THERMOCHEMISTRY
CHAPTER 5
WHY ENERGY?
•
•
•
•
•
Life requires energy
From plants – photosynthesis
To equipment – electricity
Vehicles – fossil fuels
To animals - food
ENERGY FROM WHERE?
• Most of our energy as humans
– From food
– i.e. from chemical reactions
• In the world
– From chemical reactions
WHAT IS THERMOCHEMISTRY?
Quantitative study of chemical reactions
involving heat changes
Produce
energy
Consume
energy
Examples:
Food we eat - provides us with energy
Burning coal- to produce electrical energy
Energy from the sun - responsible for
chemical reactions in plants which make
them grow
RECALL
• FORCE (F) – push/pull exerted on an object
• WORK (w) – energy used to cause an
object to move against a force (w = Fd)
• HEAT (q) – energy transfer as a result of a
temperature difference.
Heat always flows from a hotter to a
colder object until they have the same
temperature
• ENERGY (E) – capacity to do work or to
transfer heat
HOT STUFF?
• Can something transfer cold?
• Does ice make a
drink cold?
When you blow hot soup do you make it cold?
SO IN FACT:
• In theory stirring
soup long
enough and fast
enough could
make it boil
Heat always flows from a hotter to a
colder object until they have the same
temperature
BUT I KNOW IT’S COLD!!
Heat from our body is transferred,
making something
feel cold because we
lose heat from the area
touching the cold object
NATURE OF ENERGY
2 types of energy
Kinetic energy
Potential energy
Energy due to
motion of object
Energy due to
position of
object relative to
others
(Stored energy)
KINETIC ENERGY
1
Ek  mv 2
2
Note:
Ek increases as velocity increases and Ek
increases as mass increases.
Atoms and molecules have mass and are in
motion, thus they have kinetic energy!
Thermal energy
energy an object
possesses because of its
temperature.
Associated with kinetic energy of atoms and
molecules.
The higher the temperature,
 the faster the atoms and molecules move
 the more kinetic energy molecules and atoms
have
 the greater the thermal energy of the object.
The total thermal energy of the object is the
sum of the individual energies of the atoms or
molecules.
MISSING: KINETIC ENERGY,
PLEASE HELP
• What is it to have no kinetic energy?
– Absolute zero!
• Can we detect that?
– No!
– Why?!
– Because in order to detect it we would
have to interact with it which would heat it
POTENTIAL ENERGY
We know Ep = mgh
Forces other than gravity that lead to
potential energy
= ELECTROSTATIC forces
Attractions and repulsions due
to oppositely charged objects
(e.g. positive and negative ions)
Recall:
kQ1Q 2
Eel 
d
Describes the
electrostatic energy
between 2 charges.
Eel = electrostatic energy
k = Coulomb constant
Q = charge
d = distance between
charges
UNITS OF ENERGY
SI unit = J (Joule)
1 J = 1 kg.m2.s-2
Ek=0.5mv2
=(kg)(m.s-1)2
(Others 1 cal (calorie) = 4.184 J)
(The amount of heat energy to raise temp of 1g H2O by 1oC)
A BRAINTEASER
• If everything has kinetic energy
• And chemical potential energy
• Why is everything not moving around
constantly?
• Why are solids not felt to vibrate?
SYSTEM AND SURROUNDINGS
Portion singled
out for study
System
Everything
else
Surroundings
System + surroundings = universe
Open system – can exchange both
energy and matter with the
surroundings
Closed system – exchanges energy
but NOT matter with the surroundings
Isolated system – neither energy nor
matter can be exchanged with the
surroundings
A QUOTE NOT TO LIVE BY
“Thermodynamics is a funny subject. The
first time you go through it, you don't
understand it at all. The second time
you go through it, you think you
understand it, except for one or two
small points. The third time you go
through it, you know you don't
understand it, but by that time you are
so used to it, it doesn't bother you
anymore.”
http://www.eoht.info/page/Thermodynamics+humor
FIRST LAW OF THERMODYNAMICS
Energy is neither created nor destroyed, it is
merely converted from one form into another.
Energy is transferred between the system
and the surroundings in the form of work and
heat.
The total energy of the universe remains
constant.
