Chemistry – A Molecular Approach, 1st Edition

Nivaldo Tro

Chapter 6
Thermochemistry

Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
 Heating Your Home
• most homes burn fossil fuels to generate heat
• the amount the temperature of your home
increases depends on several factors
how much fuel is burned
the volume of the house
the amount of heat loss
the efficiency of the burning process
can you think of any others?
Tro, Chemistry: A Molecular Approach 2
 Nature of Energy
• even though Chemistry is the study of
matter, energy effects matter
• energy is anything that has the capacity to
do work
• work is a force acting over a distance
Energy = Work = Force x Distance
• energy can be exchanged between objects
through contact
collisions
Tro, Chemistry: A Molecular Approach 3
 Classification of
Energy
• Kinetic energy is
energy of motion or
energy that is being
transferred
thermal energy is
kinetic

Tro, Chemistry: A Molecular Approach 4

 Classification of Energy
• Potential energy is energy that is stored in
an object, or energy associated with the
composition and position of the object
energy stored in the structure of a compound is
potential

Tro, Chemistry: A Molecular Approach 5

 Law of Conservation of Energy
• energy cannot be created or
destroyed
 First Law of
Thermodynamics
• energy can be transferred
between objects
• energy can be transformed
from one form to another
 heat → light → sound
Tro, Chemistry: A Molecular Approach 6
 Some Forms of Energy
• Electrical
 kinetic energy associated with the flow of electrical charge
• Heat or Thermal Energy
 kinetic energy associated with molecular motion
• Light or Radiant Energy
 kinetic energy associated with energy transitions in an atom
• Nuclear
 potential energy in the nucleus of atoms
• Chemical
 potential energy in the attachment of atoms or because of their
position

Tro, Chemistry: A Molecular Approach 7

 Units of Energy
• the amount of kinetic energy an
object has is directly
proportional to its mass and
velocity
• when
 the
KE mass
= ½mvis2 in kg and
speed in m/s, the unit for kinetic
2
energy is 2kg  m
s
• 1 joule of energy is the amount of
energy needed to move a 1 kg mass
at a speed of 1 m/s
2
 1 J = 12 kg  m
8
s
 Units of Energy
• joule (J) is the amount of energy needed to move
a 1 kg mass a distance of 1 meter
1 J = 1 N∙m = 1 kg∙m2/s2
• calorie (cal) is the amount of energy needed to
raise one gram of water by 1°C
kcal = energy needed to raise 1000 g of water 1°C
food Calories = kcals
Energy Conversion Factors
1 calorie (cal) = 4.184 joules (J) (exact)
1 Calorie (Cal) = 1000 calories (cal)
1 kilowatt-hour (kWh) = 3.60 x 106 joules (J)
Tro, Chemistry: A Molecular Approach 9
 Energy Use
Energy Energy Energy
Required to Energy used to Used by
Raise Required to Run 1 Average
Unit
Temperature Light 100-W Mile U.S.
of 1 g of Bulb for 1 hr Citizen in
Water by 1°C (approx) 1 Day
joule (J) 4.18 3.60 x 105 4.2 x 105 9.0 x 108

calorie (cal) 1.00 8.60 x 104 1.0 x 105 2.2 x 108

Calorie (Cal) 0.00100 86.0 100. 2.2 x 105

kWh 1.16 x 10-6 0.100 0.12 2.5 x 102

Tro, Chemistry: A Molecular Approach 10
 Energy Flow and
Conservation of Energy
• we define the system as the material or process we are
studying the energy changes within
• we define the surroundings as everything else in the
universe
• Conservation of Energy requires that the total energy
change in the system and the surrounding must be zero
 Energyuniverse = 0 = Energysystem + Energysurroundings
  is the symbol that is used to mean change
 final amount – initial amount

11
 Internal Energy
• the internal energy is the total amount of
kinetic and potential energy a system possesses
• the change in the internal energy of a system
only depends on the amount of energy in the
system at the beginning and end
a state function is a mathematical function whose
result only depends on the initial and final
conditions, not on the process used
E = Efinal – Einitial
Ereaction = Eproducts - Ereactants
Tro, Chemistry: A Molecular Approach 12
 State Function

