Chapter 5
Thermochemistry
Definitions
• Thermochemistry - the study of how energy
in the form of heat is consumed and
produced by chemical reactions.
• Thermochemical Reaction: Example
H2(g) + O2(g)
2 H2O (l) + heat
• Energy is anything having the capacity to do
work or to transfer heat.
• Work is Force X Distance
• Thermodynamics The study of energy and
its transformation from one form to another.
Energy Examples
Definition-Energy is anything with the capacity to do
work, or create heat.
• Food
• Gasoline
• Electricity
• An apple on a tree
• A baseball moving
Energy Units
Units of Energy
 Joules = kg(m/s)2
 Calories, an older unit; the energy to
increase one gram of water one deg. C
 Calories, unit of food energy = kcal
Definitions Continued
• Heat is energy transferred between
objects because of a difference in their
temperatures.
• Thermodynamics is the study of
relationship between chemical reactions
and changes in heat energy.
• Heat transfer is the process of heat
energy flowing from one object into
another
Two Types of Energy
Potential: due to position or composition - can be
converted to work PE = mgh (m = mass, kg, g =
force of gravity(9.8m/s2), and h = vertical distance
in meters)
• These units multiplied together = joule
• (chemical energy is a form of potential energy)
• Kinetic: due to motion of the object
KE = 1/2 mv 2 (m = mass(kg), v = velocity(m/s))
• These units multiplied together also equal a joule
• Joule = kg(m/s)2
Kinetic vs. Potential Energy
Potential Energy: A State Function
• Depends only on the present state of
the system - not how it arrived there.
• It is independent of pathway.
Other State Functions?
 Heat (a form of energy)?
 Work? Altitude?
 Altitude?
Other State Functions?
 Heat (a form of energy)? Yes, not path dependent
 Work?
 Altitude?
Other State Functions?
 Heat (a form of energy)? Yes, not path dependent
 Work? No, work depends on path
 Altitude?
Other State Functions?




Heat (a form of energy)? Yes, not path dependent
Work? No, work depends on path
Altitude? Yes, does not depend on path
Enthalpy ΔH,?
Other State Functions?




Heat (a form of energy)? Yes, not path dependent
Work? No, work depends on path
Altitude? Yes, does not depend on path
Enthalpy ΔH,? Yes, does not depend on path
The Nature of Energy
•
Energy is conserved: Law of Conservation of
Energy states that energy cannot be created nor
destroyed, but converted from one form to another
Energy at the Molecular Level
Kinetic energy at the molecular level depends on
the mass and velocity of the particle but because its
velocity depends on temperature KE does too.
• As temperature increases then the KE of the same will
increase.
One of the most important forms of potential energy
at the atomic-molecular level arises from electrostatic
interactions.
Potential can be converted into kinetic
• Example: an apple falling from a tree
Electrostatic Potential Energy
Coulombic attraction, not gravitational force,
determines the potential energy of matter at the
atomic level.
Q1 x Q2
Eel 
d
is the electrostatic potential energy
• Eel
• Q is the charge in coulombs
• d is the distance between particles

Electrostatic Potential Energy
Bond length
Energy of an
ionic bond
Bonds contain potential energy
Energy required to break bonds
Energy released when bonds are created
Crystal Lattice of NaCl
Ionic compounds do not exist as discrete molecules. Instead they
exist as crystals where ions of opposite charges occupy
positions known as lattice sites.
In an ionic compound the
ions organize in such a way
as to minimize repulsive
and maximize attractive
forces.
-
+
+
-
The arrangement of charged particles in a covalent bond
organized in such a way as to minimize repulsive and maximize
attractive forces to give the lowest potential energy.
Terms Describing Energy Transfer
• System: the part of the universe that is
the focus of a thermodynamic study.
 Example a beaker or test tube in the lab
• Surroundings: everything in the
universe that is not part of the system.
• Universe = System + Surroundings
• An isolated system exchanges neither
energy nor matter with the surroundings.
Examples
Heat Flow
• In an exothermic process, heat flows from a
system into its surroundings.
• In an endothermic process, heat flows from
the surroundings into the system
Phase Changes
Internal Energy
• The internal energy of a system is the sum of
all the KE and PE of all of the components of
the system.
