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Principles of Chemistry: A Molecular Approach, 1st Ed.
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Nature of Energy
• Even though Chemistry is the study of
matter, energy affects matter.
Chapter 6
Thermochemistry
• Energy is anything that has the
capacity to do work.
• Work is a force acting over a distance.
9energy = work = force × distance
• Energy can be exchanged between
objects through contact.
9collisions
http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/06thermo
page/thermochemistry.htm
Tro, Principles of Chemistry: A Molecular Approach
Energy, Heat, and Work
• You can think of energy as a quantity an
object can possess.
9or collection of objects
• You can think of heat and work as the two
different ways that an object can exchange
energy with other objects.
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Classification of
Energy
• Kinetic energy is
energy of motion or
energy that is being
transferred.
9Thermal energy is
kinetic.
9either out of it, or into it
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Classification of Energy
• Potential energy is energy that is stored in
an object, or energy associated with the
composition and position of the object.
9Energy stored in the structure of a compound is
potential.
Exchanging Energy
• Energy cannot be created
•
•
or destroyed.
Energy can be
exchanged between
objects.
Energy can also be
transformed from one
form to another.
9 heat → light → sound
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Some Forms of Energy
• electrical
9 kinetic energy associated with the flow of electrical charge
• heat or thermal energy
9 kinetic energy associated with molecular motion
• light or radiant energy
9 kinetic energy associated with energy transitions in an atom
• nuclear
9 potential energy in the nucleus of atoms
• chemical
9 potential energy due to the structure of the atoms, the
attachment between atoms, the atoms’ relative position to
each other in the molecule, or the molecules’ relative
positions in the structure
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System and Surroundings
• We define the system as the material or
process that contains the energy changes
we are studying.
• We define the surroundings as everything
else in the universe.
• What we study is the exchange of energy
between the system and the surroundings.
Surroundings
Surroundings
System
System
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Units of Energy
• One joule (J) is the amount of energy needed to
move a 1-kg mass a distance of 1 meter.
9 1 J = 1 N·m = 1 kg·m2/s2
• One calorie (cal) is the amount of energy needed
to raise one gram of water by 1 °C.
9 kcal = energy needed to raise 1000 g of water 1 °C
9 food Calories = kcals
=
=
9
• The internal energy is the total amount of kinetic
and potential energy a system possesses.
The change in the internal energy of a system
depends only on the amount of energy in the
system at the beginning and end.
9 A state function is a mathematical function whose
result depends only on the initial and final conditions, not
on the process used.
9 ΔE = Efinal – Einitial
9 ΔEreaction = Eproducts – Ereactants
Tro, Principles of Chemistry: A Molecular Approach
energy change in the system and the surroundings
must be zero.
9 ΔEnergyuniverse = 0 = ΔEnergysystem + ΔEnergysurroundings
9 Δ is the symbol that is used to mean change.
4.184 joules (J) (exact)
1000 calories (cal)
Internal Energy
•
• Conservation of energy requires that the total
¾ final amount – initial amount
Energy Conversion Factors
1 calorie (cal)
1 kilo calorie (kcal)
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Energy Flow and
Conservation of Energy
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State Function
You can travel either of
two trails to reach the top
of the mountain. One is
long and winding; the
other is shorter but
steeper. Regardless of
which trail you take,
when you reach the top
you will be 10,000 ft.
above the base.
The height of the mountain is a state function. It depends
only on the distance from the base to the peak, not on how
you arrive there!
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Energy Flow
Energy Diagrams
“graphical” way of showing
the direction of energy flow
during a process.
• If the final condition has a
larger amount of internal
energy than the initial
condition, the change in the
internal energy will be +.
initial
energy added
ΔE = +
Internal Energy
•
initial
final
energy removed
ΔE = –
13
•
•
Surroundings
ΔE +
System
ΔE –
14
Energy Flow in a Chemical Reaction
• When energy flows into a
system, it must all come from the
surroundings.
When energy flows into a
system, ΔEsystem is +.
When energy flows out of the
surroundings, ΔEsurroundings is –.
therefore:
ΔEsystem = –ΔEsurroundings
system, it must all flow into the
surroundings.
