Lecture Presentation
Chapter 5
Thermochemistry
Okan Esentürk, Chemistry Dept., METU
© 2015 Pearson Education, Inc.
Thermochemistry
This chapter is about thermodynamics,
which is the study of energy transformations,
and
thermochemistry, which applies the field to
chemical reactions, specifically.
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Thermochemistry
• With the exception of the energy from the
Sun, most of the energy used in our daily
lives comes from chemical reactions.
- the combustion of gasoline,
- the production of electricity from coal,
- the heating of homes by natural gas,
- the use of batteries to power
electronic devices are
all examples of how chemistry is used to
produce energy.
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Chapter Objectives
• Define work and heat using the standard sign conventions.
• Define state functions and explain their importance.
• State the first law of thermodynamics in words and as an
equation.
• Use calorimetric data to obtain values of E and H for
chemical reactions.
• Define Hfo and write formation reactions for compounds.
• Explain Hess’s law in your own words.
• Calculate Ho for chemical reactions from tabulated data.
• Describe some important design considerations in choosing
a battery for a specific application.
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5
Energy
What kinds of energy can we think of?
1.
2.
3.
4.
5.
Kinetic Energy
Nuclear Energy
Radiant Energy
Chemical Energy
Potential Energy
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Energy
The ability to do work or transfer heat.
Work: Energy used to cause an object that has
mass to move.
Heat: Energy used to cause the temperature of an
object to rise.
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Kinetic Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
the energy
• Same mass higher speed,
higher kinetic energy
• Same speed higher mass,
higher kinetic energy
of motion.
At 80 km/h
a truck has more KE
than a car.
Thermochemistry
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Kinetic Energy
• Chemistry is interested in
the KE of atoms and
molecules.
• Although too small to be
seen, these particles have
mass and are in motion
and, therefore, possess KE.
• KE vs interaction
determines the phase.
Gas
Liquid
Solid
chem.libretexts.org
Thermochemistry
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Thermochemistry
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Potential Energy
Energy an object possesses by virtue of its
position or chemical composition.
the “stored” energy that arises from the
attractions and repulsions an object experiences
in relation to other objects.
the energy
of position.
Thermochemistry
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Potential Energy
• Potential energy is
energy an object
possesses by virtue of
its position or chemical
composition.
• The most important
form of potential energy
in molecules is
electrostatic potential
energy, Eel:
Thermochemistry
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Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1 ⎯⎯
s2
• An older, non-SI unit is still in
widespread use: The calorie (cal).
1 cal = 4.184 J
(Note: this is not the same as the calorie of
foods; the food calorie is 1 kcal!)
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Thermochemistry
System and Surroundings
– System - the part of the universe
being considered.
(here, the hydrogen and oxygen molecules)
– Surroundings - the remainder of
the universe.
(here, the cylinder and piston)
– System + surroundings = Universe
System and surroundings
are separated by a boundary.
Thermochemistry
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Definitions: System and Surroundings
Systems may be
• open
• closed
• isolated
https://lawofthermodynamicsinfo.com/whatThermochemistry
is-thermodynamic-system/
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Is a human being an isolated, closed, or open
system?
a. Closed system
b. Isolated system
c. Open system
Thermochemistry
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Definitions: Work
Energy used to move an object
over some distance is work:
w=Fd
where w is work,
F is the force, and
d is the distance over
which the force is exerted.
Lifting an object against the force of gravity.
If we define the object as the system, then we—as part of the
surroundings—are performing work on that system,
Thermochemistry
transferring energy to it.
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Heat
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
Thermochemistry
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Heat
• Heat is the energy transferred from
a hotter object to a colder one.
• A combustion reaction, such as the burning of
natural gas releases the chemical energy stored
in the molecules of the fuel. If we define the
substances involved in the reaction as the system
and everything else as the surroundings, we find
that the released energy causes the temperature
of the system to increase, then the surroundings.
Thermochemistry
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Conversion of Energy
• Energy can be converted from one type to another.
• The cyclist has potential energy as she sits on top of
the hill.
• As she coasts down the hill, her potential energy is
converted to kinetic energy until the bottom, where
the energy is converted to kinetic energy.
Thermochemistry
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First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
Thermochemistry
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Internal Energy
The internal energy, E, of a system is the sum of
all kinetic and potential energies of all components
(atoms and subatomic particles) of the system.
Thermochemistry
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Internal Energy
By definition, the change in internal energy, E,
is the final energy of the system minus the initial
energy of the system:
E = Efinal − Einitial
We generally cannot determine the
actual values of E and E for any
system of practical interest.
