Unformatted text preview: Module 21 — Energy Calculations Calculations
In Chemistry
*****
Modules 21 and 22
Phase Changes, Energy, and (∆H)
Module 21 – Phases Changes and Energy..................................................................559
Lesson 21A:
Lesson 21B:
Lesson 21C:
Lesson 21D:
Lesson 21E: Phases and Phase Changes............................................................................... 559
Specific Heat Capacity and Equations............................................................ 572
Water, Energy, and Consistent Units.............................................................. 580
Calculating Joules Using Unit Cancellation................................................... 585
Calorimetry......................................................................................................... 591 Module 22 – Heats Of Reaction (ΔH)..........................................................................599
Lesson 22A:
Lesson 22B:
Lesson 22C:
Lesson 22D:
Lesson 22E: Energy, Heat, and Work .................................................................................. 599
Exo And Endothermic Reactions .................................................................. 607
Adding ΔH Equations (Hess’s Law) ............................................................... 612
Heats of Formation and Element Formulas ................................................... 618
Using Summation to Find ΔH.......................................................................... 626
For additional modules, visit www.ChemReview.Net Page i Module 21 — Energy Calculations Module 21 — Energy Calculations
Timing: Begin this module when you are assigned problems which involve phase changes
or calculations involving joules or calories.
Prerequisites: This module for the most part does not require many prior topics. You
should be able to do most of the calculations if you have completed Modules 2, 4, 5, and 8. Lesson 21A: Phases and Energy
Three Phases
In firstyear chemistry, we are initially concerned with three phases for pure substances:
solid, liquid, and gas.
In covalently bonded molecules, the forces that hold the atoms together within the
molecule are relatively strong, but there are also relatively weak forces of attraction
between molecules. These weak attractions mean that molecules are a bit “sticky:” they
tend to attract each other somewhat like two weak but attracting magnets (these weak
molecular attractions are electrical, but the behavior is similar).
In the solid phase, molecules vibrate, but the weak attractions between molecules hold the
molecules in a crystal structure where they are limited in the extent to which they can
rotate, and they can translate (move from place to place) only very slightly.
In their liquid phase, the molecules gain some freedom: they can vibrate, rotate, and
translate. However, the liquid phase molecules are still very close together: they have
minimal space between them. This is why solids and liquids do not compress (or compress
only very slightly) when pressure is applied.
In the gas phase, molecules are separated by a considerable distance. In a gas at room
temperature, the distance between molecules is typically about 10 times the diameter of the
molecule. This means that 99.9% of the gas is empty space. Gases can be compressed
because the empty space between the molecules can be reduced.
In a gas, the molecules remain weakly attractive. If a gas is highly compressed, or if its
temperature is lowered (which slows down the speed at which the molecules move), the
stickiness of the molecules becomes a larger factor. All gases condense into a liquid or a
solid at some point as pressure is increased and temperature is decreased. Three Phase Changes
In chemistry, a chemical change is defined as a process in which atoms rearrange to form
new substances. A physical change is one in which characteristics of a substance change,
but the substance does not change: it keeps the same chemical formula. A phase change is
one example of a physical change.
There are six types of phase changes among the three phases, but most of these are familiar.
The following terms are used to describe phase changes.
• For solid/liquid changes: Solids melt (or fuse) to become liquids; liquids freeze
(or solidify) to become solids. ©2011 ChemReview.net v. e4 Page 559 Module 21 — Energy Calculations • For liquid/gas changes: Liquids boil or evaporate to form gases; gases condense to
become liquids. • For solid/gas changes: Solids sublime to become gases. In the reverse process,
gases can undergo deposition to form solids.
Sublimation is a phase change that is less commonly encountered at room
temperature and pressure, but you may be familiar with dry ice (solid carbon
dioxide) or moth crystals (paradichlorobenzene). At room conditions, these solids
do not pass through a liquid phase as they convert from the solid to the gas phase.
Vapor deposition can be observed when water vapor forms ice crystals on a
windshield that is below 0°Celsius. Melting and Freezing
For a pure substance, the temperature at which it melts (its melting point) will equal the
temperature at which it solidifies.
For pure substances: Melting Point ≡ Freezing Point For pure substances, melting occurs at a characteristic temperature (at or near standard
pressure) that can be used to identify the substance. However, even small amounts of
impurity in a substance will weaken its crystal structure and cause it to melt and freeze at a
less sharp as well as a lower temperature. Vapor Pressure
Nearly every liquid or solid has a measurable tendency for its particles to become gas
particles. At the surface of a liquid or solid, the vibrating molecules can break free and
become part of the vapor above the liquid or solid. In liquids, vapor pressure is generally
higher, and the evaporation of liquids at temperatures far below their boiling points is one
example of the effect of vapor pressure.
Vapor pressure can be measured by placing a liquid or solid in a vacuum, allowing the
system to reach equilibrium at a given temperature, then measuring the gas pressure in the
container. However, the vapor pressure of a liquid or solid an intrinsic property. The
vapor pressure of a given substance at a given temperature is the same whether it is in a
sealed container or not.
Vapor pressure is measured in pressure units, and vapor pressure increases with increasing
temperature. A liquid will boil at any temperature at which its vapor pressure equals the
atmospheric pressure above it. Boiling Temperature and Pressure
Boiling points are characteristic temperatures which can be used to identify a substance. A
normal boiling point is recorded at standard pressure (one atmosphere).
Boiling points must be recorded at a known pressure, because liquids boil at a temperature
that depends on the surrounding atmospheric pressure. The vapor pressure of a substance
increases with increasing temperature. A liquid will boil at a temperature higher than its
normal boiling point if the atmospheric pressure above it is higher than standard pressure.
It will boil at a lower than normal boiling point if atmospheric pressure is lower than
standard pressure.
©2011 ChemReview.net v. e4 Page 560 Module 21 — Energy Calculations This means that there are three ways to boil a liquid: you can heat the liquid so that its
vapor pressure rises to equal atmospheric pressure, or you can lower the lower pressure
above the liquid, such as by using a bell jar and a vacuum pump, or you can do both. Boiling Water
Atmospheric pressure is generally lower at a high altitude than at sea level. This means
that at high elevations, when you heat a liquid, its vapor pressure will equal atmospheric
pressure at a lower temperature. The liquid will therefore boil at a lower temperature at
high altitude than at sea level.
• Water boils at 100° C if the pressure above the water is 101 kPa = 760 torr = 1 atm. =
standard pressure, which is about the average atmospheric pressure on a fair
weather day at sea level. However, water boils at about 95° C under the lower
atmospheric pressure typically found in locations one mile above sea level (such as
Denver, Colorado). At high altitude, it takes more time to “hard boil” an egg than
at sea level because the boiling water around the egg is not as hot. • At 20.° C, water has a vapor pressure of 17.5 torr (see Lesson 18D). A vacuum
pump can reduce the atmospheric pressure in a bell jar to below 17 torr, and water
in such a vacuum will boil at room temperature. • In a pressure cooker, boiling water is at higher temperature than boiling water at
room pressure, and the changes required to “cook” food occur more quickly. Boiling temperatures are affected by relatively small changes in the surrounding air
pressure, such as those caused by altitude changes. Melting points are substantially
changed only by much larger changes in pressure. Boiling versus Evaporating
Boiling is not the same as evaporating. Evaporation is a surface phenomenon. Measurable
evaporation will occur from all liquids (and many solids) at any temperature. A liquid boils
only when gas bubbles can form below the surface of the liquid and not just at its edges. Practice A: Answer these questions, then, before going on to the next section, practice
until you can answer the questions from memory.
1. Name the three phases. Name six different types of phase changes.
2. Which phases of matter can be significantly compressed in volume? Why?
3. Which has a higher temperature:
a. The melting point or the freezing point of a pure substance?
b. The melting point of a substance that is pure, or one that has impurities?
4. By definition, when does a liquid boil?
5. State three ways to boil a liquid.
6. At what temperature does water boil at an atmospheric pressure of 101 kPa?
7. At approximately what temperature does water boil in a city that is one mile above sea
level? What explains the difference from the boiling temperature at sea level?
©2011 ChemReview.net v. e4 Page 561 Module 21 — Energy Calculations 8. Why does it take longer to hardboil an egg at a high altitude? Energy
Chemistry is primarily concerned with matter and energy. In chemistry, matter and energy
can be considered to be separate entities. Matter has mass, and can be described in terms of
particles such as protons, neutrons, and electrons. Energy has no mass. Sunlight, heat, and
radio waves are a few examples of the many forms of energy.
A fundamental principle of chemistry is the Law of Conservation of Energy: Energy can
neither be created nor destroyed. However, during chemical or physical processes, energy
can be transferred between substances, and to and from the environment. Energy can also
change its form.
Two forms of energy that are important in chemistry are
• kinetic energy, defined as energy of motion, and • potential energy, defined as stored energy. When a substance loses energy, the energy can do work as defined in physics (such as
moving a piston against resistance), or energy can be can transferred as heat to the
environment around the substance. A substance can gain energy when work adds energy
to the substance, such as by compressing a gas, or when the environment supplies heat to
the substance. Kinetic Energy
Kinetic energy is energy of motion. The kinetic energy of an object is calculated by the
equation
KE = ½ (mass)(velocity)2.
This equation means that if particle B has twice the mass of particle A but is moving at the
same speed, Particle B has twice as much kinetic energy. If Particle C has the same mass as
particle A but is moving twice as fast as Particle A, it has four times as much kinetic energy.
Temperature is a measure of the average kinetic energy of particles. When the temperature
of particles goes up, their average kinetic energy increases. For this to occur, since the
particles of a substance cannot change their mass, they must, on average, move faster. Practice B: Answer, and be able to answer from memory, these questions. 1. Define kinetic energy, in words, then using symbols in an equation.
2. Batter #2 hits a baseball with a bat twice as heavy as Batter #1, swinging at the same
speed. How much more energy will Batter #2 impart to the ball than Batter #1?
3. Batter #3 hits the ball with the same bat as Batter #1, but swings twice as fast. How
much more energy will Batter #3 impart to the ball than Batter #1?
4. Define temperature. ©2011 ChemReview.net v. e4 Page 562 Module 21 — Energy Calculations Potential Energy
Potential energy is stored energy. Lifting an object against gravity is one way to add stored
energy to the object. If the object returns to its former lower position, it must release that
added energy.
• To raise a hammer, you must add energy. The energy is stored in the raised hammer
as energy of position. If the hammer fall down to its original position, it must
release the energy used to raise it. It can do so by creating heat where it hits. The
hammer can also do work, such as driving nails. Heat, work, and energy of position
are simply different forms of energy. • The evaporation of water due to the absorption of radiation from the sun can store
energy. One mechanism for this storage of energy is the transfer of water out of
oceans (by evaporation) and into mountain streams (as rain). Lifting the water to a
higher position stores energy from the sun in the water. As the water falls back
toward sea level, the energy released in falling can be harnessed to spin turbines
that create electricity. This electrical energy can be converted to light or heat, or can
be used by electric motors to do work. • In chemical processes, forms of energy that can be stored and released include heat
energy (including the energy stored in plants during photosynthesis) and electrical
energy (as in rechargeable batteries). Compressing a gas can also store energy in a
chemical system. When a compressed gas moves a piston against resistance, the
potential energy stored in the gas is converted to work, another form of energy. Energy and Particle Attractions
Atoms, molecules, or particles can be held together by the attractions arising from the
protons and electrons within those particles. A “chemical bond” is a relatively strong
attraction. The attractions between molecules that cause them to be a liquid or solid at a
given temperature and pressure, rather than a gas, are generally weaker attractions.
Energy always must be added to break chemical bonds, or to change a solid to a liquid and
then to a gas. This added energy is needed to “unstick” the attracting particles. If the
“unstuck” particles return to the state they were in before they were separated, the same
amount of energy that was added and stored during their separation must be released. Energy, Reactions, and Phase Changes
Changes in the potential energy of a chemical system can be the result of chemical reactions
or phase changes.
In a chemical reaction, bonds between atoms break and new bonds form. As a result of
chemical reactions, there is nearly always a characteristic net change in the energy stored in
the substances. Energy must be added to break a bond, but more or less energy will be
released when a different bond forms. This means that in a chemical reaction, energy is
stored or released (to a major or minor extent).
In a phase change, the bonds between the atoms in a molecule do not change, and the
formulas for substances therefore do not change. However, as a substance changes phase
from solid to liquid to gas, the weak attractions between the molecules must be overcome
during each phase change, so energy must be added. As the substance changes from gas to ©2011 ChemReview.net v. e4 Page 563 Module 21 — Energy Calculations liquid to solid, that same amount of energy must be removed from the substance during each
phase change.
• When a substance is melted, a characteristic amount of energy must be added per
molecule to “unstick” the molecules so they can rotate and translate. The energy
added to unstick the molecules is stored in the molecules that change from solid to
liquid. If the liquid molecules change back to the solid state, the same amount of
energy added to melt the solid must be released in order for the liquid molecules to
solidify. • When a substance boils, a characteristic amount of energy must be added per
molecule. For that substance to be condensed from gas to liquid, that same amount
of energy must be released. Practice C: Answer, and be able to answer from memory, these questions. 1. Define potential energy.
2. Name two types of chemical processes that can change the energy stored in molecules.
3. How does the heat of fusion of a substance (the heat/mole required to melt a certain
mass) differ from the heat of freezing (the heat/mole released when the liquid changes
to a solid)? Energy and Phases
When energy is added to or removed from a pure substance, whether its kinetic or its
potential energy changes depends on whether the substance is present in one phase or two.
Recall that by our definitions, a substance is composed of particles that all have the same
chemical formula.
• When a substance is present in only one phase (all solid, all liquid, or all gas),
adding or removing energy (such as by heating or cooling) changes the average
kinetic energy of its particles (their temperature) but does not change the potential
energy stored in the substance. • During a phase change (such as melting or boiling), two phases must be present. If
energy is added to or removed from a substance during a phase change, the potential
energy stored in a substance changes, but the average kinetic energy of its particles
does not change, and its temperature therefore stays constant.
During a phase change, if the two phases present are well mixed or in close contact,
the temperature will be the same in both phases. If two phases are mixed and both
phases are present after mixing, the temperature of the particles will have adjusted
to become the same in both phases. A mixture of the solid and liquid phases of a
substance will always be at the temperature that is its melting point. Potential Energy and Phases
The solid phase of a substance will always have less stored (potential) energy than its liquid
phase, which will always have less potential energy than its gas phase. ©2011 ChemReview.net v. e4 Page 564 Module 21 — Energy Calculations For a given substance: PEsolid < PEliquid < PEgas
Changes in potential energy may not be as apparent as changes in kinetic energy, which are
evident as temperature changes. Let us therefore examine some examples of energy
changes during phase changes.
Examples of LiquidGas Phase Changes
Boiling Water
Consider a tea kettle, in a kitchen at standard pressure, partially filled with cold water and
placed on a lit gas stove. As long as the water in the kettle is below its boiling temperature, as
it is heated by the flame its temperature rises. This increase in kinetic energy is observable,
and it means that the water molecules, on average, are moving faster. What is not
observable, but is true, is that the potential energy stored in the water is not changing.
Boiling will begin when the water temperature reaches 100°C. A thermometer will show
that once the water begins a steady boil, both the liquid water and the steam above the
boiling water have the same temperature (if precautions are taken to prevent “superheating”). At standard pressure (101 kPa), for pure boiling water, that temperature will
always be 100°Celsius by definition.
During any phase change for any substance, both phases will have the same temperature as
long as they are in close contact. The particles in both phases have the same average kinetic
energy.
After 5 minutes of boiling, quite a bit of heat has been added by the flame to the water in
the tea kettle. However, a thermometer will show that both the water and steam remain at
100˚Celsius as long as any liquid water remains in the kettle.
Energy can neither be created nor destroyed. Where is all the energy being supplied by the
flame going? The flame’s energy is being stored in the gas particles (steam) that form during
the phase change.
A characteristic amount of energy must be stored in any molecules to change them from
being nearly as close as possible in their liquid phase to being far apart in their gas phase.
If the flame remains lit beneath the kettle, the water will continue to boil until the last bit of
liquid water is converted to steam. At that point, instead of two phases inside the tea
kettle, there is only one phase (steam). Adding energy with one phase present will increase
the temperature of the steam: its kinetic energy instead of its potential energy. If all of the
water is allowed to boil to steam, there is no longer a phase change to absorb the energy
supplied by the flame, and the temperature of the steam in the kettle (and of the kettle
itself) will increase very quickly.
However, as long as some liquid water remains in the kettle, the highest temperature
possible for the water or the steam is 100° Celsius: much cooler than the flame below.
Warming Leftovers
On a practical note, this is why a little water should be added when heating leftovers in a
loosely covered pan. As long as there is some liquid water between the heat and the food,
the maximum temperature of the water and the food will be 100° Celsius, enough to warm
but not to burn most foods. If all of the water boils away, the food can burn quickly. ©2011 ChemReview.net v. e4 Page 565 Module 21 — Energy Calculations When water boils in the pan, it forms steam. When the steam reaches a cooler surface, it
can condense to form water. When the steam condenses on cool food in a pan, the same
amount of potential energy which was stored in the steam as it formed from water must be
lost from the steam. As the steam turns to water, energy is transferred to the food, and the
“steamed food” heats quickly.
Thunderclouds  Water Condensing
On a humid summer day, clear water vapor (a gas) in the atmosphere can condense to tiny
drops of liquid water (clouds). As the water vapor condenses, the considerable amount of
heat required to change liquid water to water vapor must be released, and the condensation
of vapor to water heats the air around the water droplets. Since warmer air is less dense
than colder air, it rises, creating an updraft that lifts both the moist air and the water
droplets. Because the atmosphere generally cools with increasing altitude, more water
vapor in the humid air forms more liquid water and more heat as it rises. As this cycle
repeats, the cloud becomes a fast rising “thunderhead.”
As the liquid water droplets become larger with increased condensation, the drops become
too heavy to be lifted by the updraft. The result is a thunderstorm. The falling raindrops
create a powerful “downdraft” of air that strikes the ground and fans out ahead of and
with the rain. The downdraft reverses the updraft feeding the thunderhead, eventually
causing the thunderstorm to dissipate. Practice D: Answer these questions, then practice until you can answer the questions
from memory before going on to the next section.
