Download Chapter 6 Thermodynamics: The First Law

Chapter 6 Thermodynamics:
The First Law
Systems, States, and Energy (Sections 6.1–6.8)
Key Concepts
thermodynamics, statistical thermodynamics, system, surroundings, open system,
closed system, isolated system, work, internal energy, translational kinetic energy,
rotational kinetic energy, vibrational kinetic energy, mode of motion (degree of
freedom), equipartition theorem, internal energy of an ideal gas, heat, calorie, adiabatic
wall, diathermic (nonadiabatic) wall, first law of thermodynamics, state function,
expansion work, free expansion, nonexpansion work, reversible isothermal expansion,
reversible process, irreversible process, exothermic process, endothermic process,
calorimetry, calorimeter, heat capacity, specific heat capacity, molar heat capacity
6.1 Systems
Example 6.1a
Suggest a system, boundary, and surroundings for 10 moles of propane gas in a rigid metal
cylinder.
Answer
One choice is that the 10 moles of propane gas constitute the system, the metal cylinder and
everything outside of it the surroundings, and the inside walls of the cylinder a real boundary.
This is an appropriate way to study the properties of the gas alone.
Example 6.1b
Suggest a system, boundary, and surroundings for 500 mL of water in an open beaker.
Answer
To study the properties of water alone, it should constitute the system. The walls of the beaker
constitute a real, physical boundary and the phase boundary between water and air constitutes
an imaginary boundary. The surroundings consist of everything outside the boundaries
including the beaker, the platform on which it rests, and the atmosphere.
6.2 Work and Energy
Example 6.2a
Describe three commonplace examples of how work is done on or by a system.
Answer
Compression of a spring: If the spring is the system, work is done on the system by the
surroundings to compress the spring.
Compression of a gas mixture as in an automobile engine: If the gas mixture is the system,
work is done on the system by the surroundings (piston) to compress the gases.
Muscle contraction: As the muscle contracts, work is done on the surroundings.
Both mechanical and electrical work are involved as well as chemical processes.
Example 6.2b
A plumber of mass 65 kg carries a toolbox of mass 15 kg to a fifth floor walk-up apartment
15 m above ground level. Calculate the work required for this process. Recall that
(difference in potential energy) = mg∆h from Table 6.1 in the text.
Answer
The difference in potential energy between the fifth and ground floors is equal to the work
done by the plumber. The force is mg and the distance is ∆h. Both the plumber’s mass and
that of the toolbox must be elevated.
m = (65 + 15) kg = 80 kg
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Chapter 6
g = 9.81 m⋅s−2
∆h = 15 m
w = (80 kg)(9.81 m⋅s−2)(15 m) = 1.18 × 104 J = 11.8 kJ
Example 6.2c
Describe the internal energy change and work performed when a spring is compressed or
expanded.
Answer
Work is done on the system by the surroundings to compress the spring. The compressed
spring has a greater capacity to do work than the uncompressed spring. In each case, the
internal energy of the system is increased by the amount of work done on it.
∆U > 0 and ∆U = Ufinal − Uinitial = w
Therefore, w > 0 when work is done on the system.
Work is done on the surroundings by the system to expand the spring. The expanded spring
has a lesser capacity to do work than the unexpanded spring. The internal energy of the
system is decreased by the amount of work done by it.
∆U < 0 and ∆U = Ufinal − Uinitial = w
Therefore, w < 0 when work is done by the system.
Example 6.2d
Describe the internal energy change and work performed when a battery is recharged.
Answer
Electrical work is done on the system by the surroundings in recharging a battery. The
charged battery has a greater capacity to do work than the discharged battery. The internal
energy of the system is increased by the amount of work done on it. As in the case of
mechanical work, w > 0 when electrical work is done on the system.
6.3 Expansion Work
Example 6.3a
The pressure exerted on an ideal gas at 2.00 atm and 300 K is reduced suddenly to 1.00 atm
while heat is transferred to maintain the initial temperature of 300 K. Calculate q, w, and
∆U in joules for this process.
Solution
For an ideal gas undergoing an isothermal process, ∆U = 0. The work done at constant
external pressure is w = − Pex ∆V, where ∆V = Vfinal − Vinitial .
