Chapt. 15 – Energy & Chemical Change
15.1
15.2
15.3
15.4
15.5
Energy (& Modes of Heat Transfer - NIB)
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Section 15.1 Energy
Energy can change form and flow, but it
is always conserved.
• Define energy.
• Distinguish between potential and kinetic energy.
• Relate chemical potential energy to the heat lost or
gained in chemical reactions.
• Calculate the amount of heat absorbed or released by
a substance as its temperature changes.
• Describe the three ways that heat can be transferred.
Section 15.1 Energy
Key Concepts
• Energy is the capacity to do work or produce heat.
• Chemical potential energy is energy stored in the
chemical bonds of a substance by virtue of the
arrangement of the atoms and molecules.
• Chemical potential energy is released or absorbed as
heat during chemical processes or reactions.
• The amount of heat required to change the
temperature of a given mass of a substance can be
computed using
q = c × m × ∆T
• Heat can be transferred through convection,
conduction or radiation.
Nature of Energy
Energy: Ability to do work or produce heat
Two forms
Potential
• Due to composition or
position of material or object
• Water stored behind dam
Kinetic
• Energy of motion: ½ m v2
• Water falling over dam
Conservation of Energy
Water flowing over dam
• Potential energy converted to kinetic energy
Wood burning
• Chemical potential energy converted to heat
Law of conservation of energy
In any chemical reaction or physical
process, energy can be converted from one
form to another, but is neither created or
destroyed.
Conservation of Energy
Water flowing through turbine: potential
energy converted to kinetic energy which is
then converted into electrical energy
Potential Energy (PE)
Energy stored by virtue of position
Gravitational
• Higher up object is, higher the PE
• Plays very little role in chem. processes
Electrical
• Two + charges close together - high PE
• Ions involved in storing electrical PE
Magnetic
• Two N poles close together - high PE
PE in Chemical Systems
Chemical Potential energy – Energy stored
in a substance because of its composition
• Types of atoms
• Number, type (strength) of bonds
• Atomic/molecular arrangement in space
Chemical potential energy
converted to heat in many
reactions (e.g., combustion)
Heat and Temperature
Name of transfer process is heat
• Energy gets transferred
• Heat NOT a substance
Temperature: property which
• is directly proportional to KE of substance
under examination
• determines direction heat will flow when two
objects brought into contact
Temperature and heat are not the same
Heat & Heat Transfer
Heat (q) – energy in the process of flowing
from a warmer object to a cooler one
(textbook definition)
• Raises T of cooler and lowers T of hotter
Transfer of KE from one medium or object
to another, or from an energy source to a
medium or object (alternate definition)
Heat transfer can occur in 3 ways:
radiation, conduction, and convection
Conduction, Convection, & Radiation
Heat Transfer by Radiation
Heat transferred by
electromagnetic
radiation – no contact
Any form: visible light,
infrared, microwaves,
radio waves, etc.
Examples – heat lamp
for fast foods, sun
warming Earth
Thermal Conduction
Process by which energy transferred as
heat through material between two
points at different temperatures (without
material moving)
Metal rod conducting heat
Thermal Conduction
Atoms nearest higher temperature
jostle less energetic neighbors and
transfer energy in process
Rate of thermal conduction depends on
material – thermal conductivity (k)
Heat Conduction – Electric Stove
Atoms in pan vibrate about
their equilibrium positions
Those near stove coil
vibrate with larger
amplitudes
These collide with
adjacent atoms and
transfer some energy
Eventually, energy travels
entirely through pan & its
handle
Thermal Conduction
Materials giving high rates of heat
transfer are thermal conductors
• Metals
Those giving low rates are thermal
insulators
• Ceramics, anything that’s mostly air
(cork, fiberglass, etc)
Dependence of Thermal Conductivity
on Material
Best – metals
Middle – nonmetals
Lowest - Gases
Range of Thermal Conductivity of Various
Materials at Room Temperature
Styrofoam –
Good Thermal
Insulator
Conduction
When metal block
and wooden block
(both at same
temperature) picked
up, metal block feels
cooler due to faster
conduction of heat
away from hand
Convection
Energy literally carried by fluid
• Forced hot air heating
• Antifreeze cooling of engine block
Convection
Heat transport due to
movement of fluid (gas
or liquid)
Natural convection
occurs due to
circulation driven by
differing densities caused by uneven
heating
Warm air expands and
rises
Convection Currents in
Boiling Water
Convection – Room Heating
Natural vs Forced Convection
Sometimes natural convection
inadequate or inappropriate
Use forced convection with pump or fan
Thermos Bottle
Minimizes convection,
conduction, and radiation heat
transfer
Double-walled glass vessel with
evacuated space between walls
minimizes energy losses due to
conduction and convection
Silvered surfaces reflect most
radiant energy that would
otherwise enter or leave liquid
in thermos
Cooling Coffee
What modes of heat transfer involved?