1. If substantial energy is added to a substance, and it remains the same substance but its
temperature does not change, what does this tell you about the substance?
2. When does adding energy to a substance cause its temperature to rise?
3. For a given substance, which phase has the lowest amount of stored energy: solid,
liquid, or gas?
4. In a kitchen where the atmospheric pressure is close to standard pressure, water is
placed in a tea kettle and heated on a gas stove. At the point where the water first starts
to boil,
a. what is the temperature of the liquid water in the kettle?
b. What is the temperature of the steam above the water in the kettle?
5. After 5 minutes of heating, about half of the water in the kettle has boiled away.
a. What is now the temperature of the liquid water in the kettle?
b. What is the temperature of the steam above the water in the kettle?
6. During the above five minutes of boiling, the gas stove adds considerable energy to the
water in the tea kettle.
a. Has the kinetic energy of the water or steam changed?
b. Has the potential energy of the molecules that are still liquid water changed?
c. Has the potential energy of the molecules that were converted from water to steam
changed? ©2011 ChemReview.net v. e4 Page 566 Module 21 — Energy Calculations d. Where has the energy gone that was supplied by the stove in those 5 minutes?
What kind of energy has it become? LiquidSolid Phase Changes
Mixing the Solid and Liquid Phase
A stirred mixture of the solid and liquid phases of a substance will always adjust to the
temperature that is the melting point of the substance.
For example:
• H2O melts and freezes at 0° Celsius at pressures at or near typical atmospheric
pressure. It is a characteristic of water molecules that a stirred mixture containing
water and ice will always adjust to a temperature of 0° Celsius.
A mixture of ice and water is a good constant temperature bath or cold pack. It will stay
at 0°C for as long as both ice and liquid water are present. • When warm water is added to ice, two phases are present, and the temperature of
this mixture must adjust toward the melting (= freezing) point of water. As ice
melts, the warm water molecules become colder. The kinetic energy lost by the
warm water is equal to the potential energy stored in the molecules of ice that
become liquid. The warm water continues to cool, and ice continues to melt, until
either the mixture reaches its melting point (0°C), or all of the ice melts. Melting Ice
When ice melts, a solid becomes a liquid. To change a solid substance to its liquid, energy
must be added. The liquid particles have a characteristic higher amount of stored energy,
per particle, than the solid particles.
While ice is melting, its temperature does not change, but heat must be added from the
environment. This is why a mound of packed snow can take quite a while to melt even
when air temperatures are well above freezing. Considerable heat from the environment
must be absorbed by the solid ice molecules that become liquid water.
Freezing Water In an Ice Tray
The liquid phase of a substance has inherently more stored energy than its solid phase. To
convert liquid molecules to solid molecules, stored energy must be removed.
To change water into ice, the same amount of energy, per molecule, must be taken out of the
water that is put into the ice to melt it.
• When warm water in an ice tray is placed in the freezer, the temperature of the
water drops rapidly as its heat transfers to the freezer environment. When the
water’s temperature reaches 0°C, it begins to freeze. • Unless potential energy leaves a liquid, the liquid cannot become solid. To freeze
water, the air in a freezer must be colder than 0° Celsius, so that heat energy will
flow out of the 0°C water. To provide an air temperature below 0°C, the freezer
compressor pumps heat out until the air is about ―20.°C inside most household ©2011 ChemReview.net v. e4 Page 567 Module 21 — Energy Calculations freezers. You can feel this heat being pumped out if you place your hand in the
space above the coils on the back or underside of a freezer while the compressor is
running.
• After freezing begins, the water/ice mixture in an ice tray will stay at 0°C until all of
the water freezes. During this time, the water and ice mixture will be the warmest
spot in the freezer; warmer than the material already frozen and at ―20°C. • Once the icetray water is completely frozen, one phase is present, and the
temperature of the now solid ice cubes drops relatively quickly to the freezer’s air
temperature.
Ice cubes just removed from the freezer, at about ―20°C, are cold enough to both
cool and then freeze the moisture on your skin, which can cause the ice cubes to
stick to your fingers. However, at room temperature, ―20°C ice warms quickly.
When it reaches 0°C, the ice begins to melt. Ice at 0°C is not cold enough to freeze
skin moisture: melting ice will not stick to your skin. Practice E: Answer, and be able to answer from memory, these questions. 1. A mixture of crushed ice and water is added to an insulated container. After a minute
of stirring, the temperature of the mixture no longer changes, and both ice and water
remain.
a. What is the temperature of the ice? What is the temperature of the water?
b. Which phase has higher kinetic energy?
c. Which phase has higher potential energy?
2. Warm water is added to an icewater mixture in an insulated cup. After stirring for one
minute, the temperature is stable, and ice and water remain.
a. What is the temperature of the water in the cup?
b. What is the temperature of the ice?
3. During the one minute of stirring, the warm water lost some of its energy.
a. What kind of energy did it lose?
b. As the warm water lost its energy, what other change occurred?
c. Where is the energy that was lost by the warm water, and what kind of energy is it? Summary: Phases, Phase Changes, and Energy
You may want to organize the following information into charts, numbered lists, and
flashcards that will help with learning and retention in memory.
1. The three phases and three phase changes:
• Solids melt (or fuse) to become liquids; liquids freeze (or solidify) to become solids. • Liquids boil or evaporate to form gases; gases condense to become liquids. ©2011 ChemReview.net v. e4 Page 568 Module 21 — Energy Calculations • Solids sublimate to become gases directly; gases that undergo deposition form solids. 2. The Law of Conservation of Energy: Energy can neither be created nor destroyed (except
in nuclear reactions). However, energy can be transferred between substances and to
and from the environment. Energy can also change its form during chemical or
physical processes.
3. Two forms of energy are potential energy, defined as stored energy, and kinetic energy,
defined as energy of motion. Kinetic Energy = ½ (mass) (velocity)2
4. Chemical substances can store energy in the attractions between atoms, molecules, and
particles. During chemical reactions and phase changes, energy is stored or released.
5. One way to store energy in a substance is to change its phase. The solid phase of a
substance always has less stored (potential) energy than its liquid phase, which always
has less potential energy than its gas phase.
Potential energy of a substance: solid < liquid < gas
6. When a substance is in one phase (all solid, liquid, or gas), adding or removing energy
will change the average kinetic energy of its particles (its temperature), but not its
potential energy.
7. During a phase change, when two phases are present, adding or removing energy
changes the potential energy, but not the average kinetic energy (temperature) of the
particles.
8. During a phase change, temperature remains constant, and the temperature is the same
in both phases as long as they are in close contact.
9. The temperature at which a substance melts (its melting point) will equal the
temperature at which it solidifies (melting point ≡ freezing point).
10. When a mixture of the solid and liquid phases of a substance is stirred, the temperature
will adjust to the melting point of the substance.
11. The melting point is a characteristic of a substance. The melting point will be the same
no matter how the substance is formed. The melting point can be used as evidence to
identify a substance.
12. A liquid will boil at any temperature at which its vapor pressure equals the
atmospheric pressure above it. If the atmospheric pressure on a liquid is lowered, the
liquid will boil at a lower temperature. If the atmospheric pressure is raised, the liquid
will boil at a higher temperature.
13. Boiling points are a characteristic which can be used to identify a substance, but only if
the atmospheric pressure is known. Boiling points are far more sensitive to
atmospheric pressure than melting points.
14. Evaporation is a surface phenomenon; measurable evaporation will occur from all
liquids (and some solids) at any temperature. However, a liquid is boiling only when
gas bubbles of the liquid can form below the liquid surface and not just at its edges.
***** ©2011 ChemReview.net v. e4 Page 569 Module 21 — Energy Calculations Practice F: At standard pressure, small cubes of ice are removed from a freezer and
placed in a tea kettle. A thermometer is inserted into the ice cubes and the kettle is placed
on a lit gas stove. The kettle is heated until one minute after all of the water has boiled
away.
The graph below charts the changes in the temperature of the H2O molecules as they
change from ice to water to steam.
The Tea Kettle Problem
140
120 Degrees Celsius 100
80
60
40
20
0
0 5 10 15 20 25 30 20
40 Minutes Questions
1. How many phase changes occur during the above process?
2. How many phases will have been present by the time the above process is completed?
3. Which segment of the graph represents water boiling to steam?
4. How can a change in the kinetic energy of the H2O be recognized during the process?
5. How can a change in the potential energy of the system be recognized?
6. In which lettered segments of the graph does potential energy remain constant?
7. In which segments of the graph does average kinetic energy remain constant?
8. In which portions of the graph do the H2O molecules have the largest amount of stored
energy?
9. Which portions of the graph show energy from the flame being converted into potential
energy? ©2011 ChemReview.net v. e4 Page 570 Module 21 — Energy Calculations ANSWERS
Practice A
1. Phases: solid, liquid, gas. Changes: melting, freezing, boiling, condensing, sublimation, deposition.
2. Only the gas phase. There is substantial distance between particles only in the gas phase.
3. a. Melting point = Freezing point, by definition.
b. A pure substance melts at a higher temperature than the same substance with impurities.
4. When its vapor pressure equals the atmospheric pressure above it.
5. Raise the liquid’s vapor pressure by raising its temperature, or lower the atmospheric pressure above the
liquid, such as by moving to higher altitude or into a partial vacuum.
6. 101 kPa is standard pressure, so water boils at 100° C by definition.
7. Approximately 95 degree Celsius. At high altitude, atmospheric pressure is lower, and the water’s vapor
pressure will equal atmospheric pressure at a lower temperature.
8. The water boils at a lower temperature, and at a lower temperature the changes needed to “cook” food take
longer to occur.
Practice B
1. Kinetic energy is energy of motion. KE = ½ (mass)(velocity)2.
2. Batter #2 hits with twice as much energy. 3. Batter #3 hits with four times more energy. 4. Temperature is a measure of the average kinetic energy of chemical particles.
Practice C
1. Stored energy 2. Chemical reactions and phase changes. 3. Heat of melting = heat of solidification. The heat added in melting must be released when a liquid
solidifies.
Practice D
1. Two phases are present, and the substance is undergoing a change to a phase with more stored energy.
2. When only one phase is present, which means the substance is not undergoing a phase change.
3. Solid. 4a and 4b. Both the liquid water and the steam are at 100° C. 5a and 5b. Both the liquid water and the steam are still at 100° C. 6a. No. 6b. No. 6c. Yes 6d. The energy is now potential energy stored in those molecules that changed phase from liquid to gas.
Practice E
1a. Both are at 0° C. 1b. Both have the same KE. 1c. The liquid water has higher PE.
2a and 2b. Water and ice are both at 0°. If both solid and liquid are present, both must be at the melting point.
3a. The warm water lost kinetic energy: its temperature fell. 3b. The warm water melted some ice. 3c. The kinetic energy lost by the water in cooling to 0° is now potential energy that is stored in the ice
molecules that were previously ice, but the warm water melted.
Practice F
1. Two (melting and boiling) 2. Three (solid, liquid, and gas) 3. D 4. The temperature changes (this occurs when the line is not in a “plateau” region.) ©2011 ChemReview.net v. e4 Page 571 Module 21 — Energy Calculations 5. Where the graph has a “plateau” region, heat is being added from the stove for several minutes but the
temperature (average kinetic energy) remains constant, so potential energy must be increasing.
6. A,C,E – when the kinetic energy is changing. 7. B and D – the temperature stays constant. 8. D and E – the gas phase (the steam) which forms during D, and then heats during E, has the most
potential energy.
9. B and D – during the two phase changes.
***** Lesson 21B: Specific Heat Capacity and Equations
Timing: If you have not already done so, before doing Lesson 21B, complete Lesson 17C
on cancellation of complex units.
If your class assignments require calculations that include PV work before specific heat and
calorimetry calculations, do Lesson 22A, then return here.
***** Units That Measure Energy
In chemistry, energy is usually measured in joules or calories.
1. The joule (abbreviated J) is the SI unit measuring energy.
A joule is defined in physics in terms of work: as the amount of energy needed to
accelerate 1 kg by 1 meter/sec2 in 1 meter.
work = (force)(distance) = (mass)(acceleration)(distance) .
Though joules are defined in terms of work, all forms of energy are equivalent and can
be measured in any energy units. In chemistry, joules is the unit most often used to
measure the heat energy lost or gained in a chemical process.
2. Calories are a metric unit that is also used to measure energy.
A chemical calorie is defined as the amount of heat needed to raise the temperature of
one gram of liquid water by one degree (Celsius or Kelvin).
3. Because all forms of energy are equivalent, all energy units can be related by equalities.
The conversion between calories and joules is: 1 calorie = 4.184 joules
4. In studies of nutrition, a food Calorie is the unit used to measure the heat released
when food is burned completely and/or is metabolized in cells.
1 food Calorie = 1,000 chemical calories = 1 kilocalorie (kcal) = 4.184 kilojoules (kJ)
The calories listed on nutritional labels are food Calories (chemical kilocalories).
In chemistry texts, food Calories are written as Calories with a capital C, whereas
chemical calories are written with a lower case c.
Nearly all chemistry courses assign problems using joules to measure energy. Some may
also assign problems using chemical and/or food calories. You should learn the
relationships needed for your course.
©2011 ChemReview.net v. e4 Page 572 Module 21 — Energy Calculations Energy and Heat
An energy transfer between a system and its surroundings in a chemical or physical
process may involve heat (symbol q) and/or work (w).
In science, the symbol ∆ (delta) means change in. When chemical reactions and processes
take place, the change in the energy of the molecules involved (the chemical system) can be
represented by
∆Esystem = q + w
For example: Boiling water adds heat to the molecules. Burning gasoline releases heat.
Work is required to compress a gas. When a compressed gas moves a piston against
resistance, the molecules lose energy as the system does work.
In Lesson 22A, we will consider calculations that involve mechanical work. In this module,
our focus will be on calculating heat energy (symbol q ). Specific Heat Capacity
In these lessons, we will define the amount of heat required to raise one gram of a substance
by one degree Celsius or kelvin as the specific heat capacity (symbol small c) of the
substance. (Some textbooks use different terms and symbols for this quantity.)
In most calculations, specific heat capacity will be used to calculate the total energy change
in a process. We will therefore learn the equation using specific heat capacity in this form:
q = c · m · ∆t
This equation means:
The heat energy gained or lost by a substance =
= (specific heat capacity of the substance) x (its mass) x (its change in temperature)
When heat is added to a system of chemical particles, q is defined as positive. When
chemical particles lose heat, q is defined as negative.
The units of c are joules per (gram · degree) or calories per (gram · degree).
For example, the specific heat value for liquid water is written cwater = 4.184 J/g · K
Recall that the dot between gram and K means that the two units are multiplied together in
the numerator or denominator. These three notations are equivalent:
4.184 joule/gram•K = 4.184 J
g•K = 4.184 J • g─1• K─1 (If your course or textbook uses the unit─1 or unit─# notation in calculations, you may
want to complete Lesson 27F after this lesson.)
Specific heat capacity may also be measured in joules/kilogram · degree. You will learn
below how to convert data to the consistent units needed to solve equations.
Note the difference indicated by the equation between heat and temperature.
• Temperature is an intensive property: when you measure a temperature, the value
does not depend on the amount of matter or space around the thermometer. ©2011 ChemReview.net v. e4 Page 573 Module 21 — Energy Calculations • Heat is an extensive property. When calculating the heat transferred in a process,
what is being heated, the amount being heated, and how much it is being heated, all
matter. On a gas stove, to bring water at room temperature to the point that it
begins to boil, for a large amount of water you must supply more heat than for a
small amount. Change in Temperature
In science, the symbol ∆ (delta) means change in. The symbol ∆t (read “delta t”) means
change in temperature. ∆t is defined as ∆t ≡ tfinal — tinitial
This definition means that change in temperature must be labeled as positive when
temperature increases and negative when it decreases.
A change in temperature will be the same number of degrees when measured in the Celsius
or Kelvin scales. Why? The Kelvin and Celsius scales have the same size degree.
If, in Celsius, ∆t = 25°C ― 0°C = 25°C , the same measurements recorded in kelvins will
result in the same number for the change: ∆T = 298 K ― 273 K = 25 K .
This means that if the temperature units in a problem are based on ∆t measurements, the
word degree and the symbols °C and K are all equivalent.
Since specific heat capacity (c) is defined in the words and equation above as based on a
change in temperature (∆t), a value for c of “4.184 joules/gram·degree” can also be written
as “4.184 joules/gram·K.”
When units are equivalent, they can cancel. For ∆t values,
joules • °C = joules • °C = joules
K
K and calories • K = calories • K = calories
degree
degree Practice A: Assume that the temperature units below are all measurements of ∆t. Do
the unit cancellation, write the final unit.. If you need a review of the rules for unit
cancellation, see Lesson 17C.
1. 3. joules • g • °C =
gram · K
calories
=
• °C
calorie
gram · degree 2. joules
•g
joules
gram · K
4. joules
grams • °C = = 5. Write two metric units that can be used to measure q . ©2011 ChemReview.net v. e4 Page 574 Module 21 — Energy Calculations Values for Specific Heat Capacity (c)
Chemical substances have a characteristic specific heat
capacity in each phase: a fixed amount of heat
changes one gram of substance by one degree. Some
values for specific heat capacity are in the table on
the right.
As in the case of H2O in the table, for each substance,
each phase has a different c value.
Heat capacity values apply only while a substance is
in a single phase. A different way to calculate heat
changes will be needed when a substance is changing
phase. Substance Specific Heat
Capacity
In J/g·K H2O liquid 4.184 H2O solid 2.09 Cu solid 0.385 Fe solid 0.444 Specific Heat Capacity (The c Prompt)
To solve calculations that include specific heat capacity, we will use the memorized equation
that includes specific heat capacity. Our rule will be:
The c Prompt: If you see the term “specific heat capacity” or its symbol c in a
problem, write at the top of your DATA table the equation which uses c:
q = c · m · ∆t
To use this equation, we will need to use an equation method rather than a conversion
method to solve problems. Solving Problems Which Require Equations
In these lessons, we will refer to equalities that use symbols, such as q = c · m · ∆t , as
equations rather than formulas to distinguish them from chemical formulas such as H2O.