Vinitial = (nRT )/Pinitial = (1 mol)(0.082 06 L⋅atm⋅K−1⋅mol−1)(300 K)/(2.00 atm) = 12.3 L
Vfinal = (nRT )/Pfinal = (1 mol)(0.082 06 L⋅atm⋅K−1⋅mol−1)(300 K)/(1.00 atm) = 24.6 L
∆V = Vfinal − Vinitial = 24.6 L − 12.3 L = 12.3 L (system expands)
w = − Pex ∆V = −(1.00 atm)(12.3 L) = −12.3 L⋅atm
= (−12.3 L⋅atm)(101.325 J⋅L−1⋅atm−1) = −1.25 × 103 J
Work is done by the system on the surroundings.
q = ∆U − w = 0 − (−1.25 × 103 J) = +1.25 × 103 J
Heat is absorbed by the system from the surroundings.
2
Example 6.3b
Suppose the pressure change in Example 6.3a is carried out reversibly as well as
isothermally. Calculate q, w, and ∆U in joules for this process and compare the answers
to those in the irreversible expansion in one step.
Answer
The internal energy of an ideal gas depends only on temperature, and ∆U = 0.
Check the two formulas for isothermal, reversible work.
V
 24.6 L 
w = − nRT ln final = −(1.00 mol)( 8.314 47 J⋅K −1⋅ mol−1 )(300 K) ln 

Vinitial
 12.3 L 
Thermodynamics: The First Law
or w = − nRT ln
Pinitial
 2.00 atm 
= −(1.00 mol)(8.314 47 J⋅K −1⋅ mol−1 )(300 K) ln 

Pfinal
 1.00 atm 
w = −1.73 × 103 J
Note:
and
q = 1.73 × 103 J
The absolute value of the work done by the system in a reversible expansion is greater than the
value obtained in an irreversible expansion. It is possible to show that the maximum amount
of work is always done by the system in a reversible expansion.
Example 6.3c
Suppose the pressure change in Example 6.3a is carried out in the opposite direction
(compression). Calculate q, w, and ∆U in joules for this process in both a reversible and
irreversible way.
Answer
The labels for initial and final states are interchanged. The calculation for reversible
work results simply in a change in sign.
w = − nRT ln
Vfinal
 12.3 L 
= −(1.00 mol)(8.314 47 J⋅K −1⋅ mol−1 )(300 K) ln 

Vinitial
 24.6 L 
= +1.73 × 103 J
and
q = −1.73 × 103 J
The calculation for irreversible work results in a quite different value.
w = − Pex ∆V = −(2.00 atm)(−12.3 L) = +24.6 L⋅atm
= (24.6 L⋅atm)(101.325 J⋅L−1⋅atm−1) = +2.50 × 103 J and q = −2.50 × 103 J
Note:
The value of the work done on the system in a reversible compression is smaller than the value
obtained in an irreversible compression. It is possible to show that the minimum amount of
work is always done on the system in a reversible compression.
6.4 Heat
Example 6.4a
Calculate the heat flow and change in internal energy if the temperature of 1 mol of Ar is
increased from 300 K to 900 K. Assume the atoms behave as an ideal gas and no work is
done.
Solution
From Example 6.8a, Um = 3.74 kJ⋅mol−1 at 300 K. Tripling the temperature leads to
Um = 11.22 kJ⋅mol−1 at 900 K.
q = ∆Um = Um(900 K) − Um(300 K) = (11.22 − 3.74) kJ⋅mol−1 = 7.48 kJ⋅mol−1
Example 6.4b
Calculate the heat flow and change in internal energy if the temperature of 1 mol of H2 is
increased from 300 K to 900 K. Assume the molecules behave as an ideal gas and no work is
done.
Solution
In Example 6.8b, Um = 6.24 kJ⋅mol−1 at 300 K. Tripling the temperature leads to
Um = 18.72 kJ⋅mol−1 at 900 K.
q = ∆Um = Um(900 K) − Um(300 K) = (18.72 − 6.24) kJ⋅mol−1 = 12.48 kJ⋅mol−1
Example 6.4c
Calculate the heat flow and change in internal energy if the temperature of 1 mol of H2O is
increased from 300 K to 900 K. Assume the molecules behave as an ideal gas and no work is
done.
Solution
In Example 6.8c, Um = 7.48 kJ⋅mol−1 at 300 K. Tripling the temperature leads to
Um = 22.44 kJ⋅mol−1 at 900 K.
q = ∆Um = Um(900 K) − Um(300 K) = (22.44 − 7.48) kJ⋅mol−1 = 14.96 kJ⋅mol−1
Note:
In all three examples, q and the change in internal energy are positive values, because heat
flows from the surroundings into the system. The amount of heat required to increase the
temperature (and internal energy) is smaller for atoms than for molecules, and is smaller for
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Chapter 6
linear molecules than for nonlinear molecules. The greater the number of modes of motion
(degrees of freedom), the greater is the capacity of a species to store energy.
