Ch

emis

try

XX

IUnit 5

How do we predict chemical change?

M3. Measuring Rates Analyzing the factors that affect reaction rate.

M2. Comparing Free EnergiesDetermining the directionality and

extent of a chemical reaction.

M1. Analyzing Structure Comparing the relative stability of different substances

M4. Understanding Mechanism Identifying the steps that determine reaction rates.

The central goal of this unit is to help you identify and apply the different factors that help predict the

likelihood of chemical reactions.

Ch

emis

try

XX

IUnit 5

How do we predict chemical change?

Module 4: Understanding Mechanism

Central goal:

To use reaction mechanisms to make

predictions about reaction rate and vice versa.

Ch

emis

try

XX

I

The Challenge

Imagine that you were interested in understanding

why certain types of substances and processes

appeared in our planet.

How can we use reaction mechanisms to make predictions?

How can we deduce reaction mechanisms based on reaction outcomes?

TransformationHow do I change it?

Ch

emis

try

XX

I

Reaction PathwaysOur ability to predict the most likely outcomes of a chemical reaction improves considerably when we

understand the mechanism that leads from reactants to possible products.

Most reaction mechanisms involve

several steps. However, some steps play a more

central role than others in determining the overall

rate of reaction. CH4(g) + O2(g)

CO2(g) + 2 H2O(g)

Ch

emis

try

XX

IReaction Steps

In the mechanistic model, the overall reaction is viewed as the result of multiple elementary reactions

or steps occurring simultaneously in the system.

A + A B + C Bimolecular

2A 2C + E

For example, the overall reaction: 2A 2C + E may involve two elementary steps:

UnimolecularB C + E

B is an intermediate

Detecting intermediates is an important means of investigating reaction mechanisms.

Ch

emis

try

XX

IOverall Rate Order

In the mechanistic model, the overall rate of the reaction is an “emergent property” of the rates of the

individual steps.

For many reactions, one step is slow enough to limit the rate of the overall reaction:

A + A B + C Slow

2A 2C + E

FastB C + E

Overall Rate = k [A][A] = k [A]2

Ch

emis

try

XX

I

Mirror ImagesConsider the following problem of central relevance

for our understanding of the origin of life:

Most amino acids found on Earth appear in only

one of two possible mirror-image forms,

called enantiomers or optical isomers.

L D

These isomers have most of the same properties, but react differently with L and D

isomers of other “chiral” substances.

Non-superimposable

Ch

emis

try

XX

I

Chiral Centers

Molecular chirality is commonly caused by the presence of carbon atoms in a molecule attached

to four different groups:

Chiral carbonNon-Chiral

carbons

Let′s think!

Identify the chiral carbons in this molecule:

L-Glucose

Ch

emis

try

XX

I

Left or Right HandedMany biologically active molecules are chiral, including the naturally occurring amino acids,

which tend to be “left-handed” (L).

Chirality is of central importance for many biological functions.

Different enantiomers interact differently with

the chiral molecules (proteins, DNA)

in our body.

How this preferred chirality emerged on our planet?

Ch

emis

try

XX

I

RacemizationPure samples of L or D amino acids eventually

convert into a mixture of both forms. This process is called racemization. How do we explain it?

backward

Reaction Coordinate

G

HOOC

C H

H2N

H3C

L

NH2

CH3

COOH

CH

D Gorxn = 0

Mechanism

1. Unimolecular Step

Ea = 124 kJ/mol

H+

C

NH2

H3C COOH

forward2. Bimolecular Step

Ch

emis

try

XX

I

Mechanism

The racemization process can be explained by this two-step mechanism:

L C- + H+ Unimolecular (Slow) Rate = k[L]

H+ + C- D Bimolecular (Fast) Rate = k[C-][H+]

where H+ and C- are intermediates.

The rate of the overall reaction is determined by the slowest step (Rate Determining Step), thus:

L D Rate = k [L] (First Order Reaction)

The same ideas apply to the backward reaction D L.

Ch

emis

try

XX

I

According to this mechanism, the interconversion between the L and D forms should occur

with the same probability.

Let’s Think

Imagine that you start with 1 M solution of L-Ala and 30% of it transforms to D-Ala every second.

The same percentage of D-Ala in the system

transforms to L-Ala in that time.

Follow the evolution of the system. When does

the process “stop”?