INTERNAL ENERGY (U)
Also given
as E
Internal energy (of the system)
= sum of all Ek and Ep of all components
of the system
Symbol for change = 
 Change in internal energy of the system
 U
U = UFinal - Uinitial
Number & Unit  Magnitude
Sign  Direction (in which
energy is transferred)
U = UFinal - Uinitial
Usyst > 0
 Uf > Ui
 GAINED energy from surroundings
Usyst < 0
 Uf < Ui
 LOST energy to surroundings
RELATING U TO WORK AND HEAT
General ways to change ENERGY of a
closed system:
HEAT
lost/gained
by system
WORK
Done by/on
system
Relationship: When a system undergoes a
chemical/physical change
U = q + w
We can’t measure U, but we can determine U.
SIGN CONVENTION
q<0
Heat
transferred
FROM
system
w<0
Work done
BY system
SYSTEM
q>0
Heat
transferred TO
system
w>0
Work done ON
system
The sign of U will depend on the sign and
magnitude of q and w since U = q + w.
Example:
Calculate the change in internal energy for a
process in which the system absorbs 120 J of
heat from the surroundings and does 64 J of
work on the surroundings.
System
120 J heat
64 J work
U = q + w = (120 J) + (- 64 J)
= 56 J
SOME DEFINITIONS
Extensive property – dependent on the
amount of matter in the system.
E.g. mass, volume etc
Intensive Property – NOT dependent on the
amount of matter in the system.
E.g. density, temperature etc
State function – a function that depends on
the state or conditions of the system
and NOT on the details of how it came
to be in that state.
A
B
C
DEFINITIONS APPLIED TO INTERNAL
ENERGY
Recall:
Total internal energy of a system = sum of all
Ek and Ep of all components of the system
Thus total internal energy of a system
 total quantity of matter in system
U is an extensive property
Internal energy depends on the state or
conditions of the system (e.g. pressure,
temperature, location)
Does not depend on how it came to be in that
state.
 state function
U only depends on Ui and Uf and not how
the change occurred.
e.g. if a gas sample undergoes:
A
pressure
A
heat
Y
X
heat
pressure
B, or
B
U is the same in both cases.
NOTE: Heat and work are not state functions!
ENTHALPY (H)
We cannot measure enthalpy (H), we can only
measure change in enthalpy (H).
Change in enthalpy (H) is the heat gained or
lost by the system when a process occurs
under constant pressure.
E.g. atm pressure
H = Hfinal - Hinitial = qp
H is a state function and an extensive
property
NOTE:
At constant pressure, most of the energy
lost / gained is in the form of heat.
Very little work is done for the expansion /
contraction against the atmospheric force,
especially if the reaction does not involve
gases.
qp<0
H<0
Exothermic
process – system
evolves heat
SYSTEM
qp>0
H>0
Endothermic
process - system
absorbs heat
Example of an exothermic reaction:
Example of an endothermic reaction:
FINDING THE RELATIONSHIP
BETWEEN U AND H – CONSIDER
PV WORK
We know that: U = q + w
Assume we do PV work only
(e.g. expanding gases in cylinder of a car
does PV work on the piston)
F
P
A
Gas expands and
moves piston
distance d
W = Fd
But F = PA
 W = PAd
W = PV
Sign: system is doing work on the piston
W = -PV
U = q + w
and
w = -PV
 U = q - PV
At constant volume, V = 0
 U = qv
At constant pressure:
U = qp - PV
But H = qp
 U = H - PV
The volume change in many reactions is very
small, thus PV is very small and hence the
difference between U and H is small.
ENTHALPIES OF REACTION
enthalpy change that accompanies
a reaction (heat of reaction)
Hrxn = H(products) – H(reactants)
final
Thermochemical equation:
2H2(g) + O2(g)  2H2O(g)
Balanced equation
initial
exothermic
H = - 483.6 kJ
H associated with
the reaction
Coefficients in balanced equation = no. of moles of
reactant/product producing associated H (extensive)
2H2(g) + O2(g)  2H2O(g)
H = -483.6 kJ
Note:
Since enthalpy is an extensive property, H
depends on the amount of reactant
consumed.
4H2(g) + 2O2(g)  4H2O(g)
H = -967.2 kJ
Also, H for a reaction is equal in magnitude,
but opposite in sign to H for the reverse
reaction.