6.6 The First Law
Example 6.6
An electric stirrer performs 40 kJ of work on a beaker of water and water transfers 1.5 × 103
cal of heat to the surroundings. Calculate the change in internal energy of water in kJ.
Solution
Convert calories to the units of kilojoules.
(1.5 × 103 cal)(4.184 J⋅cal−1)(10−3 kJ⋅J−1) = 6.3 kJ
Work is done on the system (water), so w = 40 kJ > 0.
Heat is transferred from the system to the surroundings, so q = −6.3 kJ < 0.
∆U = q + w = (−6.3 + 40) kJ = 34 kJ
∆U > 0 and the internal energy of the system has increased.
6.7 State Functions
Example 6.7a
Describe the state of a system composed of 1 mol of Ar(g) at 1 bar and 300 K.
Solution
The pressure and temperature are uniform throughout the gas. The ideal gas equation uniquely
determines the volume of the gas, V = nRT/P. All other properties applicable to a noble gas
under these conditions, such as density, refractive index, and heat capacity, are fixed as well.
Example 6.7b
Describe the state of the following system: a partition that separates a sample of liquid water
at 10°C from a sample of water at 30°C is removed.
Solution
The resulting system is not in any given state, because its properties are not fixed. The
temperature and density are not uniform throughout. The system will eventually approach a
uniform temperature and density as a result of diffusion. The state of the system can then be
defined.
Example 6.7c
For spring break, two students decide to fly from Philadelphia (initial state) to Denver (final
state). Two of their classmates choose to drive from Philadelphia to Denver by way of New
Orleans (intermediate state). Describe one variable whose change is the same for the two
groups (state function) and another that is not.
Solution
The destination (final state) and origin (initial state) are the same for both groups. The
distances traveled depend on the path taken, and, presumably, are not the same for the plane
and car.
Example 6.7d
A laboratory technician assigned to stir an insulated beaker of water inadvertently leaves the
stirrer on overnight. In the morning, the technician notes that the temperature of the water has
risen 3°C. The same temperature rise can also be accomplished by heating the water using a
heating mantle. Compare the two processes in terms of the change in internal energy, and the
heat and work done on the system.
Solution
For the two processes, the initial and final states are the same, so ∆U is also the same. In the
first process, mechanical work is done on the system, q = 0, and ∆U = w. In the second
process, w = 0, and ∆U = q. Both q and w are clearly path dependent.
Note:
4
Because the changes in state functions are path independent, the most convenient path may be
chosen for a given change in state both in the laboratory and in calculations.
Thermodynamics: The First Law
6.8 A Molecular Interlude: The Origin of Internal Energy
Example 6.8a
Calculate the internal energy of 1 mol of Ar at 300 K. Assume the atoms behave as an ideal
gas.
Solution
An atom has only three degrees of freedom in translational motion. The internal energy per
mole is then
Um = Um(translation) = 23 RT = (1.5)(8.314 47 J⋅K−1⋅mol−1)(300 K)
= 3.74 × 103 J⋅mol−1
= 3.74 kJ⋅mol−1
Example 6.8b
Calculate the internal energy of 1 mol of H2 at 300 K. Assume the molecules behave as an
ideal gas.
Solution
All diatomic molecules are linear. The H2 molecule has three degrees of freedom in
translational motion and two degrees of freedom in rotational motion. The internal energy per
mole is then
Um = Um(translation) + Um(rotation, linear) = 23 RT + RT = 52 RT
= (2.5)(8.314 47 J⋅K−1⋅mol−1)(300 K) = 6.24 × 103 J⋅mol−1 = 6.24 kJ⋅mol−1
Example 6.8c
Calculate the internal energy of 1 mol of H2O vapor at 300 K. Assume the molecules behave
as an ideal gas.
Solution
If we use the VSEPR model (Section 3.2), the H2O molecule is predicted to be nonlinear.
