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Thermochemistry
Wade Baxter, Ph.D.
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Printed: April 25, 2014
AUTHOR
Wade Baxter, Ph.D.
EDITORS
Donald Calbreath, Ph.D.
Max Helix
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Chapter 1. Thermochemistry
C HAPTER
1
Thermochemistry
C HAPTER O UTLINE
1.1
Heat Flow
1.2
Thermochemical Equations
1.3
Heat and Changes of State
1.4
Hess’s Law
1.5
References
A fire in a fireplace on a cold winter night is welcome and comforting. The organic compounds that make up wood
react with oxygen at an elevated temperature to produce carbon dioxide, water vapor, and large amounts of heat.
Combustion is an example of an exothermic chemical reaction, meaning that heat is released in the course of the
reaction. When you sit in front of this fire, the heat energy flows from the fire to you. In this chapter, you will learn
about heat flow and how some chemical reactions naturally release energy while some other reactions absorb energy.
You will learn how to show heat changes in chemical equations and perform calculations in order to determine the
heat being absorbed or released in all sorts of chemical reactions, changes of state, and other physical processes.
Tristan Ferne. www. f lickr.com/photos/tristan f /2149343932/. CC BY 2.0.
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1.1. Heat Flow
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1.1 Heat Flow
Lesson Objectives
• Describe how chemical potential energy relates to heat and work.
• Describe the law of conservation of energy and how heat flows between system and surroundings during both
endothermic and exothermic processes.
• Use the specific heat equation to perform calculations that relate mass, specific heat, change in temperature,
and the amount of heat absorbed or released.
Lesson Vocabulary
•
•
•
•
•
•
•
•
•
•
•
calorie
chemical potential energy
endothermic
exothermic
heat
heat capacity
law of conservation of energy
specific heat
surroundings
system
thermochemistry
Check Your Understanding
Recalling Prior Knowledge
• What is energy?
• What is temperature, and what is its relationship to energy?
Chemical reactions are accompanied by transfers of energy. Keeping track of heat flow and energy requirements is
important for a full understanding of chemical processes. In this lesson, you are introduced to heat, energy, and the
specific heat of substances.
Heat and Work
Energy is the capacity for doing work or supplying heat. When you fill your car with gasoline, you are providing it
with potential energy. Chemical potential energy is the energy stored in the chemical bonds of a substance. The
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Chapter 1. Thermochemistry
various chemicals in gasoline contain a large amount of chemical potential energy that is released when the gasoline
is burned in a controlled way in the engine of the car. The release of that energy does two things. Some of the
potential energy is transformed into work, which is used to move the car ( Figure 1.1). At the same time, some of
the potential energy is converted to heat, making the car’s engine very hot. The energy changes of a system occur as
either heat or work, or some combination of both.
FIGURE 1.1
A dragster is able to accelerate because
of the chemical potential energy of its fuel.
The burning of the fuel also produces
large amounts of heat.
Heat is energy that is transferred from one object or substance to another because of a difference in temperature
between them. Heat always flows from an object at a higher temperature to an object at a lower temperature ( Figure
1.2). The flow of heat will continue until the two objects are at the same temperature.
FIGURE 1.2
(A) Object A starts with a higher temperature than object B. No heat flows when
the objects are isolated from each other.
(B) When brought into contact, heat flows
from A to B until the temperatures of the
two objects are the same.
Thermochemistry is the study of energy changes that occur during chemical reactions and during changes of state.
When chemical reactions occur, some chemical bonds are broken, while new chemical bonds form. As a result of
the rearrangement of atoms, the total chemical potential energy of the system either increases or decreases.
Exothermic and Endothermic Processes
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of
conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed.
In other words, the entire energy in the universe is conserved. In order to better understand the energy changes
taking place during a reaction, we need to define two parts of the universe, called the system and the surroundings.
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1.1. Heat Flow
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The system is the specific portion of matter in a given space that is being studied during an experiment or an
observation. The surroundings is everything in the universe that is not part of the system. In practical terms
for a laboratory chemist, the system is generally the reaction being investigated, while the surroundings include
the immediate vicinity within the room. During most processes, energy is exchanged between the system and
the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the
surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
In the study of thermochemical processes, things are viewed from the point of view of the system. A chemical
reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course
of an endothermic process, the system gains heat from the surroundings, so the temperature of the surroundings
decreases. The quantity of heat for a process is represented by the letter q. The sign of q for an endothermic
process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat
is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the
temperature of the surroundings increases. The sign of q for an exothermic process is negative because the system
is losing heat. The difference between an endothermic reaction and an exothermic reaction is illustrated below (
Figure 1.3).
FIGURE 1.3
(A) In an endothermic reaction, heat flows
from the surroundings into the system,
decreasing the temperature of the surroundings. (B) In an exothermic reaction,
heat flows from the system into the surroundings, increasing the temperature of
the surroundings.
Units of Heat
Heat flow is measured in one of two common units: the calorie and the joule. The joule (J), introduced in the chapter
Measurements, is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the
amount of energy contained within food. A calorie (cal) is the quantity of heat required to raise the temperature of 1
gram of water by 1°C. For example, raising the temperature of 100 g of water from 20°C to 22°C would require 100
× 2 = 200 cal.
Calories contained within food are actually kilocalories (kcal). In other words, if a certain snack contains 85 food
calories, it actually contains 85 kcal or 85,000 cal. In order to make the distinction, the dietary calorie is written with
a capital C.
1 kilocalorie = 1 Calorie = 1000 calories
To say that the snack “contains” 85 calories means that 85 kcal of energy are released when that snack is processed
by your body.
Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a
joule and a calorie is shown below.
1 J = 0.2390 cal or 1 cal = 4.184 J
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Chapter 1. Thermochemistry
We can calculate the amount of heat released in kilojoules when a 400. calorie hamburger is digested.
400. Cal = 400. kcal ×
4.184 kJ
= 1.67 × 103 kJ
1 kcal
Heat Capacity and Specific Heat
If a swimming pool and a bucket, both full of water at the same temperature, were subjected to the same input of
heat energy, the bucket of water would certainly rise in temperature more quickly than the swimming pool. Heat
capacity is the amount of heat required to raise the temperature of an object by 1°C. The heat capacity of an object
depends both on its mass and its chemical composition. Because of its much larger mass, the swimming pool of
water has a larger heat capacity than the bucket of water.
Different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may
become quite hot to the touch. An equal mass of water in the same sun will not become nearly as hot. Water is very
resistant to changes in temperature, while metals in general are not. The specific heat of a substance is the amount
of energy required to raise the temperature of 1 gram of the substance by 1°C. The table below ( Table 1.1) lists the
specific heats of some common substances. The symbol for specific heat is cp, with the p subscript referring to the
fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram
per degree (J/g•°C) or calories per gram per degree (cal/g•°C). This text will use J/g•°C for specific heat. Note that
the specific heat of a substance depends not only on its identity but also its state. For example, ice, liquid water, and
steam all have different specific heat values.