Conduction: (through cup walls)
Convection: in coffee and in air
Radiation:from all exterior surfaces (IR)
Heat Pipe
A Heat absorbed in
evaporator section
B Liquid boils to
produce vapor
C Heat released to
environment as vapor
condenses
D Liquid returns by
wicking or gravity
Heat Pipe Characteristics
Able to transport large
amounts of heat across
very small temperature
differences
• Thermal “super
conductors”
• 1000 x or more effective
than solid Cu
Energy, Heat, and Work
KE and PE (including chemical PE) can do work
and be converted to heat
Energy, heat, work: have same units
• SI unit – joule (J)
• Energy expended when force of one newton
is applied over displacement of one meter
Metric unit (not SI) of heat – calorie (cal)
• 1 cal = 4.184 J (exactly, by definition)
• 1 nutritional Calorie = 1 kcal = 4184 J
• Must be able to convert between units
Practice
Conversion of energy units
Problems 1- 3 page 519
Problems 62 - 66 page 552
Problems 1 - 2 page 986
The calorie (cal) & Specific Heat
One calorie = 4.184 J
(exactly, by definition)
calorie – approximately equal to
amount of heat needed to raise T of 1 g
pure H2O 1 C
Quantity 4.184 J/(gC) is the specific
heat of water – measure of ability to
absorb heat
Specific Heat
Quantity 4.184 J/(gC) is the specific
heat of water – measure of ability to
absorb heat
Alternate term: specific heat capacity
Specific Heat
Each substance has its own specific heat
Symbol c used for specific heat (specific
heat capacity)
c depends on T, but generally very weakly
• Water at 0 C – 4.218 J/(gC)
• Water at 40 C – 4.179 J/(gC)
c depends strongly on phase of substance
• c for water very different than for ice – see
following slide
Specific Heat and Liquid Water
Water has highest specific heat capacity of
common liquids
Water’s ability to store and release large
quantities of heat plays important role in
many natural phenomena and engineering
applications
• Large bodies of water moderate
temperatures (absorbs sun during day)
• Water used for cooling systems

Thermal conductivity also plays a role
Heat Release/Absorption Calculations
Heat (q) depends not only on specific heat
(c) but also upon mass (m) of substance
and the size of the temperature change (T)
q = c  m  T
m in grams, T = Tfinal – Tinitial in C
Heat Release/Absorption Calculation
Problem: Solar pond made of 14,500 kg of
granite and contains 22,500 kg of water. T rises
22 C during daylight and decreases by same
amount at night. q stored during day?
q = c  m  T
qwater = 4.184 J/(gC)  2.25x107 g  22 C
= 2.1x109 J
qgranite = 0.803 J/(gC)  1.45x107 g  22 C
= 2.6x108 J
Heat Release/Absorption Calculation
qtotal = qwater+ qgranite = 2.1x109 J + 0.26x109 J
= 2.4x109 J = 2.4x106 kJ = 2.4x103 MJ
= 2.4 GJ
2.4 GJ absorbed during day and released at
night
Specific Heat Calculation
Example problem 15-2
Temperature of sample of iron with mass of
10.0 g changed from 50.4 C to 25.0 C with
release of 114 J of heat. Calculate the
specific heat of iron.
ciron =
q
m  T
=
114 J
10.0 g  25.4 C
= 0.449 J/(gC)
Practice
Specific heat
Problems 4 - 6 page 521
Problem 10 page 522
Problem 67 page 552
Problems 3 - 5 page 986
Chapt. 15 – Energy & Chemical Change
15.1
15.2
15.3
15.4
15.5
Energy (& Modes of Heat Transfer - NIB)
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Section 15.2 Heat
The enthalpy change for a reaction is
the enthalpy of the products minus
the enthalpy of the reactants.
• Describe how a calorimeter is used to measure energy
that is absorbed or released
• Calculate the various quantities that are involved in a
calorimetry experiment, especially the specific heat of
an unknown substance
Section 15.2 Heat
• Explain the meaning of enthalpy and enthalpy change
in chemical reactions and processes and identify a
reaction as endo- or exothermic.
• Calculate the heat for a process given an associated
enthalpy change for that process and the amount of
substance.
Section 15.2 Heat
Key Concepts
• In thermochemistry, the universe is defined as the
system plus the surroundings.
• The heat lost or gained by a system during a reaction
or process carried out at constant pressure is called
the change in enthalpy (∆H).
• When ∆H is positive, the reaction is endothermic.