Thus, it has three degrees of freedom in translational motion and three degrees of freedom in
rotational motion. The internal energy per mole is then
Um = Um(translation) + Um(rotation, nonlinear) = 23 RT + 23 RT = 3RT
= (3)(8.314 47 J⋅K−1⋅mol−1)(300 K) = 7.48 × 103 J⋅mol−1 = 7.48 kJ⋅mol−1
Notes: In all three examples, the internal energy is linearly proportional to the temperature of the
ideal gas. The vibrational contribution to the internal energy of an ideal gas is small at room
temperature and can usually be ignored. Its temperature dependence is, however, decidedly
nonlinear at normal temperatures. At very high temperature, each vibrational degree of
freedom of a molecule has a contribution of of 12 kT from kinetic energy and 12 kT from
potential energy if vibration is treated as a harmonic oscillator. The total contribution to the
internal energy is then kT per degree of freedom for vibrational motion.
Enthalpy (Sections 6.9–6.13)
Key Concepts
enthalpy, heat transfer at constant pressure, heat capacity of gases, heat capacity at
constant volume, heat capacity at constant pressure, molecular origin of heat capacity,
physical change (phase transition), enthalpy of vaporization, enthalpy of fusion,
enthalpy of freezing, enthalpy of sublimation, heating curves, temperature of a
substance at its melting or boiling point
Note: Study Guide notation (not used in the text) – qV and qP (heat transfer at
constant volume and constant pressure, respectively)
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Chapter 6
6.9 Heat Transfers at Constant Pressure
Example 6.9
A gas expands against a constant external pressure and does 25 kJ of expansion work on the
surroundings. During the process, 60 kJ of heat is absorbed by the system. Determine the
values of ∆H and ∆U.
Solution
Because this is a constant pressure process, ∆H = qP = 60 kJ.
∆U = ∆H − P∆V = ∆H + w
w = −25 kJ (work done by the system is negative)
∆U = 60 kJ − 25 kJ = 35 kJ
6.10 Heat Capacities at Constant Volume and Constant Pressure
Example 6.10
The heat capacity, CP , of 1.500 moles of F2 is 46.95 J⋅K−1 at 25°C. Assuming the gas is
ideal, calculate the molar heat capacity at constant volume, CV,m , for F2 at 25°C.
Solution
CP,m = CP /n = (46.95 J⋅K−1)/(1.500 mol) = 31.30 J⋅K−1⋅mol−1
CV,m = CP,m − R = (31.30 − 8.314 51) J⋅K−1⋅mol−1 = 22.99 J⋅K−1⋅mol−1
6.11 A Molecular Interlude: The Origin of the Heat Capacities of Gases
Example 6.11a Calculate the molar heat capacity at constant volume, CV,m , for F2 at 25°C. Compare the
calculated value to the experimental one determined in Example 6.10.
Solution
Diatomic molecules are linear.
CV,m = 52 R = 20.786 28 J⋅K−1⋅mol−1
The experimental value of 22.99 J⋅K−1⋅mol−1 is slightly larger than the calculated one. The
difference is associated with vibrational motion of the F2 molecule.
Example 6.11b A 100-L vessel containing 6.00 mol of H2(g) at 2.00 atm is cooled to 203.0 K. Assuming the
gas behaves ideally, calculate ∆U and ∆H for this process.
Solution
Calculate the initial temperature, Tinitial .
Tinitial = PV/nR = [(2.00 atm)(100 L)]/[(6.00 mol)(0.082 06 L⋅atm⋅K−1⋅mol−1)]
Tinitial = 406.2 K
∆T = Tfinal − Tinitial = (203.0 − 406.2) K = −203.2 K
CV,m = 52 R = 20.786 28 J⋅K−1⋅mol−1 = 0.020 79 kJ⋅K−1⋅mol−1
CP,m = CV,m + R = 29.100 79 J⋅K−1⋅mol−1 = 0.029 10 kJ⋅K−1⋅mol−1
From Sections 6.5 and 6.10
∆U = nCV,m∆T = (6.00 mol)( 0.020 79 kJ⋅K−1⋅mol−1)(−203.2 K) = −25.3 kJ
∆H = nCP,m∆T = (6.00 mol)( 0.029 10 kJ⋅K−1⋅mol−1)(−203.2 K) = −35.5 kJ
Note: This equation applies even though pressure is not constant. For an ideal gas, enthalpy depends
only on temperature.
6
Thermodynamics: The First Law
6.12 The Enthalpy of Physical Change
Example 6.12
Considering the relative strength of intermolecular forces in the liquid phase, arrange the
following substances in the order of increasing enthalpy of vaporization at the boiling
point of the liquid: CH3OH, C6H6 , Hg, CH4 .
