Chapter 7. THERMODYNAMICS:
THE FIRST LAW
SYSTEMS, STATES, AND ENERGY
7.1 Systems
7.2 Work and Energy
7.3 Expansion Work
7.4 Heat
7.5 The Measurement of Heat
7.6 The First Law
7.7 A Molecular Interlude: The Origin of
Internal Energy
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SYSTEMS, STATES, AND ENERGY (Sections 7.1-7.7)
7.1 Systems
 Thermodynamics deals with transformation (from
one form to another) and transfer (from one place to
another) of energy.
- System means the region in which we
are interested
- Surroundings: everything else
- Universe (the system and the
surroundings
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Types of systems
- Open system: exchanging both matter and
energy with the surroundings
- Closed system: a fixed amount of matter,
but exchanging energy with the surroundings
- Isolated system: no contact with its surroundings
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Example 7.1
Identify the following systems as open, closed, or isolated:
(a) Coffee in a very high quality thermos bottle
(b) Coolant in a refrigerator coil
(c) A bomb calorimeter in which benzene is burned
(d) Gasoline burning in an automobile engine
(e) Mercury in a thermometer
(f) A living plant
Solutions
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7.2 Work and Energy
 Work: the process of achieving motion against an
opposing force.
- unit: joule (J), 1 J = 1 kg·m2·s-2 = 1 N·m
 Energy: the capacity of a system to do work:
Internal energy (U) is the total store of energy in a
system. In thermodynamics, "changes" in energy
(ΔU) are dealt with.
DU = Ufinal – Uinitial
-Symbol w: the energy transferred to a system by doing
work; in the absence of other changes, DU = w
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7.3 Expansion Work
-Expansion work: the work arising from a change in the
volume of a system. Table 7.1 shows expansion and
nonexpansion work.
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Units of work
Irreversible expansion work
Only when the external
pressure is constant during
the expansion.
The negative sign relates work to the system.
Note if Pex = 0, a vacuum, w = 0; free expansion (expansion
against vacuum)
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Self-Test 7.1B
The gases in the four cylinders of an automobile engine
expand from 0.22 L to 2.2 L during one ignition cycle.
Assuming the gear train maintains a steady pressure of
9.60 atm on the gases, how much work can the engine do
in one cycle?
Solution
Using w = -PexDV,
_
_
w = (9.60 atm) x (1.98 L) = 19.0 L atm
=
_
101.325 J
19.0 L atm x
1 L atm
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= _ 1.93 kJ
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Example 7.3
Air in a bicycle pump is compressed by pushing in the handle. If the
inner diameter of the pump is 3.0 cm and the pump is depressed 20 cm
with a pressure of 2.00 atm,
(a) How much work is done in the compression?
(b) Is the work positive or negative with respect to the air in the pump?
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Solution
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 Reversible process can be reversed by an
infinitesimal change in a variable.
E.g. reversible, isothermal expansion of
an ideal gas
PexV = nRTconst = constant
- Work done is the area beneath the ideal gas isotherm
lying between the initial and the final volumes.
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EXAMPLE 7.2
A piston confines 0.100 mol Ar(g) in a volume of 1.00 L at 25 oC. Two
experiments are performed. (a) The gas is allowed to expand through
an additional 1.00 L against a constant pressure of 1.00 atm. (b) The
gas is allowed to expand reversibly and isothermally to the same
final volume. Which process does more work?
(a) Irreversible path:
(b) Reversible path:
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Self-Test 7.2A
A cylinder of volume 2.00 L contains 0.100 mol He(g) at 30 oC.
Which process does more work on the system, compressing
the gas isothermally to 1.00 L with a constant pressure of 5.00
atm, or compressing it reversibly and isothermally to the same
final volume?