Tro, Chemistry: A Molecular Approach 13

 Energy Diagrams
• energy diagrams are a

Internal Energy
“graphical” way of showing final
the direction of energy flow energy added
during a process E = +
• if the final condition has a initial
larger amount of internal
energy than the initial
condition, the change in the

Internal Energy
internal energy will be + initial
• if the final condition has a energy removed
smaller amount of internal E = ─
final
energy than the initial
condition, the change in the
internal energy will be ─
Tro, Chemistry: A Molecular Approach 14
 Energy Flow
• when energy flows out of a
system, it must all flow into Surroundings
the surroundings E +
• when energy flows out of a System
system, Esystem is ─ E ─
• when energy flows into the
surroundings, Esurroundings is +
• therefore:
─ Esystem= Esurroundings

Tro, Chemistry: A Molecular Approach 15

 Energy Flow
• when energy flows into a
system, it must all come from Surroundings
the surroundings E ─
• when energy flows into a
System
system, Esystem is +
E +
• when energy flows out of the
surroundings, Esurroundings is ─
• therefore:
Esystem= ─ Esurroundings

Tro, Chemistry: A Molecular Approach 16

 How Is Energy Exchanged?
• energy is exchanged between the system and
surroundings through heat and work
 q = heat (thermal) energy
 w = work energy
 q and w are NOT state functions, their value depends on the
process
E = q + w
system gains heat energy system releases heat energy
q (heat) + ─
system releases energy by
system gains energy from work
w (work) +
doing work
─
system gains energy system releases energy
E + ─
Tro, Chemistry: A Molecular Approach 17
 Energy Exchange

• energy is exchanged between the system and

surroundings through either heat exchange or
work being done

Tro, Chemistry: A Molecular Approach 18

 Heat & Work
• on a smooth table, most of the kinetic energy
is transferred from the first ball to the second
– with a small amount lost through friction

Tro, Chemistry: A Molecular Approach 19

 Heat & Work
• on a rough table, most of the kinetic energy of
the first ball is lost through friction – less than
half is transferred to the second

Tro, Chemistry: A Molecular Approach 20

 Heat Exchange
• heat is the exchange of thermal energy between
the system and surroundings
• occurs when system and surroundings have a
difference in temperature
• heat flows from matter with high temperature to
matter with low temperature until both objects
reach the same temperature
thermal equilibrium

Tro, Chemistry: A Molecular Approach 21

 Quantity of Heat Energy Absorbed
Heat Capacity
• when a system absorbs heat, its temperature increases
• the increase in temperature is directly proportional to the
amount of heat absorbed
• the proportionality constant is called the heat capacity, C
 units of C are J/°C or J/K
q = C x T
• the heat capacity of an object depends on its mass
 200 g of water requires twice as much heat to raise its temperature by
1°C than 100 g of water
• the heat capacity of an object depends on the type of material
 1000 J of heat energy will raise the temperature of 100 g of sand
12°C, but only raise the temperature of 100 g of water by 2.4°C
Tro, Chemistry: A Molecular Approach 22
 Specific Heat Capacity
• measure of a substance’s intrinsic ability to
absorb heat
• the specific heat capacity is the amount of
heat energy required to raise the temperature
of one gram of a substance 1°C
 Cs
 units are J/(g∙°C)
• the molar heat capacity is the amount of heat
energy required to raise the temperature of
one mole of a substance 1°C
• the rather high specific heat of water allows it
to absorb a lot of heat energy without large
increases in temperature
 keeping ocean shore communities and beaches cool in the
summer
 allows it to be used as an effective coolant to absorb heat
Tro, Chemistry: A Molecular Approach 23
 Quantifying Heat Energy
• the heat capacity of an object is proportional to its mass
and the specific heat of the material
• so we can calculate the quantity of heat absorbed by an
object if we know the mass, the specific heat, and the
temperature change of the object
Heat = (mass) x (specific heat capacity) x (temp. change)
q = (m) x (Cs) x (T)