2H2O(g)  2H2(g) + O2(g)
H = +483.6 kJ
Notes are available on
http://hobbes.gh.wits.ac.za/chemnotes/fmain-chem1033-35.htm
Go to the section for Mrs C Billing
Lectures Chapter 7, Chapter 8 and Chapter 9 are this work
We are busy with Chapter 7 at the moment
Example:
Calculate the heat needed to convert 25 ml
water to steam at atmospheric pressure if the
enthalpy change is 44 kJ/mol.
(Assume the density of water is 1.0 kg/l)
=1.0 g/ml
H2O(l)  H2O(g)
H = 44 kJ/mol
25 ml
m = v = (1.0 g/ml)(25 ml) = 25 g
m
25 g
n=
= 1.4 mol
=
M 18.016 g/mol
qp = H = 44 kJ/mol x 1.4 mol = 62 kJ
Check - endothermic
Example:
Hydrogen peroxide can decompose to water
and oxygen by the reaction:
2H2O2 (l)  2H2O (l) + O2 (g)
H = -196 kJ
Calculate the heat produced when 2.50 g H2O2
decomposes at constant pressure.
m
2.50 g
nH2O2 =
=
= 0.0735 mol
34.02 g/mol
M
qp = H = -196 kJ for 2 mol H2O2
= -7.2 kJ for 0.0735 mol
Example – burning money:
Ethanol burns in air to give water vapour and carbon dioxide
1. Calculate the enthalpy of reaction given enthalpy of
formations of: ethanol (l) = -277.7 kJ/mol
water (l) = -285.8 kJ/mol carbon dioxide (g)= -393.5 kJ/mol
2. Calculate the minimum amount of water needed to prevent the
paper from burning after be soaked in 1g of ethanol
First: find the balanced chemical equation
C2H5OH (l) + 3O2 3H2O (l) + 2CO2 (g)
Enthalpy of reaction:
H = H°f(products) - H°f(reagents)
= 2(-393.5) + 3(-285.8) – (-277.7 + 3x0) kJ/mol
= -1365.8 kJ/mol
Example – burning money:
Ethanol burns in air to give water vapour and carbon dioxide
2. Calculate the minimum amount of water needed to prevent the
paper from burning after be soaked in 1.00g of ethanol
Find the heat released by burning 1.00 g of ethanol
m
1.00 g
n=
M = 46.07 g/mol
= 0.0217 mol
qp = H = -1365.8 kJ/mol x 0.0217 mol = -29.6 kJ
Recall: H2O(l)  H2O(g)
qp
-29.6 kJ
n=
H = 44 kJ/mol
H = 44 kJ/mol
= 0.67 mol
m = nM = (0.67 mol)(18.0148 g/mol) = 12 g
CALORIMETRY
The experimental determination of heat
flow associated with a chemical reaction
by measuring the temperature changes it
produces.
Heat Capacity
 The amount of heat required to raise the
temperature by 1K.
Recall: a change of 1K = a change of 1oC
Units: J.K-1
Extensive property
Molar Heat Capacity
 The heat capacity of 1 mol of substance.
Units: J.K-1mol-1
Specific Heat Capacity (C)
 heat capacity of 1 g of substance
Units: J.K-1g-1
 determined by measuring the change in
temperature that a known mass of substance
undergoes when it gains/loses a specific
quantity of heat.
q
C
m  T
Specific Heats for Some Subst’s at 298K
H2O (l)
4.18 J.K-1g-1
N2 (g)
1.04 J.K-1g-1
Al (s)
0.90 J.K-1g-1
Fe (s)
0.45 J.K-1g-1
Hg (l)
0.14 J.K-1g-1
Example:
a) How much heat is needed to warm 1.00 L of
water from 23.0oC to 98.0oC?
(Assume the density of water is 1.00 kg/L).
b) What is the molar heat capacity of water?
a) How much heat is needed to warm 1.00 L of water
from 23.0oC to 98.0oC?
(Assume the density of water is 1.00 kg/L).
q
C
m  T
m = v = (1.00 L)(1.00 kg.L-1) = 1.00x103 g
T = 98.0 – 23.0 = 75.0oC = 75.0 K
q  CmT
q = (4.18 J.g-1.K-1)(1.00x103 g)(75.0 K)
q = 3.14x105 J
b) What is the molar heat capacity of water?
C = 4.18 J.g-1.K-1
MH2O = 18.016 g.mol-1
 4.18 J
g.K
x 18.016 g
mol
= 75.3 J.mol-1.K-1
CONSTANT PRESSURE CALORIMETRY
Any calorimeter at
atmospheric
pressure has a
constant pressure,
e.g. “coffee-cup”
calorimeter.
