate equation between two substances A and B looks like this…

Rate = k[A]m[B]n

The equation shows the effect of changing the concentrations of the reactants on the rate of the reaction. What about all the other things like temperature and catalysts, for example which also change rates of reaction? Where do these fit into this equation?These are all included in the so-called rate constant – which is only actually constant if all you are changing is the concentration or the reactants. If you change the temperature or the catalyst the rate constant changes. These changes of the rate constant are shown mathematically in the Arrhenius Equation.

Svante August Arrhenius was a Swedish physical chemist best known for his theory that electrolytes, certain substances that dissolve in water to yield a solution that conducts electricity, are separated, or dissociated, into electrically charged particles, or ions, even when there is no current flowing through the solution.

In 1903 he was awarded the Nobel Prize for Chemistry.

1859-1927

Svante August Arrhenius

RTaE

Aek

K = rate constant

A = frequency factor

Ea = activation energy

R = the gas constant (8.314 J/mol·K)T = temperature in Kelvin

The Arrhenius Equation

RT

Eactivation

Aek

The Arrhenius Equation

RT

EAk activationlnln

Or the logarithmic form…

RTEactivation

e

The expression…

For reasons that are beyond the scope of any course at this level, this expression counts the fraction of the molecules present in a gas which have energies equal to or in excess of the activation energy at a particular temperature.

The frequency factor, A is sometimes called the pre-exponential factor of the steric factor.

It is a term which includes factors like the frequency of collisions and their orientation. It varies slightly with temperature, although not much. It is often taken as constant across small temperature ranges.

RT

Eactivation

Aek

Using the Arrhenius Equation: The effect of a change of temperature.

You can use the Arrhenius equation to show the effect of a change of temperature on the rate constant – and therefore on the rate of the reaction.

If the rate constant doubles, for example, so will the rate of the reaction.

RT

Eactivation

Aek

Example: What happens to the rate of a reaction if the temperature increases from 20oC to 30oC?

The frequency factor, a, is approximately constant for such a small temperature change. We need to look at how e-

(Ea/RT) changes --- the fraction of molecules with energies equal to or in excess of the activation energy.

9

29331.8000,50

102.1

ee RTEactivation

Assume an activation energy of 50 kJ/mol. In the equation, we have to write that as 50,000 J/mol. The value of the gas constant R is 8.31 J/K·mol

Now raise the temperature just a little bit to 30oC (303k)

9

30331.8000,50

104.2

ee RTEactivation

9

29331.8000,50

102.1

ee RTEactivation

9

30331.8000,50

104.2

ee RTEactivation

At 20oC At 30oC

The fraction of the molecules able to react has almost doubled by increasing the temperature by 10oC. That causes the rate of the reaction to almost double.

This is the basis for the old rule of thumb that a reaction rate doubles for every 10oC rise in temperature.

At the higher temperature, more molecules have energy greater than the activation energy Ea.

Catalysts: A substance that changes the rate of a reaction without being consumed in the reaction.

•Provides an easier way to react.

•Lowers activation energy

•Enzymes are biological catalysts.

•Inhibitor: A substance that decreases the rate of reaction (a negative catalyst)

Cattleist?

The effect of a catalyst is to lower the activation energy.

Energy Energy

Uncatalyzed Catalyzed

Lower activation energy means more molecules will react.

The effect of a catalystA catalyst provides an alternate reaction mechanism, or

route for the reaction. This alternate route necessarily has a lower activation energy. The overall effect of a catalyst is to lower the activation energy.

Continuing our example of a reaction with an activation energy of 50,000 J/mol… What is the effect on the rate of lowering the activation to 25,000 J/mol?

5

29331.8000,25

105.3

ee RTEactivation

9

29331.8000,50

102.1

ee RTEactivation

With Catalyst Without Catalyst

No wonder catalysts speed up reactions!

The Collision Model

Molecules must collide to react. The greater the number of collisions occurring per second, the greater the reaction rate.

Only a small fraction of collisions actually lead to a reaction.

In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy in order to react. In order to react, colliding molecules must have a total kinetic energy equal to or greater than some minimum value. This minimum energy is called the activation energy (Ea). The value of Ea varies from reaction to reaction.