First Law of Thermodynamics
• The first law of thermodynamics states that
the energy gained or lost by a system must
equal the energy lost or gained by
surroundings. ΔE = q + w (mathematical
statement)
• The calorie (cal) is the amount of heat
necessary to raise the temperature of 1 g of
water 1oC.
• The joule (J) is the SI unit of energy; 4.184 J
= 1 cal.
Energy Flow Diagrams
Change in Internal Energy
E = q + w
 E = change in system’s internal energy
 q = heat
Endothermic +q
Exothermic -q
Expansion –w (since system is losing energy to do work)
Compression +w
First Law Problem
Find the change in energy of a system when 12 j of
energy flows into the system while the system is
doing 8 j of work on the surroundings.
First Law Problem
Find the change in energy of a system when 12 j of
energy flows into the system while the system is
doing 8 j of work on the surroundings.
ΔE = q + w ↔ ΔE = + 12 – 8 = 4 j
PV Work
Expansion
Atm = 14
Facts
lb/in2
A
A
h
W=Fxd
P = F/A
F = PA
W = PA x d
ΔV = A x d
W = P ΔV
ΔV = Vfinal – Vinitial
PV Work
Facts
Expansion
A
ΔV = Vfinal – Vinitial
h
ΔV = A x d
W=Fxd
P = F/A
Since expansion is defined
as negative work and ΔV is
positive for expansion, then
we change the sign of PΔV
to –PΔV,
ΔE = q - PΔV
F = PA
Substituting
W = PxAxd
W = PΔV
PV Work
Facts
Expansion
A
h
ΔV = Vfinal – Vinitial
ΔV = A x d
W=Fxd
P = F/A
Since expansion is defined as
negative work and ΔV is
positive for expansion, then we
change the sign of PΔV to –
PΔV,
ΔE = q - PΔV (L-atm = 101.3 j)
F = PA
Substituting
W = PxAxd
W = PΔV
Calculation of Work
Calculate the work in L•atm and joules associated
with the expansion of a gas in a cylinder from 54 L
to 72 L at a constant external pressure of 18 atm
(Note that as the gas expands, it does work on its
surroundings. Energy flows out of the gas, so it is a
negative quantity)
Calculation of Work
Calculate the work in L•atm and joules associated
with the expansion of a gas in a cylinder from 54 L
to 72 L at a constant external pressure of 18 atm
(Note that as the gas expands, it does work on its
surroundings. Energy flows out of the gas, so it is a
negative quantity)
W = -pΔV
W = -18atm(72-54)L = -320L-atm
-320L-atm j
101.3 L-atm
= -3.2 j
Calculation of Work
Calculate the work in L•atm and joules associated
with the expansion of a gas in a cylinder from 54 L
to 72 L at a constant external pressure of 18 atm
(Note that as the gas expands, it does work on its
surroundings. Energy flows out of the gas, so it is a
negative quantity)
Note: The result is
W = -pΔV
W = -18atm(72-54)L = -320L-atm
-320L-atm j
101.3 L-atm
= -3.2 j
negative, which we
would predict relative to
expansion. This is the
reason that work is pΔV, to give the correct
sign for expansion.
Enthalpy and Change in Enthalpy
•
•
•
•
•
Enthalpy (H) = E + PV ( mathematical definition)
Change in Enthalpy (H) = E + PV
At constant P, qP = E + PV, therefore qP = H
H = change in enthalpy: an energy flow as heat (at
constant pressure)
H > 0, Endothermic; H < 0, Exothermic
Heating Curves
Heat Capacities
• Molar heat capacity (cp) is the heat required
to raise the temperature of 1 mole of a
substance by 1oC at constant pressure.
• q = ncpT
• Specific heat (cs) is the heat required to
raise the temperature of 1 gram of a
substance by 1oC at constant pressure.
• Heat capacity (Cp) is the quantity of heat
needed to raise the temperature of some
specific object by 1oC at constant pressure.
Phase Change and Energy
• Molar heat of fusion (Hfus) - the heat
required to convert 1 mole of a solid
substance at its melting point to 1 mole of
liquid.
• q = nHfus
• Molar heat of vaporization (Hvap) - the heat
required to convert 1 mole of a substance at
its boiling point to 1 mole of vapor.
• q = nHvap
WATER THERMO VALUES
•
•
•
•
•
Ice H2O (s) 2.06 j/g-°C
Water H2O (l) 4.184 j/g-°C
Steam H2O (g)
1.86 j/g-°C
Heat of Fusion (melting) 334.0 j/g
Heat of vaporization (evaporation) 2257 j/g
Practice
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
heat?