When ∆H is negative, the reaction is exothermic.
Measuring Heat
Calorimeter
Insulated device for
measuring heat
absorbed/released
during a chemical or
physical process
Coffee cup
calorimeter with
stirrer and
thermometer
Using Calorimeter to Determine c
Lead
Shot
Heat 150.0 g
Pb to 100 C
28.8 C 
 22.0 C
Measure Ti of
50.0 g H2O
Measure Tf of
Pb + H2O
Determining c from Calorimeter
150.0 g Pb, Ti = 100 C, Tf = 28.8 C
50.0 g H2O, Ti = 22.0 C, Tf = 28.8 C
qlead = - qwater (heat loss = heat gain)
qwater = mH2O  cH2O  TH2O =
50.0 g  4.18 J/(g °C)  6.8 °C = 1.14 x103 J
clead = qlead / (mlead  Tlead) =
-1.14 x103 J/(50.0 g  -71.2 °C) = 0.13 J/(g°C)
Temperature and Heat
c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)
Have 1.00 g each of Li (lithium) & Pb (lead)
Li is at 20.0 C Pb at 80.0 C
Which metal is hotter (has higher
temperature)?
Pb
Temperature and Heat
c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)
Have 1.00 g each of Li (lithium) & Pb (lead)
Li @ 20.0 C
Pb @ 80.0 C
Each metal put into separate beakers
containing 1.00 g of water at 10.0 C and
allowed to equilibrate.
Which one raises the temperature of the
water more (provides more heat)?
Temperature and Heat
Note: the following 3 slides show a general
algebraic solution to the question asked.
This is done to illustrate this type of
approach. However, the problem could
have been solved by doing the calculation
for each metal separately.
Temperature and Heat
m = metal w = water i = initial f = final
Heat for metal: qm = mmcmT T = Tf – Tim
Heat for H2O: qw = mwcwT T = Tf – Tiw
Energy conserved: – qm = qw
– mmcm(Tf – Tim) = mwcw(Tf - Tiw)
mm = mw (both 1.00 g)
Tf –Tim = – (cw/cm)(Tf - Tiw)
Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))
Temperature and Heat
m = metal w = water i = initial f = final
c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)
Li @ 20.0 C=Tim1
Pb @ 80.0 C=Tim2
cw = 4.184 J/(gC)
Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))
Li: cw/cm = 4.184/3.58 = 1.17
Pb: cw/cm = 4.184/0.128 = 32.7
Temperature and Heat
c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)
Li @ 20.0 C=Tim1
Pb @ 80.0 C=Tim2
Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))
Tf(Li) =(20.0 C + 1.1710.0 C) / 2.17 = 14.6 C
Tf(Pb) =(80.0 C + 32.710.0 C) / 33.7 = 12.1 C
Li raises water temperature 4.6 C
Pb raises water temperature 2.1 C
Colder metal (Li) provides much more heat
Temperature & heat are different quantities
Practice
Using specific heat / calorimetry
Problems 12 - 15 page 525
Problems 74 - 78, page 552
Problem 6 page 986
Thermochemistry
Thermochemistry is the
study of heat changes
that accompany
chemical reactions and
phase changes
Chemical Energy and the Universe
System: Specific part of universe containing
the reaction or process you wish to study
Surroundings: Everything else in the
universe except the system
Universe = System + Surroundings
Chemical Energy and the Universe
Heat pack – iron reacts with O2 in air
4Fe(s) + 3O2(g)  2Fe2O3(s) + 1625 kJ
System - Heat pack
Surroundings?
Hands, air, etc
Exothermic reaction
• Heat flows from system to
the surroundings
Chemical Energy and the Universe
Cold pack – dissolution of ammonium nitrate
27 kJ + NH4NO3(s)  NH4+(aq) + NO3-(aq)
System - Cold pack
Surroundings?
