Transcript Chapter 5 Thermochemistry

Chapter 5
Thermochemistry
© 2012 Pearson Education, Inc.
Why is a pitcher able to throw a
baseball faster than he could throw a
bowling ball?
A. It is harder to hold a baseball ball than a bowling
ball.
B. For the same input of energy, an object of smaller
mass will have the larger speed.
C. For the same input of energy, an object of larger
mass will have the larger speed.
D. The spinning of the bowling ball causes it to drop to
the ground faster.
Thermochemistry
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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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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Kinetic Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
1
Ek =  mv2
2
Thermochemistry
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Conversion of Energy
• Energy can be converted from one type to
another.
• For example, the cyclist in the figure 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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A. Yes
B. No
Thermochemistry
A.
B.
C.
D.
Increases as potential energy is converted to kinetic energy.
Increases as the kinetic energy is converted to potential energy.
Decreases as potential energy is converted to kinetic energy.
Decreases as the kinetic energy is converted to potential energy.
Thermochemistry
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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A.
B.
C.
D.
Becomes more positive and decreasing in absolute magnitude
Becomes more negative and increasing in absolute magnitude.
Becomes more positive and increasing in absolute magnitude
Becomes more negative and decreasing in absolute magnitude.
Thermochemistry
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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A. Yes, because matter is not transferred to the surroundings.
B. No, because matter is transferred to the surroundings.
Thermochemistry
A. Closed system
B. Isolated system
C. Open system
Thermochemistry
Sample Exercise 5.1 Describing and Calculating Energy Changes
A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it.
(a) What happens to the potential energy of the ball as it is raised? (b) What quantity of work, in J, is used
to raise the ball? (c) After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has
been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just
before it hits the ground? (Note The force due to gravity is F = m  g, where m is the mass of the object
and g is the gravitational constant; g = 9.8 m/s2.)
Solution
Practice Exercise
What is the kinetic energy, in J, of (a) an Ar atom moving at a speed of 650 m/s, (b) a mole of Ar atoms
moving at 650 m/s? (Hint: 1 amu = 1.66  10–27 kg.)
Thermochemistry
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
Thermochemistry
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A. Positive number
B. Zero
C. Negative number
Thermochemistry
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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A.
B.
C.
D.
No, because we are not given specific values for w and q.
No, because it an open system.
Yes, because we know the signs of w and q.
No, because we do not know the temperature of the system.
Thermochemistry
E, q, w, and Their Signs
Thermochemistry
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Sample Exercise 5.2 Relating Heat and Work to Changes of Internal
Energy
Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4
and react to form a solid product C(s): A(g) + B(g)
C(s). As the reaction occurs,
the system loses 1150 J of heat to the surroundings. The piston moves downward as the
gases react to form a solid. As the volume of the gas decreases under the constant pressure
of the atmosphere, the surroundings do 480 J of work on the system. What is the change
in the internal energy of the system?
Solution
Practice Exercise
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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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
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 the figure, 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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A. It is determined by many factors and involves
data such as amount of check, 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
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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A.
B.
C.
D.
w>0
w=0
w<0
Cannot be determined
Thermochemistry
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
A. H is italicized in the equation.
B. E is a state function in the equation and this is
the minimum requirement.
C. P and V are state functions in the equation and
this is the minimum requirement.
D. All other terms in the equation, E, P and V, are
state functions.
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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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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A. If Zn(s) is the limiting reactant then more gas is formed and more
work will be done.
B. If Zn(s) is the limiting reactant then less gas is produced and less
work will be done.
C. If HCl(aq) is the limiting reactant then addition of Zn(s) increases
the amount of gas formed and more work is done.
D. If HCl(aq) is the limiting reactant then addition of Zn(s) decreases
the amount of gas formed and less work is done.
Thermochemistry
A. Yes, because pressure may change the value
of work in w = –PΔV.
B. No, because ΔV = 0.
Thermochemistry
Endothermicity and Exothermicity
• A process is
endothermic
when H is
positive.
• A process is
exothermic when
H is negative.
Thermochemistry
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A.
B.
C.
D.
Ruler
Graduated cylinder
Barometer
Thermometer
Thermochemistry
Sample Exercise 5.3 Determining the Sign of H
Indicate the sign of the enthalpy change, H, in these processes carried out under atmospheric pressure
and indicate whether each process is endothermic or exothermic: (a) An ice cube melts; (b) 1 g of butane
(C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O.
Solution
Practice Exercise
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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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.
Thermochemistry
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.
3. H for a reaction depends on the state
of the products and the state of the
reactants.
Thermochemistry
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A.
B.
C.
D.
Exothermic, because the temperature of the system increases
Exothermic, because the temperature of the system decreases
Endothermic, because the temperature of the system increases
Endothermic, because the temperature of the system decreases
Thermochemistry
Sample Exercise 5.4 Relating H to Quantities of Reactants and Products
How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? (Use
the information given in Equation 5.18.)
Solution
Practice Exercise
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
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
s=
q
m  T
Thermochemistry
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A. Hg(l)
C. Al(s)
B. Fe(s)
D. H2O(l)
Thermochemistry
Heat Capacity and Specific Heat
The amount of energy required to raise the
temperature of an object by 1 K (1C) is its
heat capacity.
heat capacity, then, is
heat capacity = Cp =
heat transferred
temperature change
or, heat capacity = mass x specific heat
Thermochemistry
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Sample Exercise 5.5 Relating Heat, Temperature Change, and Heat
Capacity
(a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 C (about room temperature)
to 98 C (near its boiling point)? (b) What is the molar heat capacity of water?
