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Lecture Presentation
Chapter 5
Thermochemistry
John D. Bookstaver
St. Charles Community College
Cottleville, MO
© 2012 Pearson Education, Inc.
Energy
• Energy is the ability to do work or
transfer heat.
– Energy used to cause an object that has
mass to move is called work.
– Energy used to cause the temperature of
an object to rise is called heat.
Thermochemistry
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Kinetic Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
1
Ek =  mv2
2
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:
K Q 1Q 2
Eel =
d
Thermochemistry
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Conversion of Energy
• Energy can be converted from one type to
another.
• For example, the cyclist in Figure 5.2 has
potential energy as she sits on top of the hill.
Thermochemistry
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Conversion of Energy
• As she coasts down the hill, her potential
energy is converted to kinetic energy.
• At the bottom, all the potential energy she
had at the top of the hill is now kinetic energy.
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
Thermochemistry
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Definitions: System and Surroundings
• The system includes the
molecules we want to
study (here, the hydrogen
and oxygen molecules).
• The surroundings are
everything else (here, the
cylinder and piston).
Thermochemistry
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Types of Systems
• Open – matter and energy are exchanged
with surroundings
• Closed – only energy is exchanged
• Isolated – neither matter nor energy are
exchanged
• What type of system is the human body?
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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.
Thermochemistry
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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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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 of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
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
Efinal = products
Einitial = reactants
Thermochemistry
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Changes in Internal Energy
• If E > 0, Efinal > Einitial
– Therefore, the system absorbed energy
from the surroundings.
– This energy change is called endergonic.
Thermochemistry
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Changes in Internal Energy
• If E < 0, Efinal < Einitial
– Therefore, the system released energy to
the surroundings.
– This energy change is called exergonic.
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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Sample Exercise 5.2 Relating Heat and Work to Changes of Internal
Energy
Calculate the change in the internal energy for a process in
which a system absorbs 140 J of heat from the
surroundings and does 85 J of work on the surroundings.
Thermochemistry
Exchange of Heat between System and
Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
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.
• When heat is released by the system into the
surroundings, the process is exothermic.
Thermochemistry
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State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.
Thermochemistry
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State Functions
• However, we do know that the internal energy
of a system is independent of the path by
which the system achieved that state.
– In the system depicted in Figure 5.9, 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
• However, 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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Work
Usually in an open
container the only work
done is by a gas
pushing on the
surroundings (or by
the surroundings
pushing on the gas).
Thermochemistry
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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
Thermochemistry
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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
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Thermochemistry
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
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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Endothermicity and Exothermicity
• A process is
endothermic
when H is
positive.
Thermochemistry
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Endothermicity and Exothermicity
• A process is
endothermic
when H is
positive.
• A process is
exothermic when
H is negative.
Thermochemistry
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Sample Exercise 5.3 Determining the Sign of H
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?
Thermochemistry
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.
Thermochemistry
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The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. H for a reaction in the forward
direction is equal in size, but opposite
in sign, to H for the reverse reaction.
Thermochemistry
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The Truth about Enthalpy
3. H for a reaction depends on the state
of the products and the state of the
reactants.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H= -890 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H= -802 kJ
2H2O(l)  2H2O(g) H= +88 kJ
Thermochemistry
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Sample Exercise 5.4 Relating H to Quantities of Reactants and Products
Hydrogen peroxide can decompose to water and
oxygen by the reaction
2 H2O2(l)
2 H2O(l) + O2(g)
H = –196 kJ
Calculate the quantity of heat released when 5.00 g of
H2O2(l) decomposes at constant pressure.
Thermochemistry
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we
measure H through
calorimetry, the
measurement of
heat flow.
Thermochemistry
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Heat Capacity and Specific Heat
The amount of energy required to raise the
temperature of a substance by 1 K (1C) is its
heat capacity.
Thermochemistry
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Heat Capacity and Specific Heat
We define specific
heat capacity (or
simply specific heat)
as the amount of
energy required to
raise the temperature
of 1 g of a substance
by 1 K (or 1 C).
Thermochemistry
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Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
Cs =
q
m  T
q = Cs x m x T
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Thermochemistry
A.
B.
C.
D.
Hg(l)
Fe(s)
Al(s)
H2O(l)
Thermochemistry
Sample Exercise 5.5 Relating Heat, Temperature Change, and Heat
Capacity
(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?
Thermochemistry
Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, one can indirectly
measure the heat
change for the system
by measuring the heat
change for the water in
the calorimeter.