Tro, Chemistry: A Molecular Approach 24

 Example 6.2 – How much heat is absorbed by a copper
penny with mass 3.10 g whose temperature rises from
-8.0°C to 37.0°C?
• Sort Given: T1= -8.0°C, T2= 37.0°C, m=3.10 g
Information
Find: q, J
• Strategize Concept Plan: Cs m, T q

q  m  C s  T
q = m ∙ Cs ∙ T
Relationships:
Cs = 0.385 J/g (Table 6.4)
• Follow the Solution:
q  m  C s  T
Concept
Plan to
T  T2  T1
T  37.0 C - - 8.0C   
 3.10 g  0.385 gJ C  45.0 C 
Solve the
 45.0 C  53.7 J
problem
• Check Check:
the unit and sign are correct
 Pressure -Volume Work
• PV work is work that is the result of a volume change
against an external pressure
• when gases expand, V is +, but the system is doing work
on the surroundings so w is ─
• as long as the external pressure is kept constant
─Work = External Pressure x Change in Volume
w = ─PV
 to convert the units to joules use 101.3 J = 1 atm∙L

Tro, Chemistry: A Molecular Approach 26

 Example 6.3 – If a balloon is inflated from 0.100 L to
1.85 L against an external pressure of 1.00 atm, how
much work is done?
Given: V1=0.100 L, V2=1.85 L, P=1.00 atm

Find: w, J
Concept Plan:
P, V w

w  - P  V
Relationships: 101.3 J = 1 atm L

Solution:
V  V2  V1 w   P  V 101.3 J
 1.75 atm  L 
V  1.85 L - 0.100 L  1.00 atm  1.75 L 
1 atm  L
 1.75 atm  L  - 177 J
 1.75 L
Check:
the unit and sign are correct
 Exchanging Energy Between
System and Surroundings
• exchange of heat energy
q = mass x specific heat x Temperature
• exchange of work
w = −Pressure x Volume

Tro, Chemistry: A Molecular Approach 28

 Measuring E,
Calorimetry at Constant Volume
• since E = q + w, we can determine E by measuring q and w
• in practice, it is easiest to do a process in such a way that there is
no change in volume, w = 0
 at constant volume, Esystem = qsystem
• in practice, it is not possible to observe the temperature changes
of the individual chemicals involved in a reaction – so instead,
we use an insulated, controlled surroundings and measure the
temperature change in it
• the surroundings is called a bomb calorimeter and is usually
made of a sealed, insulated container filled with water
qsurroundings = qcalorimeter = ─qsystem
─Ereaction = qcal = Ccal x T
Tro, Chemistry: A Molecular Approach 29
 Bomb Calorimeter
• used to measure E
because it is a
constant volume
system

Tro, Chemistry: A Molecular Approach 30

 Example 6.4 – When 1.010 g of sugar is burned in a
bomb calorimeter, the temperature rises from 24.92°C to
28.33°C. If Ccal = 4.90 kJ/°C, find E for burning 1 mole
Given: 1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C

Find: Erxn, kJ/mol

Concept Plan:
Ccal, T qcal qcal qrxn
qcal  Ccal  T qrxn  - qcal
qcal = Ccal x T = -qrxn qrxn
Relationships: E 
MM C12H22O11 = 342.3 g/mol mol C12 H 22 O11
Solution:
q1.010 C   T 1 mol C12 H 22 O11 qrxn  16.7 kJ
cal gC 12 H 22 O11 
cal 
E 06 10 mol
2.95 -3

kJ
342.3 g mol C12 H 22O11 2.5906  10-3 mol
T4.90
28.33  3.41C  16.7 kJ
C C  24.92C
qrxn
T 3.41
 qcal
C  16.7 kJ  - 5.66  103 kJ/mol
Check: the units and sign are correct
31
 Enthalpy
• the enthalpy, H, of a system is the sum of the internal
energy of the system and the product of pressure and
volume
 H is a state function
H = E + PV
• the enthalpy change, H, of a reaction is the heat
evolved in a reaction at constant pressure
Hreaction = qreaction at constant pressure
• usually H and E are similar in value, the difference
is largest for reactions that produce or use large
quantities of gas
Tro, Chemistry: A Molecular Approach 32
Endothermic and Exothermic Reactions
• when H is ─, heat is being released by the system
• reactions that release heat are called exothermic reactions
• when H is +, heat is being absorbed by the system
• reactions that release heat are called endothermic reactions
• chemical heat packs contain iron filings that are oxidized in
an exothermic reaction ─ your hands get warm because the
released heat of the reaction is absorbed by your hands
• chemical cold packs contain NH4NO3 that dissolves in water
in an endothermic process ─ your hands get cold because
they are giving away your heat to the reaction 33
 Molecular View of
Exothermic Reactions
• in an exothermic reaction, the
temperature rises due to release of
thermal energy
• this extra thermal energy comes from
the conversion of some of the chemical
potential energy in the reactants into
kinetic energy in the form of heat
• during the course of a reaction, old
bonds are broken and new bonds made
• the products of the reaction have less
chemical potential energy than the
reactants
• the difference in energy is released as
heat 34
 Molecular View of
Endothermic Reactions
• in an endothermic reaction, the temperature drops due
to absorption of thermal energy
• the required thermal energy comes from the
surroundings
• during the course of a reaction, old bonds are broken
and new bonds made
• the products of the reaction have more chemical
potential energy than the reactants
• to acquire this extra energy, some of the thermal energy
of the surroundings is converted into chemical potential
energy stored in the products