Solution
Isothermal irreversible compression
_
w = PexDV = _ (5.00 atm) x ( 1.00 L) = +5.00 L atm
= +507 J (1 L atm = 101.325 J)
Isothermal reversible compression
w = _ nRT ln(Vfinal/Vintitial)
_ (0.100 mol) (8.315 J K-1 mol -1) x (303 K) x ln 1.00L
x
=
2.00 L
= +175 J.
Thus the irreversible compression does more work.
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7.4 Heat
 Heat is the energy transferred as a result of a
temperature difference.
- Thermal energy of a system: the sum of the potential and
kinetic energies arising from the chaotic thermal motion of
atoms, ions, and molecules
-The energy transferred to a system as heat is q
Internal energy of the system changed by transferring energy as heat
DU = q
- Unit: cal, the energy needed to raise T of 1 g of water by 1 oC.
1 cal = 4.184 J (exactly)
1 Cal (nutritional calorie) = 1 kcal
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 Exothermic process:
releasing heat into the
surroundings
 Endothermic process:
absorbing heat from the
surroundings
- Thermite reaction:
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7.5 The Measurement of Heat
-Adiabatic Process: one performed in isolated system; no
heat exchanged with surroundings
 Heat capacity, C is the ratio of the heat supplied to the
rise in temperature observed.
Heat capacity =
heat supplied
temperature rise produced
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Some Specific and Molar Heat Capacities
(Table 7.2)
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Self-Test 7.3A
Potassium perchlorate, KClO4, is used as an oxidizer
in fireworks. Calculate the heat required to raise the
temperature of 10.0 g of KClO4 from 25 oC to an ignition
temperature of 900. o C. The specific heat capacity of KClO4
is 0.8111 J K-1 g-1.
Solution
From q = mCsDT,
Heat required = (10.0 g) x (0.8111 J K-1 g-1) x (875 K)
= 7097 J or 7.10 kJ
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- Calorimeter: a device in which heat transfer is
monitored by recording the change in temperature
that it produces
- Styrofoam calorimeter:
q at constant pressure
(qp)
- Bomb calorimeter:
q at constant volume (qV)
Types of
Calorimeters
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Self-Test 7.4A
A small piece of calcium carbonate was placed in a
bomb calorimeter of 488 J (o C-1) heat capacity (Ccal) and
0.100 L of dilute hydrochloric acid was poured over it.
The temperature of the calorimeter rose by 3.57 oC.
What is the value of DU for the reaction of hydrochloric
acid with calcium carbonate?
Solution
Using qcal = CcalDT = (488 J (o C-1)) x (3.57 oC)
= 1742 J = 1.74 kJ
Hence DU = _ 1.74 kJ
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Example 7.17
(a) Calculate the heat that must be supplied to a 500.0 g copper kettle
containing 400.0 g of water to raise its temperature from 22.0 oC to the
boiling point of water, 100.0 oC (See Table 7.2).
(b) What percentage of the heat is used to raise the temperature of the
water?
Solution
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7.6 The First Law
 The first law of thermodynamics (closed system)
states that the change in internal energy (DU) is
the sum of the work and heat changes: it is applicable
to any process that begins and ends in equilibrium
states.
All the energies received are turned into the energy
of the system: this is a form of the energy
conservation law.
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 U is a state function: a property that depends only
on the current state of the system and is
independent of how that state was prepared.
Other state functions are P, V, T, H, S, G
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- For any ideal gas, ΔU = 0 for an isothermal process.
No changes in the kinetic energy of an ideal gas if ΔT = 0
No intermolecular forces for ideal gas molecules
No changes in potential energy during expansion
or compression
No changes in the total energy; ΔU = 0
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Self-Test 7.5B
A system was heated by using 300. J of heat, yet it was
found that its internal energy decreased by 150. J (so
DU = -150. J). Calculate w. Was work done on the system
or did the system do work?
Solution
Since DU = q + w
( _150 J) = (+300 J) + w
w=
_
450 J
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Work was done by the system
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EXAMPLE 7.5
Suppose that 1.00 mol of ideal gas molecules maintained at 292 K and
3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm
by two different paths. (a) Path A is an isothermal, reversible expansion.