Nevertheless, we can determine the
value of E experimentally by applying
the first law of thermodynamics.
final
initial
Thermochemistry
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Changes in Internal Energy
If E > 0, Efinal > Einitial
- energy change is
called endergonic.
- the system absorbed
energy from the
surroundings.
Thermochemistry
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Changes in Internal Energy
If E < 0, Efinal < Einitial
– energy change is
called exergonic.
– the system released
energy to the
surroundings.
Thermochemistry
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Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is,
E = q + w
Thermochemistry
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E, q, w, and Their Signs
Thermochemistry
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Relating Heat and Work to Changes of Internal Energy
Practice Exercise
Consider the following four cases:
(i) A chemical process in which heat is absorbed. E > 0
(ii) A change in which q = 30 J, w = 44 J. E > 0
(iii) A process in which a system does work on its surroundings with no
change in q. E < 0
(iv) A process in which work is done on a system and an equal amount
of heat is withdrawn. E = 0
In how many of these cases does the internal energy of the system
decrease?
a) 0
b) 1
c) 2
d) 3
e) 4
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Thermochemistry
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Exchange of Heat between System and
Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
endo- means ‘into’
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Thermochemistry
Exchange of Heat between System and
Surroundings
• When heat is released by the system into the
surroundings, the process is exothermic.
exo- means ‘out of’
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Thermochemistry
When H2(g) and O2(g) react to form H2O(l), heat is
released to the surroundings. Consider the reverse
reaction, namely, the formation of H2(g) and O2(g) from
H2O(l): 2 H2O(l) ⟶ 2 H2(g) + O2(g).
Is this reaction exothermic or endothermic?
a. Endothermic, because heat is released to the surroundings
b. Exothermic, because heat is released to the surroundings
c. Endothermic, because heat is absorbed from the
surroundings
d. Exothermic, because heat is absorbed from the
surroundings
Thermochemistry
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Thermochemistry
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When H2(g) and O2(g) react to form H2O(l), heat is
released to the surroundings. Consider the reverse
reaction, namely, the formation of H2(g) and O2(g) from
H2O(l): 2 H2O(l) ⟶ 2 H2(g) + O2(g).
Is this reaction exothermic or endothermic?
a. Endothermic, because heat is released to the surroundings
b. Exothermic, because heat is released to the surroundings
c. Endothermic, because heat is absorbed from the
surroundings
d. Exothermic, because heat is absorbed from the
surroundings
Thermochemistry
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State Functions
• Usually finding that value of internal energy is simply
too complex a problem.
• However, it is known that the internal energy of a
system is independent of the path by which the system
achieved that state.
– In the system below, the water could have reached room
temperature from either direction.
Thermochemistry
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State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
• And so, E depends only on Einitial and Efinal.
Thermochemistry
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State Functions
• q and w are not state
functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
– But q and w are different
in the two cases.
Thermochemistry
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In what ways is the balance in your checkbook a
state function?
a. It is determined by many factors and involves data such as
the number of checks, date, and payee.
b. It is a numerical value calculated from known amounts of
checks.
c. It is calculated from multiple transactions.
d. It only depends on the net total of all transactions,
and not on the ways money is transferred into or
out of the account.
Thermochemistry
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Work (Pressure-Volume)
E involves not only
the heat q added to
or removed from the
system but also the
work w done by or
on the system.
Most commonly, the only kind of work produced by
chemical or physical changes open to the
atmosphere is the mechanical work
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associated with a change in volume.
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PV Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston:
w = −PV
For example, a
reaction of Zn metal
with HCl solution rises
the piston with H2 gas
formation since internal
pressure increses.
Thermochemistry
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Thermochemistry
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Practice Exercise 1: P-V Work
If a balloon is expanded from 0.055 to 1.403 L against an external
pressure of 1.02 atm, how many L-atm of work is done?
(a) −0.056 L-atm
(b) −1.37 L-atm
(c) 1.43 L-atm
(d) 1.49 L-atm
(e) 139 L-atm
Solution:
The volume change is:
ΔV = Vfinal − Vinitial = 1.403 L − 0.055 L = 1.348 L
Thus, the quantity of work is:
w = −PΔV = −(1.02 atm)(1.348 L) = −1.37 L-atm
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If a system does not change its volume during
the course of a process, does it do pressure–
volume work?
a. Yes, because pressure may change the value of work
in w = –PΔV.
b. No, because ΔV = 0.
Thermochemistry
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Practice – A certain process results in a gas system releasing 68.3 kJ of energy.
During the process, 15.8 kcal of heat is released by the system. If the external
pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0
L, what is the final volume of the gas?