TABLE 1.1: Specific Heats of Some Common Substances
Substance
Water (l)
Water (s)
Water (g)
Ammonia (g)
Ethanol (l)
Aluminum (s)
Carbon, graphite (s)
Copper (s)
Gold (s)
Iron (s)
Lead (s)
Mercury (l)
Silver (s)
Specific Heat (J/g•°C)
4.18
2.06
1.87
2.09
2.44
0.897
0.709
0.385
0.129
0.449
0.129
0.140
0.233
Notice that water has a very high specific heat compared to most other substances. Water is commonly used as a
coolant for machinery because it is able to absorb large quantities of heat ( Figure 1.4). Coastal climates are much
more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat
from the air on hot days and releases it back into the air on cool days.
Specific Heat Calculations
The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo
when it is either heated or cooled. The equation that relates heat (q) to specific heat (c p ), mass (m), and temperature
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1.1. Heat Flow
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FIGURE 1.4
This power plant in West Virginia, like
many others, is located next to a large
lake so that the water from the lake can
be used as a coolant. Cool water from
the lake is pumped into the plant, while
warmer water is pumped out of the plant
and back into the lake.
change (∆T) is shown below.
q = m × c p × ∆T
The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in
temperature is given by ∆T = T f –Ti , where T f is the final temperature and Ti is the initial temperature.
Sample Problem 17.1: Calculating Specific Heat
A 15.0 g piece of cadmium metal absorbs 134 J of heat as its temperature is increased from 24.0°C to 62.7°C.
Calculate the specific heat of cadmium.
Step 1: List the known quantities and plan the problem.
Known
• heat = q = 134 J
• mass = m = 15.0 g
• ∆T = 62.7°C –24.0°C = 38.7°C
Unknown
• c p of cadmium = ? J/g•°C
The specific heat equation can be rearranged to solve for the specific heat.
Step 2: Solve.
cp =
q
134 J
=
= 0.231 J/g ·◦ C
m × ∆T 15.0 g × 38.7◦ C
Step 3: Think about your result.
The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals in the table above ( Table
1.1). The result has three significant figures.
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Chapter 1. Thermochemistry
Practice Problems
1. How much heat is required to raise the temperature of 13.7 g of aluminum from 25.2°C to 61.9°C?
2. A 274 g sample of air is heated with 2250 J of heat, and its temperature rises by 8.11°C. What is the specific
heat of air at these conditions?
Since most specific heats are known, they can be used to determine the final temperature attained by a substance
when it is either heated or cooled. Suppose that a 60.0 g sample of water at 23.52°C was cooled by the removal of
813 J of heat. The change in temperature can be calculated using the specific heat equation.
∆T =
q
−813 J
=
= −3.24◦ C
cp × m 4.18 J/g ·◦ C × 60.0 g
Since the water was being cooled, heat is removed from the system. Therefore, q is negative, and the temperature
decreases. The final temperature is:
T f = 23.52°C - 3.24°C = 20.28°C
Lesson Summary
• Chemical potential energy is stored within the chemical bonds of substances and can be converted into work
and heat as chemical reactions occur.
• Heat flows between system and surroundings due to a difference in temperature. Processes in which the
system absorbs heat are called endothermic, while processes in which the system releases heat are referred to
as exothermic.
• Heat capacity depends on the mass, identity, and physical state of a substance. Specific heat is the amount of
heat required to raise the temperature of 1 gram of a substance by 1°C. Water has a very high specific heat
compared to other substances.
• One equation is used to relate the heat absorbed or released by a substance to its specific heat, mass, and
change in temperature.
Lesson Review Questions
Reviewing Concepts
1. What is one potential use for substances that have a large amount of chemical potential energy? What happens
to that energy?
2. Describe what happens when two objects that have different temperatures come into contact with one another.
3. Describe the difference between an endothermic and an exothermic reaction.
4. Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very
warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and
which is exothermic? Explain.
5. What is the sign of q for an endothermic process? For an exothermic process?
6. Classify the following as endothermic or exothermic processes.
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1.1. Heat Flow
a.
b.
c.
d.
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boiling water
running a race
burning paper
water freezing
Problems
7. Make the following energy conversions.
a.
b.
c.
d.
345 cal to Cal
86.8 cal to J
217 J to cal
1.39 × 104 cal to kJ
8. 98.3 J of heat is supplied to 12.28 g of a substance, and its temperature rises by 5.42°C. What is the specific
heat of the substance?
9. 755 J of heat is supplied to 34.0 g of water, and an additional 755 J of heat is supplied to 34.0 g of iron. If
both samples are originally at 20.0°C, calculate the final temperature of the water and the iron. Comment on
the difference in your answers, and explain why water is used as a coolant in a car radiator.
10. A quantity of ethanol is cooled from 47.9°C to 12.3°C and releases 3.12 kJ of heat. What is the mass of the
ethanol sample?
11. How much heat is absorbed as 7.56 g of ice is heated from −30.0°C to its normal melting point?
12. A 25.70 g bar of gold at 23.20°C is brought into direct contact with a 19.87 g bar of silver at 81.93°C. Heat is
allowed to flow from the silver to the gold until both are at the same temperature. Assuming no heat is lost to
the surroundings, calculate the final temperature of the metals.
Further Reading / Supplemental Links
•
•
•
•
Energy in Chemical Reactions, http://www.wisc-online.com/Objects/ViewObject.aspx?ID=GCH8705
Endothermic and Exothermic Processes, http://www.kentchemistry.com/links/Matter/EndoExo.htm
Heat Flow, http://www.kentchemistry.com/links/Energy/heatflow.htm
Specific Heat, http://www.kentchemistry.com/links/Energy/SpecificHeat.htm
Points to Consider
A complete understanding of a chemical reaction requires knowledge of the amount of heat that is absorbed or
released during the reaction.
• How can heat be shown in the chemical equation?
• How can the heat involved in a reaction be used in stoichiometry calculations?
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Chapter 1. Thermochemistry
1.2 Thermochemical Equations
Lesson Objectives
• Define enthalpy, and know the conditions under which the enthalpy change in a reaction is equal to the heat
absorbed or released.
• Describe the principles behind calorimetry, and be able to calculate the heat absorbed or released during a
process that occurs in a calorimeter.
• Write and solve problems with thermochemical equations.
Lesson Vocabulary
•
•
•
•
•
calorimeter
calorimetry
enthalpy
heat of reaction
thermochemical equation
Check Your Understanding
Recalling Prior Knowledge
• How is specific heat used in calculating the heat absorbed or released in a process?
• What relationships are shown in a balanced chemical equation?
Heat is either absorbed or released during chemical reactions. In this lesson, you will learn about how heat changes
are measured in the lab and how quantities of heat are represented in a chemical equation.
Enthalpy
Heat changes in chemical reactions are most often measured in the laboratory under conditions in which the reacting
system is open to the atmosphere. In that case, the system is at a constant pressure. Enthalpy (H) is the heat
content of a system at constant pressure. Chemists routinely measure changes in the enthalpy of a chemical system
as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is
the same as the enthalpy change, which is given the symbol ∆H. Unless otherwise specified, all reactions in this text
are assumed to take place at constant pressure. Therefore, heat and enthalpy change will be used interchangeably,
and q = ∆H.