When energy flows out of a
system, ΔEsystem is –.
When energy flows into the
surroundings, ΔEsurroundings is +.
therefore:
–ΔEsystem = ΔEsurroundings
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Energy Flow
•
•
•
• If the final condition has a
smaller amount of internal
energy than the initial
condition, the change in the
internal energy will be –.
Tro, Principles of Chemistry: A Molecular Approach
• When energy flows out of a
final
• The total amount of internal energy
Surroundings
ΔE –
System
ΔE +
in 1 mol of C(s) and 1 mole of O2(g)
is greater than the internal energy in
1 mole of CO2(g).
9 at the same temperature and pressure
• In the reaction C(s) + O2(g) → CO2(g),
there will be a net release of energy
into the surroundings.
Surroundings
9 −ΔEreaction = ΔEsurroundings
9 ΔEreaction = −ΔEsurroundings
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C(s), O2(g)
energy
energy
absorbed
released
CO2(g) ΔE
ΔErxn
rxn==–+
• In the reaction CO2(g) → C(s) + O2(g),
there will be an absorption of
energy from the surroundings into
the reaction.
Tro, Principles of Chemistry: A Molecular Approach
Internal Energy
Internal Energy
• Energy diagrams are a
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System
C+O
CO
C
+ 2O→
2 → CO2
Energy Exchange
Energy Exchange
• Energy is exchanged between the system and
surroundings through heat and work.
9 q = heat (thermal) energy
9 w = work energy
9 q and w are NOT state functions; their value depends
on the process.
ΔE = q + w
• Energy is exchanged between the system
and surroundings through either heat
exchange or work being done.
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Heat and Work
• On a smooth table, most of the kinetic
energy is transferred from the first ball to
the second—with a small amount lost
through friction.
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Example 6.1 If burning fuel in a potato cannon gives 855 J of
kinetic energy to the potato and releases 1422 J of heat from
the cannon, what is the change in internal energy of the fuel?
Given:
Find:
qpotato = 855 J, wpotato = 1422 J
ΔEfuel, J
Concept Plan:
Relationships:
qsystem = −qsurroundings, wsystem = −wsurroundings, ΔE = q + w
Solution:
Check:
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The unit and sign are correct.
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Heat Exchange
• Heat is the exchange of thermal energy
between the system and surroundings.
• It occurs when the system and surroundings
have a difference in temperature.
• Heat flows from matter with high
temperature to matter with low temperature
until both objects reach the same
temperature.
9thermal equilibrium
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Quantity of Heat Energy Absorbed
“Heat Capacity”
• When a system absorbs heat, its temperature increases.
• The increase in temperature is directly proportional to the
amount of heat absorbed.
• The proportionality constant is called the heat capacity, C.
9 Units of C are J/°C .
q = C × ΔT
• The heat capacity of an object depends on its mass.
9 200 g of water requires twice as much heat to raise its temperature
by 1 °C as does 100 g of water.
• The heat capacity of an object depends on the type of
material.
9 1000 J of heat energy will raise the temperature of 100 g of sand
12 °C, but only raise the temperature of 100 g of water by 2.4 °C.
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“Specific Heat Capacity”
• measure of a substance’s
intrinsic ability to absorb heat
• The specific heat capacity
(Cs)
is the amount of heat energy
required to raise the
temperature of one gram of a
substance 1 °C.
9units are J/(g·°C)
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Quantifying Heat Energy
• The heat capacity of an object is proportional to its
mass and the specific heat of the material.
• So we can calculate the quantity of heat absorbed
by an object if we know the mass, the specific
heat, and the temperature change of the object.
heat = (mass) × (specific heat capacity) × (temp. change)
q = (m) × (Cs) × (ΔT)
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Example 6.2 How much heat is absorbed by a copper
penny with mass 3.10 g whose temperature rises from
–8.0 °C to 37.0 °C?
Sort
information
Given:
T1 = –8.0 °C, T2 = 37.0 °C, m = 3.10 g
Practice—Calculate the amount of heat
released when 7.40 g of water cools from
49° to 29 °C.
q, J
Find:
Conceptual
Plan:
Strategize
Relationships: q = m · Cs · ΔT
Cs = 0.385 J/g (Table 6.4)
Follow the
conceptual plan
to solve the
problem.