(1 cal = 4.18 J, 101.3 J = 1.00 atm•L)
Solution:
∆𝐸 = 𝑞 + 𝑤  𝑤 = ∆𝐸 − 𝑞
∆𝐸 = −6.83𝑥104𝐽
𝑞 = −1.58𝑥104 𝑐𝑎𝑙 ×
𝑤 = −𝑃∆𝑉  ∆𝑉 =
4.18𝐽
1 𝑐𝑎𝑙
= −6.60𝑥104 𝐽
𝑤 = ∆𝐸 − 𝑞 = −6.83𝑥104𝐽 − −6.60𝑥104 𝐽
𝑤 = −2.3𝑥103 𝐽
1 𝑎𝑡𝑚𝐿
𝑤 = −2.3𝑥10 𝐽 ×
= −22.7 𝑎𝑡𝑚𝐿
101.3 𝐽
∆𝑉 =
𝑤
−𝑃
−22.7 𝑎𝑡𝑚𝐿
≅ 23 𝐿
−1 𝑎𝑡𝑚
𝑉2 = ∆𝑉 + 𝑉1
𝑉2 = 23 𝐿 + 10 𝐿 = 𝟑𝟑 𝑳
3
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Enthalpy
• If a process takes place at constant pressure (as
the majority of processes we study do) and the
only work done is this pressure–volume work, we
can account for heat flow during the process by
measuring the enthalpy of the system.
• Enthalpy: is the internal energy plus the
product of pressure and volume:
H = E + PV
Thermochemistry
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Enthalpy
• Thus the change in enthalpy is
H = (E + PV)
• When the system changes at constant
pressure, the change in enthalpy, H,
becomes
H = E + PV
Thermochemistry
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Enthalpy
• Since E = q + w and w = −PV, we
can substitute these into the enthalpy
expression:
H = E + PV
H = (q + w) − w
H = q
• So, at constant pressure, the change in
enthalpy is the heat gained or lost.
Thermochemistry
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Endothermic and Exothermic
• A process is endothermic
when H is positive.
• A process is exothermic
when H is negative.
Thermochemistry
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Sample Exercise 5.4 Determining the Sign of ΔH
Indicate the sign of the enthalpy change, ΔH, in the following
processes carried out under atmospheric pressure and indicate
whether each process is endothermic or exothermic:
(a) An ice cube melts H > 0, ENDOTHERMIC
(b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give
complete combustion to CO2 and H2O H < 0, EXOTHERMIC
Practice Exercise 1
A chemical reaction that gives off heat to its surroundings is said
negative
exothermic
to be ____________
and has a ____________
value of ΔH.
(a) endothermic, positive
(c) exothermic, positive
(b) endothermic, negative
(d) exothermic, negative
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Question: Molten gold poured into a mold solidifies at
atmospheric pressure. With the gold defined as the system, is
the solidification an exothermic or endothermic process?
EXOTHERMIC
Question: If a system does not change its volume during the
course of a process, does it do pressure–volume work?
a. Yes, because pressure may change the value of work in w = PΔV.
b. No, because ΔV = 0.
Question: When a reaction occurs in a flask, you notice that the
flask gets colder. What is the sign of ΔH?
a. +
b. –
c. Neither; ΔH = 0
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Enthalpy of Reaction
The change in enthalpy,
H, is the enthalpy of
the products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
Thermochemistry
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Enthalpy of Reaction
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction (at constant Pext).
Thermochemistry
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The Truth about Enthalpy
1. Enthalpy is an extensive property.
(Proportional to the reactant amount used.)
A + B → C H = +10 kJ
2A + 2B → 2C H = +20 kJ
2. H for a reaction in the forward direction is
equal in size, but opposite in sign, to H for the
reverse reaction. C → A + B
H = –10 kJ
3. H for a reaction depends on the state of the
products and the state of the reactants. Thermochemistry
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When a reaction occurs in a flask, you notice that
the flask gets colder. What is the sign of ΔH?
a. +
b. –
c. Neither; ΔH = 0
Thermochemistry
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If the reaction to form water were written
H2(g) + O2(g) ⟶ H2O(g), would you expect the
same value of ΔH as in Equation 5.17? Why or
why not?
a. Yes, because the reactants and products are the same.
b. No, because only half as much matter is involved.
c. Yes, because mass does not affect enthalpy change.
d. No, because enthalpy is a state function.
2 H2(g) + O2(g) ⟶ 2 H2O(g)
ΔH = –483.6 kJ
[5.17]
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Practice Exercise 1
The complete combustion of ethanol, C 2H5OH (FW = 46.0 g ⁄ mol),
proceeds as follows:
C2H5OH(l) + 3→O2(g) 2CO2(g) + 3H2O(l)
ΔH = −555 kJ
What is the enthalpy change for combustion of 15.0 g of ethanol?