When solving problems using equations, we will add steps to our conversion method of
problem solving.
If you have already completed the Ideal Gas module in these lessons, you have learned the
following method for solving problems with equations. However, since many chemistry
courses cover energy calculations before gas laws, we will describe the steps of the method
in detail as it applies to energy problems as well. Solving With Equations
Problems in physical science can generally be put into three categories: those that can be
solved with unit cancellation (conversions), those that require equations, and those
requiring both.
Solving problems with equations requires a small amount of algebra.
We will use the specific heat equation to illustrate a consistent method for solving
calculations that require equations. Starting with relatively easy examples, we will
develop a method that also works for more difficult problems. ©2011 ChemReview.net v. e4 Page 575 Module 21 — Energy Calculations Easy problems can be solved in other ways, but to learn the system which is especially
useful for the more difficult calculations that lie ahead, try the method used here.
Let us start with a example:
Q. When 832 joules of heat is added to a sample of solid copper (Cu), the
temperature rises from 15.0°C to 33.0°C. Based on the specific heat capacity in the
table above, how many grams of copper were in the sample? To solve, complete the following steps in your notebook.
1. As always, begin by writing “WANTED: ?” and the unit you are looking for.
2. This problem mentions “specific heat capacity.” That’s the c prompt. In the DATA,
write the memorized equation which includes specific heat capacity (c).
3. Below the equation, make a data table that lists each symbol in the equation.
*****
For this problem, the data section should look like this:
DATA: q = c · m · ∆t
q=
c=
m=
Δt = 4. After each symbol, based on units, write each number and unit.
• In this problem, q is the symbol for heat energy, and energy is measured in joules, so
q = 832 joules. • For c = , from the table value for copper: c = 0.385 J/g·K 5. Put a ? after the symbol in the table that you are looking for in the problem. Add the unit
you are looking for. Circle this line in the DATA table.
Fill in the data table completely, and then check below.
***** ( * * * mean: write your answer, then check the answer below.) At this point, your paper should look like this:
WANTED: ? g Cu DATA: q = c · m · ∆t
q = 832 joules
c = 0.385 J/g·K
m = ? g Cu
Δt = 33.0°C ― 15.0°C = + 18.0°C ©2011 ChemReview.net v. e4 Page 576 Module 21 — Energy Calculations 6. SOLVE the fundamental memorized equation, using algebra, for the symbol that you
WANT. Do not plug in numbers until you have solved for the WANTED symbol.
(Symbols move more quickly than numbers and their units.)
Try that step, then check below.
*****
SOLVE: q = c · m · ∆t . Solving for the symbol m WANTED,
q ?=m= c · Δt
7. After solving in symbols, plug in the numbers and solve. Cancel units that cancel, but
leave the units that do not cancel, and include them after the calculated number.
Do that step, then check below.
*****
On your paper should be
SOLVE: q = c · m · ∆t ?=m= q =
832 joules
= 832 J ●
1
= 832 J ●
g·K = 120. g Cu
c · Δt 0.385 J ● 18.0°C 18.0°C
0.385 J
18.0°C
0.385 J
g·K
g·K Note that in solving, the term with a fraction in the denominator was separated and then
simplified using the rules for reciprocals. This step will often help with unit cancellation in
specific heat capacity (c) calculations.
Double check the cancellation of the units for the last step above. Since degrees and °C and
K are all equivalent when they measure a Δt, they can cancel as units.
The above problem involved finding grams of Cu, but you did not need the molar mass to
solve. This is because specific heat capacity is one of the rare quantities in chemistry that is
defined based on grams rather than moles. Other heat problems will involve moles. Our
rule will be
In heat problems, if you see both grams and moles of a substance as units in a problem,
write the molar mass in your data, because you will likely need the molar mass to solve. ©2011 ChemReview.net v. e4 Page 577 Module 21 — Energy Calculations Summary
If you need a mathematical equation to solve a problem, use these steps.
1. Write the fundamental, memorized equation in your data.
2. Make a data table with each of the symbols in the equation.
3. Use units to place each item of data after a symbol in the data table.
4. Memorize equations in one format, then use algebra to solve for symbols WANTED.
This will minimize what you need to memorize.
For example, memorize: q = c · m · ∆t . Then if you WANT c or m or ∆t , use algebra to solve the equation for that symbol.
Don’t memorize: m= q and c= c •Δ t q and Δt = m •Δ t q c •m 5. Solve the fundamental equation for the WANTED symbol before you plug in numbers.
6. Plug both numbers and units into equations. Use unit cancellation to check your
work. Practice B: Use the steps above for these problems. Answers are below. 1. When 681 joules of heat are added to 240. grams of a pure solid, the temperature of the
solid rises by 22.0 degrees. What is the specific heat capacity of the solid?
2. If 361 joules are added to a 32.5 gram sample of iron (Fe) at 20.0°C, use the value for c
from the table above and solve for the final temperature of the sample. ANSWERS
Practice A You may use other methods of unit cancellation (see Lesson 17C) as long as you arrive at the
same answers as these. Degrees and °C and K are all equivalent when they are used to measure a change in
temperature (Δt). If one is on top and one is on the bottom when you multiply terms, they can cancel.
1. joules • g • °C = joules
gram · K 3. calories
calorie • °C
gram· degree = 1 = grams
1
gram 2. = 1
1
K joules
grams • °C = joules
joules • g
gram · K
4. =K joules
grams • °C (in Problem 4, nothing cancels.) 5. Joules and calories. ©2011 ChemReview.net v. e4 Page 578 Module 21 — Energy Calculations Practice B
WANTED: c=? (Strategy: When “specific heat capacity” is mentioned, that calls the “c prompt.” Write:) DATA: 1. q = c · m · ∆t (and make a data table to match all those symbols.) q = 681 J
c=?
m = 240. g
Δt = 22.0°C
SOLVE: Since q = c · m · ∆t
?=c= q
=
m · Δt 681 J
= 0.129
J
240. g • 22.0°C
g • degree (Solve in symbols before plugging in numbers and units. Do the math for both numbers and units.
Make sure the answer unit matches what the unit should be for the symbol WANTED.)
2. WANTED: Final temperature Since Δt = ∆t ≡ tfinal — tinitial then tfinal = tinitial + Δt Strategy: When c is mentioned, that’s the prompt to write DATA: q = c · m · ∆t Make a data table to match those symbols. q = 361 J
c = 0.444 J/g•K for Fe
m = 32.5 g Fe
Δt = ?
WANTED Final t = 20.0°C + Δt
SOLVE: Since q = c · m · ∆t , and we want Δt,
? = Δt = q
c•m = 361 J
= 25.0 K = Δt
0.444 J • 32.5 g
g•K Done? WANTED = tfinal = tinitial + Δt
Final t = 20.0°C + Δt = 20.0°C + 25.0°C or K = 45.0°C final temperature
A Δt is the same number of degrees in the Celsius and Kelvin temperature scales. ***** ©2011 ChemReview.net v. e4 Page 579 Module 21 — Energy Calculations Lesson 21C: Energy, Water, and Consistent Units
The substance most often used to supply or absorb heat in a chemical process is liquid water.
Most courses require that the value for the specific heat capacity of water be memorized. In
heat calculations that involve liquid water, apply the
c Water Prompt: If a problem mentions energy or heat or joules or calories  and liquid
water, write in your data table the equation using specific heat capacity,
q = c · m · ∆t . In the data table with those 4 symbols, write
c = cwater = 4.184 J/g •K (or cwater = 1 calorie/g •K) Using this prompt, problems involving heat and water can be solved in the same way as
the specific heat problems in the previous lesson.
In calculations involving liquid water, assume that 1 mL liquid H2O = 1.00 gram liquid H2O
unless a more precise density is noted.
Note that though the common name for H2O is water, in problems dealing with energy it is
important to distinguish between ice, water, and steam. These three phases for H2O have
different values for c. However, unless ice or steam is specified, in heat problems you
should assume that water means liquid water. Consistent Units
For an equation to work, the units must match the requirements of the equation.
For example, the equation for specific heat capacity requires mass (usually grams,
occasionally kg). If the data is given in moles, you must convert moles to grams or kg
before you solve.
In addition, when solving equations, units must be consistent.
For example, in an equation involving mass, grams or kilograms may be used, but not
both. You must choose a mass unit, and then convert the other masses to that unit.
Which unit should you choose?
In most cases, you may solve equations in any consistent units, but some ways of choosing
consistent units will solve more quickly than others. To solve heat problems, we will use
this rule:
Convert DATA to the units used in the most complex unit.
For example: If a heat calculation includes a unit of “joules/kg·K ,” convert the units in
the DATA to joules and kilograms.
The best time to convert to consistent units is early in a problem. The best place to convert
to consistent units is in the DATA table.
Try this example in your notebook. If you get stuck, peek at a bit of the answer on the next
page, then try again.
Q. 16.0 moles of water at 25.0°C is supplied with 28.0 kJ from a bunsen burner. If all of
the heat is absorbed by the water, what will be the water's final temperature? ©2011 ChemReview.net v. e4 Page 580 Module 21 — Energy Calculations *****
Answer
When you see joules and liquid water, that’s the c water prompt. Write the equation that
includes specific heat capacity, a data table to match its symbols, and fill in c for liquid
water.
DATA: q = c · m · ∆t
q=
c = 4.184 joules/gram •K
m=
Δt = Since the units supplied in the problem do not match the units of the complex unit,
write a ? and a unit beside each symbol that is consistent with the complex unit.
DATA: q = ? joules =
c = 4.184 joules/gram •K
m = ? grams =
Δ t = ? K or ° C WANTED final temp = 25.0°C + Δt In heat problems, temperature change conversions can often be done by inspection,
since a Δt value in degrees Celsius and kelvins is the same.
Add the remaining data from the problem to the table, converting to the consistent
units as you go.
*****
DATA: q = ? joules = 28.0 kJ = 28.0 x 103 J (done by inspection) c = 4.184 J/g •K
m = ? g = 16.0 mol H2O • 18.0 g H2O = 288 g H2O
1 mol H2O
Δ t = ? K or ° C WANTED final temp. = 25.0°C + Δt *****
SOLVE: ? = Δt = q = c •m 28.0 x 103 J
4.184 J • 288 g = g·K
= 28,000 J • g•K • 1
=
4.184 J
288 g 23.2 K increase = Δt Since heat is being added to water, the temperature will increase.
The final temperature is 25.0°C + 23.2 °C or K = 48.2 °C
***** ©2011 ChemReview.net v. e4 Page 581 Module 21 — Energy Calculations Summary: Consistent Units
1. If an equation requires certain units, convert the DATA to those units (such as
converting temperature to K when T is used in gas equations.)
2. Convert to consistent units in the DATA table.
3. If an equation does not require certain units, but the units in the problem are not
consistent,
•
• 4. Choose consistent units based on the most complex unit in the problem.
Write the consistent unit after each symbol in the DATA table. If the WANTED unit is not consistent with the most complex unit, solve using the
units used in the complex unit, then convert to the WANTED unit. Practice: Do as many as you need to feel confident. More difficult problems are toward the bottom.
1. Convert these to the units WANTED. Try to do so without writing conversions, by
writing steps by inspection (in your head).
a. ? J = 0.25 kJ = _________________ = _____________________= ______________
In exponential notation
scientific notation
fixed notation
b. ? g H2O(l) = 75 mL H2O(l) = _______________________ in fixed notation. c. In any notation: ? g H2O(l) = 8.9 L H2O(l) =
2. A 36.0 mL sample of water is raised in temperature by 15.0°C. How many joules are
needed?
3. A 15.0 mole sample of liquid water loses 6.70 kJ of heat. At the end of the process, the
water temperature is 21.4° C. What was the original temperature of the water?
4. A quantity of water gives off 54 kilocalories as it cools from 75° to 5° C. How much
water is cooling?
5. How much heat (in joules) would be required to raise 4.50 moles of ice from ―20.0° C to
the temperature at which it begins to melt? (c for ice = 2.09 J/g·degree.) ©2011 ChemReview.net v. e4 Page 582 Module 21 — Energy Calculations ANSWERS
1. a. ? J = 0.25 kJ = 0.25 x 103 J = 2.5 x 102 J = 250 J
b. ? g H2O(l) = 75 mL H2O(l) = 75 g H2O(l) (1.00 g liquid water = 1 mL) c. ? g H2O(l) = 8.9 L H2O(l) = 8,900 mL H2O(l) = 8,900 g H2O(l) = = 8.9 x 103 g H2O(l)
2. WANTED:
(Strategy: ?J
When you see joules and liquid water, write: )
q = c · m · ∆t , make a data table with those symbols, and put c for water in the table. DATA: q = ? J WANTED
c = 4.184 J/g·K for liquid water (so find mass in grams) m = ? g = 36.0 mL = 36.0 g for liquid water (1 mL water = 1.00 g water) Δt = ? K or °C = 15.0°C
(The mL of water are converted above to the mass units that match the c unit.)
? = q = c · m · Δt = 4.184 J • 36.0 g • 15.0°C = 2,260 J
g ·K SOLVE:
3. WANTED:
(Strategy: Initial temp. Since ∆t ≡ tfinal — tinitial then tinitial = tfinal — Δt = WANTED When you see joules and liquid water, write)
q = c · m · ∆t and cwater (c water prompt) q = ? J = —6.70 kJ = —6.70 x 103 J = — 6,700 J (convert to units of c) c = 4.184 J/g·K DATA: (c water prompt) m = ? g = 15.0 mol H2O • 18.0 g H2O = 270. g H2O
1 mol H2O
When grams are needed, but moles are given, do the conversion in the data table.)
Δt = ∆t ≡ tfinal — tinitial so tinitial = tfinal — Δt = WANTED SOLVE: ? = Δt = q
=
— 6,700 J
c·m
4.184 J • 270. g
g•K = —5.93 K = Δt WANTED = tinitial = tfinal — Δt = 21.4°C —(— 5.93 °C or K) = 27.3 °C
(Since the water is losing heat, its initial must be warmer than the final temperature, and it is.)
4. WANTED:
(Strategy: ? = mass of water
When you see joules and liquid water, write: )
q = c · m · ∆t ©2011 ChemReview.net v. e4 (the “c water” prompt) Page 583 Module 21 — Energy Calculations DATA: q = ? calories = 54 kilocalories = 54,000 calories (convert to units of c) c = 1 calorie/g • K (the “c water” prompt) for liquid water (choosing a value for c in calories will simplify solving, since q is in calories and units must be
consistent. However, the calories in q could also be converted to joules, and the value for c
using joules used. Both methods give the same answer.)
(SF: because the c for water is also the definition of a calorie, the 1 is exact.)
m = ? g (WANTED)
Δt = ∆t ≡ tfinal — tinitial = 5°C ― 75°C = ― 70.°C (subtracting, the place determines doubt)
SOLVE: q = c · m · ∆t ,
?=m=?g = 5. WANTED:
(Strategy: (solve in symbols before numbers) q
=
c · Δt 54,000 calories
1 calorie • 70.°C
g•K = 770 g water q in joules
Note that although both are H2O, the c value for ice is not the same as for liquid water.
Since the problem mentions c, that’s our “c prompt.” Write the equation that uses c.) DATA: q = c · m · ∆t
q = ? J = WANTED
c = 2.09 J/g•K for ice (The equation requires mass (m), the c unit uses grams, but the ice data is in moles. The
relationship between grams and moles is the molar mass.)
m = ? g = 4.50 mol H2O • 18.0 g H2O = 81.0 g H2O(s)
1 mol H2O
Ice begins to melt when it reaches 0.0°C.
Δt = ? K or °C = ―20.0°C to 0.0°C = 20.0°C
(The data has now been converted to the 3 units used by c. The equation can work because
the units cancel properly.)
SOLVE: ? = q = c · m · Δt = 2.09 J • 81.0 g • 20.0°C = 3,390 J
g• K ***** ©2011 ChemReview.net v. e4 Page 584 Module 21 — Energy Calculations Lesson 21D: Calculating Joules Using Unit Cancellation
In solving heat problems, a key step is often to calculate the energy lost or gained in a
chemical process. However, there are a number of equations that solve for energy. How
do we know which to use when?
Rather than memorizing a large number of equations, unit cancellation can be used to
calculate energy. Unit cancellation is simply the conversionfactor method with the rules
loosened a bit to accommodate complex units. The principle is the same: if you arrange the
units to cancel to result the WANTED unit, the numbers attached to the units will be in the
right place to produce the correct answer.
Unit cancellation can also be use to “design formulas,” or as hints to help to remember
complex formulas. However, unit cancellation must be used carefully.
Using heat calculations, let us compare solving with equations to unit cancellation. Equations To Calculate Energy
1. The equation for calculating heat or energy using specific heat capacity (c) is
Heat energy lost or gained = q = c · m · ∆t In words, this equation means The heat energy gained or lost = (specific heat capacity) x (mass) x (change in °C or K )
Other equations can be used to calculate energy. Three of those equations are:
2. The molar heat capacity (symbol capital C) of a substance is defined as the amount of
heat required to raise one mole of the substance by one degree.
The equation is: change in heat energy = q = C · moles · ∆t
3. For heats of combustion or phase changes, the equation for heat lost or gained is
q = (heat of process/mass or moles) · (mass or moles)
4. For bomb calorimeter calculations, the equation for heat energy lost or gained is
q = (heat capacity of calorimeter/degree) · ( ∆t )
Problems involving items 2, 3, and 4 are frequently encountered, but we would prefer not
to memorize those formulas. Calculations using those quantities can instead be solved
using the units supplied in the problems.
Note that
• All four equations above solve for energy (q), which may be in joules or calories. • Not all of the equations involve temperature, but those that do always a change in
temperature, so degrees and °C and K will all be equivalent. • None of the equations include proportionality constants; q is calculated by a
straightforward multiplication of numbers and their units. ©2011 ChemReview.net v. e4 Page 585 Module 21 — Energy Calculations Under these special circumstances, the units in a problem can be used, without needing a
memorized equation, to calculate energy. Let’s learn the method by example.
Q1. If a problem’s data includes the unit joules
K
and the energy in joules is WANTED, what must you do to solve? *****
? joules = joules times (kelvins) .
K
To get joules from joules/K, multiply by the data whose units are kelvins.