Solution
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?
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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A.
B.
C.
D.
Styrofoam cups are very fragile.
To create a vacuum between the two cups
Gives greater support to the equipment setup
Provides more thermal insulation
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
Thermochemistry
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Sample Exercise 5.6 Measuring H Using a Coffee-Cup Calorimeter
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the
temperature of the resultant solution increases from 21.0 C to 27.5 C. Calculate the enthalpy change
for the reaction in kj/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat,
that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is
4.18 J/g–K.
Solution
Practice Exercise
When 50.0 mL of 0.100 MgNO3 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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A.
B.
C.
D.
Ensures a heterogeneous mixture
Keeps particles slowly moving
Ensures solution is at same temperature
Permits titration of the solution
Thermochemistry
Sample Exercise 5.7 Measuring qrxn Using a Bomb Calorimeter
The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2(g), and H2O(l):
2 CH6N2(l) + 5 O2(g)
2 N2(g) + 2 CO2(g) + 6 H2O(l)
When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter
increases from 25.00 C to 39.50 C. In a separate experiment the heat capacity of the calorimeter is
measured to be 7.794 kJ/ C. Calculate the heat of reaction for the combustion of a mole of CH6N2.
Solution
Practice Exercise
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.
B.
C.
D.
Condensation of 2 H2O(g) to 2 H2O(l)
Vaporization of 2 H2O(l) to 2 H2O(g)
Conversion of 2 H2O(l) to 2 O2(g)
Conversion of CO2(g) to CH4(g)
Thermochemistry
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
The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the
combustion of CO to CO2 is –283.0 kJ/mol C:
1.
2.
Using these data, calculate the enthalpy for the combustion of C to CO:
3.
Solution
Practice Exercise
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 = ?
Answer: +1.9 kJ
Thermochemistry
Sample Exercise 5.9 Using Three Equations with Hess’s Law to
Calculate H
Calculate H for the reaction
2 C(s) + H2(g)
C2H2(g)
given the following chemical equations and their respective enthalpy changes:
Solution
Practice Exercise
Calculate H for the reaction
NO(g) + O(g)
NO2(g)
given the following information
Thermochemistry
A.
B.
C.
D.
All would change.
None would change.
ΔH3 does not change.
ΔH1 does not change.
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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A. Yes, because it is just another elemental form
of oxygen.
B. No, because it is not the most stable form of
the element oxygen at the given conditions.
C. Yes, because changing the subscripts of an
elemental formula does not change standard
heat of formation.
D. No, because there is a temperature change
when ozone is formed.
Thermochemistry
Sample Exercise 5.10 Equations Associated with Enthalpies of
Formation
For which of these reactions at 25 C does the enthalpy change represent a standard enthalpy of
formation? For each that does not, what changes are needed to make it an equation whose H is an
enthalpy of formation?
(a)
(b)
(c)
Solution
Practice Exercise
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)
• 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
(a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g)
and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane with that
produced by 1.00 g benzene.
Solution
Practice Exercise
Comment Both propane and benzene are hydrocarbons. As a rule, the energy obtained from the
combustion of a gram of hydrocarbon is between 40 and 50 kJ.
Thermochemistry
Sample Exercise 5.12 Calculating an Enthalpy of Formation from
Enthalpies of Reaction
The standard enthalpy change for the reaction CaCO3(s)
CaO(s) + CO2(g) is 178.1 kJ. Use Table
5.3 to calculate the standard enthalpy of formation of CaCO3(s).
Solution
Practice Exercise
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
Energy in Foods
Most of the fuel in the food we eat comes
from carbohydrates and fats.
Thermochemistry
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A.
B.
C.
Carbohydrates
Proteins
Fats
Thermochemistry
A. Grams of fat
B. Grams of total carbohydrate
C. Grams of protein
Thermochemistry
Sample Exercise 5.13 Comparing Fuel Values
Celery contains carbohydrates in the form of starch and cellulose, which have essentially the same fuel
values when combusted in a bomb calorimeter. When we eat celery, however, our bodies receive fuel
value from the starch only. What can we conclude about the difference between starch and cellulose as
foods?
Solution
Practice Exercise
The nutrition label on a bottle of canola oil indicates that 10 g of the oil has a fuel value of 86 kcal. A
similar label on a bottle of pancake syrup indicates that 60 mL (about 60 g) has a fuel value of 200
kcal. Account for the difference.
Thermochemistry
Sample Exercise 5.14 Estimating the Fuel Value of a Food from Its
Composition
(a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g
protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these substances, estimate the
fuel value (caloric content) of this serving. (b) A person of average weight uses about 100 Cal/mi when
running or jogging. How many servings of this cereal provide the fuel value requirements to run 3 mi?
Solution
Practice Exercise
Dry red beans contain 62% carbohydrate, 22% protein, and 1.5% fat. Estimate the fuel value of
these beans. (b) During a very light activity, such as reading or watching television, the average
adult expends about 7 kJ/min. How many minutes of such activity can be sustained by the energy
provided by a serving of chicken noodle soup containing 13 g protein, 15 g carbohydrate, and 5 g
fat?
Thermochemistry
Energy in Fuels
The vast
majority of the
energy
consumed in
this country
comes from
fossil fuels.
Thermochemistry
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A. It is easily stored as a gas for a long time.
B. Its product of combustion is only H2O(g).
C. Safety considerations using hydrogen gas are
minimal.
D. It is easily found in nature as an element.
Thermochemistry