Factors Influencing Reaction Rates

The rate of a given reaction can be affected by:

•The physical states of the reactants

•The concentration of the reactants (for gases, pressure is equivalent to concentration)

•The reaction temperature

•The presence of a catalyst

The greater the concentration, the greater the frequency of collisions.

The Greater the Temperature the Greater the Frequency of Collisions.

An increase in temperature generally increases the rate of a reaction.

According to kinetic theory, at higher temperature the atoms and molecules move faster.

1. They collide more frequently

2. They collide with greater energy and each collision has a greater chance of being successful.

Note: The activation energy does NOT get any lower!

Not all collisions produce a reaction

Reactants

Products

Intermediaries – Activated Complexes

Energy Diagrams

Molecularity

Chemical reactions take place as a result of collisions between molecules. Each type of collision in a given reaction is called a single event or an elementary step.

The number of molecules that participate as reactants in an elementary step defines the molecularity of the step.

Unimolecular: A single molecule is involved in the elementary step

Bimolecular: Two molecules are involved

Termolecular: Three molecules are involved. Termolecular steps are far less probable than unimolecular or bimolecular processes and are rarely encountered.

The chances that four or more molecules will collide simultaneously

with any regularity is extremely remote, consequently, such collisions

are never proposed as a part of a reaction mechanism.

The reaction between NO2 and CO has the overall reaction:

NO2 + CO NO + CO2

A study of the kinetics of this reaction revealed the rate law for the reaction is.

Rate = k[NO2]2

This Rate Law requires that the slow step of the reaction involves a collision between two NO2 molecules. How can this be a step in the seemingly simple reaction above?

Example:

NO2 + CO NO + CO2

NO2 + NO2 NO3 + NO

NO3 + CO NO2 + CO2

2NO2 + NO3 + CO NO2 + NO3 + NO + CO2

NO2 + CO NO + CO2

Further study of this reaction established that two NO2 molecules can react as follows…

NO3 is a highly reactive material which is capable of transferring an oxygen atom.

NO2 + CO NO + CO2

NO2 + NO2 NO3 + NO

NO3 + CO NO2 + CO2

NO2 + CO NO + CO2

slow

fast

The first equations sets the rate law, so it must be the slow one.

Remember, the rate law only provides information about the slowest reaction in the mechanism.

H-C-C-O-C-H + H2O H-C-C-OH + H-C-OH

H

H H

H

H

H

H

HO

Methyl Acetate + water Acetic Acid + Methanol

Experimental evidence for reaction mechanisms:

Consider the reaction of methyl acetate with water….

The methyl acetate molecule can break in one of two places to complete the reaction…

H-C-C-O-C-H + H2O H-C-C-OH + H-C-OH

H

H H

H

H

H

H

HO

Methyl Acetate + water Acetic Acid + Methanol

H-C-C- O-C-H

O-H

HH

H O H

H

-OR-

H-C-C-O-C-H + H2O H-C-C-OH + H-C-OH

H

H H

H

H

H

H

HO

Methyl Acetate + water Acetic Acid + Methanol

H-C-C- O-C-H

O-H

HH

H O H

H

H-C-C-O- -C-H

H

H

H

H

O

O-H

H

-OR-

How can we determine experimentally which is correct?

H-C-C-O18-HH-O18

H

H

H

Use a radioactive isotope of oxygen, O18 To make up the water

O

H-C-C-O-C-H + H2O H-C-C-OH + H-C-OH

H

H H

H

H

H

H

HO

Methyl Acetate + water Acetic Acid + Methanol

H-C-C- O-C-H

O-H

HH

H O H

H

H-C-C-O- -C-H

H

H

H

H

O

O-H

H

-OR-

Elementary Steps

Experimental studies of reaction mechanisms begin with rate measurements. Next this data is analyzed to determine the rate constant and order of the reaction. The rate law is then written, and finally a plausible mechanism for the reaction in terms of elementary steps is formulated.

The elementary steps must satisfy two requirements…

1. The sum of the elementary steps must give the overall balanced equation for the reaction.

2. The rate-determining step, which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law as is determined experimentally.

Rate Laws of Multistep Mechanisms

•The slowest elementary step is the rate determining step.

•The rate determining step governs the rate law for the overall reaction.