(a) −12.1 kJ (b) −181 kJ (c) −422 kJ (d) −555 kJ (e) −1700 kJ
Solution:
∆𝐻𝑟𝑥𝑛 =
15 𝑔
46 𝑔/𝑚𝑜𝑙
× −555
𝑘𝐽
𝑚𝑜𝑙
= −181 𝑘𝐽
Practice Exercise 2
Hydrogen peroxide can decompose to water and oxygen by the reaction
2 H2O2(l) → 2 H2O(l) + O2(g)
ΔH = −196 kJ
Calculate the quantity of heat released when 5.00 g of H 2O2(l) decomposes
at constant pressure.
Solution:
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∆𝐻𝑟𝑥𝑛 =
5𝑔
34 𝑔/𝑚𝑜𝑙
× −196
𝑘𝐽
𝑚𝑜𝑙
= −28.8 𝑘𝐽
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Calorimetry
• Not possible to know the exact
enthalpy of reactants and
products
• The value of H can be
determined experimentally by
measuring the heat flow
accompanying a reaction at
constant pressure.
• calorimetry, the measurement
of heat flow.
• The instrument used to measure
heat flow is called a calorimeter.
Thermochemistry
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Heat Capacity
The amount of energy required to raise the
temperature of a substance by 1K (1C) is its heat
capacity, usually given for one mole of the substance.
Thermochemistry
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Heat Capacity and Specific Heat
Specific heat capacity
(or simply specific heat)
the amount of energy required
to raise the temperature of 1 g
of a substance by 1K (or 1C).
Thermochemistry
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Heat Capacity and Specific Heat
Specific heat, then, is
Thermochemistry
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Which substance in Table 5.2 undergoes the
greatest temperature change when the same
mass of each substance absorbs the same
quantity of heat?
a. Hg(l)
b. Fe(s)
c. Al(s)
d. H2O(l)
Thermochemistry
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Thermochemistry
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Practice Exercise
Suppose you have equal masses of two substances, A and B. When the
same amount of heat is added to samples of each, the temperature of
A increases by 14 C whereas that of B increases by 22 C. Which of
the following statements is true?
(a) The heat capacity of B is greater than that of A.
(b) The specific heat of A is greater than that of B.
(c) The molar heat capacity of B is greater than that of A.
(d) The volume of A is greater than that of B.
(e) The molar mass of A is greater than that of B.
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Practice Exercise
(a) Large beds of rocks are used in some solar-heated homes to store
heat. Assume that the specific heat of the rocks is 0.82 J ⁄ g-K. Calculate
the quantity of heat absorbed by 50.0 kg of rocks if their temperature
increases by 12.0 C.
(b) What temperature change would these rocks undergo if they
emitted 450 kJ of heat?
Solution:
(a) 𝑞 = 𝐶𝑠 × 𝑚 × ∆𝑇 = 0.82
(b) ∆𝑇 =
𝑞
𝐶𝑠×𝑚
=
𝐽
𝑔.𝐾
450000 𝐽
𝐽
0.82 ×50000 𝑔
𝑔.𝐾
× 50𝑥103 𝑔 × 12𝐾 = 4.9𝑥105 𝐽
= 11°𝐶 (𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒)
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Question: You have a Styrofoam cup with 50.0 g of water at 10 C.
You add a 50.0 g iron ball at 90 C to the water (Cwater= 4.18 J/C·g
and CFe = 0.45 J/C·g).
i. The final temperature of the water is:
a) Between 50 C and 90 C
b) 50 C
c) Between 10 C and 50 C
ii. Calculate the final temperature of the water.
Solution:
𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 = 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑖𝑟𝑜𝑛
𝐶𝑤𝑎𝑡𝑒𝑟 × 𝑚 × 𝑇𝑓𝑖𝑛𝑎𝑙 − 10°𝐶 = 𝐶𝐹𝑒 × 𝑚 × 90°𝐶 − 𝑇𝑓𝑖𝑛𝑎𝑙
𝐽
𝐽
4.18 ൗ℃. 𝑔 × 50 𝑔 × 𝑇𝑓𝑖𝑛𝑎𝑙 − 10°𝐶 = 0.45 ൗ℃. 𝑔 × 50 𝑔 × 90°𝐶 − 𝑇𝑓𝑖𝑛𝑎𝑙
𝑻𝒇𝒊𝒏𝒂𝒍 = 𝟏𝟖℃
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Constant Pressure Calorimetry
• By carrying out a reaction in
aqueous solution in a simple
calorimeter, the heat change for
the system can be found by
measuring the heat change for
the water in the calorimeter.
