Thermochemistry
HEATS OF REACTION AND
CHEMICAL CHANGE
The Common Misconception
• To start off this unit we need to address a common
misconception about heat and temperature.
• Why don’t you ask your parents or a sibling in
college the difference between the two. It will be
harder than you think.
Heat vs Temperature
• Heat is a form of energy that always flows from warm
areas to cool areas
Ø
From areas of high KE to areas of low KE
• Temperature is a measure of the average kinetic
energy of particles
Ø
As the average kinetic energy (i.e speed) of the particles
increases, the temperature increases
Ø
The “hotness” or “coldness” of an object
Heat is not temperature! Temperature is not heat!
Study of Energy
• What is energy?
Ø
Energy is defined as the ability to do work or produce
heat
• So this unit is about the study of heat change in
chemical reactions and physical changes of state
• We call this study Thermochemistry
Types of Energy
• Kinetic Energy (KE) is the energy of motion. Unless at
absolute zero (0 K), all atoms and molecules are in
motion so possess kinetic energy
• Potential energy (PE) is the energy available by virtue
of an object’s arrangement or position. The object has
the “potential” to do work
Forms of Energy
• Radiant energy from electromagnetic radiation
• Thermal energy, also called heat, is the energy
associated with the random motion of atoms and
molecules
• Chemical energy is the energy stored within the bonds
of chemical substances
• Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
Forms of Energy
• Gravitational energy is the energy resulting from the
attraction of two masses to each other
• Electric energy is the energy from a static or moving
electric charge
• Magnetic energy is the energy from the attraction of
opposite magnetic fields, repulsion of like fields, or
from an associated electric field
• Mechanical energy is the sum of all KE and PE in an
object that is used to do work
Units of Energy
• Joule (J)
Ø
The SI unit for energy
Ø
A joule is not a large amount of energy so we often use the kJ
• calorie (cal)
Ø
1 calorie is defined as the amount of energy transfer necessary
to raise the temperature of 1 g of water from 14.5oC to 15.5oC
• British Thermal Unit
Ø
1 cal = 4.18J
1 Btu = 1055 J
1 BTU is the amount of energy transfer necessary to raise
the temperature of 1 lb of water from 63oF to 64oF
First Law of Thermodynamics
• Also known as the Law of Conservation of Energy,
it states that energy can neither be created nor
destroyed; energy can only be transferred or changed
from one form to another
Ø
Turning on a light would seem to produce energy; however,
it is electrical energy that is converted
Ø
Sunlight is converted into energy in green plants
• To avoid confusion, scientists reference the system
and its surroundings
System vs Surroundings
• System – our area of focus. In chemistry, the system
is the chemical reaction
• Surroundings – everything else
Types of Systems
• Open: energy and matter can be exchanged with the
surroundings
• Closed: energy can be exchanged with the surroundings,
matter cannot
• Isolated: neither energy nor matter can be exchanged
with the surroundings
Internal Energy
• The Internal Energy (E) of a system is the sum of
all KE and PE of all components (atoms and
molecules) of the system
• We generally don’t know E, only how it changes. We
will use DE to depict a change in E
Internal Energy
• Part of E is KE from molecular motion
Ø
Translational, rotational, and vibrational motion
Ø
Collectively, these are sometimes called thermal energy
• Part of E is PE from intermolecular and intramolecular
forces of attraction and the locations of atoms
Ø
Collectively these are sometimes called chemical energy
Change in Internal Energy DE
• By definition, DE is equal to the final energy of the
system minus the initial energy of the system
∆E = Efinal – Einitial
• If DE is > 0, the system absorbed energy
from the surroundings
• If DE is < 0 the system released energy
to the surroundings
Change in Internal Energy DE
• When energy is exchanged between the system and
the surroundings, it is exchanged as either heat (q)