The heat produced by a reaction is entirely
absorbed by the solution at constant
pressure.
i.e. heat does not escape the calorimeter
Recall: H = qp
Can calculate
qsoln = CmT,
but we want qrxn
Since heat given off by the reaction is
absorbed by the solution:
qrxn = -qsoln
Example:
When 4.25 g solid ammonium nitrate dissolves
in 60.0 g water in a coffee-cup calorimeter, the
temperature drops from 22.0oC to 16.9oC.
Calculate H (in kJ/mol NH4NO3) for the
dissolution process.
Assume the specific heat of solution is the
same as that for pure water.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
When 4.25 g solid ammonium nitrate dissolves in 60.0 g
water in a coffee-cup calorimeter, the temperature drops
from 22.0oC to 16.9oC. Calculate H (in kJ/mol NH4NO3)
for the dissolution process.
T = 16.9 – 22.0 = -5.1oC = -5.1 K
m = 60.0 +4.25 = 64.3 g
assumption
qrxn = -CmT
qrxn = -(4.18 J.g-1.K-1)(64.3 g)(-5.1 K)
qrxn = 1.4x103 J
endothermic
When 4.25 g solid ammonium nitrate dissolves in 60.0 g
water in a coffee-cup calorimeter, the temperature drops
from 22.0oC to 16.9oC. Calculate H (in kJ/mol NH4NO3)
for the dissolution process.
qrxn = 1.4x103 J
MNH4NO3 = 80.052 g.mol-1
m
n=
M
4.25 g
= 80.052 g.mol-1
= 0.0531 mol
1.4x103 J
H = qp =
= 26.4 kJ.mol-1
0.0531 mol
endothermic
CONSTANT VOLUME CALORIMETRY
– BOMB CALORIMETER
Recall: U = qv
 At constant volume – measure U rather
than H, but for most U  H
A bomb calorimeter is used to study
combustion reactions etc.
We calculate the heat of combustion from the
measured change in temperature.
We need to know the heat capacity of the
calorimeter (Ccal).
qrxn = -Ccal x T
NB:
This equation cannot be applied blindly!!!
Always check units for Ccal.
Always include units in calculations to
ensure they cancel.
Bomb Calorimeter
Example:
A 2.200 g sample of quinone, C6H4O2, is burned
in a bomb calorimeter whose total heat capacity
is 7.854 kJ/oC. The temperature of the
calorimeter increases from 23.44oC to 30.57oC.
What is the heat of combustion per gram of
quinone? per mole of quinone?
T = 30.57 – 23.44 = 7.13oC
qrxn = -CcalT
qrxn = -(7.854 kJ.oC-1)(7.13oC)
qrxn = -56.0 kJ
exothermic
exothermic
A 2.200 g sample of quinone, C6H4O2, is burned in a
bomb calorimeter whose total heat capacity is 7.854
kJ/oC. The temperature of the calorimeter increases
from 23.44oC to 30.57oC.
What is the heat of combustion per gram of quinone?
per mole of quinone?
qrxn = -56.0 kJ
-25.5 kJ
produced by
i.e. -25.5 kJ.g-1
-25.5 kJ.g-1 x 108.09 g.mol-1
= -2.75 x 103 kJ.mol-1
2.200 g quinone
1 g quinone
Example:
50.0 g of water at 62.5oC was poured into a
calorimeter containing 50.0 g water at 18.7oC.
The final temperature was 35.0oC. How much
heat was lost to the surroundings during this
process?
(Heat capacity of water = 4.184 J.oC-1.g-1)
Ans: -2.34 kJ
50.0 g of water at 62.5oC was poured into a calorimeter containing
50.0 g water at 18.7oC. The final temperature was 35.0oC. How
much heat was lost to the surroundings during this process?
(Heat capacity of water = 4.184 J.oC-1.g-1)
H2O 1:
m = 50.0 g
Ti = 62.5oC
Tf = 35.0oC
T = -27.5oC
H2O 2:
m = 50.0 g
Ti = 18.7oC
Tf = 35.0oC
T = 16.3oC
qsoln = CmT
q1 = (4.184)(50.0)(-27.5)
q1 = -5753 J
q2 = (4.184)(50.0)(16.3)
q2 = 3410 J
q1 + q2 = -2343 J = -2.34 kJ
Negative q  heat lost by soln to the surroundings
HESS’S LAW
If a reaction is carried out in a series of steps,
H for the reaction will be equal to the sum of
the enthalpy changes for the individual steps.