• The specific heat for water is
well known (4.184 J/g∙K).
• We can calculate H for the
reaction with this equation:
qsln = -qrxn and qsln = m  Cs  T
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Thermochemistry
Thermochemistry
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:
:
:
:
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Bomb Calorimetry
• Reactions can be carried
out in a sealed “bomb”.
• The heat absorbed (or
released) by the water is
a very good approximation
of the enthalpy change for
the reaction.
qrxn = – qcal = – Ccal × ∆T
Thermochemistry
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Bomb Calorimetry
• Because the volume in
the bomb calorimeter
is constant, what is
measured is really the
change in internal
energy, E, not H.
• For most reactions, the
difference is very
small.
E = qv
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Practice Exercise
The combustion of exactly 1.000 g of benzoic acid in a bomb calorimeter
releases 26.38 kJ of heat. If the combustion of 0.550 g of benzoic acid causes the
temperature of the calorimeter to increase from 22.01 to 24.27 C, calculate the
heat capacity of the calorimeter.
(a) 0.660 kJ/C
(b) 6.42 kJ/C
(c) 14.5 kJ/C
(d) 21.2 kJ/g-C
(e) 32.7 kJ/C
Solution:
Reaction Heat: qrxn = -26.38 kJ(0.550 g/1.000 g) = -14.51 kJ
ΔT = 24.27 C − 22.01 C = 2.26 C
qrxn = −CCal × ∆T
Ccal = -(qrxn/T) = -(-14.51 kJ/2.26 C) = 6.42 kJ/C
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Question: When a 0.50 g sample of benzoic acid is combusted
in a bomb calorimeter, the temperature of the calorimeter
raises by 1.5 C. When a 0.25 g of caffeine is burned, the
temperature of the calorimeter raises also by 1.5 C. Using the
value -24 kJ/g for the heat of combustion of benzoic acid,
calculate the heat of combustion per gram of caffeine.
Solution:
qrxn(caffeine) = qrxn(benzoic acid)
qrxn(benzoic acid) = (-24 kJ/g)(0.5g) = -12 kJ
qrxn(caffeine) = -12 kJ
qrxn(caffeine per gram) = (-12 kJ)/(0.25 g) = -48 kJ/g
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Hess’s Law
• H is well known for many reactions,
and it is inconvenient to measure H
for every reaction in which we are
interested.
• However, we can estimate H using
published H values and the
properties of enthalpy.
Thermochemistry
© 2015 Pearson Education
Hess’s Law
• Hess’s law: If a reaction is
carried out in a series of
steps, H for the overall
reaction will be equal to the
sum of the enthalpy changes
for the individual steps.
• Because H is a state
function, the total enthalpy
change depends only on the
initial state (reactants) and the
final state (products) of the
reaction.
Thermochemistry
© 2015 Pearson Education
Thermochemistry
79
© 2015 Pearson Education
Solution:
C(graphite) + O2(g) → CO2(g)
CO2(g) → C(diamond) + O2(g)
H = -393.5 kJ
H = 395.4 kJ
C(graphite) → C(diamond)
H = +1.9 kJ
Thermochemistry
80
© 2015 Pearson Education
Thermochemistry
81
© 2015 Pearson Education
Practice Exercise 1
We can calculate ΔH for the reaction
C(s) + H2O(g) → CO(g) + H2(g)
using the following thermochemical equations:
C(s) + O2(g) → CO2(g)
2 CO(g) + O2(g) → 2 CO2(g)
2 H2(g) + O2(g) → 2 H2O(g)
ΔH1 = −393.5 kJ
ΔH2 = −566.0 kJ
ΔH3 = −483.6 kJ
By what coefficient do you need to multiply ΔH2 in determining ΔH for the
target equation?
(a) −1 ⁄ 2
Solution:
x(-½)
x(-½)
(b) −1
(c) 2
(d) 1 ⁄ 2
(e) 2
C(s) + O2(g) → CO2(g)
ΔH1 = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g)
H2O(g) → H2(g) + ½ O2(g)
ΔH2 = +283.0 kJ
ΔH3 = +241.8 kJ
C(s) + H2O(g) → CO(g) + H2(g) Hrxn = +131.3 kJ
© 2015 Pearson Education
Thermochemistry
82
Solution:
NO(g) + O3(g) → NO2(g) + O2(g)
3/2 O2(g) → O3(g)
O(g) → ½ O2(g)
NO(g) + O(g) → NO2(g)
H = -198.9 kJ
H = +142.3 kJ
H = -247.5 kJ
H = -304.1 kJ
Thermochemistry
83
© 2015 Pearson Education
Excercise:
Given the following information:
2 NO(g) + O2(g) → 2 NO2(g)
2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(aq)
N2(g) + O2(g) → 2 NO(g)
H° = −116 kJ
H° = −256 kJ
H° = +183 kJ
Calculate the H for the reaction below:
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
H° = ?