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1.2. Thermochemical Equations
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Calorimetry
Calorimetry is the measurement of the transfer of heat into or out of a system during a chemical reaction or physical
process. A calorimeter is an insulated container that is used to measure heat changes. The reactions that can most
easily be analyzed in a calorimetry experiment involve only liquids or aqueous solutions. A frequently used and
inexpensive calorimeter is a set of nested foam cups fitted with a lid to limit the heat exchange between the liquid
in the cup and the air in the surroundings ( Figure 1.5). In a typical calorimetry experiment, specific volumes of
the reactants are dispensed into separate containers, and the temperature of each is measured. They are then mixed
into the calorimeter, which starts the reaction. The reactant mixture is stirred until the reaction is complete, and the
temperature of the reaction is continuously monitored.
FIGURE 1.5
In a simple constant-pressure calorimeter, the temperature of a waterbased reaction is monitored as the reaction takes place. The substances
dissolved in the water are the system, and the water itself is the surroundings.
The key to all calorimetry experiments is the assumption that there is no heat exchange between the insulated
calorimeter and the room. Consider the case of a reaction taking place between aqueous reactants. The water in
which the solids have been dissolved is the surroundings, while the dissolved substances are the system. The temperature change that is measured is the temperature change that is occurring in the surroundings. If the temperature
of the water increases as the reaction occurs, the reaction is exothermic. Heat was released by the system into the
surrounding water. An endothermic reaction absorbs heat from the surroundings, so the temperature of the water
decreases as heat leaves the surroundings to enter the system.
The temperature change of the water is measured in the experiment, and the specific heat of water can be used to
calculate the heat absorbed by the surroundings (qsurr ).
qsurr = m × c p × ∆T
In this equation, m is the mass of the water, c p is the specific heat of the water, and ∆T is T f –Ti . The heat absorbed
by the surroundings is equal, but opposite in sign, to the heat released by the system. Because the heat change is
determined at constant pressure, the heat released by the system (qsys ) is equal to the enthalpy change (∆H).
qsys = ∆H = -qsurr = -(m × c p × ∆T)
The sign of ∆H is positive for an endothermic reaction and negative for an exothermic reaction.
Sample Problem 17.2: Calorimetry and Enthalpy Changes
In an experiment, 25.0 mL of 1.00 M HCl at 25.0°C is added to 25.0 mL of 1.00 M NaOH at 25.0°C in a foam cup
calorimeter. As the reaction occurs, the temperature of the solution rises to 32.0°C. Calculate the enthalpy change
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Chapter 1. Thermochemistry
(∆H) in kJ for this reaction. Assume the densities of the solutions are 1.00 g/mL and that their specific heats are the
same as that of pure water.
Step 1: List the known quantities and plan the problem.
Known
•
•
•
•
c p = 4.18 J/g•°C
V f inal = 25.0 mL + 25.0 mL = 50.0 mL
∆T = 32.0°C –25.0°C = 7.0°
Density = 1.00 g/mL
Unknown
• ∆H = ? kJ
The volume and density can be used to find the mass of the solution after mixing. Then, calculate the change in
enthalpy by using ∆H = qsys = −qsurr = −(m × c p × ∆T).
Step 2: Solve.
1.00 g
= 50.0 g
m = 50.0 mL ×
1 mL
∆H = −(m × c p × ∆T) = -(50.0 g × 4.18 J/g•°C × 7.0°C) = -1463 J = -1.5 kJ
Step 3: Think about the result.
The enthalpy change is negative because the reaction releases heat to the surroundings, causing the temperature of
the water to rise.
Practice Problems
1. A rock is heated and dropped into a foam cup calorimeter containing 35.0 mL of water at 21.4°C. The
temperature of the water rises to 24.8°C. How many joules of heat were released by the rock?
2. A 100.0 g sample of an unknown metal is heated to 95.00°C. This hot metal is then placed into a foam cup
calorimeter containing 50.0 g of water at 20.00°C. The water and metal come to a final temperature of 31.67°C.
Calculate the specific heat of the metal, and use the table above ( Table 1.1) to identify it.
Heats of Reaction
When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of
1 mol of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced
equation.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2 O(l) + 890.4 kJ
The equation tells us that 1 mol of methane combines with 2 mol of oxygen to produce 1 mol of carbon dioxide and
2 mol of water. In the process, 890.4 kJ is released, so it is written as a product of the reaction. A thermochemical
equation is a chemical equation that includes the enthalpy change of the reaction. The process in the above
thermochemical equation can be shown visually below ( Figure 1.6 (A)).
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1.2. Thermochemical Equations
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FIGURE 1.6
(A) As reactants are converted to products in an exothermic reaction, enthalpy
is released into the surroundings. The
enthalpy change of the reaction is negative. (B) As reactants are converted to
products in an endothermic reaction, enthalpy is absorbed from the surroundings.
The enthalpy change of the reaction is
positive.
In the combustion of methane example, the enthalpy change is negative because heat is being released by the system.
Therefore, the overall enthalpy of the system decreases. The heat of reaction is the enthalpy change for a chemical
reaction. In the case above, the heat of reaction is −890.4 kJ. The thermochemical reaction can also be written in
this way:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2 O(l) ∆H = −890.4 kJ
Heats of reaction are typically measured in kilojoules. It is important to include the physical states of the reactants
and products in a thermochemical equation, because the value of ∆H depends on those states.
Endothermic reactions absorb energy from the surroundings as the reaction occurs. When 1 mol of calcium carbonate
decomposes into 1 mol of calcium oxide and 1 mol of carbon dioxide, 177.8 kJ of heat is absorbed. This process is
shown visually above ( Figure 1.6 (B)). When heat is absorbed during a reaction, it can be written as a reactant. The
thermochemical reaction is shown below.
CaCO3 (s) + 177.8 kJ → CaO(s) + CO2 (g)
The reaction is endothermic, so the sign of the enthalpy change is positive.
CaCO3 (s) → CaO(s) + CO2 (g) ∆H = 177.8 kJ
Stoichiometry and Thermochemical Equations
Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems.
Refer again to the combustion reaction of methane. Since the reaction of 1 mol of methane releases 890.4 kJ, the
reaction of 2 mol of methane would release 2 × 890.4 kJ = 1781 kJ. The reaction of 0.5 mol of methane would
release 890.4 kJ / 2 = 445.2 kJ. As with other stoichiometry problems, the moles of a reactant or product can be
linked to mass or volume.
Sample Problem 17.3: Calculating Enthalpy Changes
Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction according to the following
thermochemical equation.
2SO2 (g) + O2 (g) → 2SO3 (g) + 198 kJ
Calculate the enthalpy change that occurs when 58.0 g of sulfur dioxide is reacted with excess oxygen.
Step 1: List the known quantities and plan the problem.