•For elementary steps, the exponent in its rate are the same as the coefficients of the reactants in the chemical equation. Thus, the rate law for an elementary process can be predicted from the chemical equation for the process.

Rate Laws of Elementary Steps

A Products Rate = k[A]

A+A Products Rate = k[A]2

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

A+A+A Products Rate = k[A]3

A+A+B Products Rate = k[A]2[B]

A+B+C Products Rate = k[A][B][C]

Measuring the rate of a reaction.

Formulating the Rate Law

Postulating a reasonable

reaction mechanism

Sequence of steps in the study of a reaction mechanism

Decomposition of Hydrogen Peroxide

The decomposition of Hydrogen Peroxide is facilitated by iodide ions.

The overall reaction is…

2H2O2 2H2O + O2

By experiment, the rate law is found to be…

rate = k[H2O2][I-]

Thus the reaction is first order with respect to both H2O2 and I-.

Example:

rate = k[H2O2][I-]

You can see that H2O2 decomposition does not occur in a single elementary step corresponding to the overall balanced equation.

If it did, the reaction would be second order in H2O2 (as a result of the collision of two H2O2 molecules). What’s more, the I- ion, which is not even part of the overall equation, appears in the rate law expression.

We can account for the observed rate law by assuming that the reaction takes place in two separate elementary steps, each of which is bimolecular:

step 1: H2O2 + I- H2O + IO-

step 2: H2O2 + IO- H2O + O2 + I-

2H2O2 2H2O + O2

step 1: H2O2 + I- H2O + IO-

step 2: H2O2 + IO- H2O + O2 + I-

The first step must be the rate determining step, because it matches the rate law. Thus, the rate of the reaction can be determined from the first step alone:

Note that the IO- ion is an intermediate because it does not appear in the overall balanced equation. Although the I- ion also does not appear in the overall equation, I- differs from IO- in that the former is present at the start of the reaction and at its completion. The function of I- is to speed up the reaction --- that is, it is a catalyst.

rate = k[H2O2][I-]

Slow

Another Example:

Consider the Reaction…

2NO2Cl 2NO2 + Cl2

This is found to be a first order reaction, it’s experimentally determined rate law is…

rate = k[NO2Cl]

Could the overall reaction occur in a single step by the collision of two NO2Cl molecules? This would lead to…

rate = k[NO2Cl]2

Chemists believe the actual mechanism is…

NO2Cl NO2 + Cl

NO2Cl + Cl NO2 + Cl2

Slow

Fast

Still Another Example:

Consider the reaction…

2NO + 2H2 N2 + 2H2O

Experimentally, the rate law has been found to be…

rate = k[NO]2[H2]A chemically reasonable mechanism that yields the correct form for the rate law is…

2NO + H2 N2O + H2O (slow)

N2O + H2 N2 + H20 (fast)

However, there is a serious flaw in the proposed mechanism --- it has a elementary process that involves the simultaneous collision between three molecules, two of which must be NO and one H2.

An alternative proposal is…

2NO ↔ N2O2 (fast)

N2O2 + H2 N2O + H2O (slow)

N2O + H2 N2 + H2O (fast)

Since the second step is rate determining, the rate for the reaction should be…

rate = k[N2O2][H2]

However, the experimental rate law does not contain the species N2O2, an intermediate.

Consider the reaction…

2NO + 2H2 N2 + 2H2O

2NO ↔ N2O2 (fast)

N2O2 + H2 N2O + H2O (slow)

N2O + H2 N2 + H2O (fast)

If, however, we view the first equation as a dynamic equilibrium reaction, then the rate of the forward and reverse reactions are equal. Then…

Rate (forward reaction) = kf[NO]2

Rate (reverse reaction) = kr[N2O2]

The first elementary step is an equilibrium reaction, so consider both forward and reverse reactions.

Kf[NO]2 = kr[N2O2]

[N2O2] = kf/kr[NO]2

Kf[NO]2 = kr[N2O2]

[N2O2] = kf/kr[NO]2

Substituting into the proposed rate law (rate = k[N2O2][H2]) gives….

Or rate = k’ [NO]2[H2]

The rate law derived from the proposed mechanism matches the experimental rate law, thus the three step mechanism does appear to be reasonable on the basis of kinetics.

rate = k (kf/kr)[NO]2[H2]