Solution
The intermolecular interactions of significance for the substances listed are London forces for
C6H6 and CH4, metallic bonding for Hg, and hydrogen bonding for CH3OH. The relative
strength of intermolecular forces increases in the order London forces, hydrogen bonding, and
metallic interactions. The strength of London interactions increases with the number of
electrons in the molecule (polarizability); thus, C6H6 has stronger interactions than CH4 .
The expected order of increasing enthalpy of vaporization is CH4 , C6H6 , CH3OH, and Hg.
The values of standard enthalpies of vaporization in Table 6.3 in the text are 8.2, 30.8, 35.3,
and 59.3 kJ⋅mol−1, respectively.
The Enthalpy of Chemical Change (Sections 6.14–6.22)
Key Concepts
reaction enthalpy, thermochemical equation, combustion, standard state, standard
reaction enthalpy, Hess’s law, reaction sequence, overall reaction, standard enthalpy of
combustion, standard enthalpy of formation, lattice enthalpy, Born-Haber cycle, bond
enthalpy, mean (average) bond enthalpy, temperature dependence of reaction enthalpy,
Kirchhoff ’s law
6.14 Reaction Enthalpies
Example 6.14a Reaction with oxygen is a combustion reaction. Write the thermochemical equation for the
combustion of hydrogen gas at 25°C. For the reaction of 1 mol H2(g), qP = −285.83 kJ.
Determine the reaction enthalpy, ∆Hr .
Solution
The chemical equation H2(g) + O2(g) → H2O(l) is balanced by inspection. Note that water
at 25°C is a liquid. The balanced equation with the smallest whole integer coefficients is
2 H2(g) + O2(g) → 2 H2O(l)
For a constant pressure reaction, qP = ∆H. The heat change is given for 1 mol H2(g),
but the balanced equation requires 2 mol.
Thus, ∆H = 2(−285.83 kJ) = −571.66 kJ. The thermochemical equation is
2 H2(g) + O2(g) → 2 H2O(l)
∆H = −571.66 kJ
The reaction enthalpy is ∆Hr = −571.66 kJ⋅mol−1. Here “mol” refers to the enthalpy change
for 2 mol H2(g), 1 mol O2(g), and 2 mol H2O(l).
More explicitly, ∆Hr = −571.66 kJ⋅(2 mol H2)−1, ∆Hr = −571.66 kJ⋅(1 mol O2)−1, and
∆Hr = −571.66 kJ⋅(2 mol H2O)−1.
7
Chapter 6
Example 6.14b Use the thermochemical equation given below to determine how much heat is released, qsurr ,
when 350 g of propane, C3H8(g), is burned completely in a backyard barbecue at 25°C.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
∆H = −2220 kJ
−1
The reaction enthalpy is ∆Hr = −2220 kJ⋅mol . Here “mol” refers to the enthalpy change for
1 mol C3H8(g), 5 mol O2(g), 3 mol CO2(g), and 4 mol H2O(l). More explicitly,
∆Hr = −2220 kJ⋅(1 mol C3H8)−1. The molar mass of propane is
3(12.01) + 8(1.0079) = 44.09 g⋅mol−1. The number of moles of propane is then
(350 g)/(44.09 g⋅mol−1) = 7.938 mol. The heat change for the system is
qP = n∆Hr = (7.938 mol)(−2220 kJ⋅mol−1) = −1.76 × 104 kJ.
The heat released to the surroundings is qsurr = −qP = 1.76 × 104 kJ = 17.6 MJ.
Solution
6.15 The Relation Between ∆H and ∆U
Example 6.15a Calculate the change in internal energy, ∆U, for the combustion reaction in Example 6.14b.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
Solution
∆H = −2220 kJ
∆H ≈ ∆U + (∆ngas )RT and ∆U ≈ ∆H − (∆ngas )RT
∆ngas = n(CO2) − [n(C3H8) + n(O2)] = 3 − (1 + 5) = 3 − 6 = −3
Be careful to exclude liquid water.
(∆ngas )RT = (−3 mol)(8.314 47 J⋅K−1⋅mol−1)(298.15 K) = −7437 J = −7.437 kJ
∆U ≈ ∆H − (∆ngas )RT = −2220 kJ − (−7.437 kJ) = −2213 kJ
Note: Use consistent energy units, either J or kJ. The internal energy change is a more positive value
than the enthalpy change for a reaction that consumes gas. The difference between ∆H and
∆U is small, only about 0.3%.