2

3

1

01 0

D-Ala (M)L-Ala (M)t (s)

Ch

emis

try

XX

I

0.3

0

0.7 0.3

0.7 x 0.3 = 0.21

0.3 x 0.3 = 0.09

0.58 0.42

0.58 x 0.3 = 0.174

0.42 x 0.3 = 0.126

0.532 0.468

0.532 x 0.3 = 0.1596

0.468 x 0.3 = 0.1404

0.5128 0.4872

Let’s Think t (s) L-Ala (M) D-Ala (M)

0 1 0

L D

D L

1

L D

D L

2

L D

D L

3

L D

D L

4

Etc.

L D 40%

D L 40%

Equilibrium?

Ch

emis

try

XX

I

The reaction keeps going at equilibrium but [L] and [D] remain constant.

Kinetics and EquilibriumChemical equilibrium is reached when the rate of the forward reactions is equal to the rate of the

backward process.

L D

D L

Rate = kf[L]

Rate = kb[D]

kf[L]eq = kb[D]eq][

][

L

D

k

kK

b

fc L D

In this case, kf = kb. Thus, Kc = 1. 1

RT

G

c

orxn

eK

Ch

emis

try

XX

I

Amino Acid ChiralityThe origin of biologically active amino acids’ chirality is an unresolved problem in science.

Many hypothesis have been suggested.

Let′s think!

Imagine that once amino acids are linked into proteins, the % of an L amino

acid converted to the D form is 20% every second, versus 80% conversion from D to L in the same period of time.

What would be [D]/[L] at equilibrium?

][

][

L

D

k

kK

b

fc

L D kf

D L kb

kb = 4 kf Kc = ¼

Ch

emis

try

XX

I

Let’s Think Consider the following data derived through our work in

this unit:

2

3

+23 3

2

H2O

2

3

2 2+3

DecompositionEa = 177 kJ/mol

DimerizationEa = 88.7 kJ/mol

HOOC

C H

H2N

H3C

NH2

CH3

COOH

CHRacemization

Ea = 124 kJ/mol

Build a hypothesis about what could have prevented amino acid racemization on primitive Earth.

Ch

emis

try

XX

I

Changing Mechanism

It has been suggested that one crucial step in the origin of life was the synthesis of substances that

could speed up the rate of certain chemical reactions.

As we know, these “catalysts”

act by either reducing the

activation energy Ea or changing

the reaction mechanism.

Ch

emis

try

XX

I

Enzymes

The substances that catalyze biological processes are called enzymes. One common model to explain

their behavior is the “lock and key” model.

MechanismE + S ES Fast

ES E + P Slow

(E)

(P)

(ES)

(S)

(E)

Ch

emis

try

XX

I

Enzyme Kinetics

Go

E + SE + P

ES

E + S ES Fast

ES E + P Slow

The rate law for this process is determined by the second

step: Rate = k2 [ES]

This expression is not very useful given that we cannot easily follow [ES] as a function of time.

If we assume that equilibrium is reached

in the first step:

]][[][]][[

][SEKES

SE

ESK cc

Rate = k2 Kc [E][S]

Ch

emis

try

XX

I

Let’s Think

For a given concentration of enzyme [E]o, the

reaction rate is first order in [S], but only at low concentrations of the

substrate. At high concentrations:

Rate ~ k [S]0 = k (constant)

(Zeroth order process)

Use the lock and key model to explain this behavior.

Ch

emis

try

XX

I

Let’s ThinkThe "lock and key" model has proven inaccurate.

The induced fit model is the most currently accepted

E + S ES Fast

ES EP Slow

EP E + P Fast

Draw an energy profile for this mechanism and analyze whether the associated rate law needs to be modified.

Ch

emis

try

XX

I

The same rate law: Rate = k2 Kc [E][S]

Induced Fit Model

Ch

emis

try

XX

I

Auto-Catalysis

A + B 2 B

Rate = k [A] [B]

Processes in which the reaction is catalyzed by its own products are called auto-catalytic and may have played a central role in the origin of life.

t

[B]

Why this shape?

The presence of auto-catalytic steps in some reaction mechanisms may explain the

appearance of metabolic cycles.

Ch

emis

try

XX

I

Let’s Think

Consider this mechanism:

A + X 2 XX + Y 2 Y

Y B

Hint: Think of A as grain, X as ducks, Y as wolves, and B as “dead” wolves.

a) Write the overall reaction;

b) Identify the auto-catalytic steps and the intermediate species;

c) Predict the structure of the plots [X] and [Y] vs. t as the reaction proceeds. t

[X]

Ch

emis

try

XX

I

Oscillating Reactions

[X]

[Y]

A + X 2 X X + Y 2 Y

Y B

Ch

emis

try

XX

I Assess what you know

Let′s apply!