Practice
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
heat?
g
2257 j
Practice
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
heat?
g
2257 j
10 3 j
kj
Practice
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
heat?
g
2257 j
10 3 j
kj
2000 kj
Practice
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
heat?
g
2257 j
10 3 j
kj
2000 kj
= 886 g water
Practice
5.53 From Text
Exactly 10 mL of water at 25oC was added
to a hot iron skillet. All of the water was
converted into steam at 100oC. If the mass
of the pan was 1.20 kg and the molar heat
capacity of iron is 25.19 J/mol•oC, what was
the temperature change of the skillet?
Sample Problem Solution
mole-°C
25.19 j
Sample Problem Solution
mole-°C 55.85 g
25.19 j mole
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j 10.0 g
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j 10.0 g =22570 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Now Combine
Evaporating 10.0 mL of water 3138 j + 22570 j = 25708j
2257 j 10.0 g =22570 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25708 j
25.19 j mole
103 g 1.20 kg
= 47.5 °C
Heat of Reaction
Heat of reaction is also known as
enthalpy of reaction (Hrxn) is the heat
absorbed or released by a chemical
reaction.
A Specific Enthalpy
• The standard enthalpy of formation (Hfo)
is also called the standard heat of formation
and is the enthalpy change of the a formation
reaction.
• A formation reaction is the process of forming
1 mole of a substance in its standard state
from its component elements in their standard
states.
• H2(g) + 1/2 O2(g) ---> H2O(l) Hfo for water
• The standard state of a substance is its most
stable form under 1 bar pressure and 25oC.
Methods of Determining Hrxn
1. from calorimetry experiments
2. from enthalpies of formation
3. using Hess’s Law
Calorimetry
• Calorimetry is the measurement of the
change in heat that occurs during a
physical change or chemical process.
• A calorimeter is the device used to
measure the absorption or release of
heat by a physical or chemical process.
Measuring Heat Capacity
-qaluminum = qwater; qwater = ncpT and -qaluminum = ncsT
Calorimetry: Bomb Calorimeter
• H = -qcal = -CcalT
• A bomb calorimeter
is a constant-volume
device used to
measure the heat of a
combustion reaction.
Practice
Write the standard enthalpy of formation
reaction for nitric acid.
Practice
Write the standard enthalpy of formation
reaction for nitric acid.
Standard enthalpy, ΔH°, is the change in
energy when elements, in their standard
state (25°C and 1 atm) combine to make 1
mole of products at their standard state.
Calculating Hrxn°
• Hrxn° = npHf(products)  nrHf(reactants)
Calculating Hrxn°
• Hrxn° = npHf(products)  nrHf(reactants)
½ N2 (g) + ½ H2 (g) + 3/2 O2 (g)
HNO3 (l)
Calculating Hrxn°
• Hrxn° = npHf(products)  nrHf(reactants)
1/2(0.00)
½(0.00)
3/2(0.0)
½ N2 (g) + ½ H2 (g) + 3/2 O2 (g)
1(-135.1 kj)
HNO3 (l)
Calculating Hrxn°
• Hrxn° = npHf(products)  nrHf(reactants)
1/2(0.00)
½(0.00)
3/2(0.0)
½ N2 (g) + ½ H2 (g) + 3/2 O2 (g)
1(-135.1 kj)
HNO3 (l)
ΔH°rxn = [1(-135.1 kj)] – [1/2(0.00) + 1/2(0.00) + 3/2(0.0)]
ΔH°rxn = - 135.1 kj
Example
Use Table 5.2 to calculate an
approximate enthalpy of reaction for
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)
Practice
One step in the production of nitric acid is the
combustion of ammonia. Using the data in the
appendix to calculate the enthalpy of this reaction.
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)
Fuel Values
Molecular
Formula
Fuel Value
(kJ/g)*
Methane
CH4
50.0
Ethane
C2H6
47.6
Propane
C3H8
46.3
Butane
C4H10
45.8
Compound
* Based on the formation of H2O (g)
Hess’s Law
• Hess’s law states that the enthalpy change of
a reaction that is the sum of two or more
reactions is equal to the sum of the enthalpy
changes of the constituent reactions.