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Chapter 1. Thermochemistry
Known
• mass of SO2 = 58.0 g
• molar mass of SO2 = 64.07 g/mol
• ∆H = −198 kJ for the reaction of 2 mol SO2
Unknown
• ∆H = ? kJ
The calculation requires two steps. First, the mass of SO2 is converted to moles. Then, mol SO2 is multiplied by the
conversion factor (−198 kJ/2 mol SO2 ).
Step 2: Solve.
∆H = 58.0 g SO2 ×
−198 kJ
1 mol SO2
×
= −89.6 kJ
64.07 g SO2 2 mol SO2
Step 3: Think about your result.
The mass of sulfur dioxide is slightly less than 1 mol. Since 198 kJ is released for every 2 mol of SO2 that reacts,
the heat released when about 1 mol reacts is one half of 198. The 89.6 kJ is slightly less than half of 198. The sign
of ∆H is negative because the reaction is exothermic.
Practice Problem
3. Given the reaction below for the decomposition of mercury(II) oxide:
(a) What is ∆H when 0.750 mol of HgO fully reacts?
(b) A certain reaction produces 28.4 g of Hg. How much heat was absorbed in this reaction?
(c) What volume of oxygen gas at STP is produced in a reaction that absorbs 432 kJ of heat?
Lesson Summary
• Enthalpy is the heat content of a system. When a chemical reaction or physical process occurs at constant
pressure, the heat absorbed or released by the system is equal to the enthalpy change of the system.
• A calorimeter is an insulated device used in the laboratory to measure the enthalpy change during a reaction.
• Thermochemical equations show the heat that is either absorbed or released during a reaction. The enthalpy
change (∆H) for a reaction can be used as a conversion factor in solving problems.
Lesson Review Questions
Reviewing Concepts
1. What experimental condition is required for the heat change in a reaction to be numerically equal to the
enthalpy change (∆H)?
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1.2. Thermochemical Equations
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2. Why is a foam cup used as a calorimeter rather than a glass beaker?
3. What are some possible sources of error that would be present in an experiment where a foam cup is used as
a calorimeter?
4. When 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas, 2 mol of ammonia gas is produced and 92.6
kJ of heat is released. Write the thermochemical equation.
Problems
5. Given the following reaction for the formation of water from hydrogen and oxygen gases: What is the ∆H for
the following reactions?
a. H2 (g) + 1/2O2 (g) → H2 O(l)
b. 2H2 O(l) → 2H2 (g) + O2 (g)
6. 100. mL of 0.500 M HCl is mixed with 100. mL of 0.500 M NaOH in a foam cup calorimeter. The
initial temperatures of both solutions are 22.50°C. After the reaction occurs, the temperature rises to 26.00°C.
Calculate the enthalpy change for the reaction. Assume the densities of the solutions are 1.00 g/mL and the
specific heat is 4.18 J/g°C.
7. An ice cube is dropped into a foam cup calorimeter containing 95.0 mL of water at 20.0°C. The temperature
drops to 13.7°C as the ice cube melts. How much total heat (in kJ) was released by the water into the ice cube?
8. 66.80 g of lead is heated to 155°C and then placed into a foam cup calorimeter containing 70.0 mL of water
at 15.2°C. Assuming no heat loss, calculate the final temperature of the water and lead.
9. Given the balanced equation below for the highly exothermic reaction of aluminum with oxygen to form
aluminum oxide:
a.
b.
c.
d.
What is the value of ∆H when 1.00 mol of aluminum reacts?
What is the value of ∆H when 5.23 g of aluminum reacts?
What is the value of ∆H when 6.17 L of oxygen gas at STP reacts with excess aluminum?
A certain reaction releases 6881 kJ of heat. What mass of aluminum oxide was produced in the reaction?
10. Propane gas combusts according to the following equation. 45.1 L of propane at 0.915 atm is combusted at
513.°C. How much heat is released in the reaction? Assume the propane is an ideal gas.
Further Reading / Supplemental Links
• Enthalpy, http://www.wisc-online.com/Objects/ViewObject.aspx?ID=GCH8805
• The Thermochemical Equation, http://science.widener.edu/svb/tutorial/thermoequationscsn7.html
Points to Consider
Enthalpy changes always accompany changes in state.
• What are the heat of fusion and heat of vaporization, and how can they be used to solve problems?
• What is the heat of solution?
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Chapter 1. Thermochemistry
1.3 Heat and Changes of State
Lesson Objectives
•
•
•
•
Describe the enthalpy change that occurs as a substance changes between the solid and liquid states.
Describe the enthalpy change that occurs as a substance changes between the liquid and gas states.
Calculate the enthalpy change involved in the change of state for any amount of a given substance.
Calculate the enthalpy changes involved as substances dissolve in water.
Lesson Vocabulary
•
•
•
•
•
molar heat of condensation
molar heat of fusion
molar heat of solidification
molar heat of solution
molar heat of vaporization
Check Your Understanding
Recalling Prior Knowledge
• What happens to the temperature of a system during a change of state?
• What information is contained in a heating or cooling curve?
Enthalpy changes accompany physical processes as well as chemical reactions. In this lesson, you will examine
the heat absorbed or released during changes of state as well as the enthalpy change that occurs when a solute is
dissolved into a solvent.
Heats of Fusion and Solidification
Suppose you hold an ice cube in your hand. It feels cold because heat energy leaves your hand and enters the ice
cube. What happens to the ice cube? It melts. However, as you learned in an earlier chapter, the temperature during
a phase change remains constant. So the heat that is being lost by your hand does not raise the temperature of the ice
above its melting temperature of 0°C. Rather, all the heat goes into the change of state. Energy is absorbed during
the process of changing ice into water. The water that is produced also remains at 0°C until all of the ice has melted.
All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing
the state rather than changing the temperature. The molar heat of fusion (∆H f us ) of a substance is the heat absorbed
by one mole of that substance as it is converted from a solid to a liquid. Since the melting of any substance absorbs
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1.3. Heat and Changes of State
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heat, it follows that the freezing of any substance releases heat. The molar heat of solidification (∆Hsolid ) of a
substance is the heat released by one mole of that substance as it is converted from a liquid to a solid. Since fusion
and solidification of a given substance are the exact opposite processes, the numerical value of the molar heat of
fusion is the same as the numerical value of the molar heat of solidification, but opposite in sign. In other words,
∆H f us = −∆Hsolid . Shown below ( Figure 1.7) are all of the possible changes of state along with the direction of
heat flow during each process.
FIGURE 1.7
From left to right, heat is absorbed from
the surroundings during melting, evaporation, and sublimation.
From right to
left, heat is released to the surroundings
during freezing, condensation, and deposition.
Every substance has a unique value for its molar heat of fusion, depending on the amount of energy required to
disrupt the intermolecular forces present in the solid. When 1 mol of ice at 0°C is converted to 1 mol of liquid water
at 0°C, 6.01 kJ of heat are absorbed from the surroundings. When 1 mol of water at 0°C freezes to ice at 0°C, 6.01
kJ of heat are released into the surroundings.