The fundamental rule is: Let the units tell you what to do.
The K in the unit above does not indicate whether K is a temperature, such as 273 K, or
a change in temperature, such as ∆t = 20 K or °C. However, in all of the energy
equations above, degrees, °C, and K all represent a ∆t, and not a T. In similar energy
calculations, you can assume that degrees, °C or K measure a change in temperature.
Try another example.
Q2. If a problem’s data has a unit of joules ,
mol · K and the joules of energy is WANTED, what equation would you design?
*****
? joules = joules
mol · K times (moles) times (kelvins) . *****
Unitcancellation rules are similar to conversion rules, but expanded.
• Complex ratio units, like J/g·K , can be used rightside up or upside down. • Single units can be used on top, or inverted, as in “1/(X grams)” • Units may be multiplied in any order. If you want a single unit, you can even start
with a ratio unit as long as the units cancel to give the WANTED unit at the end. Unit cancellation is a kind of jigsaw puzzle. Arrange the data as pieces, rightside up or
upside down, so that the units on the right cancel to give the unit that you WANT on the
left. This will result in the right answer for the numbers and the units.
Using the two examples above as a guide, try this problem.
Q3. Solve for joules using these “numbers” and units: X joules/g·K , Y °C, and Z g.
***** ©2011 ChemReview.net v. e4 Page 586 Module 21 — Energy Calculations Answer: ? joules = X joules • Y °C • Z grams = X•Y•Z joules .
g·K The terms to multiply may be listed in any order. Try one more.
Q4. Solve for ? grams using: X joules/g·K , Y °C, and Z joules. *****
When complex units are involved, though no particular order is required, it will often
help in arranging the units if you start in your given with a unit that puts the answer unit
(grams in the above case) where you WANT it in the answer.
? grams = 1 grams·K • 1 • Z joules = Z
X joule
Y °C
XY grams Practice A
Using the method above, try every other problem, and more if you need more practice.
Not sure how to proceed? Peek at the answer at the end of the lesson, then try again.
1. Solve for K using X joules, Y joules/g·K, and Z grams.
2. Solve for moles of liquid H2O using X degrees, 4.184 joules/g·K, and Z joules.
3. Solve for joules using X degrees, Y calories/g·K, and Z grams. (4.184 joules = 1 cal.)
4. Solve for calories using X degrees, Y joules/g·K, and Z kilograms. Cases When Unit Cancellation Does Not Work
Unit cancellation by itself does not work for equations such as KE = 1/2 mv2 or Volume of
a sphere = 4/3 π r3 . Those relationships have constants that must be included in the
calculations to get correct answers.
Unit cancellation does work for relationships based on equations without constants, such as
q = c·m·∆t or distance = (rate)(time) or area of a rectangle = base • height, or even
E = mc2 .
It is important, therefore, to use unit cancellation in place of memorized equations only for
equations that do not have proportionality constants.
However, though the units alone do not always predict proper equations, if you are unsure
about recalling a correct equation, unit cancellation will often supply good hints about what
the symbols must be, and where they must be, in an equation that solves for the unit. Using Unit Cancellation to Calculate Energy
Use the unit cancellation method above to solve this problem.
Q. A sample of aluminum weighs 16.5 grams. How many joules are required to raise
the temperature of the aluminum from 20.0°C to 50.0°C? The molar heat capacity of
aluminum is 24.3 J/mol·K .
***** ©2011 ChemReview.net v. e4 Page 587 Module 21 — Energy Calculations Answer
WANTED : ?J Strategy: This problem uses molar not specific heat capacity, so q = c·m·Δt cannot
be used directly. Instead, try unit cancellation. The symbols used for an
equation data table are not needed. Simply list as DATA the numbers
and units, in the manner done for conversion factors. To solve, arrange
the units to cancel correctly. *****
DATA: 16.5 g Al
∆t = 50.0°C ― 20.0°C = 30.0°C
24.3 J/mol●K
27.0 g Al = 1 mol Al (See both grams and moles? Need molar mass) Adjust your work if needed, and then check the answer below.
*****
SOLVE: ? J = 24.3 J • 1 mol Al • 16.5 g Al • 30.0°C = 446 joules
mol•K 27.0 g Al The terms that are multiplied can be in any order, so long as they are rightside up
compared to those above.
You could solve this problem using the equation for molar heat capacity, in the same way
you used the equation for specific heat capacity. However, to do so, you need to memorize
the equation for molar heat capacity. With unit cancellation, you can solve without
memorizing the many equations that calculate q. Instead, let the units tell you what to do.
*****
When should you use the original conversion rule “if you want a ratio, start with a ratio?”
When should you use the looser unit cancellation rules? A good rule of thumb is
• In a problem where all of the DATA can be listed as single units or equalities, the rule
“if you want a single unit, start with a single unit” will automatically arrange your
factors rightside up. • In a problem that includes a complex unit, in a form such as x/y●z, the rule “in the
given, put the answer unit where you WANT it” will help to arrange factors. The looser unitcancellation rules work in both cases, but they can take a bit longer to
arrange. In both cases, the unitcancellation concept is the same. Summary: Solving With Unit Cancellation
For energy calculations that do not use specific heat capacity (c), try unit cancellation.
1. List WANTED and DATA as done for conversion factor problems, without symbols.
2. If you see grams and moles for a substance, write the molar mass in the DATA table.
3. Arrange the numbers and units so that the units cancel to give the WANTED units. ©2011 ChemReview.net v. e4 Page 588 Module 21 — Energy Calculations Practice B: Using the method above, try these. If you are not sure how to proceed, peek
at the answers, then try again.
1. The heat of vaporization of water is 9.7 kcal/mole. How much heat would it take to
vaporize 0.75 grams of water? (Solve in joules.)
2. A “bomb calorimeter” is used to absorb and measure the heat released when a
substance is burned. If the heat capacity of a bomb calorimeter is 6.00 kJ/K, and a
reaction releases 42,600 calories of heat, what will be the increase in the calorimeter
temperature?
3. Calculate the heat needed to raise 2.5 moles of ice from ─40.0°C to 0.0°C. (The c for ice
is 2.09 J/g●K) (Though this problem mentions c, try solving with unit cancellation.) ANSWERS
Practice A: Your factors may be in any order but must be rightside up compared to these.
1. Solve for K using X joules, Y joules/g·K, and Z grams.
?K = 1 g·K • 1 • X joules = X
Y joules
Zg
YZ K 2. Solve for moles of liquid H2O using X degrees, 4.184 joules/g·K, and Z joules.
? mol H2O = 1 mol H2O • 1 g·K
•
1
• Z joules =
Z
mol H2O
18.0 g H2O 4.184 joule X degrees
4.184 • 18.0 • X
When a heat problem includes g and mol of a substance formula, you will likely need molar mass.
Since moles was WANTED, and one of the units was complex, a given was picked with moles on top.
BUT – if the units cancel, ANY order may be used for the conversions.
3. Solve for joules using X degrees, Y calories/g·K, and Z grams.
? joules = Y calories
g·K • 4.184 joules • Z grams • X degrees = (4.184)YZX joules
1 calorie Since a joules term was not supplied in the data, the energy term calories was placed on top to start,
where the energy term needs to be in the answer, and then converted to joules, but any order is OK.
4. Solve for calories using X degrees, Y joules/g·K, and Z kilograms.
? calories = Y joules • 1 calorie • Z kg • 103 g • X degrees = (1,000)YZX calories
g·K
4.184 joule
1 kg
4.184 ©2011 ChemReview.net v. e4 Page 589 Module 21 — Energy Calculations Practice B
1. WANTED: ?J Strategy: Since specific heat capacity is not mentioned, the equation using c cannot be used. Instead,
use unit cancellation or conversionfactor rules. List the data, then arrange the units to give
the units WANTED.
DATA: 9.7 kcal/mol H2O ( or 9.7 kcal = 1 mol H2O) 0.75 g H2O
18.0 g H2O = 1 mol H2O (grams and moles of H2O are both in the data) SOLVE: Since a heat unit is wanted, you can start with a heat unit on top:
• 1 mol H2O • 103 cal • 4.184 J • 0.75 g H2O = 1,690 joules
? joules = 9.7 kcal
mol H2O 18.0 g H2O
1 kcal
1 cal
Since these units are all “simple” y/x ratios, you could also start with the single unit 0.75 g H2O using the
conversion factor rules to solve for ratios. The answer will be the same no matter what rules you choose
to help to arrange your conversions.
2. WANTED:
(Strategy:
DATA: ? ∆t
This problem mentions heat capacity, but not specific heat capacity, so we cannot use
the c prompt equation. Since the equation is not available, try unit cancellation. )
6.00 kJ/K
4,260 calories SOLVE: ? °C = 1K
• 1 kJ • 4.184 J • 42,600 cal. = 29.7 K = ∆t
6.00 kJ
1 cal.
103 J (The order of the conversions can be different, as long as the conversions are all sameside up and
you get the same answer. Note that capital K is the abbreviation for kelvins, and a lower case k is the
abbreviation for kilo.)
3. WANTED:
(Strategy:
DATA: ? joules (joules is the SI energy unit, and heat is a form of energy) Though the cprompt can be used, in this problem try unitcancellation rules.)
2.5 moles H2O
18.0 g H2O = 1 mol H2O (the chemical formula for ice is H2O)
(the problem includes moles and grams of H2O ) ∆t = 40.0°C ― 0.0°C = 40.0°C
2.09 J/g•K
SOLVE: ? J = 2.09 J • 18.0 g H2O • 2.5 mol H2O • 40.0°C = 3,800 J
g•K
1 mol H2O This problem can be solved with the equation q=cm∆t. Equations are usually safer to use if you
know the equation, since unit cancellation does not work for relationships that include proportionality
constants. However, both the cprompt equation and unit cancellation can be used to solve specific
heat capacity problems.
***** ©2011 ChemReview.net v. e4 Page 590 Module 21 — Energy Calculations Lesson 21E: Calorimetry
The heat involved in a reaction or a phase change is often measured by having that heat
absorbed by another substance, such as by water in a calorimeter (an insulated container),
or by the material in the body of a bomb calorimeter.
In calorimetry problems, there are two entities: one that loses energy, and the other that
gains the lost energy. The key relationship is the fundamental law of conservation of
energy: in a chemical process, energy can neither be created nor destroyed.
To simplify calorimetry, we will use these rules.
1. When you see that a problem has two components: one losing and one gaining heat,
write at the top of your DATA:
In calorimetry: Energy lost by one = Energy gained by the other.
2. Divide the rest of your DATA table into two columns or parts: E gainer and E loser.
The two parts will have the same value for the energy (joules or calories) lost and gained.
The strategy to solve will be to calculate the energy for the part that does not contain
the WANTED unit, then to write that energy in the part that includes the WANTED
unit, and then to solve for the WANTED unit.
The first task in a calorimetry problem is to identify and label which component is the heat
loser and which is the heat gainer. Some rules for identifying heat losers and gainers:
• If a substance burns (combustion) or explodes, it is losing its stored heat energy. • If a substance gets colder, it is losing its kinetic energy. • If a substance gets hotter without participating in a chemical reaction, it is gaining
heat. Let’s try the method on a problem.
Q. A 18.0 gram food sample is placed in a bomb calorimeter with excess oxygen and
burned. The calorimeter temperature rises by 9.3°C. The heat capacity of the
calorimeter is 3.50 kJ/K . What is the energy content of the food, in J/g?
Do these steps in your notebook.
1. This is a calorimetry problem because it has two energy parts. The food loses stored
energy. The calorimeter gains that same amount of energy as kinetic energy: its
temperature increases. Write the fundamental calorimetry relationship in your DATA.
2. Below the fundamental relationship, set up your data table in two columns: E loser and
E gainer. Include the key relationship: E lost = E gained
E loser = food E lost = E gainer = calorimeter
= E gained = 3. Enter every number and unit in the problem into the data table. Carefully separate the
E loser from the E gainer data.
Do those steps, and then check your answer below.
***** ©2011 ChemReview.net v. e4 Page 591 Module 21 — Energy Calculations 4. Your paper should look like this.
In calorimetry: Energy lost by one = Energy gained by other.
E Loser = food E Gainer = calorimeter E lost = ? = E gained = ? 18.0 grams food burned 9.3°C = ∆t for calorimeter WANTED = ? = joules/gram food 3.50 kJ/K for calorimeter Two rules are important in a twocolumn DATA table.
• As with any problem with two entities, take care not to use DATA that applies
to one part to solve for the other part.
The exception is the linked variable, which has the same value in both columns
and/or parts. • Almost always, the column with the WANTED unit will have two unknown
quantities, and the other column will have one: the linked variable (q). You will
need to solve for the column with one ? first.
In this problem, you do not know enough data to find joules lost by the food.
You do know enough to find the joules gained by the calorimeter. Do that
calculation first, and then check your answer below. *****
6. With conversions in any order,
? joules gained by calorimeter = 9.3°C • 3.50 kJ • 103 J = 32,500 joules
K
1 kJ
Enter this answer in both columns of your DATA table. The energy gained by the
calorimeter must equal the energy lost from the 18.0 grams of food as it burned.
7. Solve for the WANTED unit, then check below.
*****
8. SOLVE: WANTED = ? J
=
g food 32,500 J
18.0 g food = 1,800 J
g food ( 2 sf in Δt ) ***** Summary: Calorimetry
When a problem has a heat or energy loser and a gainer, the steps are
1. Identify which substance is the energy loser and which is the energy gainer.
2. Make two separate data columns or parts: one for the loser, one for the gainer.
3. Identify which part (loser or gainer) has the unit you WANT.
4. Use equations or units to calculate the energy (E or q) for the other part. Use the data
in the column for that part.
5. Write that energy for the other part as data for the WANTED part, and solve.
***** ©2011 ChemReview.net v. e4 Page 592 Module 21 — Energy Calculations Practice: Use the method above. If you get stuck, peek at the answer and try again. 1. Water in an insulated container is used to trap the heat released when a flask of liquid
sodium thiosulfate pentahydrate (Na2S2O3 • 5 H2O) solidifies.
a. If there are 25.0 grams of H2O in the container, and the H2O temperature rises from
10.0°C to 37.6°C as the phase change occurs, how much energy is released by the
phase change?
b. If 2.0 moles of Na2S2O3• 5 H2O is used in the experiment, what is the heat released
by the solidification, in joules per mole?
2. A 40.0 gram metal sample is heated to 100.0°C in boiling water and then immersed in
an insulated calorimeter containing 25.0 grams of water at 26.6°C. After stirring, the
metal and water are both at 30.0°C. What is the specific heat capacity of the metal?
3. Large cubes of ice are dropped into an insulated calorimeter with 100. mL of warm
water at 49.6°C. The mixture is stirred until the temperature is 0.0°C, and the ice
remaining is then removed. The volume of the water is found to be 162 mL.
a. How many grams of ice melted?
b. What is the heat of melting of the ice, in J/g?
4. The burning of methanol (CH3OH ) releases 638 kJ/mol. How much heat in kJ is
released when 12.8 grams of methanol are burned?
5. Methane (CH4) is the primary constituent of natural gas. Its heat of combustion is 890
kJ/mol. How many grams of methane would need to be burned to raise the
temperature of 55 gallons of water from 10.°C to 37°C? (1 gallon = 3.78 liters) ANSWERS
1. This problem involves two substances and heat. Use the calorimetry rules.
•
•
• Write the fundamental calorimetry relationship.
Identify the heat loser and gainer; keep two separate data tables.
Energy lost = energy gained; find the energy from the other, use that to find the WANTED unit. The liquid Na2S2O3 • 5 H2O loses potential energy as it becomes solid, and the water gains that energy.
This problem has already been divided: part (a) for the energy gainer, and (b) for the energy loser.
COMMON DATA:
Part a. WANT: In calorimetry: Energy lost by one = Energy gained by the other.
? Energy gained by water. (Solve in joules unless other units are requested.
Grams of H2O in a heat problem calls the “water c” prompt:
When you see energy and liquid water in a problem, write):
q = c · m · ∆t ©2011 ChemReview.net v. e4 (and set up a data table to match this equation) Page 593 Module 21 — Energy Calculations DATA: q=? J
c = 4.184 J/g·K
m = 25.0 g H2O SOLVE: Δt = 37.6°C ― 10.0°C = 27.6°C up
? = q = c · m · Δt = 4.184 J • 25.0 g • 27.6°C = 2,887 J gained by water
g•K Carry an extra sf until the final part. If needed, adjust your work, then complete part b.
*****
Part b. WANT: ? J
mol Na2S2O3 • 5 H2O Strategy: Use energy from first part. E lost = E gained.
DATA: 2,887 J lost in becoming solid
2.0 mol of Na2S2O3 • 5 H2O SOLVE: ? J
J
= 2,887 J
= 1,400
mol Na2S2O3• 5 H2O
2.0 mol Na2S2O3• 5 H2O
mol Na2S2O3• 5 H2O 2. The problem involves two substances and heat, so use the calorimetry rules:
• Write the fundamental calorimetry relationship. • Identify the heat loser and gainer, keep two separate data tables; • E lost = E gained. Find the energy from the not WANTED part. Use that to find the WANTED unit. In this problem, the metal loses energy as it cools, and the water gains that energy.
COMMON DATA:
WANTED:
Strategy: In calorimetry: Energy lost by one = Energy gained by the other.
specific heat capacity (c) of the metal
Solve for energy gained by the other part, the water, first. For energy gainer – water
WANTED: ? Energy gained by water. Strategy: When you see water in a heat problem, use the “c water” prompt. Write:
q = c · m · ∆t DATA: Do a data table with those symbols, add the water c value. q=?=J
c = 4.184 J/g·K for liquid water m = 25.0 g H2O
Δt = 30.0°C ― 26.6°C = 3.4°C up
SOLVE: ? = q = c · m · Δt = 4.184 J
g ·K (2 sf ) • 25.0 g • 3.4°C = 356 J gained by water For energy loser – metal
Strategy: Since you want specific heat capacity (c), you can use unit cancellation or use the
equation using c: ©2011 ChemReview.net v. e4 Page 594 Module 21 — Energy Calculations q = c · m · ∆t
DATA: and set up a data table to match the equation q = 356 J lost by metal (from other part. That’s the key relationship.) cmetal = ? = WANTED
m = 40.0 g metal
Δt = 100.0°C ― 30.0°C = 70.0°C lower
SOLVE: Since q = c · m · ∆t
?=c= q
=
m · Δt 356 J
J
= 0.13
40.0 g • 70.0°C
g• degree Note that the answer units are the units expected for a c value.