or work (w)
• That is:
DE = q + w
Ø
Heat (q) – the transfer of thermal energy due to a
temperature difference
Ø
Work (w) – In physics, we define it as force acting
over a distance
ΔE, q, w, and Their Signs
• Sign Conventions for q, w, and E
For q
+ means system gains heat
− means system loses heat
For w
+ means work done on
system
− means work done by system
For E
+ means net gain of energy
by system
− means net loss of energy by
system
• Energy lost by the system is gained by the surroundings
and visa-versa
DEsys = -DEsurroundings
-DEsys = DEsurroundings
Exothermic Reactions (-q)
• Heat is released by the system to the surroundings
The temperature of the surroundings increases
Ø The products have lower potential energy than the reactants
Ø
Endothermic Reactions (+q)
• Heat is absorbed by the system from the surroundings
The temperature of the surroundings decreases
Ø The products have more energy than the reactants
Ø
Heat Flow
• Heat flowing into or out of a system always results in
some kind of change to the system
Ø
The temperature of the system could change
Ø
There could be some other change, like a change in
physical state
Heat Flow
• When heat flowing into or out of a substance results
in a change in temperature (DT), we can calculate the
amount of heat with the equation:
q = mcDT
more on this later
• But sometimes heat flowing into or out of a substance
results in a different kind of change - without a DT
Ø
Examples include freezing, melting, chemical reactions
Ø
These are measured as a change in enthalpy or DH
What is Enthalpy (H)
• We use the symbol H to denote enthalpy
• Enthalpy is the total energy (E) of a system at constant
pressure
• Like E, total enthalpy of a system cannot be measured
directly. We have to measure its effects
Ø
When a system reacts DH = Hfinal - Hinitial
• Enthalpy is an extensive property so depends on the
amount (mols) of substance
Ø
More reactant will generate more heat
Enthalpy Change DH
• To make things a little easier, we are going to go
straight to the equation that relates heat flow and
change in enthalpy
n = # of moles
nDH = qp
For simplicity we
normally leave off
the subscript “p”
• What this states is that at constant pressure, the
change in enthalpy is the heat gained or lost by the
system
• Usually expressed in units of kJ/mol
Enthalpy Change DH
• Any energy change to a system that does not result in
a change in temperature (DT) of the system is measured
as DH
Ø Phase
changes are good examples. During the change of
state (i.e. gas to liquid) the temperature remains constant
• DH can be measured experimentally since we assume
that all the energy gained or lost is heat
• DH for a reaction depends on the states of the reactants
and the products
Types of Enthalpy Change
• Enthalpy of reaction (∆Hrxn) – the heat that is either
absorbed or released during a chemical reaction
• Enthalpy of combustion (∆Hcomb)—the heat absorbed or
released by burning (usually with O2) in kJ/mol
• Enthalpy of solution (∆Hsoln)—the heat absorbed or
released by a substance when it dissolves
• Enthalpy of formation (∆Hf) – heat absorbed or
released when one mole of compound is formed from
elements in their standard states in kJ/mol
Types of Enthalpy (Phase Changes)
• Enthalpy of fusion (∆Hfus)—heat absorbed to change
one mol of a substance from a solid to a liquid at its
melting point
Ø
Freezing is exothermic so DH is negative
Ø
Melting is endothermic so DH is positive
• Enthalpy of vaporization (∆Hvap)—heat absorbed to
change one mol of a substance from a liquid to vapor at
its boiling point
Ø
Condensing is exothermic so DH is negative
Ø
Boiling is endothermic so DH is positive
Enthalpy of Reaction DHrxn
• For a chemical reaction:
DHreaction = Hproducts - Hreactants
• DHreaction or DHrxn, is called the enthalpy of reaction,
or the heat of reaction
• If DHrxn is positive, the reaction is endothermic
• If DHrxn is negative, the reaction is exothermic
For simplicity, we normally just write DH
Enthalpy Diagrams
• Enthalpy Diagrams can be used to determine if a
reaction is endo or exothermic
A decrease in
enthalpy during the
reaction; DH is
negative.