A+BX+Y
H1
X+YC+D
H2
A+BC+D
H1+ H2
H for a reaction is calculated from H data
of other reactions.
Recall that enthalpy is a state function  H
is independent of the path followed.
It is useful to use Hess’s Law in cases where
H cannot be measured.
Example:
Carbon occurs in two forms, graphite and
diamond. The enthalpy of combustion of
graphite is –393.5 kJ/mol and that of diamond is
–395.4 kJ/mol.
Calculate H for the conversion of graphite to
diamond.
C(s,graph) + O2(g)  CO2(g)
H = –393.5 kJ/mol
C(s,diam) + O2(g)  CO2(g)
H = –395.4 kJ/mol
C(s,graph)  C(s,diam)
H = ?
C(s,graph) + O2(g)  CO2(g)
H = –393.5 kJ/mol
C(s,diam) + O2(g)  CO2(g)
H = –395.4 kJ/mol
C(s,graph)  C(s,diam) H = ?
C(s,graph) + O2(g)  CO2(g)
H = -393.5 kJ/mol
CO2(g)  C(s,diam) + O2(g)
H = +395.4 kJ/mol
C(s,graph)  C(s,diam)
H = -393.5 + -(-395.4) = 1.9 kJ/mol
Example:
From the following enthalpies of reaction:
H2(g) + F2(g)  2HF(g)
C(s) + 2F2(g)  CF4(g)
2C(s) + 2H2(g)  C2H4(g)
H = -537 kJ
H = -680 kJ
H = 52.3 kJ
Calculate H for the reaction of ethylene with
F2:
C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
H2(g) + F2(g)  2HF(g)
C(s) + 2F2(g)  CF4(g)
2C(s) + 2H2(g)  C2H4(g)
H = -537 kJ
H = -680 kJ
H = 52.3 kJ
C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
C2H4(g)  2C(s) + 2H2(g)
2C(s) + 4F2(g)  2CF4(g)
2H2(g) + 2F2(g)  4HF(g)
H = -(52.3 kJ)
H = 2(-680 kJ)
H = 2(-537 kJ)
C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
H = -(52.3 kJ) + 2(-680 kJ) + 2(-537 kJ)
H = -2.49x103 kJ
STANDARD STATES
The magnitude of H depends on conditions
of temperature, pressure, and state of
products and reactants.
In order to compare enthalpies, need same
set of conditions.
Standard state of a substance = pure form at
1 atm and at the temperature of interest
(usually 298 K (25oC))
Standard enthalpy (Ho)  when all products
and reactants are in their standard states
STANDARD ENTHALPIES OF FORMATION
Hfo change in enthalpy for the reaction
that forms 1 mol of a substance from its
elements in their standard states.
Units: kJ/mol
Note: If there is more than one form of the
element present under standard conditions,
use the most stable form.
Hfo = 0 kJ/mol for the most stable form of
any element
Examples of most stable form:
Oxygen(g) can exist as O2 or O3 (ozone) at
1 atm and 298 K
Most stable form = O2
Hfo(O2(g)) = 0 kJ/mol
Carbon(s) can exist as graphite or diamond at
1 atm and 298 K
Most stable form = graphite
Hfo(C(s,graphite)) = 0 kJ/mol
Tabulated data of enthalpies of formation can
be used to calculate enthalpies of reaction as
follows:
o
Hrxn
 mHof (products)  nHof (reac tan ts)
Stoichiometric coefficients
Table of standard
enthalpies of formation
General Knowledge
Gray tin and white tin are two
solid forms of tin. The denser
white metallic form is the most
stable phase above 13oC, and
the powdery gray form is more
stable below 13oC.
The formation of gray tin is
said to have contributed to
Napoleon’s defeat at Moscow,
when his soldier’s buttons fell
off their clothes at the low
temperatures they
encountered there.
Example:
Nitroglycerine is a powerful explosive, giving four
different gases when detonated:
2C3H5(NO3)3 (l)  3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)
Given that the enthalpy of formation of nitroglycerine
is Hfo = -364 kJ.mol-1, calculate the enthalpy
change when 10.0 g of nitroglycerine is detonated.