Knee, air, etc
Endothermic reaction
• Heat flows from
surrounding to the system
Enthalpy & Enthalpy Changes
Enthalpy (H) is the heat content of a
system at constant pressure
• “Coffee cup” style calorimeter is open to
atmosphere and is at constant P
• Many reactions of interest take place at
constant (atmospheric) pressure
Really want to know change in enthalpy
for a chemical process,  H
H for Process (reaction /phase change)
Hprocess = enthalpy change of process
= Hfinal - Hinitial
= Hfinal state – Hinitial state
Exothermic processes
• Hfinal state < Hinitial state
• Hprocess < 0
Endothermic processes
• Hfinal state > Hinitial state
• Hprocess > 0
Enthalpy & Enthalpy Changes
Reactants
Enthalpy
Enthalpy
Products
Reactants
Products
Endothermic Exothermic
Enthalpy
Enthalpy Change - Heat Pack
4Fe(s) + 3O2(g)
H = - 1625 kJ
2Fe2O3(s)
Exothermic Reaction H < 0
Enthalpy
Enthalpy Change - Cold Pack
NH4+(aq) + NO3-(aq)
H = +27 kJ
NH4NO3(s)
Endothermic Reaction H > 0
Enthalpy Change and q
q – heat gained or lost in a general
chemical reaction or process
qp – as above but for reaction or
process occurring at constant pressure
For constant pressure reactions:
qp = Hrxn
If H units are per mole, then
qp = n Hrxn n = # moles
Chapt. 15 – Energy & Chemical Change
15.1
15.2
15.3
15.4
15.5
Energy (& Modes of Heat Transfer - NIB)
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Section 15.3 Thermochemical Equations
Thermochemical equations express
the amount of heat released or
absorbed by chemical reactions.
• Write thermochemical equations for chemical
reactions and other processes.
• Describe how energy is lost or gained during changes
of state and calculate this energy.
• Calculate the heat absorbed or released in a
chemical reaction.
• Determine the enthalpy for a given phase change by
using the additivity principle for thermochemical
equations.
Section 15.3 Thermochemical Equations
Key Concepts
• A thermochemical equation includes the physical
states of the reactants and products and specifies the
change in enthalpy.
• The molar enthalpy (heat) of vaporization, ∆Hvap, is the
amount of energy required to evaporate one mole of a
liquid.
• The molar enthalpy (heat) of fusion, ∆Hfus, is the
amount of energy needed to melt one mole of a solid.
Writing Thermochemical Equations
Thermochemical equation - balanced
chemical equation that includes
physical state of all reactants and
products and the energy change
expressed as H
Nature of reaction or process written as
subscript for H
(Hvap for vaporization)
Writing Thermochemical Equations
For reactions/processes carried out
under standard conditions (1 atm and
298 K), superscript 0 is used - H0
Note: “standard conditions” are not STP
Enthalpy Change and Combustion
Combustion reaction of glucose
C6H12O6(s) + 6O2(g) 
6CO2(g) + 6H2O(l) H0comb = -2808 kJ
Standard enthalpy of combustion (H0comb)
is enthalpy change for complete burning of
one mole of a substance at 1 atmosphere
pressure and 298 K – standard conditions
H0comb values in table 15.3, page 529
Bomb Calorimeter (constant V)
Sample of
known mass
ignited by spark
and burned in
excess of O2.
Heat released
transferred to
H2O in outer
chamber.
Bomb Calorimeter (constant V)
Because they operate at constant V
and not constant P, heat change
measured in a bomb calorimeter is not
in general equal to H.
For some reactions (those with same
number of moles of gas on both sides
of the equation) the pressure is nearly
constant and so the results can be
used as a measure of H.
Example Problem 15.4
Combustion reaction of glucose
C6H12O6(s) + 6O2(g)  (excess high P O2)
6CO2(g) + 6H2O(l) H0comb = -2808 kJ
How much heat evolved when 54.0 g
glucose is burned?
Key issue: H0comb is per mole glucose
54.0 g glucose  0.300 mol glucose
0.300 mol C6H12O6  2808 kJ/mol C6H12O6
= 842 kJ
(fudging constant P & standard conditions)
Practice
Energy released in chemical reaction
Problem 25 page 532
Problem 29 page 533
Problems 85 - 86, page 553
Enthalpy Change & State Changes
Vaporization, sublimation, melting all
require energy (endothermic, H > 0)
• In thermochemistry, melting process is
called fusion
Hvap= molar enthalpy (heat) of vaporization
• Heat required to vaporize one mole of liquid
Hfus = molar enthalpy (heat) of fusion
• Heat required to melt one mole of a solid
qprocess = n Hprocess (n=# moles)