Check
Solution:
Check:
The unit and sign are correct.
Answer: 6.2 x 102 J
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Pressure–Volume Work
• PV work is work that is the result of a volume
•
•
change against an external pressure.
When gases expand, ΔV is +, but the system is
doing work on the surroundings, so wgas is –.
As long as the external pressure is kept
constant,
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Example 6.3 If a balloon is inflated from 0.100 L to
1.85 L against an external pressure of 1.00 atm,
how much work is done?
Given:
Find:
V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm
w, J
Conceptual
Plan:
Relationships: 101.3 J = 1 atm·L
Solution:
–work = external pressure × change in volume
w = –PΔV
9 To convert the units to joules, use 101.3 J = 1 atm·L.
Check:
The unit and sign are correct.
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Exchanging Energy between
System and Surroundings
• exchange of heat energy
q = mass × specific heat × ΔTemperature
• exchange of work
w = –Pressure × ΔVolume
Measuring ΔE,
Calorimetry at Constant Volume
• Since ΔE = q + w, we can determine ΔE by measuring q and w.
• In practice, it is easiest to do a process in such a way that
there is no change in volume, w = 0.
9 at constant volume, ΔEsystem = qsystem
• In practice, it is not possible to observe the temperature
•
changes of the individual chemicals involved in a reaction. So
instead, we use an insulated, controlled surroundings and
measure the temperature change in it.
The surroundings is called a bomb calorimeter and is usually
made of a sealed, insulated container filled with water.
qsurroundings = qcalorimeter = –qsystem
–ΔEreaction = qcal = Ccal × ΔT
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Bomb Calorimeter
• used to measure
Tro, Principles of Chemistry: A Molecular Approach
Example 6.4 When 1.010 g of sugar is burned in a bomb
calorimeter, the temperature rises from 24.92 °C to 28.33 °C.
If Ccal = 4.90 kJ/°C, find ΔE for burning 1 mole.
Given:
ΔE because it is a
constant volume
system
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1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C,
2.9506 × 10−3 mol C12H22O11, ΔT = 3.41 °C,
Conceptual
Plan:
Relationships:
Ccal, ΔT
qcal
qrxn
qcal = Ccal × ΔT = –qrxn
Solution:
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Ccal = 4.90 kJ/°C
qrxn, mol
ΔE
Practice—When 1.22 g of HC7H5O2 (MM 122.12) is burned
in a bomb calorimeter, the temperature rises from 20.27 °C to
22.67 °C. If ΔE for the reaction is −3.23 × 103 kJ/mol, what is
the heat capacity of the calorimeter in kJ/°C?
Enthalpy
• The enthalpy, H, of a system is the sum of the
internal energy of the system and the product of
pressure and volume.
9 H is a state function.
H = E + PV
• The enthalpy change, ΔH, of a reaction is the
heat evolved in a reaction at constant pressure.
ΔHreaction = qreaction at constant pressure
• Usually ΔH and ΔE are similar in value; the
difference is largest for reactions that produce or
use large quantities of gas.
Answer: 13.4 kJ/
oC
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Molecular View of
Exothermic Reactions
Endothermic and Exothermic Reactions
• For an exothermic reaction, the outside
• When ΔH is –, heat is being released by the system.
• Reactions that release heat are called exothermic
•
•
•
•
•
•
reactions.
When ΔH is +, heat is being absorbed by the system.
Reactions that absorb heat are called endothermic
reactions.
Chemical heat packs contain iron filings that are oxidized in
an exothermic reaction. Your hands get warm because the
released heat of the reaction is absorbed by your hands.
Chemical cold packs contain NH4NO3 that dissolves in
water in an endothermic process. Your hands get cold
because they are giving away your heat to the reaction.
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•
•
temperature rises due to release of
thermal energy.
This extra thermal energy comes from
the conversion of some of the chemical
potential energy in the reactants into
kinetic energy in the form of heat.
During the course of a reaction, old
bonds are broken and new bonds made.
The products of the reaction have less
chemical potential energy than the
reactants.
The difference in energy is released as
heat.