Hfo {C3H5(NO3)3 (l)} = -364 kJ.mol-1
Hfo {H2O (g)} = -241.8 kJ.mol-1
Hfo {CO2 (g)} = -393.5 kJ.mol-1
Hfo {O2 (g)} =
Hfo {N2 (g)} =
0 kJ.mol-1
2C3H5(NO3)3 (l)  3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)
Calculate the enthalpy change when 10.0 g of
nitroglycerine is detonated.
Hfo {C3H5(NO3)3 (l)} = -364 kJ.mol-1
Hfo {H2O (g)} = -241.8 kJ.mol-1
Hfo {CO2 (g)} = -393.5 kJ.mol-1
o
Hrxn
 mHof (products)  nHof (reac tan ts)
Horxn = [3(0) + 1/2(0) + 6(-393.5) + 5(-241.8)]
- [2(-364)]
Horxn = [-3570]-[-728]
Horxn = -2842 kJ
2C3H5(NO3)3 (l)  3N2(g) + 1/2O2(g) + 6CO2(g) + 5H2O(g)
Calculate the enthalpy change when 10.0 g of
nitroglycerine is detonated.
Horxn = -2842 kJ
nC3H5 (NO3 )3
10.0 g
m

 0.0440 mol

1
227.1 g.mol
M
-2842 kJ for 2 mol NG

x kJ for 0.0440 mol NG
x = -62.5 kJ for 10.0 g
Example:
For you to do!
Calculate Ho for the decomposition of
limestone:
CaCO3(s)  CaO(s) + CO2(g)
Hence calculate the heat required to
decompose 1 kg of limestone.
Hfo {CaCO3 (s)} = -1207.1 kJ.mol-1
Hfo {CaO (s)} = -635.5 kJ.mol-1
Hfo {CO2 (g)} = -393.5 kJ.mol-1
Answer:
Ho = 178.1 kJ for 1 mol
qp = Ho = 1779 kJ for decomposition of 1 kg CaCO3
BOND ENTHALPIES
The stability of a molecule is related to the
strengths of the covalent bonds it contains.
The strength of a covalent bond between two
atoms is determined by the energy required
to break that bond.
Bond enthalpy
= enthalpy change for breaking a
particular bond in a mole of gaseous
substance.
Can determine approximate value for Horxn if
we have the values of average bond
enthalpies.
Horxn is calculated by determining the
energy required to break all bonds minus the
energy evolved to form all bonds.
o
Hrxn
 BE(bonds broken)  BE(bonds formed)
Reactants
Products
initial
final
This is only an approximation because the
average bond enthalpies used as the bond
enthalpies are not exactly the same in all
molecules.
e.g. C-H bond energy in CH4 is slightly
different to that in C2H6 and so on.
H
H-C-H
H
H
H
H-C–C-H
H
H
You must know these:
O2
O=O
N2
NN
CO
CO
CO2
O=C=O
CN
CN
H2CO
H
C=O
H
Example:
Hydroiodic acid reacts with chlorine as follows:
2HI(aq) + Cl2(g)  2HCl(aq) + I2(s)
Approximate Ho for the reaction using
tabulated bond energies.
2HI(aq) + Cl2(g)  2HCl(aq) + I2(s)
H-I
H-I
Cl-Cl
H-Cl
H-Cl
I-I
2HI(aq) + Cl2(g)  2HCl(aq) + I2(s)
Bonds Broken
2 x H-I
1 x Cl-Cl
Bonds Formed
2 x H-Cl
1 x I-I
Bond Energy / kJ.mol-1
2(297)
239
833
2(431)
149
1011
o
Hrxn
 BE(bonds broken)  BE(bonds formed)
= 833 – 1011 = -178 kJ.mol-1
Example:
Estimate H using bond enthalpies for the
following reaction:
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(g)
H
H
O=O
H-C–C-H
H
O=C=O
O
H H
H
6 x C-H
1 x C-C
7/ x O=O
2
4 x O=C
6 x O-H
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(g)
Bonds Broken
6 x C-H
1 x C-C
7/ x O=O
2
Bonds Formed
4 x O=C
6 x O-H
Bond Energy / kJ.mol-1
6(413)
348
7/ (494)
2
4555
4(707)
6(463)
5606
o
Hrxn
 BE(bonds broken)  BE(bonds formed)
= 4555 – 5606 = -1051 kJ.mol-1