Solution:
[2 NO2(g) → 2 NO(g) + O2(g)] x 1.5
H° = 1.5(+116 kJ)
[2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(aq)] x 0.5 H° = 0.5(−256 kJ)
[2 NO(g) → N2(g) + O2(g)]
H° = −183 kJ
Thermochemistry
84
© 2015 Pearson Education
Excercise:
Given the following information:
2 NO(g) + O2(g) → 2 NO2(g)
2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(aq)
N2(g) + O2(g) → 2 NO(g)
H° = −116 kJ
H° = −256 kJ
H° = +183 kJ
Calculate the H for the reaction below:
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
H° = ?
Solution:
[3 NO2(g) → 3 NO(g) + 1.5 O2(g)]
[1 N2(g) + 2.5 O2(g) + 1 H2O(l) → 2 HNO3(aq)]
[2 NO(g) → N2(g) + O2(g)]
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
H° = (+174 kJ)
H° = (−128 kJ)
H° = (−183 kJ)
H° = − 137 kJ
Thermochemistry
85
© 2015 Pearson Education
Enthalpies of Formation
An enthalpy of formation, Hf, is
defined as the enthalpy change for the
reaction in which a compound is made
from its constituent elements in their
elemental forms.
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g)
4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
Hf =-393.5 kJ
Hf =-285.8 kJ
By definition, the standard Hf of the most stable form of
any element is zero because there is no formation reaction
Thermochemistry
needed when the element is already in its standard state.
© 2015 Pearson Education
Standard Enthalpies of Formation
Standard enthalpies of formation, ∆Hf°, are
measured under standard conditions (25 ºC
and 1.00 atm pressure).
Thermochemistry
© 2015 Pearson Education
Standard Enthalpies of Formation
Thermochemistry
88
© 2015 Pearson Education
Yes
No
No
Thermochemistry
89
© 2015 Pearson Education
Practice Exercise 1
If the heat of formation of H2O(l) is −286 kJ/mol, which of the following
thermochemical equations is correct?
(a) 2H(g) + O(g) → H2O(l)
ΔH = −286 kJ
(b) 2H2(g) + O2(g) → 2H2O(l)
ΔH = −286 kJ
(c) H2(g) + ½ O2(g) → H2O(l)
ΔH = −286 kJ
(d) H2(g) + O(g) → H2O(g)
ΔH = −286 kJ
(e) H2O(l) → H2(g) + O2(g)
ΔH = −286 kJ
Thermochemistry
90
© 2015 Pearson Education
Practice Exercise 2
Write the equation corresponding to the standard enthalpy of formation of
liquid carbon tetrachloride (CCl4) and look up ΔHf for this compound in
Appendix C.
Solution:
C(s, graphite) + 2Cl2(g) → CCl4(l)*
*ΔHf for CCl4(l) obtained from another list of heats of formation.
© 2015 Pearson Education
ΔHf = -139.3 kJ/mol
Thermochemistry
91
Calculation of H
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
• Imagine this as occurring
in three steps:
1) Decomposition of propane
to the elements:
C3H8(g) ⎯→ 3 C(graphite) + 4 H2(g)
Thermochemistry
© 2015 Pearson Education
Calculation of H
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
• Imagine this as occurring
in three steps:
2) Formation of CO2:
3 C(graphite) + 3 O2(g) ⎯→3 CO2(g)
Thermochemistry
© 2015 Pearson Education
Calculation of H
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
• Imagine this as occurring
in three steps:
3) Formation of H2O:
4 H2(g) + 2 O2(g) ⎯→ 4 H2O(l)
Thermochemistry
© 2015 Pearson Education
Calculation of H
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
• So, all steps look like this:
C3H8(g) ⎯→ 3 C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) ⎯→3 CO2(g)
4 H2(g) + 2 O2(g) ⎯→ 4 H2O(l)
Thermochemistry
© 2015 Pearson Education
Calculation of H
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
• The sum of these equations is the overall equation!
C3H8(g) ⎯→ 3 C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) ⎯→3 CO2(g)
4 H2(g) + 2 O2(g) ⎯→ 4 H2O(l)
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
Thermochemistry
© 2015 Pearson Education
Calculation of H
We can use Hess’s law in this way:
𝑜
𝑜
Δ𝐻 = σ n Δ𝐻𝑓,𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
− σ m Δ𝐻𝑓,𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
where n and m are the stoichiometric
coefficients.
