Download 06_00 AP PPT Thermochemistry

Thermochemistry
AP Chemistry
Chapters 6 & 10.8
EU 3.C – Chemical and physical transformations may be observed in
several ways and typically involve a change in energy.
EK 3.C.2 – Net changes in energy for a chemical reaction can be
endothermic or exothermic.
EU 5.A – Two systems with different temperatures that are in thermal
contact will exchange energy. The quantity of thermal energy
transferred from one system to another is called heat.
EK 5.A.2 – The process of kinetic energy transfer at the particulate
scale is referred to in this course as heat transfer, and the
spontaneous direction of the transfer is always from a hot to a cold
body.
EU 5.B – Energy is neither created nor destroyed, but only
transformed from one form to another.
EK 5.B.1 – Energy is transferred between systems either through
heat transfer or through one system doing work on the other
system.
EK 5.B.2 – When two systems are in contact with each other and
are otherwise isolated, the energy that comes out of one system is
equal to the energy that goes into the other system. The
combined energy of the two systems remains fixed. Energy
transfer can occur through either heat exchange or work.
The Nature of
Energy
Thermo 6.1
Enduring Understanding &
Essential Knowledge
LO 3.11 – The student is able to interpret observations regarding
macroscopic energy changes associated with a reaction or process to
generate a relevant symbolic and/or graphical representation of the
energy change. (See SP 1.5, 4.4; EK 3.C.2)
LO 5.3 – The student can generate explanations or make predictions
about the transfer of thermal energy between systems based on this
transfer being due to a kinetic energy transfer between systems
arising from molecular collisions. (See SP 7.1; EK 5.A.2)
LO 5.4 – The student is able to use conservation of energy to relate the
magnitudes of the energy changes occurring in two or more
interacting systems, including identification of the systems, the type
(heat versus work), or the direction of energy flow. (See SP 1.4, 2.2;
EK 5.B.1, 5.B.2)
The Nature of
Energy
Thermo 6.1
Learning Objectives
Energy
• Energy = the capacity to do work or to produce heat
 That which is needed to oppose natural attractions.
• First Law of Thermodynamics = The total energy of the universe is constant.
• Law of Conservation of Energy = Energy can be converted from one form to
another but can be neither created nor destroyed.
Energy Units
• calorie (cal) = the amount of energy (heat) required to raise the
temperature of one gram of water 1 oC.
 1000 calories = 1 kcal = 1 Calorie
 Note: “calorie” and “Calorie” (capitalized) are NOT the same thing
• Joule (J)
 1 cal = 4.184 J
Types of Energy
• Potential Energy (PE) = energy due to position or composition
 ie: water behind a dam, energy in bonds
• Kinetic Energy (KE) = energy due to the motion of an object
 depends on the mass (m) of the object and its velocity (v)
1
 KE = mv 2
(provided on the AP Equation sheet)
2
System vs. Surroundings
• System = the part of the universe we are focusing on
• Surroundings = everything else in the universe
• Heat either goes from a system into the surroundings (exothermic), or from
the surroundings into a system (endothermic)
Example 6.1.B
For each of the following, define a system and its surroundings and give the
direction of energy transfer.
• Methane is burning in a Bunsen burner in a laboratory.
System = methane & oxygen
Surroundings = everything else around the system
Direction = system to the surroundings (exothermic)
• Water drops, sitting on your skin after swimming, evaporate
System = water drops
Surroundings = everything else around the system
Direction = surroundings (your skin) to the system (endothermic)
Example 6.1.B
Hydrogen gas and oxygen gas react violently to form water. Which is lower in
energy:
a) a mixture of hydrogen and oxygen gases
b) water
Water is lower in energy because a lot of energy
was released in the process when hydrogen and
oxygen gases reacted.
Heat & Work (ΔE = q + w)
• Temperature = a measurement of average kinetic energy
• Heat (Q or q) = energy that is transferred from a hotter object to a colder
object
 If two objects stay in contact, heat will flow from the warmer
object to the cooler object until their temperatures are the same.