(b) Path B has two parts. In step 1, the gas is cooled at constant volume
until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed
to expand against a constant pressure of 1.20 atm until its volume is
20.00 L and T = 292 K. Determine for each path the work done, the heat
transferred, and the change in internal energy (w, q, and DU).
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(a) From
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(b) Step 1, w = 0
Step 2, from
Less work is done in the irreversible path and
less energy has to enter the system as heat
to maintain its temperature.
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Self-Test 7.6A
Suppose that 2.00 mol CO 2 at 2.00 atm and 300. K is
compressed isothermally and reversibly to half its
original volume before being used to produce soda
water. Calculate w, q, and DU by treating CO 2 as an
ideal gas.
Solution
Since the process is isothermal, DU = 0 and w =
From w = _ nRT ln(V
/V
)
final
_
q.
initial
_ (2.00 mol) x (8.315 J K-1 mol-1) x (300. K) x ln 1
w=
2
= +3.46 kJ
_
Hence, q = 3.46 kJ
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Example 7.13
In an adiabatic process, no energy is transferred as heat. Indicate
whether each of the following statements about an adiabatic process in a
closed system is always true, always false, or true in certain conditions
(specify the conditions):
(a) ΔU = 0
(b) Q = 0
(c) q < 0
(d) ΔU = q
(e) ΔU = w
Solution
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7.7 A Molecular Interlude: The Origin of Internal
Energy
U = K + EP
- A system at high temperature has a greater internal energy
than the same system at a lower temperature.
(Greater kinetic energy)
 Equipartition theorem: The average value of each quadratic
contribution to the energy of a molecule in a sample
at a temperature T is equal to
.
- kB = 1.381×10–23 J·K-1; Boltzmann’s constant
- R = kBNA; RT = 2.48 kJ·mol-1 at 25 oC
-Can be used for vibrational motion only when quantum effects
can be neglected, at high temperatures. (kBT >> ΔE)
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- Molecules of N atoms; 3N degrees of freedom
(a) Center of mass
3 translational degrees
of freedom
(b) Linear molecule
2 rotational degrees
of freedom
3N – 5 vibrational
degrees of freedom
(c) Nonlinear molecule
3 rotational degrees
of freedom
3N – 6 vibrational
degrees of freedom
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- At room temperature, the equipartition theorem holds for
translational and rotational motions.
(a) Monatomic ideal gas:
translational kinetic energy only (3 modes)
(b) Diatomic or linear ideal gas:
translational energy (3 modes) + rotational energy (2 modes)
(c) Nonlinear ideal gas:
translational energy (3 modes) + rotational energy (3 modes)
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Example 7.25
7.25 Which molecular substance do you expect to have the
higher molar heat capacity, NO or NO2? Why?
Solution
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Chapter 7. THERMODYNAMICS:
THE FIRST LAW
ENTHALPY
7.8 Heat Transfers at Constant Pressure
7.9 Heat Capacities at Constant Volume and
Constant Pressure
7.10 A Molecular Interlude:
The Origin of the Heat Capacities of Gases
7.11 The Enthalpy of Physical Change
7.12 Heating Curves
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ENTHALPY (Sections 7.8-7.12)
- At constant volume and no nonexpansion work:
w = -PexDV = 0; DU = w + q = q
- However, most chemical reactions take place at a
constant pressure of about 1 atm.
7.8 Heat Transfers at Constant Pressure
- At constant pressure Pex, if no work other than
pressure-volume work is done, then
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 Enthalpy is defined as:
(constant pressure, pressure-volume work only)
Since U, P, and V are state functions, H is also a
state function!
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Exothermic and endothermic reactions
-Exothermic
reaction (ΔH < 0)
-Endothermic
reaction (ΔH > 0)
DH = -208 kJ
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Self-Test 7.7A
In an exothermic reaction at constant pressure, 50. kJ of
energy left the system as heat and 20. kJ of energy left
the system as expansion work. What are the values of
(a) DH and (b) DU for this process?