An increase in
enthalpy during the
reaction; DH is
positive.
Potential Energy Diagrams
• By comparing the PE of reactants to products, you
can determine the components with the most PE and
determine whether the net energy is positive or
negative – meaning is energy (as heat) released or
absorbed by the system
Heat Capacity: A Thought Experiment
Place an empty iron pot weighing 5 lb on the burner of
a stove
Place a 1 lb iron pot containing 4 lb of water on an
identical burner (same total mass)
Turn on both burners and wait five minutes
Which pot handle can you grab with your bare hands?
Why?
Iron has a lower specific heat than does water. It takes
less heat to “warm up” iron than it does water
Heat Capacity and Specific Heat
• The heat capacity (C) of a system is the quantity of
heat required to change the temperature of a
substance by 1 oC or 1 K
• If the amount of the substance heated is one gram, it is
the specific heat
Molar Heat Capacity
• If the amount of substance is 1 mole, it is the molar
heat capacity (units are J/mol • K)
Heat Capacity Problem
Calculate the heat capacity of an aluminum block that
absorbs 629 J of heat from its surroundings in order for
its temperature to rise from 22 °C to 145 °C
q = 629 J
DT = 145-22 = 123 oC
C=
=
= 5.1 J/oC
Specific Heat
• A useful form of the specific heat equation is:
q = m c DT
Ø
If DT > 0, then q > 0 and heat is gained by the system
Ø If
DT < 0, then q < 0 and heat is lost by the system
Specific Heat Capacities of Some Substances
Substance
Water (l)
Specific Heat (J/g oC)
4.186
Water (s)
2.06
Water (g)
Ammonia (g)
Ethanol (l)
Aluminum (s)
Carbon, graphite (s)
Copper (s)
Gold (s)
Iron (s)
Lead (s)
Mercury (l)
Silver (s)
2.02
2.09
2.44
0.897
0.709
0.385
0.129
0.449
0.129
0.140
0.233
Note the difference in specific heat for the different phases (or states)
of water
More on Specific Heat
• The calorie (c), while not an SI unit, is still used to
some extent
• Water has a specific heat of 1 cal/(g oC)
• 4.184 J = 1 cal
• 1 Calorie (C) is used in nutrition and is equal to 1000 c
or 4.184 kJ
Specific Heat Problem
How much heat does it take to raise the temperature
of 225.0 g of water from 25.0 to 100.0 °C?
Mass = 225.0 g
DT = 100.0 - 25.0 = 75.0 oC
c = 4.184 J/g oC
q = m c DT
q = 225.0 (4.184)(75.0)
q = 70.6 kJ
Calorimetry
• Since we cannot know the exact
enthalpy of the reactants and
products, we measure DH through
calorimetry, the measurement of
heat flow
• The instrument used to measure
heat flow is called a calorimeter
Ø
Ø
Its normally filled with water to
absorb or release heat
The apparatus, including the water,
are part of the surroundings
Calorimetry
• For a reaction carried out in a calorimeter, all the heat
evolved by a reaction is absorbed by the calorimeter
and its contents
•
-qrxn = qcalorimeter
Ø
Since the calorimeter is made of the solution, stirrer,
thermometer, and container, we break qcal into two terms
-qrxn = qcal + qsoln
Ø
or
qrxn = Ccal DT + mcDT
Heat capacity of calorimeter
DT and Thermal Equilibrium
• We can determine the heat exchange once the system
has reached a state of thermal equilibrium
• Thermal equilibrium occurs when the system and
surroundings reach the same temperature and heat
transfer stops
Ø
“Flow” ceases at thermal equilibrium
Coffee Cup Calorimeter
• The design of the constant-pressure
calorimeter can be as simple as two nested
Styrofoam cups covered with a piece of
corrugated cardboard with a hole through
which a thermometer is inserted.