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Molecular View of
Endothermic Reactions
• In an endothermic reaction, the surrounding
•
•
•
temperature drops due to absorption of some of
its thermal energy.
During the course of a reaction, old bonds are
broken and new bonds made.
The products of the reaction have more chemical
potential energy than the reactants.
To acquire this extra energy, some of the thermal
energy of the surroundings is converted into
chemical potential energy stored in the products.
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Example 6.6 How much heat is evolved in the
complete combustion of 13.2 kg of C3H8(g)?
Given:
Find:
Enthalpy of Reaction
• The enthalpy change in a chemical reaction is an
extensive property.
9 The more reactants you use, the larger the enthalpy
change.
• By convention, we calculate the enthalpy change
for the number of moles of reactants in the reaction
as written.
• For the reaction:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
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Practice—How much heat is evolved when a
0.483-g diamond is burned?
(ΔHcombustion = –395.4 kJ/mol C)
13.2 kg C3H8
q, kJ/mol
Conceptual
Plan:
Relationships: 1 kg = 1000 g, 1 mol C3H8 = –2044 kJ, Molar Mass = 44.09 g/mol
Solution:
Check:
The sign is correct and the value is reasonable.
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ΔH = –2044 kJ
Answer: -15.9 kJ
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Measuring ΔH
Calorimetry at Constant Pressure
• Reactions done in aqueous solution are at
constant pressure.
9 open to the atmosphere
• The calorimeter is often nested foam cups
containing the solution.
Example 6.7 What is ΔHrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacting in
100.0 mL of solution changes the temperature from 25.6 °C to 32.8 °C?
(assume d sol = 1.00 g/mL and specific heat of the solution is 4.18 J/g·°C)
Given:
Find:
Conceptual
Plan:
0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C
d = 1.00 g/mL, Cs = 4.18 J/g·°C
ΔH, kJ/mol
Cs, ΔT, m
qsol’n
qrxn
qrxn, mol
ΔH
Relationships:
Solution:
qreaction = –qsolution = –(masssolution × Cs, solution × ΔT)
y ΔHreaction = qconstant pressure = qreaction
step2
9 To get ΔHreaction per mol, divide by the number
of moles.
Step 1
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Practice—What will be the final temperature of the solution in a
coffee cup calorimeter if a 50.00-mL sample of 0.250 M HCl(aq) is
added to a 50.00-mL sample of 0.250 NaOH(aq) if the initial
temperature is 19.50 °C and the ΔHrxn is −57.2 kJ/mol NaOH?
(Assume the density of the solution is 1.00 g/mL and the specific
heat of the solution is 4.18 J/g·°C.)
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Relationships Involving ΔHrxn
Rule 1 When a reaction is multiplied by a
factor, ΔHrxn is multiplied by that factor.
9because ΔHrxn is extensive
C(s) + O2(g) → CO2(g)
ΔH = −393.5 kJ
2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = 2(−393.5 kJ) = −787.0 kJ
Rule 2 If a reaction is reversed, then the sign
of ΔH is reversed.
ΔH = +393.5 kJ
CO2(g) → C(s) + O2(g)
Answer: 21.21 oC
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Relationships Involving ΔHrxn
Hess’s Law
• If a reaction can
Given the following information:
ΔH° = −206 kJ
Cu(s) + Cl2(g) → CuCl2(s)
2 Cu(s) + Cl2(g) → 2 CuCl(s) ΔH° = −36 kJ
be expressed as a
series of steps,
then the ΔHrxn for
the overall
reaction is the sum
of the heats of
reaction for each
step.
Tro, Principles of Chemistry: A Molecular Approach
Practice—Hess’s Law
Calculate the ΔH° for the reaction below:
Cu(s) + CuCl2(s) → 2 CuCl(s)
ΔH° = ? kJ
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Tro, Principles of Chemistry: A Molecular Approach
Practice—Hess’s Law
Given the following information:
Cu(s) + Cl2(g) → CuCl2(s)
2 Cu(s) + Cl2(g) → 2 CuCl(s)
ΔH° = −206 kJ
ΔH° = −36 kJ
Calculate the ΔH° for the reaction below:
Cu(s) + CuCl2(s) → 2 CuCl(s)
ΔH° = ? kJ
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Practice—Hess’s Law
Given the following information:
2 NO(g) + O2(g) → 2 NO2(g)
2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(aq)
N2(g) + O2(g) → 2 NO(g)
ΔH° = −116 kJ
ΔH° = −256 kJ
ΔH° = +183 kJ
Calculate the ΔH° for the reaction below:
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
ΔH° = ?