Thermochemical Equations for
State Changes
H2O(l)  H2O(g) Hvap = 40.7 kJ
H2O(s)  H2O(l) Hfus = 6.01 kJ
The reverse processes (condensation,
freezing) release the same amount of
energy as were absorbed in the above
Hvap = -Hcond
Hfus = -Hsolid
Temperature Change of Ice, Water,
and Steam with Added Energy
Changes for heating 10.0 g ice
Water/
Steam Steam
Ice / Water
Water
Heat (x103 J)
Relating q and H
Example for simple phase change
Calculate energy released when 64.08
g of methanol (CH3OH) freezes
From table 15.4, page 530
Hfus = 3.22 kJ/mol
Freezing (solidification) is opposite
process to melting (fusion)
Hsolid = - 3.22 kJ/mol (sign changed)
Relating q and H
Calculate energy released (in J) when
64.08 g of methanol (CH3OH = MeOH)
freezes
q = Hsolid  n
n = # moles MeOH = 64.08 g MeOH 
1 mol MeOH/32.04 g MeOH
= 2.000 mol MeOH
Relating q and H
Calculate energy released (in J) when
64.08 g of methanol (CH3OH = MeOH)
freezes
Hsolid = - 3.22 kJ/mol
n = 2.000 mol MeOH
q = H x n = - 3.22 kJ/mol  2.000 mol
 1x103 J/kJ = - 6.44x103 J
negative sign for energy release
Energy Requirements for Heating
Substance with Phase Changes
Determine energy required to raise 225 g of
water from  46.8 C to 173.0 C
225 g  1 mol H2O/18.02 g = 12.5 mol H2O
Have two phase changes and 3 states
cice, cwater, csteam – table 15.2, page 520
Enthalpies fusion, vaporization in table 15.4
Hfus = 6.01 kJ/mol
Hvap = 40.7 kJ/mol
Energy Requirements for Heating
Substance with Phase Changes
m = 225 g n = 12.5 mol H2O
T1= 46.8 C T3= 100 C T5= 73.0 C
q1 = cice  m  T1
q2 = Hfus  n (heat required to melt ice)
q3 = cwater  m  T3
q4 = Hvap  n (heat required to boil water)
q5 = csteam  m  T5
Energy Requirements for Heating
Substance with Phase Changes
Total q = q1 + q2 + q3 + q4 + q5
= 21.4 + 75.0 + 94.1+ 508 + 33.0 kJ
= 732 kJ
Note that majority of energy was used to boil
the water
State Changes: Sublimation, Deposition
H for sublimation &
deposition not
tabulated on page 530 
Hsub
Figure shows:
40.7 kJ
Hcond
 H
Hsub = Hfus + Hvap
dep
Hdep = Hcond +
Hsolid
Hdep = –Hsub

6.01 kJ
Hsolid
Adding Thermochemical Equations
Have two phase changes for water
H2O(l)  H2O(g) Hvap = 40.7 kJ
H2O(s)  H2O(l) Hfus = 6.01 kJ
Can add them together to get
H2O(s)  H2O(g) Hsubl = 46.7 kJ
The resulting H is for the phase
change from a solid to a gas =
sublimation
State Changes: Sublimation, Deposition
H for sublimation &
deposition not
tabulated on page 530
Hsub

Figure shows:
40.7 kJ
Hcond
Hsub = Hfus + Hvap

Hdep
= 6.01 kJ + 40.7 kJ
Hsub = 46.7 kJ
Hdep= Hcond + Hsolid 
Hsolid

6.01
kJ
Hdep= – Hsub
Hdep= – 46.7 kJ
Practice
Energy in changes of state
Problems 23 – 24 page 532
Problems 27, 28, 30 page 533
Problems 83, 84, 87, 88 page 553
Problems 7 – 8, page 986
Chapt. 15 – Energy & Chemical Change
15.1
15.2
15.3
15.4
15.5
Energy (& Modes of Heat Transfer - NIB)
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Section 15.4 Calculating Enthalpy Change
The enthalpy change for a reaction
can be calculated using Hess’s law.
• Apply Hess’s law to calculate the enthalpy change for
a reaction.
• Explain the basis for the table of standard enthalpies
of formation.
• Calculate ∆Hrxn using thermochemical equations.
• Determine the enthalpy change for a reaction using
standard enthalpies of formation data.
Section 15.4 Calculating Enthalpy Change
Key Concepts
• The enthalpy change for a reaction can be
calculated by adding two or more thermochemical
equations and their enthalpy changes.
• Standard enthalpies of formation of compounds are
determined relative to the assigned enthalpy of
formation of the elements in their standard states.
• The standard enthalpy change for a reaction can be
computed from the standard enthalpies of formation
using
Calculating Enthalpy Change
Calorimetry cannot be used to measure
H for some reactions, including:
• Slow reactions
C(s, graphite)  C(s, diamond)
occurs on time scale of 106 years
• Conditions difficult to reproduce in
laboratory
• Ones that produce side products
Hess’s Law
If you add 2 or more thermochemical
equations to produce a final equation
for a reaction, then the sum of the
enthalpy changes for the individual
reactions is the enthalpy change for the
final reaction
Hfinal =  Hindividual
 = summation operation
Hess’s Law
Combustion of sulfur to form sulfur trioxide
2S(s) + 3O2(g)  2SO3(g) H=?