• Work (w) = force acting over a distance
System and Energy
Endothermic
• A system can increase its internal energy, ΔE, in two ways:
 the system can absorb energy (+q) from the surroundings
 the surroundings can do work (+w) on the system
Exothermic
• A system can lose energy in two ways:
 the system can release energy (-q) to the surroundings
 the system can do work (-w) on the surroundings
Example 6.1.A
1) A system does 20 J of work on the surroundings. The change in energy of
-20 J.
the system, ΔE, is
2) A system has a change in energy of -25 J. No work is done by or on the
25
system, so
J of heat must have been (absorbed / released).
3) The E of a system is 50 J if the system absorbs 30 J of heat and has
20 J of work done on it.
4) The E of a system is -40 J if the system releases 50 J of heat and has
10 J of work done on it.
5) A system has a change in energy of +45 J. If 30 J of work was done by the
system, then
75 J of heat must have been (absorbed / released).
Exothermic
• Exothermic = system releases heat to
the surroundings
• Example: A + B  AB + heat
 Energy (heat) is a product
 Energy (heat) was released to
make the bond between A & B
• Making a bond ALWAYS releases energy
• The energy of the reactants is higher
than the energy of the products, so
energy is lost overall
• ΔH has a –q (negative) value
Endothermic
• Endothermic = system absorbs
energy from the surroundings
• Example: AB + heat  A + B
 Energy (heat) is a reactant
 Energy (heat) was absorbed to
break the bond between A & B
• Breaking a bond ALWAYS requires /
uses energy
• The energy of the reactants is lower
than the energy of the products, so
energy was gained overall
• ΔH has a +q (positive) value
Example 6.1.B
1) A container of melted paraffin wax is allowed to stand at room
temperature until the wax solidifies. What is the direction of heat flow as
the liquid wax solidifies?
2) When solid barium hydroxide octahydrate is mixed in a beaker with solid
ammonium thiocyanate, a reaction occurs. The outside of the beaker
quickly becomes very cold. Was the reaction exothermic or endothermic?
Example 6.1.C
Classify each process as exothermic or endothermic. Explain. The system
is underlined in each example.
a) Water freezes.
Exothermic
b) Your hand gets cold when you touch ice.
Exothermic
c) The ice gets warmer when you touch it.
Endothermic
d) Water boils in a kettle being heated on a stove.
Endothermic
e) Water vapor condenses on a cold pipe.
Exothermic
f) Ice cream melts.
Endothermic
EU 3.C – Chemical and physical transformations may be observed in several
ways and typically involve a change in energy.
EK 3.C.2 – Net changes in energy for a chemical reaction can be
endothermic or exothermic.
EU 5.A – Two systems with different temperatures that are in thermal contact
will exchange energy. The quantity of thermal energy transferred from one
system to another is called heat.
EK 5.A.2 – The process of kinetic energy transfer at the particulate scale is
referred to in this course as heat transfer, and the spontaneous direction of
the transfer is always from a hot to a cold body.
EU 5.B – Energy is neither created nor destroyed, but only transformed from
one form to another.
EK 5.B.1 – Energy is transferred between systems either through heat
transfer or through one system doing work on the other system.
EK 5.B.2 – When two systems are in contact with each other and are
otherwise isolated, the energy that comes out of one system is equal to
the energy that goes into the other system. The combined energy of the
two systems remains fixed. Energy transfer can occur through either heat
exchange or work.
EK 5.B.3 – Chemical systems undergo three main processes that change
their energy; heating/cooling, phase transitions, and chemical reactions.
EK 5.B.4 – Calorimetry is an experimental technique that is used to
measure the change in energy of a chemical system.
