Transcript Slide 1

Unit 7
How do we analyze a
complex chemical system?
Through this course, we have learned to use
thermodynamic and kinetic properties
at the macroscopic level, and
Chemistry XXI
electronic and steric factors
at the submicroscopic scale
to predict and control chemical reactions.
Can you apply what you have learned to
analyze a relevant complex system?
Unit 7
How do we analyze a
complex chemical system
Chemistry XXI
The central goal of this unit is to apply and extend
central concepts and ideas discussed in this course
to the analysis of a complex chemical system.
M1. Tracking Electron Transfer
Detect electron transfer among
reacting species in a system.
M2. Detecting Electron Sharing
Analyze electron. sharing among
reacting species in a system.
M3. Analyzing Coupled Processes
Analyze processes occurring
simultaneously in a system.
Context
To illustrate the power of the concepts, ideas, and
ways of thinking discussed in the course, we will
focus our attention on understanding the causes
and effects of water acidification in our environment.
NO2(g) + O2(g)  NO(g) + O3(g)
Chemistry XXI
CO2(g) + H2O(l)  H2CO3(aq)
CO32-(s) + H3O+(aq)  HCO3-(aq) + H2O(l)
C(s) + O2(g)  CO2(g)
C8H18(l) + 12.5O2(g)  8CO2(g) + 9H2O(g)
Why do we care?
In which ways what
we have learned can
help us predict and
control water
acidification in our
planet?
The Problem
Many complex systems of interest, such as our
body or our planet, are made of hundreds of
substances in constant interaction.
The central question is how to apply chemical
concepts, ideas, and ways of thinking to predict and
control relevant processes:
Chemistry XXI
Outcomes
Mechanism
Directionality
Extent
Unit 7
How do we analyze a
complex chemical system?
Chemistry XXI
Module 1: Tracking Electron Transfer
Central goal:
To analyze charge
distribution in chemical
compounds to detect
transfer of electron density
among reacting species.
The Challenge
Transformation
How do I change it?
Chemistry XXI
We live in a complex environment made of
hundreds of different substances in constant
chemical interaction. Some of these
interactions are crucial for the survival of life on
Earth; others threaten several ecosystems.
How can we analyze the
types of chemical
processes in which
these substances are
involved?
Substances
To face this challenge we need to analyze:
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a) the chemical nature of the substances that
comprise the system;
For example, identifying
whether the substances
are molecular or ionic is
very useful in predicting
their physical and
chemical properties.
Let’s Think
CH4
H 2O
NO2
N2
CO2
H2CO3
O2
FeS2
SO2
SO3
HNO3
NO
Chemistry XXI
H2SO4
CaCO3
C6H12O6
O3
Al(OH)3
This is a list of substances involved in or affected by
water acidification in our planet.
Classify them as molecular or ionic.
The Substances
Ionic: CaCO3, FeS2, Al(OH)3
Chemistry XXI
Properties determined by the
charge and size of the ions in the
ionic network.
Molecular: CH4, CO2, C6H12O6, H2CO3
N2, NO, NO2, HNO3
O2, O3, H2O
SO2, SO3, H2SO4
Properties determined by their molecular
structure and charge distribution.
Substances and Reactions
In the analysis of complex chemical systems
we also need to understand:
a) the chemical nature of the substances that
comprise the system;
Chemistry XXI
b) the characteristics of the chemical reactions
in which they participate.
Again, analysis of structure
and charge distribution in
reactants and products is
crucial to make predictions
about the types of
processes that may occur.
The Reactions
For example, this set of interrelated chemical
processes lead to water acidification in our planet:
Chemistry XXI
CO2
Production
Acid
Formation
and
Dissolution
C6H12O6(g) + 6O2(g)  6H2O(l) + 6CO2(g)
CH4(g) + 2O2(g)  2H2O(l) + CO2(g)
H2O(l) + CO2(g)  H2CO3(aq)
H2CO3(aq) + H2O(l)
HCO3-(aq) + H3O+(aq)
Acid
Al(OH)3(s) + 3H3O+(aq)
Neutralization
Al3+(aq) + 6H2O(l)
How do we differentiate these processes?
Chemical Reactions
Most chemical reactions are driven by the
interaction between positive and negative charge
centers on different particles.
Chemistry XXI
Thus, a chemical reaction
tends to result in the
transfer or redistribution of
charged particles:
Electrons
Ions
d+
d-
+
-
among the reacting
species.
How can we decide what is actually happening?
Focus on Structure
To distinguish types of reactions and make
predictions about reactivity we need to have a
good idea of the structural features of reactants
and products.
Let’s consider one of the triggering reactions for
water acidification:
Chemistry XXI
CH4(g) + 2O2(g)  2H2O(l) + CO2(g)
Let′s think!
Build the Lewis structure each
substance. Predict their molecular
geometry, and their bond and molecular
polarity.