Example 6.15b Calculate the change in internal energy, ∆U, for the dissociation of hydrogen gas into atoms at
25°C, given the following thermochemical equation:
H2(g) → 2 H(g)
∆H = +435.94 kJ
∆ngas = n(H) − n(H2 ) = 2 − 1 = +1
(∆ngas )RT = (+1 mol)(8.314 47 J⋅K−1⋅mol−1)(298.15 K) = +2478.97 J = +2.479 kJ
∆U ≈ ∆H − (∆ngas )RT = +435.94 kJ − (+2.479 kJ) = +433.46 kJ
Solution
Note: The internal energy change is a more negative value than the enthalpy change for a reaction
that produces gas. The difference between ∆H and ∆U is small, only about 0.6%.
6.16 Standard Reaction Enthalpies
Example 6.16
Solution
8
Compare the following two thermochemical reactions at 25°C. Comment on the nature of the
difference in standard reaction enthalpy between reaction (2) and reaction (1).
(1)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
∆H ° = −2220 kJ
(2)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
∆H ° = −2044 kJ
Reaction (2) is less exothermic than reaction (1). Reaction (2) produces water vapor and
reaction (1) produces liquid water. For reaction (1), ∆Hr° = −2220 kJ⋅mol−1. For reaction (2),
∆Hr° = −2044 kJ⋅mol−1. Recall that “mol” refers to 1 mol C3H8 , 5 mol O2 , 3 mol CO2 , and
4 mol H2O. The enthalpy of vaporization of water is 44.01 kJ⋅mol−1 at 25°C. The difference
Thermodynamics: The First Law
in standard reaction enthalpies between reaction (2) and reaction (1) is 176 kJ⋅(4 mol H2O)−1
or 44 kJ⋅mol−1, which is the enthalpy of vaporization of water at 25°C. Thus, the difference
corresponds to the vaporization of 4 mol of liquid water at 25°C. The thermochemical
equations may be combined by subtracting reaction (1) from reaction (2). The result is
4 H2O(l) → 4 H2O(g)
∆H ° = +176 kJ
Changing the stoichiometric coefficients to the simplest whole integers yields
H2O(l) → H2O(g)
∆Hr° = +44 kJ⋅mol−1 = ∆Hvap
Combining thermochemical equations is an application of Hess’s law, which is discussed in the
next section.
6.17 Combining Reaction Enthalpies: Hess’s Law
Example 6.17
Determine the standard reaction enthalpy for
∆H ° = ?
(3)
2 SO2(g) + O2(g) → 2 SO3(g)
Use Hess’s law to combine combustion reactions (1) and (2) given below.
∆H ° = −792 kJ
(1)
2 S(s) + 3 O2(g) → 2 SO3(g)
∆H ° = −297 kJ
(2)
S(g) + O2(g) → SO2(g)
Solution
In reaction (1), the proper number of moles of product SO3 is present, and the reaction may be
combined as written. In reaction (2), the number of moles of SO2 needs to be doubled and the
reactants and products interchanged. The thermochemical equations that combine to give
reaction (3) are then
(1)
2 S(s) + 3 O2(g)
2 SO2(g) → 2 S(s) + 2 O2(g)
−2×(2)
(3)
→ 2 SO3(g)
2 SO2(g) + O2(g)
→ 2 SO3(g)
∆H ° = −792 kJ
∆H ° = +594 kJ
____
∆H ° = −198 kJ
6.18 The Heat Output of Reactions
Example 6.18a Write a balanced thermochemical equation for the combustion of one mole of octane (use
Table 6.4 in the text). Suppose an average automobile travels 11,000 miles in a year and
averages 21.0 miles per gallon of gasoline. Approximating gasoline as octane and assuming a
gallon of gasoline has a mass of 2.66 kg, calculate the enthalpy produced from the combustion
of fossil fuel and used by an average automobile per year. Assume the complete combustion
of octane in the engine.
Solution
The thermochemical equation for one mole of octane is (Table 6.4 in the text)
C8H18(l) +
25
2
O2(g) → 8 CO2(g) + 9 H2O(l)
∆Hc° = −5471 kJ⋅mol−1
(11 000 miles / 21.0 miles⋅gallon−1) = 524 gallons of gasoline or octane
(524 gallons × 2.66 kg⋅gallon−1) = 1.39 × 103 kg = 1.39 × 106 g
molar mass of octane = [8(12.01) + 18(1.0079)] = 114.22 g⋅mol−1
(1.39 × 106 g / 114.22 g⋅mol−1) = 1.22 × 104 mol octane per year
(5471 kJ⋅mol−1 × 1.22 × 104 mol) = 6.67 × 107 kJ = 66.7 GJ per year
Example 6.18b How many grams of CO2(g) are added to the atmosphere by the average automobile in one
year? Use the results of Example 6.18a.