Solution: reverse the 1st equation and keep the second the same
CuCl2(s) → Cu(s) + Cl2(g)
2 Cu(s) + Cl2(g) → 2 CuCl(s)
Cu(s) + CuCl2(s) → 2 CuCl(s)
ΔH° = +206 kJ
ΔH° = −36 kJ
ΔH° = +170. kJ
Answer: ΔH = -137kJ
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Standard Conditions
Formation Reactions
• The standard state is the state of a material at a defined
set of conditions.
9 pure gas at exactly 1 atm pressure
9 pure solid or liquid in its most stable form at exactly 1 atm pressure
and temperature of interest
¾ usually 25 °C
9 substance in a solution with concentration 1 M
• The standard enthalpy change, ΔH°, is the enthalpy
•
change when all reactants and products are in their
standard states.
The standard enthalpy of formation, ΔHf°, is the
enthalpy change for the reaction forming 1 mole of a pure
compound from its constituent elements.
9 The elements must be in their standard states.
9 The ΔHf° for a pure element in its standard state = 0 kJ/mol.
• reactions of elements in their standard state
to form 1 mole of a pure compound
9If you are not sure what the standard state of an
element is, find the form in Appendix IIB that
has a ΔHf° = 0.
9Since the definition requires 1 mole of
compound be made, the coefficients of the
reactants may be fractions.
¾ by definition
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Writing Formation Reactions—
Write the formation reaction for CO(g).
• The formation reaction is the reaction between the
•
elements in the compound, which are C and O.
C + O → CO(g)
The elements must be in their standard state.
9 There are several forms of solid C, but the one with
ΔHf° = 0 is graphite.
9 Oxygen’s standard state is the diatomic gas.
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Write the formation reactions for the following.
CO2(g)
C(s, graphite) + O2(g) → CO2(g)
Al2(SO4)3(s)
C(s, graphite) + O2(g) → CO(g)
• The equation must be balanced, but the coefficient
of the product compound must be 1.
H2O (g)
9 Use whatever coefficient in front of the reactants is
necessary to make the atoms on both sides equal
without changing the product coefficient.
C(s, graphite) + ½ O2(g) → CO(g)
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Calculating Standard Enthalpy
Change for a Reaction
• Any reaction can be written as the sum of
formation reactions (or the reverse of formation
reactions) for the reactants and products.
• The ΔH° for the reaction is then the sum of the
ΔHf° for the component reactions.
ΔH°reaction = Σ n ΔHf°(products) – Σ n ΔHf°(reactants)
9Σ means sum.
9n is the coefficient of the reaction.
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Example—Calculate the enthalpy change in
the reaction.
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Step 2
ΔH°reaction = Σ n ΔHf°(products) – Σ n ΔHf°(reactants)
Example—Calculate the enthalpy change in
the reaction.
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Step 1: Write formation reactions for each
compound and determine the ΔHf° for each.
2 C(s, gr) + H2(g) → C2H2(g)
ΔHf° = +227.4 kJ/mol
C(s, gr) + O2(g) → CO2(g)
ΔHf° = –393.5 kJ/mol
H2(g) + ½ O2(g) → H2O(l)
ΔHf° = –285.8 kJ/mol
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Practice—Calculate the ΔH° for decomposing 10.0 g
of limestone, CaCO3, under standard conditions.
CaCO3(s) → CaO(s) + O2(g)
Material
ΔHf°, kJ/mol
CaCO3(s) −1207.6
O2(g)
0
CaO(s)
−634.9
ΔHrxn = [(4•ΔHCO2 + 2•ΔHH2O) – (2•ΔHC2H2 + 5•ΔHO2)]
ΔHrxn = [(4•(–393.5) + 2•(–285.8)) – (2•(+227.4) + 5•(0))]
ΔHrxn = −2600.4 kJ
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