• That’s why we often refer to this as “Coffee
Cup Calorimetry”
Ø
Ø
Ø
The inner cup holds the solution in which the reaction takes
place
The second cup provides insulation
The loose-fitting cardboard cover provides insulation while
allowing the pressure in the container to equilibrate with the
atmospheric pressure outside
Constant Pressure Calorimetry
• Constant-pressure calorimeters are run at atmospheric
pressure and are used to determine DH for reactions in
solution
• DH for the system can be found by measuring the heat
change for the solution
Ø
The energy given off by the reaction is gained by the solution
or visa-versa
• For simplicity, we will assume that the heat capacity of
the calorimeter (C) is negligible in our problem sets and
assessments so:
-qrxn = qsoln
Constant Pressure Calorimetry
• We calculate ΔH for the reaction with this equation:
qsoln = csoln x msoln x DT
Ø
m is the mass of the solution
ü
ü
Ø
c is the specific heat of the solution
ü
Ø
Remember that solution = solute + solvent
You can assume that the density of the original solutions and
the final solution is the same as water: 1.00 g/mL.
Unless otherwise given, use 4.184 J/g oC. We’re normally dealing
with so small amount of solute that there is no real change in c
DT = Tfinal - Tinitial
Calorimetry Problem #1
A 0.500-g sample of KCl is added to 50.0 g of water
in a calorimeter. If the temperature decreases by 1.05
°C, what’s the approximate amount of heat involved in
the dissolution of the KCl? Assume the heat capacity
of the resulting solution is 4.18 J/g °C. Is the reaction
exothermic or endothermic?
Calorimetry Problem #1
Solution
Givens: mass KCl = 0.500 g
mass H2O = 50.0 g
DT = 1.05 oC
csolution = 4.18 J/g oC
1. We have to assume the KCl and water are in thermal
equilibrium (at same temperature)
2. Mass of solution is 50.0 + 0.500 = 50.5 g
3. qsoln = mcDT = 4.18 J/g oC (50.5 g)(-1.05 oC)
4. qsoln = -222 J therefore qrxn =+222 J
The reaction is endothermic. 222 J has been absorbed
from the water and KCl for the KCl to dissolve
Problem #2
When burned in air, hydrogen releases 120 J/g of energy
and methane gives off 50.0 J/g. If a mixture of 5.00 g of
H2 and 10.0 g CH4 are burned and the heat released is
transferred to 50.0 g of water at 25.0 °C, what’s the final
temperature of the water?
q = (120. J/g H2)(5.00 g H2) + (50.0 J/g CH4)(10.0 g CH4)
q = 1100 J = m · c ·DT
1100
q = Jm =· s(4.18
· DT J/°C · g H2O) · (50.0 g H2O) · DT
DT = 5.26°C
Final temp = 25.0 °C + 5.26°C = 30.3°C
Problem #3
The addition of 3.15 g of Ba(OH)2•8H2O to a solution of 1.52
g of NH4SCN in 100 g of water in a calorimeter caused the
temperature to fall by 3.1 °C. Assuming the specific heat of
the solution and products is 4.20 J/g °C, calculate the
approximate amount of heat absorbed by the reaction,
which can be represented by the following equation:
Ba(OH)2⋅8H2O(s) + 2NH4SCN(aq) →
Ba(SCN)2(aq) + 2NH3(aq) +10H2O(l)
Thermochemical Equations
• Thermochemical Equations describe the reaction
and energy changes that occur during the chemical
reaction
• In a thermochemical reaction, ∆H for the reaction can
be written either as a product
CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ
Ø
We would normally write this equation as ...
CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ
Thermochemical Equations
• Or ∆H can be written as a reactant
2NaHCO3(s) +129 kJ → Na2CO3(s)+H2O(g)+CO2(g)
Ø
We would normally write this equation as ...