Answer
4.29 × 106 g = 4.29 Mg of CO2(g)
9
Chapter 6
6.19 Standard Enthalpies of Formation
Example 6.19a Using enthalpy of formation values in Appendix 2A, calculate the enthalpy of combustion of
two moles of octane.
Solution
The thermochemical equation for the combustion of two moles of octane is
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)
∆Hc° = ?
The standard enthalpy of formation of an element in its most stable form is 0.
Therefore, ∆Hf°[O2(g)] = 0.
∆Hr° = ∑n∆Hf°(products) − ∑n∆Hf°(reactants) = ∆Hc°
= 16∆Hf°[CO2(g)] + 18∆Hf°[H2O(l)] − 2∆Hf°[C8H18(l)]
= 16(−393.51 kJ⋅mol−1) + 18(−285.83 kJ⋅mol−1) − 2(−249.9 kJ⋅mol−1)
= −6296.15 − 5144.94 + 499.8 kJ⋅mol−1
= −10 941.3 kJ⋅mol−1 at 298.15 K
Recall that “mol” refers to 2 mol of C8H18 , 25 mol O2 , 16 mol CO2 , and 18 mol H2O.
Example 6.19b Using enthalpy of formation values in Appendix 2A, calculate the enthalpy of dissociation of
one mole of hydrogen molecules into hydrogen atoms.
H2(g) → 2 H(g)
Answer
∆Hr° = ?
∆Hr° = 2∆Hf°[H(g)] = 2(217.97 kJ⋅mol−1) = 435.94 kJ⋅mol−1 at 298.15 K
This enthalpy change is also known as the bond enthalpy, ∆HB(H−H), defined in Section 6.21.
Note:
It is possible to obtain an unknown enthalpy of formation if a reaction enthalpy and the other
enthalpies of formation in that reaction are known. See Example 6.10 in the text.
6.21 Bond Enthalpies
Example 6.21a Write a thermochemical equation for breaking one mol of C−H bonds in methane, CH4 .
The enthalpy of formation of CH3(g) is 138.9 kJ⋅mol−1. Use Appendix 2A for other data.
Solution
∆Hr° = ∆HB(C−H)
CH4(g) → CH3(g) + H(g)
∆Hr° = ∆Hf°[CH3(g)] + ∆Hf°[H(g)] − ∆Hf°[CH4(g)]
= 138.9 + 217.97 − (−74.81) kJ⋅mol−1 = 431.7 kJ⋅mol−1 = ∆HB(C−H)
The C−H bond has an average bond enthalpy of 412 kJ⋅mol−1 in polyatomic molecules (see
Table 6.8 in the text). Note that “mol” refers to the amount of C−H bonds. Breaking the first
C−H bond in methane requires more heat than subsequent ones.
Example 6.21b Write a thermochemical equation for breaking one mol of carbonyl C−H bonds in ethanal
(acetaldehyde), CH3COH . The enthalpy of formation of CH3CO(g) is −18.8 kJ⋅mol−1.
Use Appendix 2A for other data.
Solution
∆Hr° = ∆HB(C−H)
CH3COH(g) → CH3CO(g) + H(g)
∆Hr° = ∆Hf°[CH3CO(g)] + ∆Hf°[H(g)] − ∆Hf°[CH3COH(g)]
= −18.8 + 217.97 − (−192.30) kJ⋅mol−1 = 391.5 kJ⋅mol−1 = ∆HB(C−H)
The carbonyl C−H bond in ethanal is considerably weaker than the average C−H bond
enthalpy in polyatomic molecules.
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Thermodynamics: The First Law
Example 6.21c Calculate the average C−H bond enthalpy in methane, CH4 . Use the data in Appendix 2A.
Solution
∆Hr° = 4∆HB(C−H)
CH4(g) → C(g) + 4 H(g)
∆Hr° = ∆Hf°[C(g)] + 4∆Hf°[H(g)] − ∆Hf°[CH4(g)]
= 716.68 + 4(217.97) − (−74.81) kJ⋅mol−1 = 1663.37 kJ⋅mol−1
∆HB(C−H) = ∆Hr°/4 = 415.84 kJ⋅mol−1
The average C−H bond enthalpy in methane is slightly larger than the average C−H
bond enthalpy in polyatomic molecules.