H2 O(s) → H2 O(l) ∆Hfus = 6.01 kJ/mol
H2 O(l) → H2 O(s) ∆Hsolid = −6.01 kJ/mol
The molar heats of fusion and solidification of a given substance can be used to calculate the heat absorbed or
released when various amounts are melted or frozen.
Sample Problem 17.4: Heat of Fusion
Calculate the heat absorbed when 31.6 g of ice at 0°C is completely melted.
Step 1: List the known quantities and plan the problem.
Known
• mass of ice = 31.6 g
• molar mass of H2 O = 18.02 g/mol
• molar heat of fusion (H2 O) = 6.01 kJ/mol
Unknown
• ∆H = ? J
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Chapter 1. Thermochemistry
The mass of ice is first converted to moles. This is then multiplied by the conversion factor of (6.01 kJ/1 mol) in
order to find the kJ of heat absorbed.
Step 2: Solve.
31.6 g ice ×
1 mol ice
6.01 kJ
×
= 10.5 kJ
18.02 g ice 1 mol ice
Step 3: Think about your result.
The given quantity is a bit less than 2 moles of ice, so just under 12 kJ of heat is absorbed by the melting process.
Practice Problems
1. What mass of ice at 0°C can be melted by the addition of 559 J of heat?
2. What is ∆H when 250. g of water is frozen?
Heat of Vaporization and Condensation
Energy is also absorbed when a liquid is converted into a gas. As with the melting of a solid, the temperature
of a boiling liquid remains constant, and the input of energy goes into changing the state. The molar heat of
vaporization (∆Hvap ) of a substance is the heat absorbed by one mole of that substance as it is converted from
a liquid to a gas. As a gas condenses to a liquid, heat is released. The molar heat of condensation (∆Hcond )
of a substance is the heat released by one mole of that substance as it is converted from a gas to a liquid. Since
vaporization and condensation of a given substance are the exact opposite processes, the numerical value of the
molar heat of vaporization is the same as the numerical value of the molar heat of condensation, but opposite in sign.
In other words, ∆Hvap = −∆Hcond .
When 1 mol of water at 100°C and 1 atm pressure is converted to 1 mol of water vapor at 100°C, 40.7 kJ of heat are
absorbed from the surroundings. When 1 mol of water vapor at 100°C condenses to liquid water at 100°C, 40.7 kJ
of heat are released into the surroundings.
H2 O(l) → H2 O(g) ∆Hvap = 40.7 kJ/mol
H2 O(g) → H2 O(l) ∆Hcond = −40.7 kJ/mol
Molar heats of fusion and vaporization for some other substances are given in the table below ( Table 1.2).
TABLE 1.2: Molar Heats of Fusion and Vaporization
Substance
Ammonia (NH3 )
Ethanol (C2 H5 OH)
Methanol (CH3 OH)
Oxygen (O2 )
Water (H2 O)
∆H f us (kJ/mol)
5.65
4.60
3.16
0.44
6.01
∆Hvap (kJ/mol)
23.4
43.5
35.3
6.82
40.7
Notice that for all substances, the heat of vaporization is substantially higher than the heat of fusion. Much more
energy is required to change the state from a liquid to a gas than from a solid to a liquid. This is because of the
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1.3. Heat and Changes of State
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large separation of the particles in the gas state. The values of the heats of fusion and vaporization are related to the
strength of the intermolecular forces. All of the substances listed above ( Table 1.2), with the exception of oxygen,
are capable of hydrogen bonding. Consequently, the heats of fusion and vaporization of oxygen are far lower than
the others.
Sample Problem 17.5: Heat of Vaporization
What mass of methanol vapor condenses to a liquid as 20.0 kJ of heat are released?
Step 1: List the known quantities and plan the problem.
Known
• ∆H = -20.0 kJ
• ∆Hcond = −35.3 kJ/mol
• molar mass of CH3 OH = 32.05 g/mol
Unknown
• mass of methanol = ? g
First, the amount of heat released in the condensation is multiplied by the conversion factor of (1 mol/−35.3 kJ) to
find the moles of methanol that condensed. Then, moles are converted to grams.
Step 2: Solve.
−20.0 kJ ×
1 mol CH3 OH 32.05 g CH3 OH
×
= 18.2 g CH3 OH
−35.3 kJ
1 mol CH3 OH
Step 3: Think about your result.
Condensation is an exothermic process, so the enthalpy change is negative. Slightly more than one half of a mole of
methanol has condensed.
Practice Problem
3. How much heat is absorbed when 1.00 g of liquid ammonia is vaporized at its boiling point?
Multi-Step Problems with Changes of State
In the chapter States of Matter, you learned about heating curves and how they show the phase changes that a
substance undergoes as heat is continuously absorbed.
In the previous lesson, Thermochemical Equations, you learned how to use the specific heat of a substance to
calculate the heat absorbed or released as the temperature of the substance changes. It is possible to combine that
type of problem with a change of state to solve a problem involving multiple steps. The figure above ( Figure 1.8)
shows ice at −30°C being converted in a five-step process to gaseous water (steam) at 140°C. It is now possible
to calculate the heat absorbed during that entire process. The process and the required calculation is summarized
below.
1. Ice is heated from −30°C to 0°C. The heat absorbed is calculated by using the specific heat of ice and the
equation ∆H = m × c p × ∆T.
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Chapter 1. Thermochemistry
FIGURE 1.8
Heating curve of water. The process of
converting ice below its melting point to
steam above its boiling point involves five
distinct steps. The enthalpy change for
each step can be calculated separately.
2. Ice is melted at 0°C. The heat absorbed is calculated by multiplying the moles of ice by the molar heat of
fusion.
3. Water at 0°C is heated to 100°C. The heat absorbed is calculated by using the specific heat of water and the
equation ∆H = m × c p × ∆T.
4. Water is vaporized to steam at 100°C. The heat absorbed is calculated by multiplying the moles of water by
the molar heat of vaporization.
5. Steam is heated from 100°C to 140°C. The heat absorbed is calculated by using the specific heat of steam and
the equation ∆H = m × c p × ∆T.
Sample Problem 17.6: Multi-Step Problems using a Heating Curve
Calculate the total amount of heat absorbed (in kJ) when 2.00 mol of ice at −30.0°C is converted to steam at 140.0°C.
The required specific heats can be found in the table above ( Table 1.1).
Step 1: List the known quantities and plan the problem.
Known
•
•
•
•
•
•
2.00 mol ice = 36.04 g ice
c p (ice) = 2.06 J/g•°C
c p (water) = 4.18 J/g•°C
c p (steam) = 1.87 J/g•°C
∆H f us = 6.01 kJ/mol
∆Hvap = 40.7 kJ/mol
Unknown
• ∆Htotal = ? kJ
Follow the five steps outlined in the text. Note that the mass of the water is needed for the calculations that involve
the specific heat, while the moles of water is needed for the calculations that involve changes of state. All heat
quantities must be in kilojoules so that they can be added together to get a total for the five-step process.