Calculations via Hess’s Law
1. If a reaction is reversed, H sign changes.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by
an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
Example
Calculate the enthalpy change for
C2H4(g) + H2(g)
C2H6(g) using the following data.
H2(g) + 1/2O2(g)
C2H4(g) + 3O2(g)
C2H6(g) + 7/2O2(g)
H2O(l)
-285.8 kJ
2H2O(l) + 2CO2(g) -1411 kJ
3H2O(l) + 2CO2(g) -1560 kJ
Example
Calculate the enthalpy change for
C2H4(g) + H2(g)
C2H6(g) using the following data.
H2(g) + 1/2O2(g)
C2H4(g) + 3O2(g)
3H2O(l) + 2CO2(g)
H2O(l)
-285.8 kJ
2H2O(l) + 2CO2(g) -1411 kJ
C2H6(g) + 7/2O2(g) +1560 kJ
H2(g) +1/2O2(g)+C2H4(g)+3O2(g)+3H2O(l)+2CO2(g)
H2O(l)+ 2H2O(l) + 2CO2(g)+ C2H6(g) + 7/2O2(g) -136.8
simplify
C2H4(g) + H2(g)
C2H6(g) -136.8 kj
Review
ChemTour: State Functions and
Path Functions
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PC | Mac
This ChemTour defines and explores the difference
between state and path functions using a travel analogy that
leads into a discussion of energy, enthalpy, heat, and work.
ChemTour: Internal Energy
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This ChemTour explores how energy is exchanged between
a system and its surroundings as heat and/or work, and
how this transfer in turn affects the internal energy (E) of a
system.
ChemTour: Pressure-Volume
Work
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PC | Mac
An animated ChemTour of an internal combustion engine
shows how a system undergoing an exothermic reaction
can do work on its surroundings; students can explore the
relationship among pressure, volume, and work.
ChemTour: Heating Curves
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PC | Mac
In this ChemTour, students use interactive heating curve
diagrams to explore phase changes, heat of fusion, and
heat of vaporization. Macroscopic views of ice melting and
water boiling are shown in sync with the appropriate
sections of the heating curve.
ChemTour: Calorimetry
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This ChemTour demonstrates how a bomb calorimeter
works, and walks students through the equations used to
solve calorimetry problems. Includes an interactive
experiment.
ChemTour: Hess’s Law
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This ChemTour explains Hess’s law of constant heat of
summation using animated sample problems and step-bystep descriptions.
An ideal gas in a sealed piston is
allowed to expand isothermally (at a
constant temperature) against a
pressure of 1 atm. In what direction, if
at all, does heat flow for this process?
A) into the system
B) out of the system
Isothermal Expansion of Ideal Gas
C) heat does not flow
Consider the following arguments for each answer and
vote again:
A. When the gas expands isothermally, it does work
without a decrease in its energy, so heat must flow into
the system.
B. During the expansion, the gas pressure decreases,
thereby releasing heat to the surroundings.
C. The fact that the process is isothermal means that heat
does not flow.
Isothermal Expansion of Ideal Gas
When air is released from a tire does it get warmer or cooler?
When air is released from a tire does it get warmer or cooler?
Cooler
Isothermal means that the temperature is unchanged, so what
must happen for the temperature to remain the same?
When air is released from a tire does it get warmer or cooler?
Cooler
Isothermal means that the temperature is unchanged, so what
must happen for the temperature to remain the same?
Heat must flow into the tire
An ideal gas in an insulated piston is
compressed adiabatically (q = 0) by its
surroundings. What can be said of the
change in the temperature (ΔT) of the
gas for this process?
A) ΔT > 0
Adiabatic Compression of an Ideal Gas
B) ΔT = 0 C) ΔT < 0
Consider the following arguments for each answer
and vote again:
A. The surroundings are doing work on the system, and
no heat is flowing. Therefore, ΔE > 0 and so ΔT > 0.
B. The volume of the gas decreases, but the pressure
increases to keep the product of the pressure and
volume constant. Therefore, the temperature is also
constant.
C. The gas is being compressed to a more ordered state,
which corresponds to a lower temperature.
Adiabatic Compression of an Ideal Gas
Consider a stretched rubber band that is
suddenly released. What can be said of
the change in the temperature (ΔT) of
the rubber band for this process?