Enthalpy &
Calorimetry
Thermo 6.2
Essential Knowledge
LO 3.11 – The student is able to interpret observations regarding
macroscopic energy changes associated with a reaction or process to
generate a relevant symbolic and/or graphical representation of the
energy change. (See SP 1.5, 4.4; EK 3.C.2)
LO 5.3 – The student can generate explanations or make predictions
about the transfer of thermal energy between systems based on this
transfer being due to a kinetic energy transfer between systems
arising from molecular collisions. (See SP 7.1; EK 5.A.2)
Enthalpy &
Calorimetry
LO 5.5 – The student is able to use conservation of energy to relate
the magnitudes of the energy changes when two nonreacting
substances are mixed or brought into contact with one another. (See
SP 2.2; EK 5.B.1 & 5.B.2)
Thermo 6.2
LO 5.6 – The student is able to use calculations or estimations to
relate: energy changes associated with heating/cooling a substance to
the heat capacity, energy changes associated with a phase transition
to the enthalpy of fusion/vaporization, energy changes associated with
a chemical reaction to the enthalpy of the reaction, and energy
changes to PV work. (See SP 2.2 & 2.3; EK 5.B.3)
Learning Objectives
LO 5.7 – The student is able to design and/or interpret the results of
an experiment in which calorimetry is used to determine the change
in enthalpy of a chemical process (heating/cooling, phase transition,
or chemical reaction) at constant pressure. (See SP 4.2 & 5.1; EK
5.B.4)
Enthalpy
• Enthalpy (ΔH) = Heat = the total energy of a system when it is at constant
pressure
ΔH = Hfinal – Hinitial
• For a chemical reaction, the initial state refers to the reactants and the
final state refers to the products, so we can write:
ΔHreaction = Hproducts – Hreactants
• Heat of Reaction = ΔHreaction = ΔHrxn = ΔH (when written with an equation)
Enthalpy Signs
• If ΔH is negative (-), the reaction is exothermic
 – ΔH (negative) when Hproducts < Hreactants
 Energy is leaving the system and goes into the surroundings
• If ΔH is positive (+), the reaction is endothermic
 ΔH (positive) when Hproducts > Hreactants
 Note: “+” sign is not required when writing ΔH , but a “-” sign is
 Energy is being absorbed by the system from the surroundings
Different Forms of Enthalpy/Heat
• ΔH = Heat/Enthalpy
 The units of ΔH are usually kilojoules (kJ)
• ΔHrxn = ΔHreaction = ΔH (with an equation) = Enthalpy/Heat of Reaction
• ΔHo = ΔH at standard conditions
 Thermochemical reactions are generally written at standard
thermochemical temperature and pressure, 25 oC and 1 atm
• ΔHf = Enthalpy/Heat of Formation
Thermochemical Equation Example
N2 (g) + 3 H2 (g)  2 NH3 (g)
ΔHo = – 92.38 kJ
Things to note:
• All physical states (solid/liquid/gas) of the reactants and products are cited
in this equation.
• How can you tell that this reaction is exothermic?
• ΔHo represents the energy (Heat of Reaction) released when 1 mol N2
reacts with 3 mol H2 to form 2 mol NH3
• Look at the coefficients in the balanced equation
Example 6.2.A
N2 (g) + 3 H2 (g)  2 NH3 (g)
ΔHo = –92.38 kJ
What is the enthalpy of this reaction given only 1.5 mol of H2?
According to the equation, the reaction of 3 mol H2 has a ΔHo of –92.38 kJ
Conversion Factor: 3 mol H2 = –92.38 kJ
1.5 mol H2
–92.38 kJ
3 mol H2
= – 46.19 kJ = – 46 kJ
Heats of Combustion
Heat of Combustion = the amount of energy released during a combustion
reaction
Substance
HEATS OF COMBUSTION AT 25 oC
H
Formula
Substance
(kJ/mol)
Formula
H
(kJ/mol)
Hydrogen
H2 (g)
-286
Glucose
C6H12O6 (s)
-2808
Carbon
C (s), graphite
-394
Octane
C8H18 (l)
-5471
Methane
CH4 (g)
-890
Sucrose
C12H22O11 (s)
-5645
Acetylene
C2H2 (g)
-1300
Ethanol
C2H5OH (g)
-1368
Propane
C3H8 (g)
-2220
Example 6.2.A
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = –2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate
ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.
According to the equation, the reaction of 1 mol C3H8 has a ΔH of –2221 kJ
5.00 g C3H8
1 mol C3H8
–2221 kJ
= – 252 kJ
44.094 g C3H8 1 mol C3H8
Specific Heat (q = mcΔT)
q (or Q)
m
C (or c)
ΔT
Represents
Heat or energy
Mass
Specific Heat
Change (Δ) in temperature
Definition
The transfer of energy
released or absorbed
by the reaction
The amount of energy needed to
change the temperature of 1 gram
of substance by 1 degree Celsius
ΔT = Tf – Ti
(Final temperature minus initial
temperature)
Unit
kJ or J or kcal (Cal) or
cal
g
goC
Sign
(-) = heat is released
(+) = heat is absorbed
(+)
(+)
J
• Specific Heat Capacity = Specific Heat
oC
(-) = cooling down
(+) = heating up
Specific Heat & Heat Capacity
Specific Heat (AKA Specific Heat Capacity)= the amount of
energy needed to raise the temperature of one gram of
substance by one degree Celsius.