Ch

emis

try

XX

I

Let′s apply!

2

3

+2

3 3

2

H2O

Analyze

which we found to be a second order reactionRate = k [Aa]2

Consider the dimerization of amino acids (Aa):

Several possible mechanisms have been proposed for this type of reaction.

Ch

emis

try

XX

ILet′s apply!

AnalyzeGo

2AaAa-Aa+ H2O

Go

2AaAa-Aa+ H2O

Aa-Aa*

2 Aa Aa-Aa*

Aa-Aa* Aa-Aa + H2O

2 Aa Aa-Aa + H2O

Is there a way to determine which of these mechanisms is more plausible?

Could you propose a different

mechanism that leads to the same

experimental rate law

Rate = k[Aa]2?

Ch

emis

try

XX

ILet′s apply!

Both mechanisms lead to Rate = k[Aa]2. We would have to experimentally confirm the

existence of the intermediate.

Analyze

Go

2AaAa-Aa+ H2O

Aa-Aa*

Another possibility:2 Aa Aa-Aa* Fast

Aa-Aa* Aa-Aa + H2O Slow

Rate = k2 [Aa-Aa*]

Kc = [Aa-Aa*]/[Aa]2

[Aa-Aa*] = Kc [Aa]2

Rate = k2 Kc [Aa]2

Ch

emis

try

XX

I

Discuss with a partner one thing you do not fully understand about

the content of this Module.

Ch

emis

try

XX

I

Understanding Mechanism

Summary

Reaction mechanisms allow us to understand reaction kinetics and make predictions about most

likely outcomes.

Most chemical processes can be thought as occurring in a sequence of elementary steps:

A + B C + D Bimolecular Rate = k1[A][B]

C B + E Unimolecular Rate = k2[C]

A D + E

The rate law is determined by the slowest step.

Ch

emis

try

XX

I

Catalysts

Catalysts” act by either reducing the

activation energy Ea or changing

the reaction mechanism.

The reaction mechanism can be altered by the presence of substances that help create alternative

reaction paths.

Ch

emis

try

XX

I

Are You Ready?

Ch

emis

try

XX

I

The Quest for Ammonia

Ammonia, NH3, is one of the most important industrial

chemical substances.

It is widely used in the production of fertilizers,

pharmaceuticals, refrigerants, explosives, and cleaning agents.

It ranks as one of the 10 top chemicals substances produced

annually in the world.

Ch

emis

try

XX

I

The Synthesis

Ammonia is mainly produced via this simple chemical reaction:

1/2 N2(g) + 3/2 H2(g) NH3(g)

Compare the energetic and entropic stability of reactants and products.

Make a prediction of the signs of Ho

rxn and Sorxn for this process.

Energy: Horxn < 0

A-A bonds A-B bonds

Entropy: Sorxn < 0

4 mol gas 3 mol gasmixture one compound

Let′s think!

Ch

emis

try

XX

I

Reaction Horxn (kJ) So

rxn (J/K)

1/2 N2(g) + 3/2 H2(g) NH3(g) -45.9 -99.1

Directionality

Use these data to calculate Gorxn.

Write Kp for this reaction and calculate its value at 25 oC.

Let′s think!

Grxn= Hrxn–TSrxn = -45.9 – 298.15*0.0991 = -16.4 kJ

242/32/1

1047.7))15.298314.8/(1064.1exp(22

3 xxxPP

PK

HN

NHp

Ch

emis

try

XX

I

Reaction Horxn (kJ) So

rxn (J/K)

1/2 N2(g) + 3/2 H2(g) NH3(g) -45.9 -99.1

K = 1 Grxn = 0

Grxn= -45.9 – T*0.0991 = 0 T = 463 K

The reaction is favored at low temperatures.

Is the synthesis of ammonia thermodynamically favored at low or high temperatures?

Estimate the temperature at which the directionality switches (K = 1)?

Let′s think!

Ch

emis

try

XX

I

Energy Profile

Reaction Horxn (kJ) Ea (kJ/mol)

1/2 N2(g) + 3/2 H2(g) NH3(g) -45.9 325

Use the following information to build the energy profile for the reaction:

Let′s think!