Thermochemistry
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Constant Pressure Calorimetry
Because the specific
heat for water is well
known (4.184 J/g-K), we
can measure H for the
reaction with this
equation:
q = m  s  T = H
qsoln = -qrxn
Thermochemistry
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Sample Exercise 5.6 Measuring H Using a Coffee-Cup Calorimeter
When 50.0 mL of 0.100 MAgNO3 and 50.0 mL of 0.100 M HCl
are mixed in a constant-pressure calorimeter,
the temperature of the mixture increases from 22.30 C to 23.11
C. The temperature increase is caused by the following
reaction:
AgNO3(aq) + HCl(aq)
AgCl(s) + HNO3(aq)
Calculate H for this reaction in AgNO3, assuming that the
combined solution has a mass of 100.0 g and a specific heat of
4.18 J/g C.
Thermochemistry
Bomb Calorimetry
• Reactions can be
carried out in a sealed
“bomb” such as this
one.
• The heat absorbed
(or released) by the
water is a very good
approximation of the
enthalpy change for
the reaction.
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.
Thermochemistry
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Sample Exercise 5.7 Measuring qrxn Using a Bomb Calorimeter
A 0.5865-g sample of lactic acid (HC3H5O3) is burned
in a calorimeter whose heat capacity is 4.812 kJ/C.
The temperature increases from 23.10 C to 24.95 C.
Calculate the heat of combustion of lactic acid (a) per
gram and (b) per mole.
Thermochemistry
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
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Hess’s Law
Hess’s law states that
“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.”
Thermochemistry
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Hess’s Law
Because H is a state
function, the total
enthalpy change
depends only on the
initial state of the
reactants and the final
state of the products.
Thermochemistry
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a.
A.
B.
C.
D.
No change.
Sign of H changes.
Value of H increases.
Value of H decreases.
Thermochemistry
b.
A.
B.
C.
D.
No change.
Sign of H changes.
Value of H doubles.
Value of H decreases by half.
Thermochemistry
Sample Exercise 5.8 Using Hess’s Law to Calculate H
Carbon occurs in two forms, graphite and diamond. The
enthalpy of the combustion of
graphite is –393.5 kJ/mol, and that of diamond is –395.4 kJ/mol:
C(graphite) + O2(g)
CO2(g)
H = –393.5 kJ
C(diamond) + O2(g)
CO2(g)
H = –395.4 kJ
Calculate H for the conversion of graphite to diamond:
C(graphite)
C(diamond)
H = ?
Thermochemistry
Sample Exercise 5.8 Using Hess’s Law to Calculate H
Carbon occurs in two forms, graphite and diamond. The
enthalpy of the combustion of
graphite is –393.5 kJ/mol, and that of diamond is –395.4 kJ/mol:
C(graphite) + O2(g)
CO2(g)
H = –393.5 kJ
C(diamond) + O2(g)
CO2(g)
H = –395.4 kJ
Calculate H for the conversion of graphite to diamond:
C(graphite)
C(diamond)
H = ?
Thermochemistry
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.
Thermochemistry
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Standard Enthalpies of Formation
Standard enthalpies of formation, Hf°, are
measured under standard conditions (25 °C
and 1.00 atm pressure).
Thermochemistry
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Sample Exercise 5.10 Equations Associated with Enthalpies of
Formation
Write the equation corresponding to the standard
enthalpy of formation of liquid carbon tetrachloride
(CCl4).
Thermochemistry
Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
Thermochemistry
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Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
Thermochemistry
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Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
4H2(g) + 2O2(g)  4H2O(l)
Thermochemistry
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Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• The sum of these
equations is
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
4H2(g) + 2O2(g)  4H2O(l)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Thermochemistry
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Calculation of H
We can use Hess’s law in this way:
H = nHf,products – mHf°,reactants
where n and m are the stoichiometric
coefficients.
Thermochemistry
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Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(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
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Sample Exercise 5.11 Calculating an Enthalpy of Reaction from
Enthalpies of Formation
Use Table 5.3 to calculate the enthalpy change for the
combustion of 1 mol of ethanol:
C2H5OH(l) + 3 O2(g)
2 CO2(g) + 3 H2O(l)
Thermochemistry
Sample Exercise 5.12 Calculating an Enthalpy of Formation from
Enthalpies of Reaction
Given the following standard enthalpy change, use the standard
enthalpies of formation in Table 5.3 to calculate the standard
enthalpy of formation of CuO(s):
CuO(s) + H2(g)
Cu(s) + H2O(l)
H = –129.7 kJ
Thermochemistry