2NaHCO3(s) → Na2CO3(s)+H2O(g)+CO2(g) DH = +129kJ
Thermochemical Equations
• The physical state of the reactants & products must be
stated
H2O(l) → H2(g) + ½ O2(g)
∆H = 285.8 kJ
H2O(g) → H2(g) + ½ O2(g) ∆H = 241.8 kJ
• Even though the stoichiometry is the same, the state
of the reactants is different
Ø
The difference in ∆H is 44.0 kJ. You need to check that
the values you are using coincide to the state of the
substance
Manipulating Thermochemical Equations
• We can manipulate or rearrange thermochemical
equations to help us solve problems. Two primary
ways are:
Ø
Ø
Flip the equation and write it backwards. Products
become reactants and visa-versa
Multiply/divide the entire equation (including DH) by
some factor (natural number or fraction)
Manipulating Thermochemical Equations
• If you multiply or divide a balanced thermochemical
equation by a factor n, then that same factor n must be
applied to DH
C(s) + O2(g) → CO2(g) DH = -393.5 kJ
Ø
Let’s say I need 2O2. I would simply multiply everything by 2
2[C(s) + O2(g) → CO2(g) DH = -393.5 kJ]
Ø
Many students forget to multiply DH by 2 so you could write it
2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2(-393.5 kJ) = -787.0 kJ
Manipulating Thermochemical Equations
• If you reverse or flip a thermochemical equation, the
sign of DH changes
• DH for a reaction in the forward direction is equal in
size, but opposite in sign, to DH for the reverse reaction
Ø
Let’s say I need Ca(OH)2 to be a reactant vice the product . I
simply flip the entire equation
CaO(s) + H2O(l) → Ca(OH)2(s)
Ø
∆H = −65.2 kJ
So DH switches signs. Again, a common mistake by students
Ca(OH)2(s) → CaO(s) + H2O(l)
∆H = 65.2 kJ
Conversion Factors
• We can generate conversion factors involving DH
• For example, the reaction:
CaO(s) + H2O(l) → Ca(OH)2(s) ∆H = −65.2 kJ
can be used to write:
–65.2 kJ
–65.2 kJ
1 mol CaO
1 mol H2O
–65.2 kJ
1 mol Ca(OH)2
Calculating DH Using Stoichiometry
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -890kJ/mol
• How much heat is released if 8.2 g of methane gas is
burned?
Given: 8.2 g CH4 Unknown: DHrxn
8.2 g CH4
1 mol CH4
16.01 g CH4
-890 kJ
1 mol CH4
= -460 kJ
So 460 kJ was released
Problem #2
The major source of aluminum in the world is bauxite
(mostly aluminum oxide). Its thermal decomposition can
be represented by
You can have a fraction!
Al2O3(s) → 2Al(s) + 3/2O2(g)
DH = 1676 kJ
If aluminum is produced with this process, how many grams
will be produced from 1000.0 kJ of heat?
1000.0 kJ 2 mol Al
26.98 g Al
1676 kJ
The “given” determines sigfigs
1 mol Al
= 32.196 g Al
Problem #3
What is the heat evolved from the combustion of 15.5 g of
C3 H8 ?