3a. WANT: ? g ice melted.
DATA: The amount of ice that melted is 162 mL ─ 100 mL = 62 mL ice melted
For liquid water, 1.00 g = 1 mL, so 62 g of the water was made by the ice melting
= 62 g ice melted. 3b. In this part, you have a heat loser (the warm water) and a heat gainer (the ice that melted). That makes
this a calorimetry calculation. A key is to label and separate the data for the warm water and the ice
In calorimetry: Energy lost by one = Energy gained by other.
WANTED:
Strategy: specific heat capacity (c) of the ice
Solve for energy lost by the not WANTED part, the water, first. For energy gainer – water
WANTED: ? Energy lost by water. Strategy: When you see water in a heat problem, use the “c water” prompt. Write:
q = c · m · ∆t DATA: Do a data table with those symbols, add the water c value. q=?=J
c = 4.184 J/g·K for liquid water m = 100. mL H2O = 100. g H2O
Δt = 49.6°C ― 0.0°C = 49.6°C (subtraction = doubtful place = 3 sf ) SOLVE: ? = q = c · m · Δt = 4.184 J • 100. g • 49.6°C = 20,760 J lost by water
g ·K
For energy gainer – ice
(A complete label is especially helpful in this problem) WANTED: J
g ice melted DATA: 20,760 J lost by water = 20,760 J gained by ice (from other part: the key relationship.)
g ice melted = 62 g ice melted (from Part a) WANTED: J
g ice melted ©2011 ChemReview.net v. e4 = 20,760 J gained by ice = 330
J
62 g ice melted
g ice melted Page 595 Module 21 — Energy Calculations 4. WANT: heat (q) in kJ
DATA: 638 kJ released = 1 mol CH3OH burned
12.8 g CH3OH
32.0 g CH3OH = 1 mol CH3OH (grams prompt) When the data is mostly equalities, with one singleunit WANTED and given, how might you solve?
*****
SOLVE: Try conversions.
? = q in kJ = 12.8 g CH3OH • 1 mol CH3OH
32.0 g CH3OH • 638 kJ released
1 mol CH3OH burned = 255 kJ heat released 5. The problem involves two substances and heat, so use the calorimetry rules.
• Write the fundamental calorimetry relationship. • Identify the heat loser and gainer; keep two separate data tables. • E lost = E gained, find the energy from the other, use that to find the WANTED unit. In this problem, the methane loses stored energy as it burns, the water gains that energy.
COMMON DATA: In calorimetry: Energy lost by one = Energy gained by the other. a. Final WANTED: grams methane.
Strategy: Solve for energy gained by water first. Energy gainer – water:
WANT: ? joules gained by water.
Grams of water in a heat problem calls the cwater prompt:. Write:
q = c · m · ∆t
DATA: and set up a data table to match the equation q = ? = joules
c = 4.184 J/g·K for liquid water
m = ? g H2O = 55 gal H2O • 3.78 L • 1 mL • 1.00 g water = 2.08 x 105 g H2O
1 gal 10―3 L 1 mL water
Δt = 37°C ― 10.°C = 27°C SOLVE: ? = q = c · m · Δt = 4.184 J • 2.08 x 105 g • 27°C = 2.35 x 107 J
g· K If needed, adjust your work to this point, and then finish the problem.
*****
Energy loser – methane:
WANTED: ? g CH4 burned ©2011 ChemReview.net v. e4 Page 596 Module 21 — Energy Calculations DATA: 890 kJ = 1 mol CH4 burned. ( you may write x/y units as ratios or equalities) 2.35 x 107 joules (use E for other part to solve for WANTED) 16.0 g CH4 = 1 mol CH4
Strategy: (since units for CH4 use g and moles, need molar mass) If you don’t see specific heat capacity, which solves with an equation, try using units.
Since all of these units are x/y simple ratios, as opposed to x/y•z complex ratios, you
might use the rule “If you want a single unit, start with a single unit.” The squiggles
above work backwards from the answer unit through the conversions to the given
single unit. But any order of multiplying is OK. SOLVE: ? g CH4 = 2.35 x 107 J • 1 kJ • 1 mol CH4 • 16.0 g CH4 = 420 g CH4
890 kJ
1 mol CH4
103 J
The factor which limits the sf in the answer is the Δt value, which only had 2 sf.
***** Summary: Energy Calculations
A summary for phases and phase changes is listed at the end of Lesson 21A.
1. The Law of Conservation of Energy: energy can neither be created nor destroyed by
chemical processes.
2. In chemistry, energy is usually measured in joules or calories.
a. Joules are the SI unit measuring energy. Joules are defined in terms of “work.”
b. Calories are a metric unit also used in chemistry to measure energy.
A chemical calorie is defined as the amount of heat needed to raise the temperature
of one gram of liquid water by one degree (Celsius or Kelvin).
3. Because all forms of energy are equivalent, all energy units can be related by equalities.
1 calorie = 4.184 joules
4. 1 food Calorie = 1,000 chemical calories = 1 kilocalorie (kcal) = 4.184 kJ
Food Calories are abbreviated with a capital Cal., chemical calories with a cal. 5. The specific heat capacity (symbol small c) of a substance is defined as the amount of heat
required to raise one gram of the substance by one degree (Celsius or Kelvin).
The units of c are joules per (gram · degree) or calories per (gram · degree).
6. ∆ is a symbol meaning change in.
∆t ≡ tfinal — tinitial ∆t means change in temperature. A ∆t value should be labeled as positive or negative. A change in temperature is the same number of degrees whether measured in °C or K.
For ∆t measurements, the terms degree and °C and K are all equivalent.
When terms are equivalent, they can cancel. ©2011 ChemReview.net v. e4 Page 597 Module 21 — Energy Calculations 7. The c Prompt: If you see the term specific heat capacity or its symbol c in a problem, write
at the top of your data the equation which includes c :
q = c · m · ∆t
Heat energy (q) is defined as positive when energy is added to a chemical system and
negative when energy leaves.
8. IF you need an equation to solve a problem,
a. write the fundamental, memorized equation in your data.
b. Write a data table with each of the symbols in the equation.
c. Use units to match the data with the symbols in the data table.
d. Memorize equations in one format; use algebra to solve for symbols WANTED.
e. Solve the equation for the WANTED symbol before you plug in numbers.
f. Put both numbers and units into equations. Do the math for numbers and units. 9. The c water prompt: If a problem mentions energy or heat or joules or calories  and
liquid water  write the c prompt equation: q = c · m · ∆t , a data table listing those
symbols, and enter c = cwater = 4.184 J/g·K (or c = 1 calorie/g·K).
10. Units must be consistent in order to cancel. When units are not consistent, or do not
match what is needed in an equation,
•
• Write the chosen unit after each symbol in the DATA. •
11. pick appropriate units to convert DATA to (preferably those used in a complex
unit in the problem),
Convert the supplied units to the chosen units in the DATA table, then solve. Unit cancellation can be used in place of memorized equations when equation
relationships do not include constants. Most equations that calculate heat energy (q)
in chemistry can be solved with unit cancellation.
a. List WANTED and DATA as done for conversion problems, without labeling
measurements with symbols.
b. Arrange numbers and units so that the units cancel to give the WANTED units. 12. Calorimetry: When a problem has a heat or energy loser and a gainer, the steps are,
a. identify which substance is the energy loser, and which is the energy gainer.
b. Make two separate data columns or parts: one for the loser, one for the gainer.
c. Identify which part (loser or gainer) has the unit WANTED.
d. Use equations or units to calculate the heat energy (q) in the column or part that
does not have the final WANTED unit.
e. Write that heat energy value as data in the column or part that includes the final
WANTED unit, then solve for the WANTED unit.
##### ©2011 ChemReview.net v. e4 Page 598 Module 22 — Heats of Reaction (ΔH) Module 22 — Heats of Reaction (ΔH)
Prerequisites: You should be able to do most of the calculations in this module if you have
completed Lesson 10B (Balancing Equations).
***** Lesson 22A: Energy, Heat, and PV Work
Timing: Some courses address PV work at this point, before ∆H calculations; others include
PV work as part of a thermodynamics unit later in the course. Do this lesson at this time
when you are assigned calculations that involve PV work in your course. This topic is not
essential for the ∆H calculations in the remainder of this module.
***** Thermodynamics
Thermodynamics studies relationships involving heat and work. The rules for
thermdynamics help us to predict why reactions go in the direction they do, as well as the
changes in energy and in the arrangements of particles that occur during reactions and
processes. In Modules 21 and 22, our interest is primarily measuring the changes in heat
energy that occur during chemical processes. Other aspects of thermodynamics are
addressed in Module 36. System Versus Surroundings
To study thermodynamics, it is helpful to divide the universe into two parts. In chemistry,
• The system is the particles of interest, which may be atoms, molecules, or ions. • The surroundings is the environment outside of the system. Universe = system + surroundings
To explain chemical changes, the system and the surroundings are accounted for
separately.
For example, in a chemical reaction, if a collection of molecules lose energy, the same
amount of energy must be gained by the surrounding environment (as heat or work). The First Law: Conservation of Energy
Energy can be described as the capacity to do work. The first law of thermodynamics is
the law of conservation of energy:
In physical and chemical processes, energy can neither be created nor destroyed.
The exception to this law is nuclear reactions, in which energy can be created from, or
converted to, mass. Nuclear processes are the major reactions in stars, but some nuclear
reactions occur on earth, such as in nuclear reactors and radioactive decay. However,
unless nuclear processes are specified, you should assume for problems in chemistry that
the first law is obeyed. ©2011 ChemReview.net v. e4 Page 599 Module 22 — Heats of Reaction (ΔH) The first law means that in chemical or physical processes, there is no change in the total
energy of the system and its surroundings. In equation form, for a chemical change,
ΔEuniverse = Euniverse final ─ Euniverse initial = 0 (Equation 1) Internal Energy, Heat, and Work
During physical and chemical changes, though the total energy before, during, and after the
process cannot change, energy can change its form, such as from potential to kinetic. In
addition, energy can flow into and out of a system.
• The internal energy of a system (Einternal or Esystem) is the total of the potential
and kinetic energy of its particles.
Esystem = PE + KE • Energy can flow into and out of a system as heat (q) or work (w). Heat is defined as a transfer of thermal energy between two systems that are at different
temperatures. As defined in physics, there are many types of work. One type of work is
the energy that is transferred as a force acts over a distance ( w = F· d ).
The change in the internal energy of a system during a process can be calculated by
∆Einternal or ∆Esystem = q + w (Equation 2) When heat (q) is added to a system, or work (w) is done on a system (such as by
compressing a gas), the internal energy of the system increases. If a process or reaction
releases heat from the system to its surroundings, or if the expansion of gas particles under
pressure does work (such as creating a force that moves a resistant object), energy must
leave the system.
The change in energy (∆E) is a termed a state function. In state functions, only the initial
and final states matter. The steps between the initial and final states do not matter. In
equation form,
∆E = Efinal ─ Einitial (Equation 3) The values for q and w during a process do vary depending on how the change occurs
(though the total of q + w does not), so q and w are not state functions. By convention in
thermodynamics, symbols for state functions, such as ∆E, are written in upper case, and
nonstate functions such as heat (q) and work (w) are assigned lowercase symbols. Units and Signs
By the law of dimensional homogeneity (Lesson 11G), terms that are added or subtracted
on one side of an equation must have the same units, and the units must be the same on
both sides of an equation. When using SI units, energy is measured in joules. For the units
of ∆E, q, and w,
• in equation 3, since a change in energy is the difference between two values in
joules, the SI unit for ∆E must be joules. ©2011 ChemReview.net v. e4 Page 600 Module 22 — Heats of Reaction (ΔH) • In equation 2, the SI units for heat and work must be the same as ∆E: joules. Heat
and work measure transfers of energy. The variables ∆E, q, and w must be assigned signs: their values can be positive or negative.
In chemistry (but not necessarily in other sciences), we define the signs from the
perspective of the system: the chemical particles in a process. In chemistry calculations,
• if energy flows out of a system into its surroundings, ∆E is given a negative sign,
because the system loses internal energy.
o o • If the energy flowing out of a system of particles produces heat, such as
when a substance burns, the q value is given a negative sign because the
system has lost heat energy. Such a process is termed exothermic.
If a chemical process does work, such as when burning produces gases that
expand and move a piston that is under pressure, the w value for that
process is given a negative sign because the system is losing internal energy
as it converts that energy to work. If energy flows into the system, ∆E must be assigned a positive sign, because the
system gains energy.
o If heat is added to a system, such as by boiling water molecules, the q value
for the process is given a positive sign because the system gains internal
energy, and the process is termed endothermic. o If work is done on a chemical system, such as when you compress the air in
a bicycle tire with a hand pump, the w value for the process is given a
positive sign. As the air in the tire is compressed, the molecules gain
internal energy (the capacity to do work) from the work put into the system. Practice A: Chart and learn the rules above, then answer these questions from memory. 1. From the perspective of the particles in a chemical system,
a. if heat is added to the particles, is the sign of q positive or negative ?
b. If a gas expands by pushing a piston that is under pressure, is the sign of
w positive or negative ?
2. If a substance burns, but no work is done,
a. is the sign of q positive or negative ?
b. Is the sign of ∆E positive or negative ?
c. Is the reaction exothermic or endothermic?
3. If a gas is compressed by a piston, but no heat is added to the gas,
a. is the sign of w positive or negative ?
b. is the sign of q positive or negative ?
c. Is the sign of ∆E is positive or negative ?
4. If 30.0 joules of heat is released by burning, and the gases produced do 25.0 joules of
work in moving an automotive piston, what is the ∆E for the process? ©2011 ChemReview.net v. e4 Page 601 Module 22 — Heats of Reaction (ΔH) PV Work
In physics, we investigate many types of work. One type is the mechanical work done in
accelerating a mass over a distance:
Work = force times distance = f • d = m • a • d = mass times acceleration times distance.
In chemistry, we are most interested in the work done during a chemical reaction or
process. This type of work include electrical work (work done by moving charges) and
PV work (the work involved in the expansion or compression of a gas).
For the special case of the expansion or compression of a gas, the PV work involved can be
calculated by
Work content of system = ─ (external pressure) times (change in the volume of the gas)
or, in symbols, wsystem = ─ Pexternal ∆Vsystem (Equation 4) When a gas expands against external pressure, ∆V is positive. Since pressure is always
positive, based on equation 2, the value for the work must be negative. This makes sense:
when a gas expands against pressure, it does work on its surroundings, and for energy to
be conserved, the system must lose internal energy.
In calculations, to converting between the SI units of work (joules) and the units for PV
work, you will need this equality:
1 liter • atmosphere = 101 joules (Equation 5) Enthalpy
Enthalpy (H) is a property defined as Hsystem = Esystem + PsystemVsystem where Esystem is the internal energy of the system, and Psystem is the pressure exerted
by the system.
By convention, when thermodynamic equations are written in chemistry (by not in all other
sciences), if a symbol is not assigned a subscript, the subscript is assumed to be system (as
opposed to surroundings). The equation above is then written as
H = E + PV (Equation 6) In chemical processes, we are most interested in the change in enthalpy (∆H):
Hfinal ─ Hinitial = ΔH = ΔE + Δ(PV) (Equation 7) Let’s explore what ∆H means. For the special case in which the external pressure on a
system is held constant, equation 7 becomes
ΔH = ΔE + PΔV at constant pressure When we substitute our definition for “PV work,”
∆E = q + w
∆E = q ─ P ∆V (Equation 8)
w = ─ P ∆V into equation 2: , equation 2 becomes
. Substituting this result into equation 8, ∆H becomes ©2011 ChemReview.net v. e4 Page 602 Module 22 — Heats of Reaction (ΔH) Δ H = Δ E + P Δ V = ( q ─ P ∆V ) + P Δ V
ΔH = q =q or at constant pressure . (Equation 9) This last equation means:
If pressure on a system is held constant, and any work is limited to “PV work,” change
in enthalpy (∆H) represents the heat flow to or from a system: q .
The ∆H of a reaction is termed the heat of reaction. The ∆H for a phase change may be
described with terms such as heat of melting, heat of fusion or heat of vaporization. ΔH, ΔE, and ΔPE
For the special case of a chemical process in which pressure is constant and the change in
volume is small:
ΔH = ΔE + PΔV ≈ ΔE + P(~0) ΔH ≈ Δ E ≈ q ≈ ΔE + ~0 ≈ q or if pressure is constant and the volume change is small Reactions in which the change in volume is very small include
• Reactions that do not involve gases, and • Gas reactions in which the moles of gas do not change. Using the calorimetry calculations in Module 21, by calculating the heat transfer (q) during
a process at constant volume or pressure, we can calculate the ΔH value for the process.
Even for reactions in which the volume does change substantially, in most cases the size of
the PVwork term is small compared to the heat term (q), so that
ΔE = q + small w = ΔH ─ PΔV ≈ ΔH ─ small ≈ q
ΔH ≈ ΔE ≈ q or for most reactions (Equation 10) This equation means that in most (though not all) processes, ∆H measures the approximate
change in internal energy of a system.
Finally, the internal energy of a system (E) is the combination of its kinetic and potential
energy. The temperature of a system is a measure of its average kinetic energy: if
temperature does not change, ΔEkinetic = 0 . For reactions run at constant temperature and
pressure with a small work term,
ΔH = ΔE + PΔV = (ΔEpotential +ΔEkinetic) + PΔV = (ΔEpotential + ~0 ) + ~0 ≈ q or
ΔH ≈ ΔEpotential ≈ q for most reactions at constant P and T (Equation 11) For example, during a solidliquid phase change for a substance under constant
pressure, the volume change is very small, and the temperature is constant. The
change in enthalpy, which is equal to the flow of heat into or out of the system, is
approximately equal to the change in the potential energy of the system. ©2011 ChemReview.net v. e4 Page 603 Module 22 — Heats of Reaction (ΔH) Summary: Energy and Enthalpy
1. In studying the energy changes during physical or chemical processes, it is helpful to
divide the universe into the system (the chemical particles) and its surroundings.