Thermochemistry
© 2015 Pearson Education
Calculation of H using Values from the
Standard Enthalpy Table
C3H8(g) + 5 O2(g) ⎯→ 3 CO2(g) + 4 H2O(l)
H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)]
= [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)]
= (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ
Thermochemistry
© 2015 Pearson Education
Energy in Foods
Most of the fuel in the food we eat comes
from carbohydrates and fats.
Thermochemistry
© 2015 Pearson Education
Energy in Foods
• Most of the energy our bodies need comes from
carbohydrates and fats.The carbohydrates known as
starches are decomposed in the intestines into glucose,
C6H12O6.
• Glucose is soluble in blood, and in the human body it is
known as blood sugar. It is transported by the blood to cells
where it reacts with O2 in a series of steps, eventually
producing CO2, H2O, and energy:
Thermochemistry
© 2015 Pearson Education
Energy in Foods
Because carbohydrates break down rapidly, their energy is
quickly supplied to the body. However, the body stores only a
very small amount of carbohydrates. The average fuel value
of carbohydrates is 17 kJ/g (4 kcal/g).
Like carbohydrates, fats produce CO and H O when
metabolized. The reaction of tristearin, C57H110O6, a typical
fat, is
2
2
Thermochemistry
© 2015 Pearson Education
Energy in Foods
The body uses the chemical energy from foods to maintain body
temperature to contract muscles, and to construct and repair tissues.
Any excess energy is stored as fats. Fats are well suited to serve as the
body’s energy reserve for at least two reasons:
(1) They are insoluble in water, which facilitates storage in the body,
(2) they produce more energy per gram than either proteins or
carbohydrates, which makes them efficient energy sources on a
mass basis. The average fuel value of fats is 38 kJ/g 19 kcal/g.
Thermochemistry
© 2015 Pearson Education
Energy in Fuels
The vast majority of the
energy consumed in
this country comes from
fossil fuels.
Thermochemistry
© 2015 Pearson Education
Other Energy Sources
• Nuclear fission produces
8.5% of the U.S. energy
needs.
• Renewable energy
sources, like solar,
wind, geothermal,
hydroelectric, and
biomass sources
produce 7.4% of the
U.S. energy needs.
Thermochemistry
© 2015 Pearson Education
Energy in Turkey
Energy consumption in Turkey based on energy source.
• Fossil fuels supply roughly 89% of consumed energy.
o Self sufficiency in 1990: 48.1%
o Self sufficiency in 2011: 28.2%
© 2015 Pearson Education
Thermochemistry
105
Energy in Turkey
Change in energy consumption and source distribution in
Turkey between 1980 and 2011.
Thermochemistry
106
© 2015 Pearson Education
The sum of all of the kinetic and
potential energies of a system is
called the
a.
b.
c.
d.
integral energy.
dynamic energy.
internal energy.
work.
Thermochemistry
© 2015 Pearson Education
Which set of values for heat and
work will result in a decrease of
internal energy?
a.
b.
c.
d.
q = –150 J; w = +150 J
q = –150 J; w = +300 J
q = +150 J; w = –300 J
q = +300 J; w = –150 J
Thermochemistry
© 2015 Pearson Education
Which set of values for heat and
work will result in a decrease of
internal energy?
a.
b.
c.
d.
q = –150 J; w = +150 J
q = –150 J; w = +300 J
q = +150 J; w = –300 J
q = +300 J; w = –150 J
Thermochemistry
© 2015 Pearson Education
A system absorbs heat during an
_______ process.
a.
b.
c.
d.
exothermic
isothermal
endothermic
isobaric
Thermochemistry
© 2015 Pearson Education
A system absorbs heat during an
_______ process.
a.
b.
c.
d.
exothermic
isothermal
endothermic
isobaric
Thermochemistry
© 2015 Pearson Education
Which group includes only state
functions?
a.
b.
c.
d.
P, T, V
P, H, w
H, q, w
P, R, q, w
Thermochemistry
© 2015 Pearson Education
Which group includes only state
functions?
a.
b.
c.
d.
P, T, V
P, H, w
H, q, w
P, R, q, w
Thermochemistry
© 2015 Pearson Education
Which of these is not a state
function?
a.
b.
c.
d.
enthalpy
internal energy
temperature
work
Thermochemistry
© 2015 Pearson Education
Which of these is not a state
function?
a.
b.
c.
d.
enthalpy
internal energy
temperature
work
Thermochemistry
© 2015 Pearson Education
Which statement about enthalpy is
not correct?
a.
b.
c.
d.