C3 H 8 ( g ) + 5O2 ( g ) = 3CO2 ( g ) + 4 H 2O(l )...DH = -2219kJ / mol
Combustion of one mole of C3H8 yields -2219 kJ
Molar mass of C3H8 = 44.0 g/mol
No moles in 15.5 g = 15.5/44.0 = 0.352 mol
Heat produced = - 2219 x 0.352 = - 782 kJ
Hess’s Law
• Because enthalpy is a state function, the total enthalpy
change depends only on the initial state (reactants) and
the final state (products) of the reaction
• This allows us to calculate ΔH for reactions that cannot
be easily determined or to predict DH before we go to
the lab (safety issue)
Hess’s Law
• If a reaction can be expressed as a series of steps, then
the DH for the target reaction is the sum of the heats of
reaction for each step
DHtarget = SDHknowns
Ø
S
is the sigma symbol and essentially means “sum
up”
Using Hess’s Law
• To solve a Hess’s Law problem, we algebraically
manipulate equations (and enthalpies) so their sum
equals the target equation
Ø
Arrange equations so all substances are on the correct side
of the reaction arrow. You may need to flip one or more
Ø
Scale (multiply or divide) the reactions so the coefficients
are the same as the target. Whatever you do to the reaction,
you do to DH
Ø
Cancel out unnecessary substances (same on both sides of
the reaction arrow
Ø
Add the equations and DH’s together to solve the problem
Calculating DH for Combustion of Propane
C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l)
• Imagine this as occurring in three steps:
1. Decomposition of propane to the elements:
C3H8(g) ® 3C(s) + 4H2(g) DH = 103.85 kJ
2. Formation of CO2
3C(s) + 3O2(g) ® 3CO2(g) DH = -1181 kJ
3. Formation of H2O
4H2(g) +2O2(g) ® 4H2O(l) DH = -1143 kJ
Calculating DH for Combustion of Propane
C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l)
Cross out like terms from both
sides or arrow
The sum of these equations is the
overall equation
Add the DH values
C3H8(g) ® 3C(s) + 4H2(g)
3C(s) + 3O2(g) ® 3CO2(g))
4H2(g) +2O2(g) ® 4H2O(l)
C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l)
DH = 103.85 kJ + (-1181 kJ) + (-1143 kJ) = -2220 kJ
2C(graphite) + 3H2(g) ® C2H6(g) DH = ?
Calculate the enthalpy change for the target reaction
given the equations below:
1. C(graphite) + O2(g) à CO2(g)
DH = –393.5 kJ
2. 2C2H6(g) + 7O2 (g) à 4CO2(g) + 6H2O(l) DH = –3119.6 kJ
3. H2(g) + ½O2 (g) à H2O(l)
DH = –285.8 kJ
Over the next few slides, we will break down each reaction
and then add them together to make the target
2C(graphite) + 3H2(g) ® C2H6(g) DH= ?
• First, multiply equation the first equation by 2 to
match the target
2[C(graphite) + O2(g) à CO2(g) DH = –393.5 kJ]
• Equation #1 becomes:
2C(graphite) + 2O2(g) à 2CO2(g) DH = –787 kJ
2C(graphite) + 3H2(g) ® C2H6(g) DH= ?
• Next, flip the second equation to move C2H6 to the
product side and then divide by 2 (only need 1 C2H6)
1/2[2C2H6(g) + 7O2 (g) à 4CO2(g) + 6H2O(l)] 1/2DH = –3119.6 kJ
• Equation #2 becomes:
2CO2(g) + 3H2O(l) à C2H6(g) + 7/2 O2 (g) DH = +1559.8 kJ
2C(graphite) + 3H2(g) ® C2H6(g) DH= ?
• Next, multiply the third reaction by 3 since we need 3H2
3[H2(g) + ½O2 (g) à H2O(l)]
3DH = –285.8 kJ]
• Equation #3 becomes:
3H2(g) + 3/2O2 (g) à 3H2O(l) DH = –857.4 kJ
You can have fractional coefficients in thermochemistry
2C(graphite) + 3H2(g) ® C2H6(g) DH= ?
Cross off like terms and add the equations together:
1. 2C(graphite) + 2O2(g) à 2CO2(g)
DH = –787 kJ
2. 2CO2(g) + 3H2O(l) à C2H6(g) + 7/2 O2 (g) DH = +1559.8 kJ
3. 3H2(g) + 3/2O2 (g) à 3H2O(l)
2 C(graphite) + 3 H2(g) à C2H6(g
3/2 O2 + 2 O2 = 7/2 O2
DH = –857.4 kJ
DH = -84.6 kJ
N2(g) + 2O2(g) ® 2NO2(g)
ΔH° = ?