Universe = system + surroundings
2. Energy can be described as the capacity to do work.
3. The first law of thermodynamics is the law of conservation of energy: in chemical (nonnuclear) processes, energy can neither be created nor destroyed.
In equation form: ΔEuniverse = Etotal final ─ Etotal initial = 0 4. The internal energy of a system (Einternal or Esystem) is the total of the potential and
kinetic energy of its particles.
Esystem = PE + KE
5. In chemical processes, energy can change form, such as from kinetic to potential energy,
and energy can flow into and out of a system as heat (q) or work (w).
In chemical processes: ∆Esystem = q + w 6. Energy (E), heat (q), and work (w) in the SI system are measured in joules (J).
7. Measurements of heat, work, and changes in energy and enthalpy are assigned signs
from the perspective of the system.
• If heat is added to the system, q is given a positive sign. If a reaction or process
releases heat from the system to the surroundings, q is given a negative sign. • If work is done on a system, such as in compressing a gas, w is positive. If a system
does work, such as an expanding gas moving a piston that is under pressure, the
system loses energy and w is negative. 8. In the case of PV work by a gas, wsystem = ─ Pexternal ∆Vsystem 9. In chemistry, if no subscript is given after a symbol in a thermodynamic equation,
assume the subscript is system.
10. In converting from PV work units to energy units: 1 liter • atm = 101 joules
11. The definition of enthalpy (H) is
and Hfinal ─ Hinitial = H = E + PV ΔH = ΔE + Δ(PV) if work is limited to PV work. 12. For reactions in which the external pressure on a system is held constant and work is
limited to “PV work,” ∆H will measure the flow of heat energy into or out of the system.
ΔH = q .
13. If pressure is constant and the change in volume is small, and/or if the work term is
much smaller than q (true for most reactions): ΔH ≈ ΔE ≈ q ©2011 ChemReview.net v. e4 Page 604 Module 22 — Heats of Reaction (ΔH) Practice B: Learn the Summary points 5 through 12 above, then do these problems. 1. If a sample of gas is compressed from 4.5 to 2.5 liters by 606 joules of work done on the
system, what is constant external pressure on the gas, in atmospheres?
2. When hydrogen gas (H2) is burned under constant pressure, ΔH = ─284 kJ/mol . How
much heat in kJ is released when 0.12 g of H2 burns? ANSWERS
Practice A
1 a. If heat is added to the particles, the sign of q is positive. The particles gained energy.
b. If a gas expands by pushing a piston that is under pressure, the sign of w is negative . The system
has done work on its environment, and in the process the system loses internal energy. In the
equation w = ─ Pexternal ∆Vsystem , ∆V is positive, and P must be positive, so w is negative. 2. a. q is negative . In burning, the surroundings gain heat. The system must lose heat.
b. ∆E = q + w . If q is negative and w is 0, ∆E must be negative. If the system loses heat with no work
involved, the system loses energy, and ∆E for the system must be negative.
c. If the surroundings gain heat, the reaction is exothermic.
3. a. w is positive . Work is done on the system. Energy is added to the system.
b. q is zero: neither positive nor negative. c. ∆E = q + w . If w is positive and q is 0, ∆E must be positive. 4. WANTED:
DATA: ∆E
When heat is released to the surroundings, q is negative. ─25.0 J = w
SOLVE: ─30.0 J = q When work is done on the surroundings, w is negative. ∆E = q + w = ─ 30.0 J + (─ 25.0 J ) = ─55.0 J In both releasing heat and doing work, the system of particles loses internal energy.
Practice B
1. WANTED: external P in atm. The equation that relates external pressure, work, and volume change is w = ─ Pexternal ∆V
In problems that use the PVwork formula, you will likely need the conversion 1 liter • atm = 101 joules
*****
DATA:
w = + 606 J
When work is done on the system, w is positive.
∆V = final ─ initial = 2.5 L ─ 4.5 L = ─ 2.0 L
SOLVE: The system loses volume. w = ─ Pexternal ∆V . Pexternal in atm. = ©2011 ChemReview.net v. e4 ─ w = ─ 606 J
∆V
─ 2.0 L = 303 J • 1 liter • atm = 3.0 atm
L
101 joules Page 605 Module 22 — Heats of Reaction (ΔH) 2. Tip: You will need an equation plus conversions.
*****
WANTED: q in kJ The equation that relates q and ∆H is ΔH = q at constant pressure. For conversions, list the DATA as equalities and single units.
DATA: ─284 kJ = 1 mol H2 (2 equivalent amounts) 2.016 g H2 = 1 mol H2 (grams prompt)
(the singleunit given). 0.12 g H2
SOLVE: ? kJ = 0.12 g H2 • 1 mol H2 • ─ 284 kJ
2.016 g H2
1 mol H2 = ─ 17 kJ = q ***** Lesson 22B: Exo and Endothermic Reactions
Timing: Begin this lesson when you are assigned ∆H calculations.
***** The Energy Stored in Substances
Energy is stored in a substance during its formation. The amount of energy stored in a
substance is dependent on (1) its atoms and bonds, and how they are arranged, and (2) the
physical state of the substance: whether it is a gas, liquid, solid, or in solution.
Energy can be stored in the bonds of a substance. Some types of bonds have relatively
large amounts of stored energy compared to others. For example, the C─H bond stores
more energy than an O─H or a C─O bond.
Petroleum products have considerable commercial value in large part because they contain
highenergy C─H bonds that release their energy when burned to form O─H and C─O
bonds. That energy can be used to supply heat or to do work. Enthalpy and ∆H Terminology
1. In general terms, the property enthalpy (H) is often described as the “heat content” of a
substance: a measure of the energy that can be extracted from a substance.
The absolute enthalpy (H) of a system is not possible to measure, but the absolute
enthalpy (H) of a particle can be assigned a value relative to an arbitrary zero enthalpy.
In enthalpy measurements, an arbitrary zero enthalpy is assigned to elements in their
standard state, and the absolute enthalpy of other particles compared to those zero
values by measuring the ∆H for chemical reactions and processes. ©2011 ChemReview.net v. e4 Page 606 Module 22 — Heats of Reaction (ΔH) 2. ∆H is the symbol for change in enthalpy during a reaction or process. ∆H is defined as:
∆H = (enthalpy stored in products) minus (enthalpy stored in reactants)
∆H = Hfinal ─ Hinitial = Hproducts ─ Hreactants
During most chemical processes, the change in enthalpy (∆H) of a system is a measure
of the heat flow (q) into or out of the system.
3. The SI units of ΔH are joules. In equations that include energy terms, the coefficients are
understood to be moles.
If a chemical equation has one mole of one substance as a product, the units of ΔH can be
written either as joules (J) or as joules/mol (J/mol). In energy equations, the energy
term will often be expressed in kilojoules (kJ).
4. Energy and enthalpy changes in chemical processes can be classified as either
exothermic or endothermic.
a. Exothermic reactions are those in which heat is released to the environment (exo is a
prefix from ancient Greek meaning “out of”). Heat flows out of the chemical
particles in exothermic reactions.
b. Endothermic reactions are those in which heat is required. Heat must flow endo, “to
the inside,” into the chemical particles. ΔH Notation
The energy change in a reaction can be expressed in two ways: either as a reactant or
product inside the reaction equation, or written in ΔH notation after the equation.
Exothermic Reactions
The explosive burning of hydrogen gas is exothermic. The reaction can be written as
H2(gas) + 1/2 O2(g) H2O(g) + 242 kilojoules (Equation 1) This equation indicates that heat is a product of the reaction: energy is released to the
environment. When energy is a term in a reaction equation, the coefficients are understood
to be in moles. An omitted coefficient is assumed to represent one mole.
Though the energy term can be written inside the equation as in equation (1), more often
the energy factor is represented using ΔH notation. The above reaction is written as
H2(gas) + 1/2 O2(g) H2O(g) ΔH = ― 242 kJ (Equation 2) When equations include ΔH notation, the units of ∆H are always energy units (such as
joules or calories), and the coefficients of the reaction equation are read in moles.
In the above reaction, ∆H has a negative sign. For exothermic reactions, ∆H is defined as
negative. The minus sign indicates the chemical particles have lost stored energy in the
reaction. Some of the energy stored in the reactant particles is released into the
environment as the products are formed. ©2011 ChemReview.net v. e4 Page 607 Module 22 — Heats of Reaction (ΔH) Endothermic Reactions
When the products of a reaction have a higher stored energy than the reactants, energy
must be added to the chemical particles for the reaction to take place. Cooking food,
melting ice, and boiling water are examples of chemical process that require energy to be
added. For these reactions, ∆H will have a positive sign, because the chemical particles gain
stored energy.
For example, the process of boiling water can be represented as either
1 mol H2O(liquid) + 44.0 kJ 1 mol H2O(gas) (Equation 3) OR as
H2O(l) H2O(g) ΔH = + 44.0 kJ (Equation 4) For endothermic reactions, the ΔH value is always positive.
This reaction is an example of why, when representing the energy involved in a chemical
process, the phase of each particle must be shown: (s), (1), (g), or (aq). The solid, liquid, gas,
and aqueous phases of a substance have different amounts of stored energy. Writing Energy Terms Inside Reactions
The explosion of hydrogen gas in reaction (1) above was written as
H2(gas) + 1/2 O2(g) H2O(g) + 242 kilojoules By the rules of algebra, we could also write the above reaction as
H2(g) + 1/2 O2(g) ― 242 kilojoules H2O(g) However, by the conventions of chemistry, we do not write negative signs in front of
energy terms that are written inside reaction equations. An energy term written inside a
reaction equation is always assigned a positive sign: written on the right side (as a product)
for exothermic reactions, and on the left (as something which needs to be added) for
endothermic reactions.
Energy terms are always positive inside equations, but may be positive or negative as ΔH. Summary
To solve problems involving energy equations, first commit the following rules to memory. Rules for Representing Energy Changes in Chemical Processes
When an energy term is included in a chemical equation:
1. The coefficients are in moles.
2. The phase of each particle must be shown: (s), (1), (g), or (aq).
3. In exothermic reactions, energy is released into the environment. The energy term
is shown
• EITHER with a positive sign on the products side; • OR (preferred) with a negative ΔH value written after the reaction. ©2011 ChemReview.net v. e4 Page 608 Module 22 — Heats of Reaction (ΔH) 4. In endothermic reactions, energy must be added as the reaction proceeds. The
energy term is shown
• EITHER with a positive sign on the reactants side of the equation, • OR with a positive ∆H value written after the reaction. 5. The SI units of ΔH are joules. In equations that include energy terms, the
coefficients are understood to be moles. If a chemical equation has one mole of one
substance as a product, the units of ΔH can be written either as joules (J) or as
joules/mol (J/mol).
After you learn the rules, complete the problems below to cement your knowledge. Practice A: Answers are at the end of this lesson. Check your answer after each part. 1. Rewrite these equations so that the heat term appears as a product or a reactant inside
the equation.
a. C(s) + O2(g) CO2(g) b. 1/2 N2 (g) + 1/2 O2(g) ∆H = ─ 393.5 kJ
NO(g) ∆H = + 90.3 kJ 2. Label each reaction in (1) above as either exothermic or endothermic.
3. Rewrite these equations using ∆H notation.
a. H2O2(l) + 187.6 kJ H2(g) + O2(g) b. CO(g) + 110.5 kJ C(s) + 1/2 O2(g) 4. After each reaction in problem (3) above, add a label identifying the side that has more
energy stored in its particles: the reactants or the products. Reversing Equations That Include Heat Terms
In theory, any chemical reaction can go backwards. In practice, many do.
Any reaction with a heat term inside will be true if written in the reverse direction.
Example: Since, from equation (3) above,
This will also be true: H2O(liquid) + 44.0 kJ
H2O(gas) H2O(gas) H2O(liquid) + 44.0 kJ Given a reaction in which a ∆H value represents an energy change, the reaction can be
written in the reverse direction if you change the sign of ∆H.
Example: Since, from equation 4 above,
This equation is also true: H2O(l) H2O(g) ΔH = + 44.0 kJ H2O(g) H2O(l) ΔH = ─ 44.0kJ If an amount of energy must be added to change reactants to products, this same amount
will be released when products are changed back to reactants. In chemical and physical
processes, energy must be conserved. ©2011 ChemReview.net v. e4 Page 609 Module 22 — Heats of Reaction (ΔH) Multiplying Equations That Include Heat Terms
Given a known balanced equation that includes an energy term, the coefficients representing
the moles of the reactants and products may all be multiplied or divided by the same factor,
provided you do the same to the energy term.
Examples
a. Since H2(gas)+ 1/2 O2(g)
Then H2O(g) + 242 kJ 2 H2(gas)+ 1 O2(g) 2 H2O(g) + 484 kJ b. Since H2(g) + 1/2 O2(g)
Then 2 H2(g) + O2(g) is true, H2O(g) (all doubled) is also true. ΔH = ― 242 kJ 2 H2O(g) is true, ΔH = ― 484 kJ (all doubled) is also true. Summary
Memorize these additional rules for energy terms in equations.
6. Reactions that include energy terms can be reversed (written backwards). If ∆H
notation is used, change the sign of ∆H.
7. All coefficients and energy terms in a balanced equation can be multiplied or
divided by a factor. If ∆H notation is used, do the same to the ∆H.
Once you have the above seven rules firmly in memory, try the problems below. Practice B: Do every other problem. Save the rest for your next practice session. 1. Write these reactions in the reverse direction. Express the energy term as a ΔH value.
UF6(g) a. UF6(l) b. C(s) + 2 H2(g) ΔH = + 30.1 kJ
CH4(g) ∆H = ― 74.9 kJ 2 Use these 4 “known” reactions to fill in the blanks below.
H2(g) + 1/2 O2(g) H2O(g) ΔH = ― 241.8 kJ (1) H2(g) + 1/2 O2(g) H2O(l) ΔH = ― 285.8 kJ (2) 1/2 N2 (g) + O2(g) NO2(g) ΔH = + 33.8 kJ (3) ΔH = ― 812 kJ (4) H2(g) + 1/8 S8(s) + 2 O2(g)
a. 2 NO2(g) N2(g) + 2 O2(g) b. 4 H2(g) + 2 O2(g)
c. 8 H2SO4(l) 4 H2O(l) H2SO4(l) ΔH = _______
ΔH = _______ 8 H2(g) + S8(s) + 16 O2(g) ©2011 ChemReview.net v. e4 ΔH = _______ Page 610 Module 22 — Heats of Reaction (ΔH) ANSWERS
Practice A
1&2a. C(s) + O2(g) CO2(g) + 393.5 kJ 1&2b. 1/2 N2 (g) + 1/2 O2(g) + 90.3 kJ
3&4a. H2O2(l) H2(g) + O2(g) 3&4b. C(s) + 1/2 O2(g) Exothermic NO(g) Endothermic ΔH = +187.6 kJ CO(g) (Products have more PE) ΔH = ― 110.5 kJ (Reactants have more PE) Practice B
1a. UF6(g)
2a. 2 NO2 (g) UF6(l) ΔH = ― 30.1 kJ
N2 (g) + 2 O2 (g) 1b. CH4(g) C(s) + 2 H2(g) ∆H = + 74.9 kJ ΔH = ― 67.6 kJ This reaction is table reaction # 3 doubled and written backwards: double and change sign of ΔH.
2b. 4 H2 (g) + 2 O2(g) 4 H2O(l) ΔH = ― 1,143.2 kJ This reaction is table reaction #2 quadrupled – be careful to distinguish the gas/liquid/solid states.
Significant figures: Multiplying by an exact number does not change the place with doubt.
8 H2 (g) + S8(s) + 16 O2 (g) ΔH = + 6,496 kJ
This is table reaction #4 multiplied by 8: multiply ΔH by 8. Since this reaction is also written backwards
from reaction #4; change the sign of ΔH. 2c. 8 H2SO4(l) ***** ©2011 ChemReview.net v. e4 Page 611 Module 22 — Heats of Reaction (ΔH) Lesson 22C: Adding ∆H Equations (Hess’s Law)
Timing: Do this lesson when you are assigned problems that require finding a ∆H by
adding equations with ∆H values attached.
***** Hess’s Law
Hess’s law of heat summation states that the enthalpy change in an overall chemical
process can be calculated by adding its individual steps.
This means that for reactions where the heat of reaction (∆H) is not known, ∆H can be
calculated by adding reactions for which ∆H is known.
Rules For Adding Equations With ∆H Terms
All reaction equations with energy terms can be added.
• Multiple equations can be added to produce one net equation. • Like particles on opposite sides of the arrows in the added reactions can
cancel. • Like particles on the same side of the arrows of added equations can add. • All particles to the left of the arrows add to become the reactants in the
final equation. • All particles to the right of the arrows add to become the products of the
final reaction. • The final ∆H will be the added ∆H values of the reactions being added. The meaning of these rules is best explained with an example. Apply the above rules to the
following problem, and then check the answer below.
Q. Find the heat of reaction for the burning of carbon monoxide
CO(g) + 1/2 O2(g) CO2(g) ∆H = ??? by adding these two reactions.
CO(g) C(s) + 1/2 O2(g) ∆H = + 110.5 kJ CO2(g) ∆H = ─ 393.5 kJ C(s) + O2(g) ____________________________________________ ***** ©2011 ChemReview.net v. e4 Page 612 Module 22 — Heats of Reaction (ΔH) Answer
When adding, first cancel like terms on opposite sides of the arrows in the equations
being added. Writing your equations with the arrows lined up one above the other will
help keep the two sides separated.
CO(g) ∆H = + 110.5 kJ C(s) + 1/2 O2(g) CO2(g)
∆H =  393.5 kJ
C(s) + O2(g)
________________________________________________
If we add the reactions at this point, the result is:
CO(g) + 1 O2(g) CO2(g) + 1/2 O2(g) However, another cancellation can be made. The 1 O2 on the left can be split into
CO(g) + 1/2 O2(g) + 1/2 O2(g) CO2(g) + 1/2 O2(g) The 1/2 O2 on the right now cancels a 1/2 O2 on the left, leaving 1/2 O2 on the left.