• Specific heats are different for the same substance in
different phases
• ie: ice water, liquid water, and gaseous water (steam)
are all H2O but they have different specific heats.
• Substances with lower specific heats are (easier /
more difficult) to heat up or cool down than
substances with higher specific heats.
Heat Capacity = the amount of energy needed to raise
the temperature of an entire sample of substance by one
degree Celsius.
• Units: J/oC
Example 6.2.B
74.8 J of heat is required to raise the temperature of
18.69 g of silver from 10.0C to 27.0C.
a) What is the heat capacity (J/C) of the silver sample? Note: This is a useful
value only for this specific sample of silver.
b) What is the specific heat capacity (J/g·C) of silver? Note: This is a useful
value for any sample of silver that is heated or cooled. This is equivalent to
the 4.184 J·g-1·°C-1 that we use for water. This value is also called the
specific heat.
Example 6.2.C #1
Water has a specific heat capacity of 4.184 J/g·°C. This means it takes 4.184
J to heat 1.00 gram of water 1.00°C.
a) How much energy will it take to heat 10.0 grams of water 1°C?
b) How much energy is needed to heat 30.0 g H2O from 10.0 °C to 50.0 °C?
Example 6.2.C #2
A pot of water (2.5 Liters of water) initially at 25.0C is heated to boiling
(100.°C).
a) How much energy (in J) is needed to heat the water? (The density of
water is 1 g/mL.)
b) What would this amount of heat be in kJ?
Example 6.2.C #3
What amount of heat is released when 175 g of water cools from 100.°C to
room temperature, 20.0 °C?
Example 6.2.C #4
We don’t always have to warm up or cool down water. The
specific heat capacity of copper metal is 0.39 J/g·°C. How
much energy would it take to heat up a 5.20 g sample of
copper from 20.0 °C to 100.°C?
Calorimetry
• Calorimeter = a device used to determine the
heat from a chemical reaction
• The calorimeter is insulated so the energy
gained and lost are kept in a specific container.
• –Qexo = Qendo
• The energy gained and the energy lost within
a calorimeter are equal to one another in
value (but not in sign).
Example 6.2.C #5
If 300. J of heat energy were used to heat up a 5.00 gram sample of copper
metal and a 5.00 gram sample of water both starting at 10.0°C, calculate
the final temperature of each sample.
Example 6.2.C #6
Suppose we mix 90.0 grams of hot water (90.0°C) with 10.0 grams of cold
water (10.0°C). Let x = the final temperature. C = 4.184 J/g·C
a) Set up an expression for the energy released (q) by the hot water (qhot =
mhotCThot)
b) Set up an expression for the energy absorbed (q) by the cold water (qcold =
mcoldCTcold)
c) Knowing that the heat absorbed =  heat released, combine the two
expressions and solve for x.
Example 6.2.C #7
175 grams of hot aluminum (100.°C) is dropped into an insulated cup that
contains 40.0 mL of ice cold water (0.0°C). Determine the final temperature.
a) Set up an expression for the heat lost by the aluminum (C=0.900 J/g·°C)
b) Set up an expression for the heat gained by the cold water.
c) Put the two expressions together and solve for x.
EU 2.A – Matter can be described by its physical properties. The physical
properties of a substance generally depend on the spacing between the
particles (atoms, molecules, ions) that make up the substance and the forces
of attraction among them.
EK 2.A.1 – The different properties of solids and liquids can be explained by
differences in their structures, both at the particulate level and in their
supramolecular structures.
EU 2.C – The strong electrostatic forces of attraction holding atoms together in
a unit are called chemical bonds.
EK 2.C.2 – Ionic bonding results from the net attraction between oppositely
charged ions, closely packed together in a crystal lattice.
EU 5.B – Energy is neither created nor destroyed, but only transformed from
one form to another.
EK 5.B.3 – Chemical systems undergo three main processes that change
their energy; heating/cooling, phase transitions, and chemical reactions.