In lab, mostly sulfur dioxide produced
H = -297 kJ
Need to use Hess’s Law to compute H
S(s) + O2(g)  SO2(g)
Rules for Applying Hess’s Law
1. If you multiply/divide the coefficients of the
chemical equation by some factor, the
corresponding H must be multiplied/divided
by the same factor
2S(s) + 3O2(g)  2SO3(g) H = -792 kJ
S(s) + 3/2 O2(g)  SO3(g) H = -396 kJ
2. If you reverse the direction of a reaction,
the sign of H must also be reversed
SO3(g)  S(s) + 3/2 O2(g) H = +396 kJ
Hess’s Law – Sulfur Trioxide
2S(s) + 3O2(g)  2SO3(g) H = ?
Reactions with known H
a. S(s) + O2(g)  SO2(g)
H = -297 kJ
b. 2SO3(g)  2SO2(g) + O2(g) H = 198 kJ
Multiply reaction a by 2; reverse reaction b
c. 2S(s) + 2O2(g)  2SO2(g) H = -594 kJ
d. 2SO2(g) + O2(g)  2SO3(g) H = -198 kJ
Add c and d, canceling common terms
2S(s) + 3O2(g)  2SO3(g) H = -792 kJ
Hess’s Law – Sulfur Trioxide
ENTHALPY
2S(s) + 2O2(g)
H = -594 kJ
2SO2(g)
Overall
enthalpy
change
H = -792 kJ
2SO2(g) + O2(g)
H = -198 kJ
2SO3(g)
Hess’s Law – Sulfur Trioxide
2S(s) + 3O2(g)  2SO3(g)
H = -792 kJ
Often written on per mole of product basis
S(s) + 3/2 O2(g)  SO3(g)
H = -396 kJ
Hess’s Law – H2O2 Decomposition
2H2O2(l)  2H2O(l) + O2(g) H = ?
Reactions with known H
a. 2H2(g) + O2(g)  2H2O (l) H = -572 kJ
b. H2(g) + O2(g)  H2O2(l)
H = -188 kJ
Reverse reaction b and multiply by 2
a. 2H2(g) + O2(g)  2H2O (l) H = -572 kJ
c. 2H2O2(l)  2H2(g) + 2O2(g) H = 376 kJ
Add a and c, canceling common terms
2H2O2(l)  2H2O(l) + O2(g) H = -196 kJ
Practice
Hess’s Law
Problems 32 - 33, page 537
Problem 42 page 541
Problems 93 - 94, page 553
Problems 9 - 10, page 987
Standard Enthalpy of Formation
H0f = change in enthalpy that accompanies
the formation of one mole of the compound
in its standard state from its constituent
elements in their standard states
AKA Standard Heat of Formation
Standard state means the normal physical
state of the substance at 1 atm & 298 K
• Fe(s)
• Hg(l)
• O2(g)
Standard Enthalpy of Formation
Examples:
Formation of 1 mole SO3 from its elements
S(s) + 3/2 O2(g)  SO3(g)
H0f = - 396 kJ
Formation of 1 mole CO2 from its elements
C(s,graphite) + O2(g) CO2(g) H0f = -394 kJ
Note: the standard state of carbon has been
assigned to be the graphite allotrope
Table 15.5, page 538 has examples
Standard Enthalpy of Formation
Arbitrary (but key) standard
Every free element in its standard state is
assigned H0f = 0.0 kJ (exactly)
• Fe(s), C(s,graphite), S(s), Al(s)
• H2(g), O2(g), N2(g)
Using Std. Enthalpies of Formation
Key data to compute H0rxn using Hess’s Law
H2S(g) +4F2(g)  2HF(g) + SF6(g) H0rxn ?
Use a H0f for every species in the equation
that is not an element in its standard state
a. ½ H2(g) + ½ F2(g)  HF(g) H0f = -273 kJ
b. S(s) + 3F2(g)  SF6(g)
H0f = -1220 kJ
c. H2(g) + S(s)  H2S(g)
H0f = -21 kJ
Multiply eqn. a by 2 and reverse eqn. c
Using Std. Enthalpies of Formation
H2S(g) +4F2(g)  2HF(g) + SF6(g) H0rxn ?