Tro, Chemistry: A Molecular Approach 35

 Enthalpy of Reaction
• the enthalpy change in a chemical reaction is an
extensive property
 the more reactants you use, the larger the enthalpy change
• by convention, we calculate the enthalpy change for the
number of moles of reactants in the reaction as written
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) H = -2044 kJ
  2044 kJ
or
1 mol C3H 8
Hreaction for 1 mol C3H8 = -2044 kJ 1 mol C3H 8  2044 kJ
 2044 kJ 5 mol O 2
Hreaction for 5 mol O2 = -2044 kJ or
5 mol O 2  2044 kJ
Tro, Chemistry: A Molecular Approach 36
 Example 6.6 – How much heat is evolved in the
complete combustion of 13.2 kg of C 3H8(g)?
Given: 13.2 kg C3H8,

Find: q, kJ/mol
Concept Plan: kg g mol kJ
1000 g 1 mol C3H 8 - 2044 kJ
1 kg 44.09 g 1 mol C3H 8
Relationships: 1 kg = 1000 g, 1 mol C H = -2044 kJ, Molar Mass = 44.09 g/mol
3 8

Solution:
1000 g 1 mol - 2044 kJ
13.2 kg     6.12 105 kJ
1kg 44.09 g 1 mol

Check:
the sign is correct and the value is reasonable
37
 Measuring H
Calorimetry at Constant Pressure
• reactions done in aqueous solution are at
constant pressure
 open to the atmosphere
• the calorimeter is often nested foam cups
containing the solution
qreaction = ─ qsolution = ─(masssolution x Cs, solution x T)
 Hreaction = qconstant pressure = qreaction
 to get Hreaction per mol, divide by the number of
moles

Tro, Chemistry: A Molecular Approach 38

 Example 6.7 – What is Hrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
100.0 mL of solution changes the temperature from 25.6°C to 32.8°C?
Given: 0.158 g Mg, 100.0 mL,
Find: q, kJ/mol
Concept Plan: kg g mol kJ
1000 g 1 mol C3H 8 - 2044 kJ
1 kg 44.09 g 1 mol C3H 8
Relationships: 1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol

Solution:

1000 g 1 mol - 2044 kJ

13.2 kg     6.12 105 kJ
1kg 44.09 g 1 mol

Check:
the sign is correct and the value is reasonable
39
 Example 6.7 – What is Hrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
100.0 mL of solution to change the temperature from 25.6°C to 32.8°C?
Given: 0.158 g Mg, 100.0 mL sol’n, T1 = 25.6°C, T2 = 32.8°C, Cs = 4.18 J/°C, dsoln = 1.00
g/mL
Find: Hrxn, J/mol Mg

Concept Plan: m, Cs, T qsoln qsoln qrxn

qsoln  m  Cs  T qrxn  - qsoln
Relationships: qrxn
qsoln = m x Cs x T = -qrxn H 
mol Mg
Solution:
1.00 g
100.0 mL   1.00 102 g
qsoln  m  C1 smL
 T qrxn  3.0  103 J
1 mol H   -3
0.158
 1.00gMg 2 J
10 g  4.18 g6C.4994
7.210
-3
C  mol 3
3.0  10 J mol Mg 6 .4994  10 mol
24.31 g 5
3  - 4.6  10 J/mol
qrxn   q
T  32.8soln  3 .0  10
C  25.6C  7.2C J
Check: the units and sign are correct
40
 Relationships Involving Hrxn
• when reaction is multiplied by a factor, Hrxn is
multiplied by that factor
because Hrxn is extensive
C(s) + O2(g) → CO2(g) H = -393.5 kJ
2 C(s) + 2 O2(g) → 2 CO2(g) H = 2(-393.5 kJ) = 787.0 kJ