EU 5.D – Electrostatic forces exist between molecules as well as between
atoms or ions, and breaking the resultant intermolecular interactions requires
energy.
EK 5.D.2 – At the particulate scale, chemical processes can be distinguished
from physical processes because chemical bonds can be distinguished from
intermolecular interactions
Vapor Pressure
& Changes of
State
Thermo 10.8
Essential Knowledge
LO 2.1 – Students can predict properties of substances based on their
chemical formulas and provide explanations of their properties based
on particle views. (See SP 6.4; EK 2.A-2.D)
LO 2.19 – The student can create visual representations of ionic
substances that connect the microscopic structure to macroscopic
properties, and/or use representations to connect the microscopic
structure to macroscopic properties (e.g., boiling point, solubility,
hardness, brittleness, low volatility, lack of malleability, ductility, or
conductivity. (See SP 1.1, 1.4 & 7.1; EK 2.C.2, 2.D.1, 2.D.2)
LO 5.6 – The student is able to use calculations or estimations to
relate: energy changes associated with heating/cooling a substance to
the heat capacity, energy changes associated with a phase transition
to the enthalpy of fusion/vaporization, energy changes associated with
a chemical reaction to the enthalpy of the reaction, and energy
changes to PV work. (See SP 2.2 & 2.3; EK 5.B.3)
LO 5.10 – The student can support the claim about whether a process
is a chemical or physical change (or may be classified as both) based
on whether the process involves changes in intramolecular versus
intermolecular interactions (See SP 5.1; EK 5.D.2)
Vapor Pressure
& Changes of
State
Thermo 10.8
Learning Objectives
Phase Changes
Endothermic
• Exothermic phase changes:
 Deposition
 Condensation
 Freezing
Exothermic
• Endothermic phase changes:
 Sublimation
 Vaporization
 Melting
Vaporization
• Vaporization = the phase change from a liquid to a gas
• (Equilibrium) Vapor Pressure = pressure of the vapor present at equilibrium
 In a closed container, a liquid will vaporize.
 As moles of vapor molecules increases, some vapor will condense.
 Equilibrium occurs when ratecondensation = ratevaporization
Brain Check
• What is the vapor pressure of water at 100°C? How do you know?
• The vapor pressure of water at 100oC is 1 atm. You know this because
atmospheric pressure is 1 atm and this is the temperature at which we
observe water to boil.
Factors That Affect Vapor Pressure
• Liquids in which the intermolecular forces
are strong have relatively low vapor
pressures.
• Vapor pressure increases significantly with
temperature.
Boiling vs. Vaporization
• (Equilibrium) Vapor Pressure = pressure of the vapor present at equilibrium
• Boiling = when the vapor pressure of a liquid is equal to the atmospheric
pressure
Boiling and Freezing Points
• Boiling = when the vapor pressure of a liquid is equal to the atmospheric
pressure
 Normal Boiling Point = the temperature at which the vapor
pressure of the liquid is exactly 1 atm.
• Freezing Point (AKA Melting Point) = the temperature at which a
substance goes from a liquid to a solid (or from a solid to a liquid)
 Normal Melting Point = the temperature at which the solid and
liquid states have the same vapor pressure under conditions
where the total pressure is 1 atm.
Energy of Phase Changes
• When we write an equation, there is usually energy involved. Consider
the evaporation of water:
H2O(l) + energy  H2O(g)
• This energy is called the heat of vaporization with the symbol, Hvap.
• The equation represents one mole of water being vaporized, so the
energy is called the molar heat of vaporization and is reported in kJ/mol.
• If “energy / heat” is on the reactants side, the reaction is endothermic.
• If “energy / heat” is on the products side, the reaction is exothermic.