d. H2(g) + F2(g)  2HF(g) H0f = -546 kJ
b. S(s) + 3F2(g)  SF6(g)
f. H2S(g)  H2(g) + S(s)
H0f = -1220 kJ
H0f = 21 kJ
Add and cancel terms common to both sides
H2S(g) + 4F2(g)  2HF(g) + SF6(g)
H0rxn= -1745 kJ
Using Std. Enthalpies of Formation
H0rxn can be obtained from H0f in a
simpler way than the method just
illustrated by using
H0rxn =  H0f(products)
-  H0f(reactants)
H2S(g) +4F2(g)  2HF(g) + SF6(g)
H0rxn = [2 H0f(HF) + H0f(SF6)]
- [H0f(H2S) + 4 H0f(F2)]
Must use proper coefficients in formula
Using Std. Enthalpies of Formation
CH4(g) +2O2(g)  CO2(g) + 2H2O(g)
H0rxn =  H0f(products)
-  H0f(reactants)
H0rxn = [H0f(CO2) + 2 H0f(H2O)]
- [H0f(CH4) + 2 H0f(O2)]
Note that O2(g) is element in standard
state, H0f = 0; rest are from table
H0rxn = [(-394) + 2(-286)]-[(-75) + 0] kJ
= -891 kJ
Practice
Enthalpy Change from Standard
Enthalpies of Formation
Problems 34 – 37, 41, page 541
Problem 92, page 553
Problems 11-12, page 987
Chapt. 15 – Energy & Chemical Change
15.1
15.2
15.3
15.4
15.5
Energy (& Modes of Heat Transfer - NIB)
Heat
Thermochemical Equations
Calculating Enthalpy Change
Reaction Spontaneity
Section 15.5 Reaction Spontaneity
Changes in enthalpy and entropy
determine whether a process is
spontaneous.
• Differentiate between spontaneous and
nonspontaneous processes.
• Explain the meaning of entropy and of the Gibb’s free
energy.
• State the second law of thermodynamics.
Section 15.5 Reaction Spontaneity
• Predict the sign of the entropy change for various
processes.
• Explain how changes in entropy and free energy
determine the spontaneity of chemical reactions and
other processes and determine if a reaction is
spontaneous by calculating the free energy at a given
temperature.
• Predict the spontaneity of certain reactions at low and
high temperature extremes from information about the
associated enthalpy and entropy changes.
Section 15.5 Reaction Spontaneity
Key Concepts
• Entropy is a measure of the disorder or randomness
of a system.
• Spontaneous processes always result in an increase
in the entropy of the universe.
• Free energy is the energy available to do work. The
sign of the free energy change indicates whether the
reaction is spontaneous.
• The following equation relates the system free energy
change to the changes in enthalpy and entropy.
∆Gsystem = ∆Hsystem – T∆Ssystem
Spontaneous Process (SP)
SP = physical or chemical change that
once begun, occurs with (no outside
intervention)*
* except perhaps for some small amount of
energy needed to get process started
• Spark to light Bunsen burner flame
Formation of rust is spontaneous
4Fe(s) + 3O2(g)  2Fe2O3(s)
H = -1625 kJ
Spontaneous Process (SP)
Ball rolls downhill. Never spontaneously
rolls uphill.
A gas fills a container uniformly. Never
spontaneously collects at one end.
Heat flow always occurs from hot object
to a cooler one. Reverse never
spontaneously occurs.
Spontaneous Process (SP)
Wood burns spontaneously in an
exothermic reaction to form CO2 and
H2O
Wood not formed when CO2 and H2O
mixed together.
Spontaneous Processes and H
Many but not all endothermic reactions
are nonspontaneous
H2O(s)  H2O(l) H = 6.01 kJ
Spontaneous at T  0 C
Exothermic or endothermic nature of
reaction not the sole determinant of
reaction spontaneity
Entropy (S) must also be considered
Entropy
Measure of disorder or randomness of
particles that make up a system
Measure of possible ways that energy
of system can be distributed; related to
freedom of system’s particles to move
and number of ways they can be
arranged
Gas
Expansion
of a Gas
into an
Evacuated
Bulb
Vacuum
Three Possible
Arrangements (states)
of Four Molecules in a
Two-Bulbed Flask
#1 abcd : 0
#2 bcd : a
acd : b
abd : c
abc : d
#3 ab : cd x2
ac : bd x2
ad : bc x2
#1
1 way
4 ways
#2
6 ways
#3
Probability of All Molecules Being in
Left Hand Bulb of a Two-Bulbed Flask
# Molecules
Relative Probability
1
1/2
2
1/22 =1/4
3
1/23 = 1/8
100
1/2100 = 7.9x10-31
6.0x1023
0
Entropy
Low
Higher
Second Law of Thermodynamics
2d Law = Spontaneous processes
always proceed in such a way that the
entropy of the universe (Suniv) increases
Suniv > 0 for any spontaneous process
Universe = System + Surroundings
Second Law of Thermodynamics
2d Law = Spontaneous processes
always proceed in such a way that the
entropy of the universe (Suniv) increases
Equivalent statement:
Nature spontaneously proceeds
towards the states that have the
highest probability of existing.