Step 2: Solve.
1. ∆H1 = 36.04 g × 2.06 J/g ·◦ C × 30◦ C ×
1 kJ
= 2.23 kJ
1000 J
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1.3. Heat and Changes of State
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6.01 kJ
= 12.0 kJ
1 mol
1 kJ
3. ∆H3 = 36.04 g × 4.18 J/g ·◦ C × 100◦ C ×
= 15.1 kJ
1000 J
40.7 kJ
4. ∆H4 = 2.00 mol ×
= 81.4 kJ
1 mol
1 kJ
5. ∆H3 = 36.04 g × 1.87 J/g ·◦ C × 40◦ C ×
= 2.70 kJ
1000 J
2. ∆H2 = 2.00 mol ×
∆Htotal = ∆H1 + ∆H2 + ∆H3 + ∆H4 + ∆H5 = 113.4 kJ
Step 3: Think about your result.
The total heat absorbed as the ice at −30°C is heated to steam at 140°C is 113.4 kJ. By far, the largest absorption of
heat comes during the vaporization of the liquid water.
Practice Problem
4. Calculate the heat released when 100.0 g of water at 35.0°C is converted to ice at −18.0°C.
Heat of Solution
Enthalpy changes also occur when a solute undergoes the physical process of dissolving into a solvent. Hot packs
and cold packs ( Figure 1.9) use this property. Many hot packs use calcium chloride, which releases heat when it
dissolves according to the equation below.
CaCl2 (s) → Ca2+ (aq) + 2Cl− (aq) + 82.8 kJ
The molar heat of solution (∆Hsoln ) of a substance is the heat absorbed or released when one mole of the substance
is dissolved in water. For calcium chloride, ∆Hsoln = −82.8 kJ/mol.
FIGURE 1.9
Chemical hot packs and cold packs work
because of the heats of solution of the
chemicals inside them. When the bag is
squeezed, an inner pouch bursts, allowing
the chemical to dissolve in water. Heat is
released in a hot pack and absorbed by a
cold pack.
Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.
−
NH4 NO3 (s) + 25.7 kJ → NH+
4 (aq) + NO3 (aq)
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Chapter 1. Thermochemistry
Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the
affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps limit swelling. For
ammonium nitrate, ∆Hsoln = 25.7 kJ/mol.
Sample Problem 17.7: Heat of Solution
The molar heat of solution, ∆Hsoln , of NaOH is −445.1 kJ/mol. In a certain experiment, 5.00 g of NaOH is
completely dissolved in 1.000 L of water at 20.0°C in a foam cup calorimeter. Assuming no heat loss, calculate
the final temperature of the water.
Step 1: List the known quantities and plan the problem.
Known
•
•
•
•
•
•
mass of NaOH = 5.00 g
molar mass of NaOH = 40.00 g/mol
∆Hsoln (NaOH) = −445.1 kJ/mol
mass H2 O = 1.000 kg = 1000. g (assumes density = 1.000 g/mL)
Tinitial (H2 O) = 20.0°C
c p (H2 O) = 4.18 J/g•°C
Unknown
• T f inal of H2 O = ? °C
This is a multiple-step problem: 1) the mass of NaOH is converted to moles; 2) the resulting value is multiplied by
the molar heat of solution; 3) the heat released in the dissolving process is used with the specific heat equation and
the total mass of the solution to calculate ∆T; 4) T f inal is determined from ∆T.
Step 2: Solve.
−445.1 kJ
1000 J
1 mol NaOH
×
×
= −5.56 × 104 J
40.00 g NaOH 1 mol NaOH
1 kJ
q
−5.56 × 104 J
∆T =
=
= 13.2◦ C
cp × m 4.18 J/g ·◦ C × 1005 g
Tfinal = 20.0◦ C + 13.2◦ C = 33.2◦ C
5.00 g NaOH ×
Step 3: Think about your result.
The dissolving process releases a large amount of heat, which causes the temperature of the solution to rise. Care
must be taken when preparing concentrated solutions of sodium hydroxide because of the large amounts of heat
released.
Practice Problem
5. Calculate the final temperature of a solution prepared by dissolving 18.40 g of ammonium nitrate into 250. g
of water in a foam cup calorimeter. Both the ammonium nitrate and the water are initially at 22.0°C. Assume
that the heat capacity of the solution is the same as that of pure water.
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1.3. Heat and Changes of State
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Lesson Summary
• The molar heat of fusion is the heat absorbed when one mole of a solid melts and is numerically equivalent to
the molar heat of solidification, the heat released when one mole of the liquid freezes.
• The molar heat of vaporization is the heat absorbed when one mole of a liquid boils and is numerically
equivalent to the molar heat of condensation, the heat released when one mole of the gas condenses.
• Conversion factors can be used to calculate the heat absorbed or released during any change of state. Multiplestep problems, such as ice being transformed into steam, can be solved using specific heats along with heats
of fusion and vaporization.
• The molar heat of solution is the heat absorbed or released when one mole of a solute dissolves in water.
Lesson Review Questions
Reviewing Concepts
1. How does the molar heat of fusion of a substance compare to the molar heat of solidification?
2. How does the molar heat of vaporization of a substance compare to the molar heat of condensation?
3. State the names given to the following changes of state, and classify each as being endothermic or exothermic.
a.
b.
c.
d.
e.
Ar(g) → Ar(l)
KBr(s) → KBr(l)
C4 H10 (l) → C4 H10 (g)
CO2 (s) → CO2 (g)
Br2 (l) → Br2 (s)
4. Explain why the temperature of ice at 0°C does not initially rise when heat is added to it.
5. The molar heat of solution of ammonia is −30.50 kJ/mol. What happens to the temperature of water when
ammonia is added to it?
Problems
6. Heats of fusion and vaporization can also be expressed in J/g or kJ/g.
a. Use the molar mass of H2 O to convert the molar heat of fusion and the molar heat of vaporization of
water to kJ/g.
b. Calculate the energy required to melt 50.0 g of ice at 0°C and to boil 50.0 g of water at 100°C.
7. Calculate the quantity of heat that is absorbed or released (in kJ) during each of the following processes.
a.
b.
c.
d.
655 g of water vapor condenses at 100°C.
3.25 g of CaCl2 (s) is dissolved in water.
8.20 kg of water is frozen.
40.0 mL of ethanol is vaporized at 78.3°C (its boiling point). The density of ethanol is 0.789 g/mL.
8. Various systems are each supplied with 9.25 kJ of heat. Calculate the mass of each substance that will undergo
the indicated process with this input of heat.
a.
b.
c.
d.
22
melt ice at 0°C
vaporize water at 100°C
dissolve NH4 NO3 in water
vaporize water originally at 25.0°C
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Chapter 1. Thermochemistry
9. Calculate the total amount of heat (in kJ) absorbed by the process of converting 4.33 g of ice at −60.0°C to
steam at 185°C.
10. An ice cube at 0°C was dropped into a foam cup containing 100. g of water at 40.0°C. The final temperature
of the water in the cup is 21.2°C. Assuming no heat loss, what was the mass of the ice cube?