Example 10.8.A
Write an equation for the following changes and include energy:
a) The melting of ice:
b) The condensation of steam:
c) The sublimation of dry ice, CO2(s):
d) The freezing of liquid water:
Latent Heat (q = mΔHfus) (q = mΔHvap)
Represent
s
q (or Q)
m
ΔHfus
ΔHvap
Heat or
energy
Mass
Heat of Fusion
Heat of Vaporization
The amount of energy needed to
melt or freeze (l ↔ s) one gram
of substance
The amount of energy needed to
vaporize or condense (l ↔ g) one
gram of substance
J/g (kJ/g) (cal/g)
J/g (kJ/g) (cal/g)
Sometimes “g” is replaced with “mol”
and then it becomes the Molar Heat of
Fusion
Sometimes “g” is replaced with “mol” and
then it becomes the Molar Heat of
Vaporization
(+) = melting
(–) = freezing
(+) = vaporization
(–) = condensation
Definition
Unit
kJ or J or
kcal (Cal) or
cal
g
Sign
(+) = endo
(–) = exo
(+)
ΔHfus as a Conversion Factor
a) When 1.00 gram of H2O(s) melts, 334.6 Joules of energy is absorbed.
Calculate the amount of energy (in kJ/mol) required to melt 1.00 mole
of ice. (This is the molar heat of fusion for H2O.)
b) Calculate the amount of heat needed to melt 35.2 grams of ice.
c) What mass of ice can be melted with 975 J of energy?
Example 10.8.B
How many kilocalories are given off when 50.0 g of water at 0 oC freezes?
(The accepted value of the heat of fusion of water, Hfus, is 6.03 kJ/mol.)
1) What information is provided by the problem?
m = 50.0 g
T = 0 oC (T does not change)
Water freezes (l  s), so ΔHfus = - 6.01 kJ/mol
2) What equation will be used to solve this problem?
q = m ΔHfus
Example 10.8.B (Continued)
How many kilocalories are given off when 50.0 g of water at 0 oC freezes?
(The accepted value of the heat of fusion of water, Hfus, is 6.03 kJ/mol.)
3) Isolate the unknown variable from the equation.
(Not necessary because the equation already has “q” isolated.)
4) Plug in the numbers for all the known variables and solve.
q =
50.0 g H2O
- 6.03 kJ
1 mol H2O
1 mol H2O 18.02 g H2O
1 kcal
4.184 kJ
= -3.99 kcal
Answer: 3.99 kcal are given off.
* Note: Although the value for q is negative, the question asks how many kilocalories are
released. You cannot release a negative amount of energy. The negative sign for “q” indicates
that the process is exothermic, so energy is being released, rather than absorbed.
Example 10.8.C
The accepted value for Hfus is 6.03 kJ/mol.
The accepted value for Hvap is 40.67 kJ/mol.
1) Calculate the amount of heat energy needed to boil 40.0 grams of
water.
2) What mass of water can be vaporized with 55.0 kJ of energy.
3) How much heat is needed to vaporize 1.00 gram of H2O?
Example 10.8.C (Continued)
Hfus = 6.03 kJ/mol
Hvap = 40.67 kJ/mol
4) 100. grams of steam condenses to liquid water. What amount of energy
is released?
5) 775 J of energy is added to liquid water at 100°C. What mass of water is
vaporized?
Heating Curves (General)
• Energy can either be used
for a temperature change
or for a phase change, but
not both simultaneously.
• Horizontal lines indicate
phase changes
(temperature remains
constant). These are
labeled as the normal
boiling point and the
normal melting point
• Each diagonal lines
indicates a temperature
changes for a different
phase (solid, liquid or gas)
Heating Curve for Water
• Normal Freezing/Melting Point
= 0 oC (@ 1 atm)
• Normal Boiling Point = 100 oC
(@ 1 atm)
• Density
Liquid Water = 1.00 g/mL
Ice = 0.917 g/mL
Cooling Curve for Water
Things to note:
• From left to right, all
processes on this
curve are
exothermic.
• This curve is the
mirror image of the
heating curve of
water.
• The normal freezing
point is still 0 oC.
• The normal boiling
point is still 100 oC.
Brain Check
1) Explain why the Hvap is larger than the Hfus .
It takes a lot more energy to break the intermolecular forces between
the liquid molecules to completely break apart to form a gas.
2) As intermolecular forces increase, what happens to each of the following:







Boiling point
Viscosity
Surface tension
Enthalpy of fusion
Freezing point
Vapor pressure
Heat of vaporization
increases
increases
increases
increases
increases
decreases
increases
No AP Learning Objectives
Hess’ Law
Thermo 6.3
Hess’ Law
When going from a particular set of reactants to a particular set of products,
the change in enthalpy is the same whether the reaction takes place in one
step or in a series of steps.