Calculate the enthalpy for the above reaction using
the following two equations:
N2(g) + O2(g) ® 2NO(g)
ΔH° = +180 kJ
2NO2(g) ® 2NO(g) + O2(g)
ΔH° = +112 kJ
Enthalpies of Formation
An enthalpy of formation, ΔHf , is defined as the
enthalpy change for the reaction in which a
compound is made from its constituent elements in
their elemental forms
Standard Enthalpies of Formation
• The standard enthalpy of formation (ΔHf°) for
a reaction is the enthalpy change that occurs when 1
mole of a substance is formed from its component
elements in their most stable states under standard
state conditions
Ø
Also called the Standard Heat of Formation
Ø
Standard conditions are denoted by the o
• ΔHf° values are measured under standard conditions
of 25 oC (298 K) and 1.00 atm pressure
Standard Enthalpies of Formation
Important - The ΔHf° for any element in its most stable state is zero
Standard Enthapy of Reaction, 1111111
• The enthalpy change for a given reaction is calculated by
subtracting the enthalpies of formation of the reactants
from the enthalpies of formation of the products:
=
Snp
[products]
- Snr
[reactants]
Again, for simplicity we sometimes leave out the rxn
o
ΔH using the Standard Enthalpy Table
C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l)
Substance
DHof
H2 O
-285.8 kJ
CO2
-393.5 kJ
C3H8
-103.85 kJ
O2
0
DHo = [3(-393.5 kJ + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0)]
DHo = [-1180.5 kJ + (-1143.2 kJ)] - [(-103.85 kJ) + 0)]
DHo = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ
Bond Enthalpies
• Bond Energy values can be used to calculate the
approximate enthalpies for reaction
• In order to break a bond, energy must be added to the
system (endothermic)
• Why? Atoms are at their lowest potential energy levels
when bonds are formed and therefore more stable. To
separate two atoms (break a bond), energy must be
added
Bond Enthalpies
• Conversely, when bonds are formed, atoms return to
a stable state with the lowest potential energy and
the excess energy is release as heat (exothermic)
• The difference between the energy to break bonds
and the energy released when new bonds are made
determines whether the overall or net reaction is
endo- or exothermic
• Note: bond enthalpies are not as accurate as heats of
formation
Using Bond Enthalpies
• The reactants & products must be in the gaseous state
Ø
If you have liquids present, you need an extra energy term
to work out the enthalpy change when you convert from
liquid to gas. That term is the enthalpy change of
vaporization
• The more energy that is required to break a bond, the
more chemically stable the compound
• To determine the number and types of bonds that are
broken & made, you may have to draw Lewis
structures
Average Bond Enthalpies (kJ/mol)
• Average bond enthalpies are positive, because bond
breaking is an endothermic process
Note that these are averages over many different compounds
Bond Enthalpies and Enthalpy of Reaction
• The heat released or absorbed
during a chemical change is
due to differences between the
bond energies of reactants and
products
Ø
Energy is added to break bonds
and released when making
bonds
n = number of bonds
DH = SnBond Energy(reactants) - SnBond Energy(products)
Note that the equation is reactants minus products!
Calculating Enthalpy Using Bond Enthalpies
H2 (g) + F2 (g)
Type of
bonds broken
H
H
F
F
Type of
bonds formed
H
F
2HF(g)
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
1
1
432
159
432
159
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
2
-565
-1130
DH = SBE (reactants) – SBE (products)
DH= 432 + 159 – 1130 = -539 kJ
Calculate the Enthalpy of Combustion for CH4
CH4(g) + 2O2(g)
2H2O(g) + CO2(g)
• Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 (413) + 2 (498)
= 1662 + 996 = 2658 kJ
• Bonds formed = 4 x (O-H) + 2 x (C=O)
= 4 (464) + 2 (805)
= 1856 + 1610 = 3466 kJ
• DH = 2658 - 3466 = -808 kJ
Questions?