Enthalpy can be measured in joules per mole.
Enthalpy is a state function.
Enthalpy is an extensive property.
Change in enthalpy is negative for an
exothermic reaction.
Thermochemistry
© 2015 Pearson Education
Which statement about enthalpy is
not correct?
a. Enthalpy can be measured in joules per
mole.
b. Enthalpy is a state function.
c. Enthalpy is an extensive property.
d. Change in enthalpy is negative for an
exothermic reaction.
Thermochemistry
© 2015 Pearson Education
The laboratory technique used to
measure heat flow is called
a.
b.
c.
d.
calorimetry.
neutralization.
titration.
voltammetry.
Thermochemistry
© 2015 Pearson Education
The laboratory technique used to
measure heat flow is called
a.
b.
c.
d.
calorimetry.
neutralization.
titration.
voltammetry.
Thermochemistry
© 2015 Pearson Education
When a piece of iron at 356 K is
placed in water at 298 K, what
happens?
a.
b.
c.
d.
Energy
Energy
Energy
Energy
flows from iron to water.
flows from water to iron.
does not flow.
is not conserved.
Thermochemistry
© 2015 Pearson Education
When a piece of iron at 356 K is
placed in water at 298 K, what
happens?
a.
b.
c.
d.
Energy flows from iron to water.
Energy flows from water to iron.
Energy does not flow.
Energy is not conserved.
Thermochemistry
© 2015 Pearson Education
A substance’s specific heat is its heat
capacity per
a.
b.
c.
d.
mole.
gram.
joule.
kelvin.
Thermochemistry
© 2015 Pearson Education
A substance’s specific heat is its heat
capacity per
a.
b.
c.
d.
mole.
gram.
joule.
kelvin.
Thermochemistry
© 2015 Pearson Education
The specific heat of water is
relatively large. This means that in
response to a ____ amount of heat, it
shows a ____ change in
temperature.
a. large; large
c. large; small
b. small; large
d. small; small
Thermochemistry
© 2015 Pearson Education
The specific heat of water is
relatively large. This means that in
response to a ____ amount of heat, it
shows a ____ change in
temperature.
a. large; large
c. large; small
b. small; large
d. small; small
Thermochemistry
© 2015 Pearson Education
The standard enthalpy of formation of
carbon in its graphite form is ___
kJ/mole.
a.
b.
c.
d.
100
1000
1
0
© 2015 Pearson Education
Thermochemistry
The standard enthalpy of formation of
carbon in its graphite form is ___
kJ/mole.
a.
b.
c.
d.
100
1000
1
0
© 2015 Pearson Education
Thermochemistry
The standard enthalpy of
formation of carbon in its
diamond form is +1.88 kJ/mole,
which means that diamond is
__________ graphite.
a.
b.
c.
d.
as stable as
more stable than
less stable than
an isotope of
© 2015 Pearson Education
Thermochemistry
The standard enthalpy of
formation of carbon in its
diamond form is +1.88 kJ/mole,
which means that diamond is
__________ graphite.
a.
b.
c.
d.
as stable as
more stable than
less stable than
an isotope of
© 2015 Pearson Education
Thermochemistry
2 H2 + O2 → 2 H2O
If the reaction above releases 483.6
kJ, then the standard enthalpy of
formation of H2O =
a.
b.
c.
d.
–483.6 kJ/mole.
+483.6 kJ/mole.
–241.8 kJ/mole.
+241.8 kJ/mole.
© 2015 Pearson Education
Thermochemistry
2 H2 + O2 → 2 H2O
If the reaction above releases 483.6
kJ, then the standard enthalpy of
formation of H2O =
a.
b.
c.
d.
–483.6 kJ/mole.
+483.6 kJ/mole.
–241.8 kJ/mole.
+241.8 kJ/mole.
© 2015 Pearson Education
Thermochemistry
The average fuel values for proteins,
carbohydrates, and fats are,
respectively,
_______ kcal/g.
a.
b.
c.
d.
4, 4, and 4
4, 4, and 9
4, 9, and 4
9, 4, and 4
© 2015 Pearson Education
Thermochemistry
The average fuel values for proteins,
carbohydrates, and fats are,
respectively,
_______ kcal/g.
a.
b.
c.
d.
4, 4, and 4
4, 4, and 9
4, 9, and 4
9, 4, and 4
© 2015 Pearson Education
Thermochemistry
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
T
T
F
T
T
Thermochemistry
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Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
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gives off
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education
Thermochemistry
© 2015 Pearson Education