• if a reaction is reversed, then the sign of H is

reversed
CO2(g) → C(s) + O2(g) H = +393.5 kJ

Tro, Chemistry: A Molecular Approach 41

 Relationships Involving Hrxn
Hess’s Law
• if a reaction can be
expressed as a series
of steps, then the
Hrxn for the overall
reaction is the sum of
the heats of reaction
for each step

Tro, Chemistry: A Molecular Approach 42

 Sample – Hess’s Law
Given the following information:
2 NO(g) + O2(g)  2 NO2(g) H° = -173 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) H° = -255 kJ
N2(g) + O2(g)  2 NO(g) H° = +181 kJ
Calculate the H° for the reaction below:
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) H° = ?

[32 NO2(g)  32 NO(g) + 1.5 O2(g)]

O2(g)] x 1.5 H° = (+259.5
1.5(+173kJ)
kJ)
[1
[2 N2(g) + 2.5
5 OO 2(g)
2(g) + +2 1HH 2O(l)
2O(l) 4 2HNO
HNO 3(aq)]
3(aq)] x 0.5 H° = (-128 kJ) kJ)
0.5(-255
[2 NO(g)  N2(g) + O2(g)] H° = -181 kJ
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) H° = - 49 kJ

Tro, Chemistry: A Molecular Approach 43

 Standard Conditions
• the standard state is the state of a material at a defined set of
conditions
 pure gas at exactly 1 atm pressure
 pure solid or liquid in its most stable form at exactly 1 atm pressure and
temperature of interest
 usually 25°C
 substance in a solution with concentration 1 M
• the standard enthalpy change, H°, is the enthalpy change
when all reactants and products are in their standard states
• the standard enthalpy of formation, Hf°, is the enthalpy
change for the reaction forming 1 mole of a pure compound
from its constituent elements
 the elements must be in their standard states
 the Hf° for a pure element in its standard state = 0 kJ/mol
 by definition

Tro, Chemistry: A Molecular Approach 44

 Formation Reactions
• reactions of elements in their standard state to
form 1 mole of a pure compound
if you are not sure what the standard state of an
element is, find the form in Appendix IIB that has a
Hf° = 0
since the definition requires 1 mole of compound be
made, the coefficients of the reactants may be
fractions

Tro, Chemistry: A Molecular Approach 45

 Writing Formation Reactions
Write the formation reaction for CO(g)
• the formation reaction is the reaction between the
elements in the compound, which are C and O
C + O → CO(g)
• the elements must be in their standard state
 there are several forms of solid C, but the one with Hf° = 0 is
graphite
 oxygen’s standard state is the diatomic gas
C(s, graphite) + O2(g) → CO(g)
• the equation must be balanced, but the coefficient of the
product compound must be 1
 use whatever coefficient in front of the reactants is necessary to
make the atoms on both sides equal without changing the
product coefficient
C(s, graphite) + ½ O2(g) → CO(g) 46
 Calculating Standard Enthalpy Change
for a Reaction
• any reaction can be written as the sum of formation
reactions (or the reverse of formation reactions) for
the reactants and products
• the H° for the reaction is then the sum of the Hf°
for the component reactions
H°reaction =  n Hf°(products) -  n Hf°(reactants)
 means sum
n is the coefficient of the reaction
Tro, Chemistry: A Molecular Approach 47
 The Combustion of CH4

Tro, Chemistry: A Molecular Approach 48

 Sample - Calculate the Enthalpy Change in
the Reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound and
determine the Hf° for each
2 C(s, gr) + H2(g)  C2H2(g) Hf° = +227.4 kJ/mol
C(s, gr) + O2(g)  CO2(g) Hf° = -393.5 kJ/mol
H2(g) + ½ O2(g)  H2O(l) Hf° = -285.8 kJ/mol