Hess’ Law = If you add two or more thermochemical equations to get a final
equation, then you can also add the heats of reaction to give the final heat of
reaction.
For Example:
Equation 1
Equation 2
+ Equation 3
New Equation
ΔH1o
ΔH2o
ΔH3o
ΔH1o + ΔH2o + ΔH3o
Hess’ Law Visualization
Rule #1 For Manipulating Equations
When an equation is reversed (with the reactants and products switched),
the sign of ΔHo must also be reversed.
C (s, graphite) + O2 (g)  CO2 (g)
ΔHo = –393.5 kJ
When reversed becomes:
CO2 (g)  C (s, graphite) + O2 (g)
ΔHo = 393.5 kJ
Rule #2 For Manipulating Equations
If all the coefficients of an equation are multiplied or divided by a common
factor, the value of ΔHo must be likewise changed.
N2 (g) + 3 H2 (g)  2 NH3 (g)
ΔHo = –92.38 kJ
When multiplied by 3 becomes:
3 N2 (g) + 9 H2 (g)  6 NH3 (g)
ΔHo = –277.14 kJ
Rule #3 For Manipulating Equations
Cancel (cross out) identical formulas from both sides of an equation in
identical physical states.
It is okay if the formulas have different coefficients. However, you can only
cancel out the smaller coefficient and you must modify the larger coefficient
accordingly.
 C (s graphite) + O2 (g)
CO2 (g)
C (s diamond) + O2 (g)  CO2 (g)
 C (s graphite)
C (s diamond)
ΔH = 393.5 kJ
ΔH = –395.4 kJ
ΔH = –1.9 kJ
Example 6.3.A
1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
CO2(g)
2) CO(g) + ½ O2(g)
ΔHo = –26.7 kJ
ΔHo = –283.0 kJ
Calculate the ΔHo for the following reaction:
2 Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
ΔHo = ?
This is the “target equation.”
Note that thermochemical equations can have fractions for coefficients.
Example 6.3.A (Continued #1)
Step 1: Look for a reactant/product from the target equation that shows up
in only ONE of the given equations & compare its position in all the given
equations.
1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
ΔHo = –26.7 kJ
CO2(g)
2) CO(g) + ½ O2(g)
ΔHo = –283.0 kJ
2 Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
ΔHo = ?
We can focus on any of these chemicals. Let’s choose to focus on Fe (s) first.
Problem: To match the target equation correctly, Equation #1 must have 2 Fe on the left,
but it has 2 Fe on the right.
Example 6.3.A (Continued #2)
Problem: To match the target equation correctly, Equation #1 must have 2 Fe
on the left, but it has 2 Fe on the right.
Solution: Reverse Equation #1 and the ΔHo sign.
1)
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)

1*) 2 Fe(s) + 3 CO2(g)  Fe2O3(s)+ 3 CO(g)
ΔHo = –26.7 kJ
ΔHo = 26.7 kJ
Example 6.3.A (Continued #3)
Step 2: If possible, repeat Step 1 until each equation has been modified.
Focus on another reactant/product that shows up only once in the target
equation. For this example, we will focus on O2 next.
1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
CO2(g)
2) CO(g) + ½ O2(g)
2 Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
ΔHo = –26.7 kJ
ΔHo = –283.0 kJ
ΔHo = ?
Problem: We need 3/2 O2 on the left, but we only have ½ O2 on the left in
equation #2.
Example 6.3.A (Continued #4)
Problem: We need 3/2 O2 on the left, but we only have ½ O2 on the left in
equation #2.
Solution: Multiply Equation #2 by 3.
2)
CO(g) + ½ O2(g)
 CO2(g)

2) 3 [CO(g) + ½ O2(g)  CO2(g) ]

2*) 3 CO(g) + 3/2 O2(g)  3 CO2 (g)
ΔHo = –283.0 kJ
ΔHo = 3(–283.0 kJ)
ΔHo = –849.0 kJ
Example 6.3.A (Continued #5)
Step 3: If there is an equation that was not modified because none of its
reactants or products appear in the Target Equation, modify the last equation
so that it is canceled out by the reactants/products in the previously modified
equations.
• Note that an original equation does not have to be modified if the
reactant/product that you are focusing on appears in the Target Equation
with the same coefficient and on the same side of the equation.