CO(g) + 1/2 O2(g) + 1/2 O2(g) CO2(g) + 1/2 O2(g) After cancellation, the result is the reaction that was WANTED.
When reactions add to give the reaction wanted, add the ∆H’s to find the ∆H wanted.
CO(g) C(s) + 1/2 O2(g) ∆H = + 110.5 kJ CO2(g)
∆H = ─ 393.5 kJ
C(s) + 1/2 1 O2(g)
________________________________________________
CO2(g)
CO(g) + 1/2 O2(g)
∆H = ─ 283.0 kJ
This process of adding equations will become easier with practice. Practice A: Add these equations and their ∆H values. Check answers as you go.
H2O(l) 1. H2(g) + 1/2 O2(g) ∆H = + 285.8 kJ H2O(l)
∆H = ─ 44.0 kJ
H2O(g)
__________________________________________________________ 2. CH4(g) C(s) + 2 H2(g) 2 H2(g) + O2(g) 2 H2O(g) ∆H = + 74.9 kJ
∆H = ─ 483.6 kJ C(s) + O2(g)
CO2(g)
∆H = ─ 393.5 kJ
________________________________________________ ©2011 ChemReview.net v. e4 Page 613 Module 22 — Heats of Reaction (ΔH) Finding An Unknown ∆H By Adding Equations
When ∆H for an equation is not known, it can be found by adding equations for which ∆H
is known. The key is to arrange the known equations so that they add to result in the
equation with the unknown ∆H.
The following problem will illustrate a system for deciding which known equations to add,
as well as when to multiply and reverse them.
Q. Find the ∆H for the burning of methyl alcohol,
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) using these three equations for which ∆H values are supplied.
C(s) + 2 H2(g) + 1/2 O2(g)
H2(g) + 1/2 O2(g)
C(s) + O2(g) CH3OH(l) H2O(g) CO2(g) ∆H = ─ 238.6 kJ (1) ∆H = ─ 241.8 kJ (2) ∆H = ─ 393.5 kJ (3) Complete the following steps in your notebook.
1. Write the WANTED equation, and then draw a dotted line beneath it.
2. Rewrite the first term in the WANTED equation, with its coefficient, below the dotted
line.
(Putting a WANTED reactant on the side where it must be in the final added equation
will help to arrange the rest of the added equations).
*****
At this point, your paper should look like this:
WANTED: CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) ∆H = ??? CH3OH(l)
3. Find an equation that both includes the dropped particle and has a known ∆H.
*****
CH3OH is part of equation (1) in the problem, which includes a ∆H.
Your goal is to write a known equation, reversed and/or multiplied if needed, so that the
dropped particle is on the side it is dropped on and has the same coefficient as the
particle dropped.
If needed, reverse the direction of the known equation so that the term dropped below
the line is on the side of the arrow, left or right, where it was dropped. If needed,
multiply the coefficients of the known equation so that the coefficient of the dropped
particle is the coefficient WANTED. Modify and write the ∆H value after the equation.
Try that step, then check your answer below.
***** ©2011 ChemReview.net v. e4 Page 614 Module 22 — Heats of Reaction (ΔH) In this problem, 1 CH3OH is needed on the left side of the arrow, so write known
equation (1) backwards.
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) ∆H = ??? CH3OH(l) C(s) + 2 H2(g) + 1/2 O2(g) ∆H = + 238.6 kJ Because the known equation must be reversed, reverse the sign of its ∆H.
4. A TIP that works often in problems involving burning or combustion (both mean reacting
with O2 gas) is “don’t worry about O2 until the end.” If the equations have been
chosen properly, when you add the terms at the end, O2 should have the coefficient
WANTED. This will be a check that you have selected the reactions properly.
5. Now compare the first particle on the right side of the arrow below the dotted line to the
particles WANTED on the right above the dotted line.
If what you have below the dotted line does not exactly match what is WANTED above
the dotted line, get rid of the UNwanted particle. Start a new equation on the line below
by writing the unwanted particle, with its coefficient, on the side where it will cancel the
same term in the first equation.
Try that step, then check your answer below.
*****
In this problem, after the arrow and below the dotted line, is C(s). What is WANTED
on the right that contains carbon is CO2. , so we need to get rid of C(s). Write C(s)
at the start of a new equation below, on the side where it will cancel when the
equations are added:
CO2(g) + 2 H2O(g)
∆H = ???
∆H = + 238.6 kJ
C(s) + 2 H2(g) + 1/2 O2(g)
CH3OH(l)
CH3OH(l) + 3/2 O2(g) C(s)
Then find an equation with a known ∆H that both includes the particle in the bottom
equation that you set up to cancel, and places its atoms in a compound WANTED above
the dotted line.
Try that step, then check your answer below.
*****
The equation that converts the C(s) you need to get rid of on the left to the CO2 you
WANT on the right is equation (3) supplied in the problem.
CO2(g) + 2 H2O(g)
∆H = ???
CH3OH(l)
∆H = + 238.6 kJ
C(s) + 2 H2(g) + 1/2 O2(g) CH3OH(l) + 3/2 O2(g) C(s) + O2(g) ©2011 ChemReview.net v. e4 CO2(g) ∆H = ─ 393.5 kJ Page 615 Module 22 — Heats of Reaction (ΔH) 5. Ignoring O2, compare the particles below the dotted line to what is WANTED above the
dotted line. If there is any particle and coefficient below the dotted line that does not
match exactly what is WANTED above the line, put the coefficient and particle that is
not WANTED where it will cancel. Then find an equation that will convert the
unwanted particle to a WANTED particle.
Take a look: what coefficient and particle below the dotted line do you not want?
* ** * *
Ignoring O2, it is 2 H2(g) . Try step 5, and then check the answer below.
* ** * *
After the arrow and below the dotted line, is 2 H2. What is WANTED that contains
hydrogen is 2 H2O. So, write 2 H2 below in a new equation where it will cancel, and
find an equation that converts H2 to the H2O WANTED.
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) ∆H = ??? ∆H = + 238.6 kJ
C(s) + 2 H2(g) + 1/2 O2(g)
CH3OH(l)
C(s) + O2(g)
2 H2(g) + O2(g) CO2(g)
2 H2O(g) ∆H = ─ 393.5 kJ
∆H = ─ 483.6 kJ Equation (2) supplied in the problem, and its ∆H, is doubled.
6. Compare what is WANTED above the dotted line to what is written below the line.
If what is below adds to give what is above, add the equations below the dotted line.
Then add the ∆H values to get the WANTED ∆H.
* * * **
Your paper should look like this:
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) ∆H = ??? CH3OH(l)
∆H = + 238.6 kJ
C(s) + 2 H2(g) + 1/2 O2(g)
C(s) + O2(g) CO2(g) ∆H = ─ 393.5 kJ ∆H = ─ 483.6 kJ
2 H2(g) + O2(g)
2 H2O(g)
_________________________________________________________
CO2(g) + 1/2 O2(g) + 2 H2O(g) ∆H = ???
CH3OH(l) + 2 O2(g)
3/2
Note that the O2 coefficient, after cancellation, matches what is WANTED. That’s
an indication that you have probably added the proper equations.
Once the equations below the dotted line add to give the equation WANTED, add
the modified ∆H values.
* * * ** ©2011 ChemReview.net v. e4 Page 616 Module 22 — Heats of Reaction (ΔH) ∆H = + 238.6 – 393.5 – 483.6 = ─ 638.5 kJ Done! Summary: To find an unknown ∆H for a reaction equation,
• write the equation WANTED; write a dotted line below it. • Drop the first coefficient and particle below the dotted line. • Including that first coefficient and particle, write an equation with a known ∆H. • Add other equations with known ∆H values that cancel particles you don’t want,
form particles you want, and add to give the equation WANTED. • Add the ∆H values. Practice B: Use the method above on these. If you get stuck, check a part of the answer
and try again. More practice can be found in the next lesson.
1. Use the first two equations to find ∆H for the third: the heat required to melt ice.
H2(g) + 1/2 O2(g) H2O(l) ∆H = ─ 285.8 kJ H2(g) + 1/2 O2(g) H2O(s) ∆H = ─ 291.8 kJ H2O(s) H2O(l) ∆H = ? 2. The gas ethane (C2H6) is one of the constituents of natural gas, a major component of
the world energy economy. The equation for the burning of ethane is
C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g) Use the formation equations below to find ∆H for the burning of ethane.
2 C(s) + 3 H2(g) C2H6(g) ∆H = ─ 84.7 kJ (1) H2(g) + 1/2 O2(g) H2O(g) ∆H = ─ 241.8 kJ (2) ∆H = ─ 393.5 kJ (3) C(s) + O2(g) CO2(g) ANSWERS
Practice A
1. H2O(l) H2(g) + 1/2 O2(g) ∆H = + 285.8 kJ H2O(l)
∆H = ─ 44.0 kJ
H2O(g)
________________________________________________
H2O(g) H2(g) + 1/2 O2(g) ©2011 ChemReview.net v. e4 ∆H = + 241.8 kJ Page 617 Module 22 — Heats of Reaction (ΔH) 2. CH4(g) C(s) + 2 H2(g) 2 H2(g) + O2(g) 2 H2O(g) ∆H = + 74.9 kJ
∆H = ─ 483.6 kJ C(s) + O2(g)
CO2(g)
∆H = ─ 393.5 kJ
________________________________________________
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ∆H = ─ 802.2 kJ Practice B
1. WANTED: H2O(s)
H2O(l)
∆H = ?
H2(g) + 1/2 O2(g)
H2O(s)
∆H = + 291.8 kJ ∆H = ─ 285.8 kJ
H2(g) + 1/2 O2(g)
H2O(l)
______________________________________________________
H2O(s)
H2O(l)
∆H = + 6.0 kJ
2 CO2(g) + 3 H2O(g)
2. WANT: C2H6(g) + 7/2 O2(g)
2 C(s) + 3 H2(g)
∆H = + 84.7 kJ
C2H6(g)
∆H = ─ 787.0 kJ (equation (2) doubled) ∆H = ─ 725.4 kJ
3 H2(g) + 3/2 O2(g)
3 H2O(g)
______________________________________________________ (equation (3) tripled) 2 C(s) + 2 O2(g) C2H6(g) + 7/2 O2(g)
***** 2 CO2(g) (equation (1) reversed) 2 CO2(g) + 3 H2O(g) ∆H = ─ 1427.7 kJ Lesson 22D: Heats of Formation and Element Formulas
Timing: Complete this lesson when you are asked to solve problems using heats of
formation.
***** Gas Versus Thermodynamic Standard Conditions
The history of chemistry has given us multiple definitions for standard conditions.
• For gas law calculations, standard temperature and pressure (STP) is defined as 0ºC
and 1 atmosphere pressure. • For thermodynamic measurements, for substances to be in their standard state,
o elements must be at 25ºC and 1 atm pressure and (in most cases) in the phase
and/or solid structure that is most stable. o Compounds that are gases must be at one atmosphere pressure. o Substances in solutions must have a concentration of 1 mol/L. o Solid and liquid compounds are in their standard state under nearly all conditions
if they are in the form that is most stable at 25ºC and 1 atm pressure. ©2011 ChemReview.net v. e4 Page 618 Module 22 — Heats of Reaction (ΔH) You should be aware of the difference between gas and thermodynamic standard conditions,
but in most problems it will be clear from the context which definition of “standard
conditions” applies. Writing the Standard State For Elements
To work with formation equations, you will need to be able to write the standard state of
elements: both the molecular formula of the element and the state (gas, liquid, or solid) in
which it is most stable at room temperature (25ºC) and one atmosphere pressure. The
following are rules for writing formulas and standard states for most of the elements you
will encounter in firstyear chemistry.
1. Only two elements are liquids at room temperature: mercury (Hg) and bromine (Br2).
2. Over 75% of the elements are metals, and the formulas for all metals (except mercury)
are written as monatomic solids.
Examples of element formulas for metals: Na(s), Al(s), Ag(s), Hg(liquid)
3. Eight elements are diatomic: The 5 halogens, plus the gases hydrogen, oxygen, and
nitrogen:
F2(g), Cl2(g), Br2(l), I2(s), At2(s), H2(g), O2(g), N2(g)
4. Except for hydrogen, all of the 11 elements that are gases at room temperature are
toward the top and right of the periodic table. The gaseous elements are 5 of the
diatomics: H2(g), O2(g), N2(g), F2(g), Cl2(g); plus all 6 of the monatomic noble gases.
5. Some periodic tables indicate the state of each element at 25ºC by the color of the
element symbol: often solids are black, the two liquids blue, and gases red.
6. If a table of heats of formation shows a particle with a value for its heat of formation
( ΔH˚f ) of zero, the formula shown will be that of the element in its standard state.
Example: In the table at the right, for the elements
chlorine and hydrogen, the formulas in their
standard state are the formulas with the zero ΔH:
Cl2(g) and H2(g).
The singleatom forms of each element can be
formed at room temperature, but, as their higher
enthalpy indicates, they are less stable than the
molecules formed from two neutral atoms. They
therefore tend to react readily with other particles. Formula
Cl(g) ΔH˚f in kJ/mole
+ 121.0 Cl2(g) 0 H2(g) 0 H(g) + 218.0 7. Other frequently encountered nonmetal elements are S8(s) and P4(s). ©2011 ChemReview.net v. e4 Page 619 Module 22 — Heats of Reaction (ΔH) Writing Formation Equations
A formation equation is a balanced equation in which the reactants are all elements, the
product is one mole of one compound, and the elements and the compound are all in their
standard states.
Examples of formation equations are
CO2(g) For carbon dioxide: C(s) + O2(g)
For sulfuric acid: H2(g) + 1/8 S8(s) + 2 O2(g) H2SO4(l) To write a formation equation for a compound,
1. To the right of an arrow, write the compound formula with a coefficient of 1.
2. On the left, for each of the elements that make up the compound, write the formula
of the element in its standard state at 25ºC and 1 atmosphere pressure.
3. Add element coefficients (that often include fractions) to balance the equation.
Try Q. Write the formation reaction for ethanol: C2H5OH(l)
*****
Steps 1 and 2:
*****
Step 3: C(s) + H2(g) + O2(g) 2 C(s) + 3 H2(g) + 1/2 O2(g) 1 C2H5OH(l)
1 C2H5OH(l) Practice A: Learn the rules above. Then, for each numbered problem, do every other
letter. Check answers as you go. Save a few for your next practice session.
1. Which of these are not elements in their standard state?
a. H2(g) b. CO(g) c. Co(s) d. Hg(g) e. N(g) f. Cl2(l) 2. Write the formulas for these elements in their standard state at room temperature.
a. Fluorine b. Iron c. Oxygen d. Carbon e. Nitrogen 3. Write balanced formation equations for these compounds.
a. NaCl(s) b. HCl(g) c. Al2O3(s) d. C2H6(g) e. NO2(g) Defining Zero For ΔH
Every particle has a characteristic enthalpy that depends on its structural formula and its
state. The absolute enthalpy of a substance cannot be measured, but we can work around
this because in chemical processes, we are concerned with changes in enthalpy that occur
(ΔH). To measure enthalpy changes, we can assign an arbitrary zero point to the enthalpy
scale, and then measuring change relative to that zero value. For enthalpy, the zero point
that has been chosen is: all elements in their standard state are assigned a heat of formation
(H˚f) of zero kJ/mol.
©2011 ChemReview.net v. e4 Page 620 Module 22 — Heats of Reaction (ΔH) For all elements in their standard state, ΔH˚f ≡ 0 kJ/mol. Heat of Formation
A particle in a given state has a characteristic heat of formation ( ΔH˚f ) and this can be
considered to be the value for its enthalpy. For elements in their standard state, this value
is always 0 kJ/mol. For all other particles, ΔH˚f is the change in enthalpy that occurs in its
formation reaction.
In the symbol for a heat of formation, ΔH°f , the subscript f means “formed from elements
in their standard state.” The superscript degree symbol ° means that the particle formed is
also in its standard state.
Values for heats of formation are
often provided in reference tables.
The table at the right tells us that
1. Graphite (used in “pencil lead”)
and diamond are composed of
solid carbon. Name Formula Graphite C(s) 0 Diamond C(s) + 1.9 Carbon Dioxide CO2(g) ΔH°f in kJ/mol ─ 393.5 2. Graphite and diamond are
Methane
CH4(g)
─ 74.9
often represented by the same
Steam
H2O(g)
─ 241.8
molecular formula: C(s).
However, they have different
structural formulas, and graphite has slightly lower enthalpy. Graphite is therefore the
more stable of the two forms of pure carbon. The system with lower enthalpy is termed
“more stable” because more energy must be added to change it.
The most stable form at at 25ºC is designated as the standard state of an element.
Heats of formation are the ΔH values for formation equations. Formation equations may be
written with ΔH values labeled as either ΔH, ΔH° or ΔH°f .
Examples
For carbon dioxide: C(s) + O2(g) CO2(g) For sulfuric acid: H2(g) + 1/8 S8(s) + 2 O2(g) ΔH°f = ― 393.5 kJ/mol
H2SO4(l) ΔH°f = ― 812 kJ/mol A formation equation can be treated as any other equation with a ΔH attached: it can be
multiplied, reversed, and added.
If you are given a ΔH°f value for a compound in a problem or in a table, you can write a
formation equation for the compound, attach the ΔH°f, and then reverse, multiply, or add
that equation to find ΔH values for other reactions. Try this example.
Q. Using a ΔH°f in the table above, write the ΔH for
2 H2O(g) 2 H2(g) + O2(g) ∆H = _______ ***** ©2011 ChemReview.net v. e4 Page 621 Module 22 — Heats of Reaction (ΔH) Starting from a ΔH°f table value, to find ΔH for a related equation, first write the
balanced formation equation and attach its ΔH°f.
H2(g) + 1/2 O2(g) H2O(g) ΔH°f = ─ 241.8 kJ/mol Then reverse and/or multiply the formation reaction to get the WANTED reaction.
2 H2(g) + 2 H2O(g) O2(g) ∆H = ─ 2(─ 241.8 kJ) = + 483.6 kJ Note that the unit converts from kJ/mol to kJ when the product is no longer one mole
of one substance. Note also that if a formation equation is multiplied or reversed, the
ΔH°f becomes a ΔH because the equation is no longer a formation equation. Practice B: Learn the rules above. Then, for each numbered problem, do every other
letter. Check answers as you go. Save a few for your next practice session.