1/2 N2 3/2 H2

NH345.9 KJ

325 kJ

Ep

Reaction Coordinate

Ch

emis

try

XX

I

Reaction Horxn (kJ) Ea (kJ/mol)

1/2 N2(g) + 3/2 H2(g) NH3(g) -45.9 325

Reaction Conditions

Given its high Ea, the reaction is normally done at high T (~500 oC) and P (~200 atm).

Discuss how the increased T will affect the:

Rate (Calculate k500/k25)Extent (Calculate K500/K25)

of the reaction.

Let′s think!

Ch

emis

try

XX

I

Reaction Conditions

RT

Ea

Aek

35298

1

773

1

314.8

1025.311

1000.1

5

12

1

2 xeek

k xTTR

E

T

Ta

)(R

S

RT

H orxn

orxn

eK

5298

1

773

1

314.8

1059.411

1014.1

4

12

1

2

xeeK

K xTTR

H

T

T

orxn

Ch

emis

try

XX

I

Reaction Conditions

High temperature increases reaction rate, but decreases reaction extent. That is why the

reaction is carried out at high P too.

Discuss why high P favors the product side in this process:

Let′s think! N2(g) + 3 H2(g) 2 NH3(g)

The collision rate is higher in the side with more particles. The forward reaction is favored.

Ch

emis

try

XX

ICatalysts and Reaction Order

The synthesis of NH3 is carried out in the presence of catalysts. The order of the reaction depends on

the composition and structure of this catalyst.

t(s) 0 0.1 0.2 0.3 0.4 0.5 0.6

CN2(mol/L) 2.00 1.68 1.42 1.19 1.01 0.846 0.71

What is the reaction order with respect to N2(g)? What is the value of the rate

constant under these conditions?Let′s think!

For example: T = 500 oC P = 200 atm Fe catalyst

The rate does only depends on the concentration of N2(g).

Ch

emis

try

XX

IReaction Order

t(s) 0 0.1 0.2 0.3 0.4 0.5 0.6

CN2(mol/L) 2.00 1.68 1.42 1.19 1.01 0.846 0.71

ln(C) 0.693 0.521 0.349 0.177 0.005 -0.167 -0.339

Rate = k [N2(g)]

First order with respect to

[N2(g)]

k = 1.72 s-1

y = -1.72x + 0.6931

-0.4

-0.2

0

0.2

0.4

0.6

0.8

0 0.2 0.4 0.6 0.8

t(s)

Ln

(C)

ln(C)= -kt + ln(Co)

Ch

emis

try

XX

IReaction Mechanism

In the presence of a solid catalyst, the reaction takes place on the surface of the solid.

N2(g) N2(ad)

N2(ad) 2 N(ad)

N(ad) +H(ad) NH(ad)

Let′s think!

What other mechanistic steps are involved in the synthesis of NH3? Which step can be expected to be

the slowest given that Rate = k[N2]?

Ch

emis

try

XX

IReaction

Mechanism

H2(g) H2(ad)

N2(g) N2(ad)

N2(ad) 2 N(ad)

H2(ad) 2 H(ad)

N(ad) +H(ad) NH(ad)

NH(ad) + H(ad) NH2(ad)

NH2(ad) + H(ad) NH3(ad)

NH3(ad) NH3(g)Slowest:

N2(g) N2(ad)

Rate = k [N2(g)]

Rate = k’ [N2(ad)]

Kc = [N2(ad)]/[N2(g)]

or N2(ad) 2 N(ad)

Rate = k’ Kc [N2(g)]

Ch

emis

try

XX

IActivation Energy

The presence of the catalyst reduces Ea considerably.

Use the following data to derive Ea for an Fe-based catalyst.

How many times faster is the reaction at T = 500 oC in the presence of the catalyst?

(Ea = 325 kJ/mol without it)

Let′s think!

T(oC) 25 100 200 300 400 500

k(s-1) 3.24x10-9 2.32x10-6 5.81x10-4 0.0211 0.264 1.72

Ch

emis

try

XX

I

y = -9743x + 13.146

-25

-20

-15

-10

-5

0

5

0 0.001 0.002 0.003 0.004

1/T

ln(k

)

T(oC) 25 100 200 300 400 500

k(s-1) 3.24x10-9 2.32x10-6 5.81x10-4 0.0211 0.264 1.72

Activation Energy

)ln(1

)ln( ATR

Ek a Ea = 9743R

Ea = 81 kJ/mol

12

1

2

1aa

a

aEE

RT

E

Ee

k

k

161009.31

2 xk

k

a

a

E

E