2nd Law of Thermodynamics
No cyclic process that converts heat
entirely into work is possible (engine)
No cyclic process can transfer energy
as heat from a low-T body to a high-T
body without work being done (fridge)
2nd Law of Thermodynamics
A cyclic process must transfer heat
from a hot to cold reservoir if it is to
convert heat into work
Work must be done to transfer heat
from a cold to a hot reservoir
A useful perpetual motion machine
does not exist
Second Law of Thermodynamics
Conservation of energy tells us that
total amount of energy in the universe
is constant
• Strictly speaking: mass + energy
Second law of thermodynamics says
that the entropy of the universe is
constantly increasing
Predicting Entropy Change, S
Entropy changes associated with
changes in state, with creating certain
types of solutions, with reactions of
gases, and with increasing the
temperature can be predicted
Predicting S
Ssystem = Sproducts – Sreactants
1.State Changes
Entropy increases as a substance
changes from a solid to a liquid to a gas
• Increased mobility allows more states
(positions of molecules) to be
accessible
S H2O(s) < S H2O(l) << S H2O(g)
Increase
in
entropy
from
solid to
liquid to
gas
Large
changes
in S occur
at phase
transitions
Predicting S
Ssystem = Sproducts – Sreactants
2. Dissolving gas in a liquid
Entropy decreases – gas molecules
more limited in movements
CO2(g)  CO2(aq) Ssystem < 0
Gas dissolving in liquid: Ssystem < 0
O2 gas
O2
molecule
confined
by solvent
“cage”
O2
dissolved
Predicting S
Ssystem = Sproducts – Sreactants
3. Reactions with gaseous reactants
and products
Entropy increases if # of product
particles > # of reactant particles
2SO3(g)  2SO2(g) + O2(g) Ssystem>0
(2 mol of gas generates 3 mol of gas)
Predicting S
Ssystem = Sproducts – Sreactants
4. Solids/liquids dissolving to form
solutions
Entropy generally increases
NaCl(s)  Na+(aq) + Cl-(aq) Ssystem>0
(S of water decreases slightly due to
ordering of H2O around hydrated ions)
NaCl(s) + H2O(l)  Na+(aq) + Cl-(aq)
Ssystem>0
Small Increase in S When Ethanol
Dissolves in Water
Freedom of movement remains ~ unchanged;
S increase due solely to random mixing
ethanol
water
solution of
ethanol and
water
Predicting S
Ssystem = Sproducts – Sreactants
5. Temperature increase
T increase  KE increase
Ssystem > 0 (increased disorder)
Practice
Predicting sign of S
Problems 44 (a-d), 45, page 545
Problems 96 - 97, page 554
Problems 13-14, page 987
S, the Universe, Spontaneity
Suniverse > 0 for spontaneous process
Suniverse = Ssystem + Ssurroundings
Suniverse tends to be positive when
a. Hsystem<0
• Heat released by exothermic process
raises T of surroundings and makes
Ssurroundings > 0
b. Ssystem > 0
Exothermic reactions with Ssystem >0
always spontaneous
S, the Universe, and Free Energy
Gibbs Free Energy G
• Combines enthalpy and entropy
• Commonly called Free Energy
• For processes that occur at constant
P and T, G is energy available to do
work
Gsystem= Hsystem - T Ssystem
T in K
For standard conditions (1 atm, 298 K)
G0system= H0system - T S0system
Free Energy and Spontaneity
Gsystem = Hsystem - T Ssystem
Sign of Gsystem indicates if reaction is
spontaneous at specified T and P
Allows information about system only to
predict entropy change of universe
Reaction/process Gsystem
Spontaneous
<0
Nonspontaneous
>0
Suniverse
>0
<0
Free Energy
G used in equilibrium (chapter 17)
concepts in higher level courses
Calculating Free Energy Change
N2(g) + 3H2(g)  2NH3(g)
H0system= - 91.8 kJ S0system= -197 J/K
Gaseous reaction – system S decreases
because # moles gas decreases
G0system determines spontaneity
G0system= H0system - T S0system
= -9.18x104 J – 298 K  (-197 J/K)
= -9.18x104 J + 5.87x104 J = -3.31x104 J
G0system< 0 so reaction is spontaneous
Reaction Spontaneity and the Signs of
Ho, So and Go (= Ho – T So)
All values for system See also table 15.6, page 547
Ho So -TSo Go
Reaction or process
-
+
-
-
Spont. all T
+
-
+
+
Nonspont. all T
+
+
-
+ or -
-
-
+
Spont. at higher T
Nonspont. at lower T
+ or Spont. at lower T
Nonspont. at higher T
Practice
Determine Reaction Spontaneity
Problems 46 (a-c), 47, 51, page 548
Problems 95, 97 (a-c), 98 - 103 page
554
Problems 15-16, page 987