1. In your notebook, based on the table data at the right,
write the formation equation for each compound and
attach a ∆H value to each equation. Formula ΔH°f in kJ/mol NH3(g) ─ 45.9 AgCl(s) ─ 127.0 NO(g)
2. Using the table values from Problem 1, write ∆H
values for these reactions.
a. AgCl(s) Ag(s) + 1/2 Cl2(g) b. N2(g) + 3 H2(g)
c. 3 H2O(l) 2 NH3(g) 3 H2(g) + 3/2 O2(g) + 90.3 H2O(l) ─ 285.8 ∆H = _______
∆H = _______
∆H = _______ 3. Which of the reactions in Problem 2 are endothermic?
4. Circle the compounds involved in each of these reactions, then write the value for the
heat of formation (ΔH°f) for each compound.
a. 1/8 S8(s) + O2(g) SO2(g) b. 3/8 S8(s) + 9/2 O2(g)
c. 2 NO(g)
d. 4 H2O(g) 3 SO3(g) N2(g) + O2(g)
4 H2(g) + 2 O2(g) ∆H = ─ 296.8 kJ
∆H = ─ 1,188 kJ
∆H = ─ 180.6 kJ
∆H = + 967.2 kJ 5. Which equation(s) in Problem 4 are formation equations? ©2011 ChemReview.net v. e4 Page 622 Module 22 — Heats of Reaction (ΔH) 6. Use this table data to find ∆H values for the two
reactions below, then add the two reactions and their
∆H values. 1/2 N2(g) + O2(g) ΔH°f NO2(g) + 33.8 kJ/mol NO(g)
NO2(g) Formula + 90.3 kJ/mol ∆H = _____________ NO(g)
∆H = _____________
1/2 N2(g) + 1/2 O2(g)
______________________________________________________ 7. Acetylene gas can be burned to produce the extremely hot flame used in a “cutting
torch.” The equation is
C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g) Use the above equation plus the table values to
find the heat of formation for acetylene. ∆H = ─ 1,255.8 kJ
Formula ΔH°f in kJ/mole H2O(g) ─ 241.8 CO2(g) ─ 393.5 ANSWERS
Practice A
1. b, d, e, and f. b. CO(g) is a compound. d. Hg(g) . Mercury is a liquid at room temperature.
e. N(g) . Nitrogen as an element is N2 . f. Cl2(l) . Chlorine is a gas at room temperature.
2. a. Fluorine F2(g) Fluorine is a halogen. All halogen elements are diatomic. Fluorine is at the top
right of the periodic table, where several elements are gases at room temperature.
b. Iron Fe(s) c. Oxygen Iron is a metal. All metal elements are monatomic solids (except mercury). O2(g) Oxygen is a gas at room temperature. The fact that the formula for the element oxygen is O2 must be memorized.
C(s) The elemental form of carbon is graphite (“pencil lead”).
e. Nitrogen N2(g) The air we breathe is about 80% nitrogen gas. The diatomic formula for
d. Carbon elemental nitrogen gas is used frequently and must be memorized.
3. a. NaCl(s) In formation equations, elements in their standard state are the reactants, one mole of
the compound is the product, and coefficients must be added that balance the equation:
Na(s) + 1/2 Cl2(g) NaCl(s) Sodium is a metal. For metals, the formula is written as one atom; all are solids except Hg. ©2011 ChemReview.net v. e4 Page 623 Module 22 — Heats of Reaction (ΔH) Chlorine is a halogen: all halogen elements are diatomic. In the halogen column, the elements above
bromine are gases, and below bromine are solids, at room temperature.
NaCl is a compound: it has more than one kind of atom. You must add coefficients that keep the same
particle formulas, but make one mole of the compound in a balanced equation.
b. HCl(g) 1/2 H2(g) + 1/2 Cl2(g) HCl(g) Hydrogen atoms are stable at room temperature as H2 gas; this element formula should be
memorized.
c. Al2O3(s) 2 Al(s) + 3/2 O2(g) d. C2H6(g) 2 C(s) + 3 H2(g) Al2O3(s) Aluminum is a metal. C2H6(g) Some texts write C(graphite) to distinguish graphite, diamonds, buckyballs, and other forms of pure
carbon. Graphite is the “standard state” form of carbon.
e. NO2(g) 1/2 N2(g) + O2(g) NO2(g) Practice B
1. a. 1/2 N2(g) + 3/2 H2(g) NH3(g) ∆H = ─ 45.9 kJ/mol When translating heats of formation into formation equations, elements are added as reactants, one
mole of the compound is the product, and the heat of formation is the ∆H for the reaction.
b. Ag(s) + 1/2 Cl2(g)
c. 1/2 N2(g) + 1/2 O2(g)
d. H2(g) + 1/2 O2(g) AgCl(s)
NO(g)
H2O(l) ∆H = ─ 127.0 kJ/mol
∆H = + 90.3 kJ/mol ∆H = ─ 285.8 kJ/mol When a reaction has one product and its coefficient is one, the units of ∆H may be written as either kJ/mol
or as kJ. In equations with energy terms, coefficients are in moles.
2. a. AgCl(s) Ag(s) + 1/2 Cl2(g) ∆H = + 127.0 kJ When starting from the table, first write out the formation reaction, as in answer 3b. Then, since this
problem asks for the reverse of the formation reaction, change the sign of ∆H.
b. N2(g) + 3 H2(g) 2 NH3(g) ∆H = ─ 91.8 kJ Starting from the table, first write the formation reaction, as in answer 3a. Then, since this problem
doubles the formation reaction, double the ∆H.
c. 3 H2O(l) 3 H2(g) + 3/2 O2(g) ∆H = + 857.4 kJ This problem is table equation 3d, tripled and reversed.
When a reaction does not have one particle as a product, the units of ∆H are kJ.
3. 2a and 2c. Endothermic reactions have a positive ∆H. ©2011 ChemReview.net v. e4 Page 624 Module 22 — Heats of Reaction (ΔH) 4. Compounds have more than one kind of atom.
a. 1/8 S8(s) + O2(g) SO2(g) ∆H = ─ 296.8 kJ or kJ/mol Assuming that all of the formulas in the reactants are elements in their “standard state” (and they are),
this is a formation reaction for one mole of SO2(g). The heat of this reaction (∆H) is therefore the
heat of formation of the compound: ΔH°f for SO2(g) = ─ 296.8 kJ/mol
b. 3/8 S8(s) + 9/2 O2(g) 3 SO3(g) ∆H = ─ 1188 kJ In this reaction, the elements are on the left, the compound is on the right, and 3 moles of compound is
formed. For a formation reaction involving heat, we must make one mole of the compound. To get
the formation reaction, multiply all terms of the given reaction by 1/3.
ΔH˚f = 1/3 x (─ 1188 kJ) = ─ 396.0 kJ .
c. 2 NO(g) N2(g) + O2(g) ∆H = ─ 180.6 kJ In a formation reaction, elements must be on the left, so this reaction must be reversed.
Reversing this reaction, the sign of ∆H is reversed. This is double a formation reaction, so the ∆H
would be cut in half for the heat of formation. Heats for formation must be for one mole of compound.
ΔH°f for NO(g) = 1/2 x (─ 180.6 kJ), reversed= + 90.3 kJ .
d. 4 H2O(g) 2 H2(g) + O2(g) ∆H = + 967.2 kJ This is the formation reaction for water vapor, quadruped and reversed. To find to the formation
reaction, write the reaction backwards, changing the sign of ∆H. Then multiply the coefficients and ∆H
by 1/4, to make one mole of water vapor.
ΔHf for water vapor = 1/4 x (+ 967.2 kJ) reversed= ─ 241.8 kJ
5. Only 4a. Formation equations must have elements in their standard state on the left and one mole of a
nonelement on the right.
6. NO2(g) 1/2 N2(g) + 1/2 1 O2(g) ∆H = ─ 33.8 kJ (reverse of formation) NO(g)
∆H = + 90.3 kJ
(this is the formation equation)
1/2 N2(g) + 1/2 O2(g)
______________________________________________________
NO2(g)
NO(g) + 1/2 O2(g)
∆H = + 56.5 kJ
When you add the equations, 1/2 O2 cancels on both sides. When the equations are written so that the
arrows do not line up in a column, be careful to cancel substances on opposite sides of the arrows.
7. The heat of formation of acetylene is WANTED. That equation would be:
C2H2(g)
∆H = ?
2 C(s) + H2(g)
*****
Arrange the given equations to add and cancel to result in the equation wanted. The equations may
be added in any order. In burning (combustion) reactions, the O2 coefficients should add correctly if
the equations to be added are correct. Here, the 5/2 O2 on both sides will cancel. ©2011 ChemReview.net v. e4 Page 625 Module 22 — Heats of Reaction (ΔH) 2 C(s) + H2(g)
C2H2(g)
∆H = ?
 (wanted equation) 2 C(s) + 2 O2(g) 2 CO2(g) ∆H = ─ 787.0 kJ (table rxn. doubled) 2 CO2(g) + H2O(g) C2H2(g) + 5/2 O2(g) ∆H = + 1255.8 kJ (given rxn reversed) ∆H = ─ 241.8 kJ
H2(g) + 1/2 O2(g)
H2O(g)
______________________________________________________
2 C(s) + H2(g) C2H2(g) ∆H = + 227.0 kJ (table reaction)
(wanted equation) ***** Lesson 22E: Using Summation To Find ∆H
Enthalpy is a state function: the change in enthalpy in a process (ΔH) is simply the
difference between the total enthalpy of the particles at the end (the products) and at the
beginning (the reactants). This means that one way to measure the enthalpy change of a
reaction is to add up the enthalpy values of all of the products, then subtract the enthalpy
values of all of the reactants.
ΔHreaction = sum of ΔH°f for final particles ─ sum of ΔH°f for initial particles
ΔH°f values measure the enthalpy in one mole of a particle. Two moles of the particles will
have double that value for their total enthalpy.
Apply the logic of the above to this
Q. If we needed to find ∆H for the reaction in problem 7 above
C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g) ∆H = ? and we know the ΔH°f for each of the particles in
the equation, as given in the table at the right, Formula a. calculate the total ΔH°f of the products of the
above reaction, based on the table and
reaction coefficients. H2O(g) ─ 241.8 kJ/mol CO2(g) ─ 393.5 kJ/mol C2H2(g) + 227.0 kJ/mol ***** ΔH°f Total ΔH°f of the products = 2 (ΔH°f of CO2(g)) + 1 (ΔH°f of H2O(g)) =
= 2 (─ 393.5 kJ) + 1 (─ 241.8 kJ) = ─ 1,028.8 kJ b. Find the total ΔH°f of the reactants.
*****
Sum of ΔH°f of the reactants = 1 (ΔH°f of C2H2(g) + 5/2 (ΔH°f of O2(g)) =
= 1 (+ 227.0 kJ) + 5/2 ( 0 kJ) = + 227.0 kJ O2(g) is an element, and the ΔH°f of all elements is 0 kJ. ©2011 ChemReview.net v. e4 Page 626 Module 22 — Heats of Reaction (ΔH) c. Using the equation above with sums, find ∆H for the reaction.
*****
ΔHreaction = sum of ΔH°f of products ─ sum of ΔH°f of reactants
= ─ 1,028.8 kJ ─ (+ 227.0 kJ) = ─ 1,255.8 kJ Compare this answer to the ΔH supplied after the reaction in problem 7 above.
What is the difference between the above Question Part c and problem 7?
• In this question, we knew all of the heats of formation for the particles in the
reaction. • In problem 7, we WANTED one of the heats of formation, and had data for a
reaction that was not a formation reaction, but could be used to find the ΔH°f . To find a reaction ΔH, if the ΔH°f values for all of the compounds in an equation are
known, this summation method is likely to be quicker than writing out formation equations
and adding using Hess’s law. However, in problems with data that mix formation and
nonformation reactions, such as problem 7, the summation method does not work, but
writing out the formation reactions then adding both types of reactions using Hess’s law
does work.
Hess’s law is slower, but it will solve using heats of formation equations, nonformation
equations, or a mixture of the two.
To find ΔHreaction , a suggested strategy that may speed your work is
• use the summation method if you know all of the needed ΔH°f values. • If not, write out the formation equations and add equations using Hess’s Law. Let’s summarize:
To find ΔHreaction :
1. If standard enthalpy of formation (ΔH°f) values are known for all of the particles in a
chemical reaction, the ΔH of the reaction can be calculated by substituting those values
into the state function equation that defines ΔH:
ΔHreaction = ΔHfinal ─ ΔHinitial
= sum of ΔH°f of products ─ sum of ΔH°f of reactants
= [sum of (coefficient x ΔH°f) of products] ─ [sum of (coefficient x ΔH°f) of reactants]
Abbreviated in symbols, we will call this the heat of formation summation equation.
ΔHreaction = ∑ c ΔH˚f of products ─ ∑ c ΔH˚f of reactants
2. If not all of the needed ΔH°f values are known, write out the formation equations and
other supplied equations, adding the equations using Hess’s Law. ©2011 ChemReview.net v. e4 Page 627 Module 22 — Heats of Reaction (ΔH) Try Q. Use the table and the summation equation to
find ΔH for this reaction. Formula 2 SO2(g) + O2(g) SO2(g) ─ 296.8 SO3(g) ─ 396.0 2 SO3(g) ΔH°f in kJ/mole *****
ΔH = [sum of (coefficient x ΔH°f) of products] ─ [sum of (coefficient x ΔH°f) of reactants]
= [ 2 (─ 396.0 kJ) ] ─ [ 2 (─ 296.8 kJ) + 0 kJ ] = ─ 198.4 kJ Practice. On these, use the summation equation and table values to find ∆H. 1. Steam can be added to hot coal (which is primarily carbon) to produce a burnable
mixture:
C(s) + H2O(g) CO(g) + H2(g) ∆H = ? Formula ΔH˚f in kJ/mol CO(g) ─ 110.5 H2O(g) ─ 241.8 C2H6(g) + O2(g) ©2011 ChemReview.net v. e4 CO2(g) + H2O(g) Formula ΔH˚f in kJ/mol C2H6(g) ─ 84.7 CO2(g) ─ 393.5 H2O(g) 2. For the burning of ethane, a component of natural
gas, balance the equation, then find ∆H. ─ 241.8 Page 628 Module 22 — Heats of Reaction (ΔH) ANSWERS
Practice
1. C(s) + H2O(g) CO(g) + H2(g) ΔHreaction = (sum of ΔH°f values of products) ─ (sum of ΔH°f values of reactants)
= ( ─ 110.5 + 0 ) ─ ( 0 + ─ 241.8 ) = 241.8 ─ 110.5 = + 131.3 kJ
2. C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g) ΔHreaction = ∑ c ΔH˚f of products ─ ∑ c ΔH˚f of reactants
= [ 2 (─ 393.5 kJ) + 3 (─ 241.8) ] ─ [ 1 (─ 84.7 + 7/2 (0) ] = ─ 1427.7 kJ
***** SUMMARY: Heats of Reaction (∆H)
A summary for initial thermodynamic and PV work equations is given at the end of Lesson
22A. The summary below covers calculations that involve ∆H.
If you have not already done so, you may want to organize this summary into charts,
numbered lists, and flashcards that help with longterm memory.
1. Energy (E), heat (q), and work (w) in the SI system are measured in joules (J).
2. If the external pressure on a system is held constant and work is limited to “PV work” ,
then the heat flow in our out of a system of particles equals its change in enthalpy.
ΔH = q at constant P and PV work. 3. In equations that include energy terms, the coefficients are in moles.
4. In energy equations, the phase of each particle must be shown: (s), (1), (g), or (aq).
5. In exothermic reactions, energy is released into the environment, and the energy term is
shown
• EITHER with a positive sign on the products side; • OR (preferred) with a negative ΔH value written after the equation. 6. In endothermic reactions, energy must be added, and the energy term is shown
• EITHER with a positive sign on the reactants side of the equation, • OR with a positive ∆H written after the reaction. 7. Reactions involving energy or heat can be reversed (the equation can be written
backwards). If a ∆H is attached to the equation, change the sign of ∆H.
8. All reaction coefficients and energy terms can be multiplied or divided by a number. If
∆H notation is used, do the same to the value of ∆H. ©2011 ChemReview.net v. e4 Page 629 Module 22 — Heats of Reaction (ΔH) 9. Equations with energy terms can be added to produce a new equation.
• Like particles on the same side in different equations can add, and on opposite
sides can cancel. • When equations are added, ∆H values add according to their signs. 10. Hess’s Law. When ∆H for an equation is not known, it can be found by adding
together equations for which ∆H is known.
• Write the reaction equation WANTED, then a dotted line below it. • Write the first coefficient and substance formula below the dotted line. • Find an equation with a known ∆H that includes that first formula. Adjust the
direction and coefficients of the known equation to put the dropped particle and
its coefficient on the side where it is WANTED. Modify and include a ∆H. • Arrange other equations and their known ∆H to cancel particles not wanted, and
add to result in the equation WANTED. 11. In a formation equation, the reactants are all elements in their standard state at 25°C and
1 atm pressure, and the product is one mole of a compound or other particle that is not
an element.
12. The heat of formation (ΔH°f) of a compound is the amount of heat required or released
when one mole of the compound is formed from its elements in their standard state.
13. All elements in their standard state are assigned a heat of formation of zero kJ/mole.
14. The standard enthalpy value for a compound (ΔH°) is its ΔH°f.
15. If heats of formation are known for all of the substances in a reaction, the ΔH of the
reaction can be found by substituting ΔH°f values into:
ΔH = ΔHfinal ─ ΔHinitial
ΔH = [sum of (coefficient x ΔH°) of products] ─ [sum of (coefficient x ΔH°) of reactants]
or ΔHreaction = ∑ c ΔH˚f of products ─ ∑ c ΔH˚f of reactants 16. To find a ΔHreaction , Hess’s law is slower, but it will solve using heats of formation
equations, nonformation equations, or a mixture of both. ##### ©2011 ChemReview.net v. e4 Page 630 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.
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