11. The molar heat of solution of NaCl is 3.88 kJ/mol. 125 g of NaCl is dissolved in 500. g of water at 20.0°C
in a calorimeter. What is the temperature of the solution when the salt has completely dissolved? Assume the
specific heat of the solution is the same as that of water.
Further Reading / Supplemental Links
• Heat of Fusion and Heat of Vaporization, http://www.wisc-online.com/Objects/ViewObject.aspx?ID=GCH460
4
• Heat of Fusion of Water, http://www.kentchemistry.com/links/Energy/HeatFusion.htm
• Heat of Vaporization of Water, http://www.kentchemistry.com/links/Energy/HeatVaporization.htm
• Complex Calorimetry Problems, http://www.kentchemistry.com/links/Energy/ComplexCalProblems.htm
Points to Consider
A formation reaction is a reaction in which elements in their standard states are combined to form a compound.
• How can knowledge of heats of formation be used to determine the enthalpy change for any reaction?
• What does it mean to say that heats of reaction are additive?
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1.4. Hess’s Law
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1.4 Hess’s Law
Lesson Objectives
• Use Hess’s law of heat summation to add chemical reactions together in such a way as to produce a desired
final equation. Calculate the enthalpy change for that final reaction.
• Define the standard heat of formation of a compound.
• Use known values for standard heats of formation to calculate a previously unknown standard heat of reaction.
Lesson Vocabulary
• heat of combustion
• Hess’s law of heat summation
• standard heat of formation
Check Your Understanding
Recalling Prior Knowledge
• What is a combustion reaction?
• How does a compound compare to an element?
Calorimetry is an experimental technique used to directly measure a heat of reaction in a laboratory setting. However,
many reactions for which it may be desirable to know the heat of reaction are too difficult to perform in a controlled
manner. In this lesson, you will learn about two indirect methods that will allow you to find the enthalpy change for
almost any chemical reaction.
Adding Heats of Reaction
It is sometimes very difficult or even impossible to measure the enthalpy change for a reaction directly in the
laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a
given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because
multiple side reactions may occur at the same time. Fortunately, it is possible to measure the enthalpy change for a
reaction by an indirect method. Hess’s law of heat summation states that if two or more thermochemical equations
are added together to give a final equation, then the heats of reaction for those equations can also be added together
to give a heat of reaction for the final equation.
An example will illustrate how Hess’s law can be used. Acetylene (C2 H2 ) is a gas that burns at an extremely high
temperature (3300°C) and is used in welding ( Figure 1.10). On paper, acetylene gas can be produced by the reaction
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Chapter 1. Thermochemistry
of solid carbon (graphite) with hydrogen gas. However, this is not an especially easy reaction for which to measure
the total enthalpy change. In contrast, enthalpy changes for combustion reactions are relatively easy to measure. The
heat of combustion is the heat released when one mole of a substance is completely reacted with oxygen gas.
FIGURE 1.10
The heats of combustion for carbon, hydrogen, and acetylene are shown below along with the balanced equation for
each process.
1. C(s, graphite) + O2 (g) → CO2 (g)
∆H = −393.5 kJ
1
2. H2 (g) + O2 (g) → H2 O(l)
∆H = −285.8 kJ
2
5
3. C2 H2 (g) + O2 (g) → 2CO2 (g) + H2 O(l) ∆H = −1301.1 kJ
2
The combustion reactions are written with fractional coefficients for O2 because the heats of combustion that are
found in a table are for the combustion of 1 mol of the given substance. To use Hess’s law, we need to determine
how the three equations above can be manipulated so that they can be added together to result in the desired equation
(the formation of acetylene from carbon and hydrogen).
In order to do this, we will go through the desired equation, one substance at a time—choosing the combustion
reaction from the equations numbered 1-3 above that contains that substance. It may be necessary to either reverse a
combustion reaction or multiply it by some factor in order to make it “fit” to the desired equation. The first reactant
is carbon, and the in the equation for the desired reaction, the coefficient of the carbon is a 2. We will therefore start
by writing the first combustion reaction with all of its coefficients doubled. Because we are doubling the coefficients,
we also need to double the value of ∆H, since we are now looking at the heat released when two moles of graphite
are combusted instead of just one.
2C(s, graphite) + 2O2 (g) → 2CO2 (g)
∆H = 2(−393.5) = −787.0 kJ
The second reactant is hydrogen, and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that
reaction will be used as written.
1
H2 (g) + O2 (g) → H2 O(l)
2
∆H = −285.8 kJ
The product of the reaction is C2 H2 , and its coefficient is also a 1. In combustion reaction #3, the acetylene is a
reactant. Therefore, we will reverse reaction 3. When the reactants and products are reversed, ∆H for the resulting
equation has the same numerical value but the opposite sign.
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1.4. Hess’s Law
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5
2CO2 (g) + H2 O(l) → C2 H2 (g) + O2 (g) ∆H = 1301.1 kJ
2
Now, these three reactions can be summed together. Any substance that appears in equal quantities as a reactant in
one equation and a product in another equation, cancels out algebraically. The values for the enthalpy changes are
likewise added.
2CO
2C(s, graphite) + 2O
2 (g) → 2 (g)
1 H2 (g) + O
H2
O(l)
2 (g) → 2
∆H = 2(−393.5) = −787.0 kJ
∆H = −285.8 kJ
→ C H (g) + 5 O 2CO
H2
O(l)
2 2
2 (g) + 2 (g)
2
2C(s, graphite) + H2 (g) → C2 H2 (g)
∆H = 1301.1 kJ
∆H = 228.3 kJ
So the heat of reaction for the combination of carbon with hydrogen to produce acetylene is 228.3 kJ. When one
mole of acetylene is produced, 228.3 kJ of heat are absorbed, making the reaction endothermic.
Standard Heat of Formation
A relatively straightforward chemical reaction is one in which elements are combined to form a compound. For
example, sodium and chlorine react to form sodium chloride (as seen in the video below), and hydrogen and oxygen
combine to form water. Like other reactions, these are accompanied by either the absorption or release of heat.
The standard heat of formation (∆H f °) is the enthalpy change associated with the formation of one mole of a
compound from its elements in their standard states. The standard conditions for thermochemistry are 25°C and
101.3 kPa. Therefore, the standard state of an element is its state at 25°C and 101.3 kPa. For example, iron is a
solid, bromine is a liquid, and oxygen is a gas under those conditions. The standard heat of formation of an element
in its standard state is by definition equal to zero. ∆H f ° = 0 for H2 (g), N2 (g), O2 (g), F2 (g), Cl2 (g), Br2 (l), and I2 (g),
because these elements are most commonly found in their pure form as diatomic molecules. The graphite form of
solid carbon is defined as its standard state, so it has a ∆H f ° of 0. Because diamond is not the standard state of
carbon, it has a non-zero value for ∆H f °. Some standard heats of formation are listed below ( Table 1.3).