Focus on Structure
CH4
O2
d+
CO2
H2O
d+
d-
d-
d+
Chemistry XXI
d-
Tetrahedral
Linear
Bent
Linear
Non Polar
Non Polar
Polar
Non Polar
We can gain insights about chemical reactions
by analyzing the changes they induce in the
electron density around each atom.
Charge Transfer
CH4(g)
d+
+
2O2(g)

2H2O(l)
d+
d-
+
CO2(g)
d-
d+
Chemistry XXI
d-
Combustion reactions belong to an important type
of chemical processes characterized by the transfer
of electron density from one atom to another.
C goes from having d- to d+ (loses electron density).
O goes from neutral to d- (gains electron density).
How can we better characterize this charge transfer?
Oxidation Number
Chemistry XXI
The extent to which a reaction leads to
electron density transfer can be assessed by
analyzing changes in the
oxidation number (or state) of each atom.
The oxidation number is defined
as the partial charge that an atom
in a molecule would have if all of
the bonding electrons were
assigned to the most
electronegative atom in the bond
(molecule seen as fully ionic).
H+
H+ C-4 H+
H+
Oxidation Number
How do we determine the oxidation number?
Consider this distribution of
electrons in the molecule of CO2.
Chemistry XXI
If the bonding electrons are assigned
to the most electronegative atom:
C has 0 valence e- in the molecule.
O has 8 valence e- in the molecule.
Central Question:
How does these numbers compare with what
they would have in their elemental form?
Oxidation Number
To calculate the oxidation number (ON) we compare
the number of valence electron each atom has with
those that it would have in its elemental form:
ON = # of valence e- of the elemental atom –
# valence e- in fully ionic molecule.
Chemistry XXI
FC(O) = 6 – 8 = -2
FC(C) = 4 – 0 = +4
Notice that SON =
charge of molecule
We say carbon in the molecule is in a highly
oxidized state (largest positive ON), while oxygen
is in a highly reduced state (largest negative ON).
Let’s Think
CH4(g)
2O2(g) 
+
2H2O(l)
+
CO2(g)
d+
d+
Chemistry XXI
d-
d-
d+
d-
d+ d-
Assign ONs to all of the atoms in CH4, O2, and H2O.
Identify which atoms are “oxidized” (its oxidation
number increases) and which ones are “reduced” (its
oxidation number decreases) in this process.
+1
0
0
+1
-2
+1
-2
-4
C is oxidized (loses e- density),
O is reduced (gains e- density).
+4
-2
Redox Reactions
Combustion reactions are typical examples of
oxidation-reduction (redox) reactions in which the
oxidation number of the atoms involved changes,
signaling a transfer of electron density.
CH4(g)
+
2O2(g) 
2H2O(l)
+
CO2(g)
Chemistry XXI
Assigning oxidation numbers is useful in:
 Identifying electron-rich and electron-poor
centers in molecules;
 Tracking electron transfer or redistribution
during a chemical reaction;
 Making predictions about reaction directionality.
Redox Reactions
In general, we may expect that compounds with
highly electronegative atoms in high oxidation
states (  0 ) will be good oxidizing agents
(they can oxidize other substances).
Chemistry XXI
0
On the other hand,
compounds with
weakly eletronegative
atoms in low oxidation
states (  0 ) will be
good reducing agents
(they can reduce other
substances).
0
O2 is a good
oxidizing agent.
+
+1
-4
DG < 0
Hydrocarbons
are good
reducing agents.
Favored
Processes
Simple Rules
The assignation of the oxidation numbers of atoms in
chemical compounds can be greatly facilitated by
applying these basic rules:
1) ON = 0 for all atoms in elemental substances:
O2, O3, Al, Cu.
Chemistry XXI
2) ON = charge for monoatomic ions:
Cl-1 ON = -1
O-2 ON = -2
3) Some atoms, when combined, USUALLY have the
same oxidation number:
O usually is -2
F always is -1
H usually is +1
Simple Rules
4) SON = 0 for a neutral polyatomic formula.
CH4  ON(C) + 4ON(H) = 0
If ON(H) = +1  ON(C) = - 4
CO2  ON(C) + 2ON(O) = 0
If ON(O) = -2  ON(C) = + 4
Chemistry XXI
5) SON = charge for a charged polyatomic ions.
NH4+  ON(N) + 4ON(H) = + 1
If ON(H) = +1  ON(N) = - 3
SO4-2 
ON(S) + 4ON(O) = - 2
If ON(O) = -2  ON(S) = + 6
Let’s Think
Chemistry XXI
Most of the reactions that generate the substances
that are ultimately responsible for water
acidification in our planet are redox reactions:
The
relevant
substances
are oxides
of non
metallic
elements.