• This step does not apply to this example because we modified all the
equations that were originally given.
Example 6.3.A (Continued #6)
Step 4: Cancel (cross out) identical formulas in identical physical states from
both sides of the modified equations (or unmodified equations, if
modification was not necessary). (Rule #3 for Modifying Equations)
1*) 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g)
2*) 3 CO(g) + 3/2 O2(g)  3 CO2(g)
2 Fe(s) + 3/2 O2(g)  Fe2O3(s)
ΔHo = 26.7 kJ
ΔHo = –849.0 kJ
ΔHo = –822.3 kJ
EU 5.C – Breaking bonds requires energy, and making bonds releases
energy.
EK 5.C.2 – The net energy change during a reaction is the sum of
the energy required to break the bonds in the reactant molecules
and the energy released in forming the bonds of the product
molecules. The net change in energy may be positive for
endothermic reactions where energy is required, or negative for
exothermic reactions where energy is released.
Standard
Enthalpies of
Formation
Thermo 6.4
Essential Knowledge
LO 5.8 – The student is able to draw qualitative and quantitative
connections between the reaction enthalpy and the energies involved
in the breaking and formation of chemical bonds. (See SP 2.3, 7.1 &
7.2; EK 5.C.2)
Standard
Enthalpies of
Formation
Thermo 6.4
Learning Objectives
Heats of Formation (ΔHf)
• ΔHo = (sum of ΔHfo of all the products) – (sum of ΔHfo of all the reactants)
• Standard heat (enthalpy) of formation (ΔHfo ) = the change in enthalpy
when one mole of substance is formed at standard thermochemical
conditions from its elements and in its standard form (solid/liquid/gas).
 Standard States for a compound = 1 atm (gas), 1 M (soln)
 Standard States for a gas = 1 atm & 25 oC
• Tables for heats of formation are provided.
• You are not required to memorize the different heats of formation, except
for pure elements in their standard states (because ΔHf under these
conditions is zero)
Sample ΔHf Table
• ΔHfo for all pure elements in their standard states = 0 (zero) kJ/mol
• Most ΔHfo are exothermic, but some are endothermic.
• Different ΔHfo for the same compound in different states (gas, liquid or
solid)
Substance
ΔHfo
(kJ/mol)
Substance
ΔHfo
(kJ/mol)
Al (s)
0
Cl2 (g)
0
BaCO3 (s)
- 1219
H2O (l)
- 286
CO (g)
- 110.5
H2O (g)
- 241.8
CO2 (g)
- 393.5
NaHCO3 (s)
- 947.7
CS2 (g)
+ 117
Na2CO3 (s)
- 1131
Heats of Formation (ΔHf) Visualization
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
• ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ
Example 6.6.B
2 NaHCO3 (s)  Na2CO3 (s) + H2O (l) + CO2 (g)
Some chefs keep baking soda, NaHCO3 , handy to put out grease fires. When
thrown on the fire, baking soda partly smothers the fire and the heat
decomposes it to give CO2 , which further smothers the flame. The equation
for the decomposition of NaHCO3 is shown above. What is the ΔHo of this
reaction?
Example 6.6.B (Continued)
2 NaHCO3 (s)  Na2CO3 (s) + H2O (l) + CO2 (g)
ΔHo = (sum of ΔHfo of all the products) – (sum of ΔHfo of all the reactants)
Use a reference table to find the heats of formation for each of the reactants
and products
ΔHfo of NaHCO3 (s)
=
– 947.7 kJ/mol
ΔHfo of Na2CO3 (s)
=
– 1131 kJ/mol
ΔHfo of H2O (l)
=
– 286 kJ/mol
ΔHfo of CO2 (g)
=
– 394 kJ/mol
Example 6.6.B (Continued)
2 NaHCO3 (s)  Na2CO3 (s) + H2O (l) + CO2 (g)
ΔHo = (ΣΔHfo products) – (ΣΔHfo reactants)
ΔHo = (ΔHfo of Na2CO3 + ΔHfo of H2O + ΔHfo of CO2) – (ΔHfo of NaHCO3)
ΔHo = [(-1131) + (-286) + (-394)] – [2 x (-947.7)]
The ΔHfo of NaHCO3 must be multiplied by 2 because it has a coefficient of 2.
ΔHo = 84 kJ