Elemental sodium (Na) and chlorine gas (Cl2 ) react to form sodium chloride, releasing 411 kJ of heat for every mole
of NaCl produced. This spectacular reaction can be seen in the following video: http://www.youtube.com/watch
?v=Ftw7a5ccubs (0:42).
MEDIA
Click image to the left for more content.
TABLE 1.3: Standard Heats of Formation of Selected Substances
Substance
Al2 O3 (s)
BaCl2 (s)
26
∆H f ° (kJ/mol)
−1669.8
−860.1
Substance
H2 O2 (l)
KCl(s)
∆H f ° (kJ/mol)
−187.6
−435.87
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Chapter 1. Thermochemistry
TABLE 1.3: (continued)
Substance
Br2 (g)
C (s, graphite)
C (s, diamond)
CH4 (g)
C2 H5 OH(l)
CO(g)
CO2 (g)
CaO(s)
CaCO3 (s)
HCl(g)
CuO(s)
CuSO4 (s)
Fe2 O3 (s)
H2 O(g)
H2 O(l)
∆H f ° (kJ/mol)
30.91
0
1.90
−74.85
−276.98
−110.5
−393.5
−635.6
−1206.9
−92.3
−155.2
−769.86
−822.2
−241.8
−285.8
Substance
NH3 (g)
NO(g)
NO2 (g)
NaCl(s)
O3 (g)
P(s, white)
P(s, red)
PbO(s)
S(rhombic)
S(monoclinic)
SO2 (g)
SO3 (g)
H2 S(g)
SiO2 (s, quartz)
ZnCl2 (s)
∆H f ° (kJ/mol)
−46.3
90.4
33.85
−411.0
142.2
0
−18.4
−217.86
0
0.30
−296.1
−395.2
−20.15
−910.7
−415.89
An application of Hess’s law allows us to use standard heats of formation to indirectly calculate the heat of reaction
for any reaction that occurs at standard conditions. An enthalpy change that occurs specifically under standard
conditions is called the standard enthalpy (or heat) of reaction and is given the symbol ∆H°. The standard heat of
reaction can be calculated by using the following equation.
∆H◦ = Σn∆H◦f (products) − Σn∆H◦f (reactants)
The symbol Σ is the Greek letter sigma and means “the sum of.” The standard heat of reaction is equal to the sum
of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the
reactants. The symbol “n” signifies that each heat of formation must first be multiplied by its coefficient in the
balanced equation. Sample Problem 17.8 illustrates the use of this equation.
Sample Problem 17.8: Calculating Standard Heat of Reaction
Calculate the standard heat of reaction (∆H°) for the reaction of nitrogen monoxide gas with oxygen to form nitrogen
dioxide gas.
Step 1: List the known quantities and plan the problem.
Known
• ∆H f ° for NO(g) = 90.4 kJ/mol
• ∆H f ° for O2 (g) = 0 (pure element in its standard state)
• ∆H f ° for NO2 (g) = 33.85 kJ/mol
Unknown
• ∆H° = ? kJ
First, write the balanced equation for the reaction. Then, apply the equation to calculate the standard heat of reaction
from the standard heats of formation.
Step 2: Solve.
The balanced equation is: 2NO(g) + O2 (g) → 2NO2 (g)
27
1.4. Hess’s Law
www.ck12.org
Applying the equation from the text:
∆H f ° = [2 mol NO2 (33.85 kJ/mol)] - [2 mol NO (90.4 kJ/mol) + 1 mol O2 (0 kJ/mol)] = -113 kJ
The standard heat of reaction is −113 kJ.
Step 3: Think about your result.
The reaction is exothermic, which makes sense, because combustion reactions typically release heat.
Practice Problem
1. Calculate ∆H° for the following reactions.
a. CaCO3 (s) → CaO(s) + CO2 (g)
b. 7O2 (g) + 4NH3 (g) → 4NO2 (g) + 6H2 O(l)
Lesson Summary
• Hess’s law of heat summation allows chemical reactions to be added together in such a way as to produce
a final desired equation. The heats of reaction for each individual reaction are also added, resulting in the
enthalpy change for the final reaction.
• The standard heat of formation is the enthalpy change that occurs when a compound is formed from its
elements in their standard states.
• The standard enthalpy change for any reaction is equal to the sum of the heats of formation of the products
minus the sum of the heats of formation of the reactants.
Lesson Review Questions
Reviewing Concepts
1.
2.
3.
4.
What are two reasons why the heat of reaction may not be able to be determined directly?
What are standard conditions for thermochemical problems?
What is the standard heat of formation of an element in its standard state?
For which of the following is the standard heat of formation equal to zero?
a.
b.
c.
d.
e.
f.
F(g)
F2 (g)
He(l)
Hg(l)
CO(g)
Co(s)
5. In general, the more negative the value of the ∆H f ° for a compound, the more stable the compound. Explain
why this is true. Use the table above ( Table 1.3) to choose the more stable substance from each pair below.
a.
b.
c.
d.
28
CO(g) or CO2 (g)
graphite or diamond
Al2 O3 or Fe2 O3
O2 (g) or O3 (g)
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Chapter 1. Thermochemistry
Problems
6. Write the balanced chemical equation for the formation of one mole of solid sodium nitrite, NaNO2 , from its
elements in their standard states.
7. Determine ∆H for the reaction below.
8. Given the heats of reaction for the following combustion reactions: Calculate the enthalpy of formation of
methanol from its elements, according to the following equation:
9. Find the enthalpy change for the reaction below for the formation of phosphorus pentachloride from its
elements. Use the thermochemical equations below.
10. Use standard heats of formation from the table above ( Table 1.3) to calculate the standard heat of reaction for
each of the following.
a. 2H2 O2 (l) → 2H2 O(l) + O2 (g)
b. Fe2 O3 (s) + 3CO(g) → 2Fe(s) + 3CO2 (g)
c. C2 H5 OH(l) + 3O2 (g) → 2CO2 (g) + 3H2 O(l)
Further Reading / Supplemental Links
• Hess’s Law, http://www.kentchemistry.com/aplinks/chapters/6thermochem/Hess.htm
• Hess’s Law: The Principle of the Conservation of Energy, http://www.science.uwaterloo.ca/~cchieh/cact/c12
0/hess.html
Points to Consider
Chemical reactions proceed at a wide variety of speeds or rates. Some reactions happen almost instantly, as soon as
the reactants come into contact with one another, while other reactions may take years to reach completion.
• How do chemical reactions occur on the molecular level?
• What factors influence the rate of a chemical reaction?
29
1.5. References
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1.5 References
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
30
Brett Levin. http://www.flickr.com/photos/scubabrett22/8380510248/ . CC BY 2.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
User:Raeky/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Mount_Storm_Power_Plant,_Areial.jpg . Public Domain
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Christopher Auyeung. CK-12 Foundation . CC BY-NC 3.0
Courtesy of Cpl. Bryson K. Jones, US Marine Corps. http://commons.wikimedia.org/wiki/File:Cutting_torch
.jpg . Public Domain