C6H12O6(g) + 6O2(g)  6H2O(l) + 6CO2(g)
N2(g) + O2(g)  2NO(g)
2NO(g) + O2(g)  2NO2(g)
2SO2(g) + O2(g)  2SO3(g)
Identify the oxidized and reduced atoms as well
as the oxidizing and reducing species.
Let’s Think
0 +1 -2
0
C6H12O6(g) +
Reducing
Oxidizing
+
2NO(g)
Chemistry XXI
+2 -2
 2NO(g)
O2(g)
+2 -2
Agents
0
+
Reducing
O2(g)
Oxidizing
+4 -2
Reducing
Agents
Oxidizing
0
+
O2(g)
Oxidizing
+4 -2
6H2O(l) + 6CO2(g)
0
Reducing
2SO2(g)

6O2(g)
0
N2(g)
+1 -2
+4 -2
 2NO2(g)
Agents
+6 -2
 2SO3(g)
Agents
Chemistry XXI
Let′s apply!
Assess what you know
Natural Cycles
Chemistry XXI
Compounds of sulfur and nitrogen play a
central role at various levels in our planet.
They participate in natural cycles that have
been altered by human activities.
Compounds of sulfur
and nitrogen, for
example, not only are
the cause of “acid rain”
but also contribute to
Global Warming and to
the depletion of the
Ozone Layer.
N 2O
NO2
HNO3
SO2
H2SO4
C 2H 6S
(DMS)
Sulfur Cycle
Chemistry XXI
SO2, H2S,
DMS, and
SO42- are the
main
chemical
species in
this cycle.
72
SO2 leads
to the
formation
of acids.
In million of
tons/year
9
FeS2
Main Sources
Biogenic Sources:
SO42-(aq) + 2CH2O(aq) + 2H3O+(aq)
H2S(g) + 4H2O(l) + 2CO2(g)
Anaerobic bacteria
CH3
S
H 3C
DMSP
+
CH2
CH2
O
Enzyme
H 3C
S
C
OH
DMS
H 3C
+
O
H 2C
C
CH
OH
+ H+
Chemistry XXI
Plankton
Anthropogenic Sources:
Combustion of coal, which typically contains 1%
to 3% sulfur in the form of pyrite (FeS2):
4FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)
Analyze
Let′s apply!
SO42-(aq) + 2CH2O(aq) + 2H3O+(aq)
CH3
S
H 3C
DMSP
+
CH2
CH2
O
DMS
S
C
OH
Chemistry XXI
Enzyme
H 3C
H2S(g) + 4H2O(l) + 2CO2(g)
H 3C
+
O
H 2C
C
CH
OH
+ H+
4FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)
Are these redox reactions?
If they are, identify the oxidized and reduced atoms,
and the oxidizing and reducing agents.
Further Oxidations
Further oxidation of sulfur compounds in the gas
phase in the atmosphere tends to follow reaction
mechanisms involving free radicals.
Chemistry XXI
hn
The hydroxyl radical
HO, constantly formed
and destroyed in the
atmosphere, plays a
central role in these
processes.
O3  O2 + O
O + H2O  2HO
H
ON = +1
O
ON = -1
Let′s apply!
Analyze
Consider the transformations of H2S to SO2, and
of SO2 to H2SO4:
H2S to SO2
1) H2S + HO HS + H2O
Chemistry XXI
2) HS + O3  O2 + HSO
SO2 to H2SO4
1) SO2 + HO HOSO2
3) HSO + O3  HSO2 + O2
2) HOSO2 + O2  HO2 + SO3
4) HSO2 + O2 HO2 + SO2
3) SO3 + H2O  H2SO4
a) In which of these steps are sulfur atoms being?
Oxidized; b) How is their ON changing in each step?
c) Is this change due to the actual transfer of e-?.
Chemistry XXI
Working in pairs, summarize the
important chemical information that
can be derived from the analysis of
the oxidation numbers of atoms in
reactants and products.
Tracking Electron Transfer
Summary
Oxidation-Reduction (Redox) are an important type
of chemical processes characterized by the transfer
of electron density from one atom to another.
Chemistry XXI
The extent to which a reaction leads to electron density
transfer can be assessed by analyzing changes in the
oxidation number (or state) of each atom.
+1
0
0
+1
-2
+1
-2
-4
C is oxidized (loses e- density),
O is reduced (gains e- density).
+4
-2
Oxidation Numbers
The oxidation number is defined as the partial
charge that an atom in a molecule would have if all
of the bonding electrons were assigned to the
most electronegative atom in the bond
(molecule seen as fully ionic).
Chemistry XXI
Assigning oxidation numbers is useful in:
 Identifying electron-rich and electron-poor
centers in molecules;
 Tracking electron transfer or redistribution
during a chemical reaction;
 Making predictions about reaction directionality.
Chemistry XXI
For next class,
Investigate what is a Lewis acid and a
Lewis base.
Why is that CO2 is considered a Lewis acid
while H2O is a Lewis base?