Tro, Chemistry: A Molecular Approach 49

 Sample - Calculate the Enthalpy Change in
the Reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
2. Arrange equations so they add up to desired reaction
2 C2H2(g)  4 C(s) + 2 H2(g) H° = 2(-227.4) kJ
4 C(s) + 4 O2(g)  4CO2(g) H° = 4(-393.5) kJ
2 H2(g) + O2(g)  2 H2O(l) H° = 2(-285.8) kJ
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l) H = -2600.4 kJ

Tro, Chemistry: A Molecular Approach 50

 Sample - Calculate the Enthalpy Change in
the Reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
H°reaction = n Hf°(products) - n Hf°(reactants)

Hrxn = [(4•HCO2 + 2•HH2O) – (2•HC2H2 + 5•HO2)]

Hrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]

Hrxn = -2600.4 kJ

Tro, Chemistry: A Molecular Approach 51

 Example 6.11 – How many kg of octane must be
combusted to supply 1.0 x 1011 kJ of energy?
Given: 1.0 x 1011 kJ
Find: mass octane, kg

Concept Plan: Write the balanced equation per mole of octane

Hf°’s Hrxn° Hrxn  nH f products - nH f reactants

kJ mol C8H18 g C8H18 kg C8H18

from 114.2 g
1 kg
above 1 mol 1000 g

Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g

Solution: C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
11 1 mol C8H18
H rxn 114.22
nH f products - ngH f reactants1 kg
Material Hf°, kJ/mol
- 1.0  10 kJ   
 8H f kJ
- 5074.1 CO 2  1
9mol
Look up the
H f HC
for each material
O 
H
H

2 8 18 °
fH 1000
C
f 8 18CH H g

8 18 2
25
(l) H f O 2 

-250.1
8 393.5 in
 2.3  106 kg C8 H18 kJ Appendix 8 kJ  O250
 9 241.IIB 2(g) 
.1 kJ   25
2
00 kJ  
CO2(g) -393.5
 5074.1 kJ
H2O(g) -241.8
Check: the units and sign are correct
the large value is expected 52
 Energy Use and the Environment
• in the U.S., each person uses over 105 kWh of energy per year
• most comes from the combustion of fossil fuels
 combustible materials that originate from ancient life
C(s) + O2(g) → CO2(g) H°rxn = -393.5 kJ
CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) H°rxn = -802.3 kJ
C8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g) H°rxn = -5074.1 kJ
• fossil fuels cannot be replenished
• at current rates of consumption, oil and natural gas
supplies will be depleted in 50 – 100 yrs.

Tro, Chemistry: A Molecular Approach 53

 Energy Consumption

• the increase in energy

consumption in the US

• the distribution of energy consumption in the US

Tro, Chemistry: A Molecular Approach 54

 The Effect of Combustion Products
on Our Environment
• because of additives and impurities in the fossil
fuel, incomplete combustion and side reactions,
harmful materials are added to the atmosphere
when fossil fuels are burned for energy
• therefore fossil fuel emissions contribute to air
pollution, acid rain, and global warming

Tro, Chemistry: A Molecular Approach 55

 Global Warming
• CO2 is a greenhouse gas
 it allows light from the sun to reach the earth, but does not
allow the heat (infrared light) reflected off the earth to escape
into outer space
 it acts like a blanket
• CO2 levels in the atmosphere have been steadily
increasing
• current observations suggest that the average global air
temperature has risen 0.6°C in the past 100 yrs.
• atmospheric models suggest that the warming effect
could worsen if CO2 levels are not curbed
• some models predict that the result will be more severe
storms, more floods and droughts, shifts in agricultural
zones, rising sea levels, and changes in habitats
Tro, Chemistry: A Molecular Approach 56
 CO2 Levels

Tro, Chemistry: A Molecular Approach 57

 Renewable Energy
• our greatest unlimited supply of energy is the sun
• new technologies are being developed to capture
the energy of sunlight
parabolic troughs, solar power towers, and dish engines
concentrate the sun’s light to generate electricity
solar energy used to decompose water into H2(g) and
O2(g); the H2 can then be used by fuel cells to generate
electricity
H2(g) + ½ O2(g) → H2O(l) H°rxn = -285.8 kJ
• hydroelectric power
• wind power
Tro, Chemistry: A Molecular Approach 58