Example 6.21d Calculate the average C=O bond enthalpy in carbon dioxide, CO2 . The enthalpy of
formation of O(g) is 249.17 kJ⋅mol−1. Use Appendix 2A for other data.
Solution
∆Hr° = 2∆HB(C=O)
CO2(g) → C(g) + 2 O(g)
∆Hr° = ∆Hf°[C(g)] + 2∆Hf°[O(g)] − ∆Hf°[CO2(g)]
= 716.68 + 2(249.17) − (−393.51) kJ⋅mol−1 = 1608.53 kJ⋅mol−1
∆HB(C=O) = ∆Hr°/2 = 804.27 kJ⋅mol−1
The C=O double bond in carbon dioxide is considerably stronger than the average C=O bond
enthalpy of 743 kJ⋅mol−1 in polyatomic molecules.
6.22 The Variation of Reaction Enthalpy with Temperature
Example 6.22a Calculate a value for the enthalpy of vaporization of liquid water at 298.15 K. Use
Kirchhoff ’s law to estimate the enthalpy of vaporization of liquid water at the normal
boiling point of 373.15 K. Compare the value to the one in Table 6.3 in the text. Use the
data in Appendix 2A.
Solution
H2O(l) → H2O(g)
∆Hr° = ∆Hvap°
∆Hr° = ∆Hf°[H2O(g)] − ∆Hf°[H2O(l)]
= −241.82 − (−285.83) kJ⋅mol−1 = +44.01 kJ⋅mol−1 = ∆Hvap° at 298.15 K
∆CP = CP,m[H2O(g)] − CP,m[H2O(l)] = 33.58 − 75.29 J⋅K−1⋅mol−1
= −41.71 J⋅K−1⋅mol−1 = −0.041 71 kJ⋅K−1⋅mol−1
∆Hvap° (373.15 K) = ∆Hvap°(298.15 K) + ∆CP (373.15 − 298.15 K)
= 44.01 + (−0.041 71)(75.00) = 44.01 − 3.13 kJ⋅mol−1
= 40.88 kJ⋅mol−1
The value in Table 6.3 in the text is 40.7 kJ⋅mol−1. The approximation is an excellent one
in this case.
Example 6.22b The bond enthalpy of H−H is 436 kJ⋅mol−1 at 298.15 K. Use Kirchhoff ’s law to estimate the
bond enthalpy at 0 K. Use the data in Appendix 2A.
Solution
H2(g) → 2 H(g)
∆Hr° = ∆HB(H−H) = 436 kJ⋅mol−1 at 298.15 K
∆CP = 2CP,m[H(g)] − CP,m[H2(g)] = 2(20.78) − 28.82 J⋅K−1⋅mol−1
= +12.74 J⋅K−1⋅mol−1 = +0.012 74 kJ⋅K−1⋅mol−1
∆HB(0 K) = ∆HB(298.15 K) + ∆CP (0 − 298.15 K)
= 436 + (0.012 74)(−298.15) = 436 − 3.8 kJ⋅mol−1
= 432 kJ⋅mol−1
The spectroscopic dissociation energy of H2(g) is 432.07 kJ⋅mol−1, which corresponds to a
thermodynamic temperature of absolute zero. Any physical change of state that occurs at
low temperature is deliberately ignored.
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Chapter 6
Example 6.22c If ∆H ° = −571.66 kJ for reaction (1) at 25°C, calculate ∆H ° for reaction (2) at 0°C.
(1)
(2)
Solution
2 H2(g) + O2(g) → 2 H2O(l)
2 H2(g) + O2(g) → 2 H2O(s)
Solve the problem in two parts. First use Kirchoff ’s law to calculate ∆H ° for Reaction (1)
at 0°C. Then consider the phase change from liquid water to solid water (Hess’s law).
The following information is taken from Table 6.3 in the text and Appendix 2A.
∆Hfus° [H2O] = 6.01 kJ⋅mol−1
CP,m [H2(g)] = 28.82 J⋅K−1⋅mol−1
Answer
CP,m [H2O(l)] = 75.29 J⋅K−1⋅mol−1
CP,m [O2(g)] = 29.36 J⋅K−1⋅mol−1
∆H ° = −585.27 kJ for reaction (2) at 0°C
Note: This example illustrates how to modify Kirchoff ’s law when phase changes occur.
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