A) ΔT > 0
ΔT of a Released Rubber Band
B) ΔT = 0 C) ΔT < 0
Consider the following arguments for each answer
and vote again:
A. The stretched rubber band is at a higher energy state
than the unstretched rubber band. Releasing the
stretched rubber band causes the energy to be
released.
B. Because the recoil of the rubber band is rapid, this
process is essentially adiabatic.
Therefore, the
temperature of the rubber band will not change.
C. As the rubber band contracts, it does work and its
energy decreases, resulting in a decrease in its
temperature.
ΔT of a Released Rubber Band
A 1.0 gram block of Al (cs = 0.9 J/·°C1·g) at 100 °C and a 1.0 gram block of
Fe (cs = 0.4 J/·°C-1·g) at 0 °C are
added to 10 mL of water (cs =
4.2 J·/°C-1·g) at 50 °C. What will be
the final temperature of the water?
A) < 50 °C
Specific Heat Capacity of Al and Fe
B) 50 °C C) > 50 °C
Consider the following arguments for each answer and
vote again:
A. The specific heat capacity of Fe(s) is smaller than that
of Al(s), so heat from both the Al(s) and the water will
be required to warm the Fe(s).
B. The average initial temperature of the three
components is 50 °C. Therefore, the final temperature
of the water will be 50 °C.
C. The specific heat capacity of Al(s) is greater than that
of Fe(s), so the Al block at 100 °C will heat the water
more than the Fe block will cool it.
Specific Heat Capacity of Al and Fe
This is called a first law problem (energy cannot be created nor
destroyed), and a mathematical statement would be – q = q.
Remember q is measured in joules, so – j = + j
This is called a first law problem (energy cannot be created nor
destroyed), and a mathematical statement would be – q = q.
Remember q is measured in joules, so – j = + j
Since the heat capacity of aluminum is twice that of carbon,
then it will lose more heat than iron will gain. This means
more heat is lost by the aluminum than gained by the iron.
Therefore, the water will increase in temperature and the
resultant temperature will be greater than 50°C.
Cyclooctatetraene, C8H8, can undergo a
transformation between two possible states, A and B,
by rearranging its 4 double bonds.
Which of the following graphs depicts the
dependence of the equilibrium constant (K) on
temperature for the conversion from state A to state
B?
A)
B)
Equilibium Constant of Cyclooctatetraene
C)
Consider the following arguments for each answer
and vote again:
A. The intermediate is at a higher energy state than
either states A or B, so at high temperatures, the
reaction will favor the intermediate and K will
decrease.
B. The enthalpies of formation for states A and B are
equal, so ΔH° = 0 and K is not temperature
dependent.
C. At high temperatures, the conversion from state A to
state B will occur at a much faster rate, thus
increasing the value of K.
Equilibium Constant of Cyclooctatetraene
The reaction of N2(g) and O2(g) to form
N2O(g) is an endothermic process. The
reaction of N2(g) and H2(g) to form
NH3(g) is an exothermic process.
Given this information, which of the
following species has the lowest
enthalpy of formation, ?
A) N2(g)
Enthalpies of N , NH , and N O
B) NH3(g)
C) N2O(g)
Consider the following arguments for each answer
and vote again:
A. N2(g) is the elemental form of nitrogen, which by
definition will have H f= 0, the lowest possible
enthalpy of formation.
B. N2O(g) has a higher H f than N2(g), O2(g), and
H2(g), whereas NH3(g) has a lower H f than N2(g),
O2(g), and H2(g).
C. The formation of an N−N double bond and a N−O
double bond, as found in N2O, releases more energy
than does the creation of 3 N−H bonds to form
NH3(g).
Enthalpies of N , NH , and N 0
Which of the following is true of ΔH° for the
polymerization of ethylene to form polyethylene?
Note: the C-C single bond enthalpy is ~350 kJ/mole
and the C-C double bond enthalpy is ~600 kJ/mole.
A) ΔH° > 0
Polymerization of Ethylene
B) ΔH° = 0
C) ΔH° < 0
Consider the following arguments for each answer
and vote again:
A. The energy required to break double bonds is more
than the energy released by forming new single
bonds.
B. The total number of C-C bonds (if we count double
bonds twice) does not change with polymerization.
Therefore, there can be no change in ΔH°.
C. For each C-C double bond that breaks (~600
kJ/mole), two single bonds form, (2×~350 = ~700
kJ/mole).
Polymerization of Ethylene
THE END