Solution
(a) Since the reaction occurs at constant P, and is
exothermic, DH = _ 50. kJ
(b) Since DH = DU + PDV, and PDV = +20. kJ,
_
D U = ( _ 50. kJ) _ (+20. kJ) = 70. kJ
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7.9 Heat Capacities at Constant Volume and
Constant Pressure
for any ideal gas
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- If CV and CP do not change with temperature,
qV = nCV,m ΔT
qP = nCP,m ΔT
qV < qP
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7.10 A Molecular Interlude: The Origin of the
Heat Capacities of Gases
- For monatomic ideal gases (see 7.7),
- For linear molecules (see 7.7),
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Variation of molar heat
capacity of iodine vapor
at constant volume
(Fig. 7.20)
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Example 7.29
Predict the contribution to the heat capacity CV,m made by molecular
motions for each of the following atoms and molecules:
(a) HCN; (b) C2H6; (c) Ar; (d) HBr
Solutions
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EXAMPLE 7.6
Calculate the final temperature and the change in internal energy
when 500 J of energy is transferred as heat to 0.900 mol O2(g) at
298 K and 1.00 atm at (a) constant volume; (b) constant pressure.
Treat the gas as ideal.
(a)
= +26.7 K
(b)
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Self-Test 7.8A
Calculate the final temperature and the change in internal
energy when 500. J of energy is transferred as heat to 0.900
mol Ne(g) at 298 K and 1.00 atm (a) at constant volume and
(b) at constant pressure. Treat the gas as ideal.
Solution
(a) CV,m = 3/2R =
DT =
q
nCV,m
=
3
2
(8.315 J K-1 mol-1) = 12.47 J K-1 mol -1
(500. J)
(0.900 mol) x (12.47 J K-1 mol-1)
= 44.6 K
Hence, final temperature is 342.6 K or 343 K
D U = q at constant volume = 500. J
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(b) CP,m = 5/2R = 5 (8.315 J K-1 mol -1) = 20.79 J K-1 mol-1
2
q
(500. J)
DT =
=
nCV,m
(0.900 mol) x (20.79 J K-1 mol -1)
= 26.7 K
Hence, final temperature is 324.7 K or 325 K
DU = q = nCV,mDT = (0.900 mol) x (12.47 J K-1 mol-1) x (26.7 K)
= 300. J
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7.11 The Enthalpy of Physical Change
 Enthalpy of vaporization is the
difference in molar enthalpy
between the vapor and the
liquid states (> 0, always).
- Energy needed to separate
liquid molecules
- Temperature dependence:
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 Enthalpy of fusion is the molar enthalpy change
that accompanies melting (fusion).
 Enthalpy of freezing is the change in molar enthalpy
change of a liquid when it solidifies.
DHfreez = Hm(solid) – Hm(liquid)
In general:
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Self-Test 7.9A
A sample of benzene, C6H6, was heated to 80 o C, its
normal boiling point. The heating was continued until
15.4 kJ had been supplied; as a result, 39.1 g of boiling
benzene was vaporized. What is the enthalpy of
vaporization of benzene at its boiling point?
Solution
The molar mass of benzene is 78.2 g/mol, hence 39.1 g
corresponds to 0.50 mol
It takes 15.4 kJ of heat to vaporize 0.50 mol of benzene,
hence DHvap = (15.4 kJ) / (0.50 mol) = 30.8 kJ mol-1
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 Enthalpy of sublimation is
the molar enthalpy change
when a solid sublimes
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Standard Enthalpies of Physical Change*
Table 7.3)
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Self-Test 7.10B
The enthalpy of vaporization of methanol is 38 kJ mol -1 at
25 o C and the enthapy of fusion is 3 kJ mol-1 at the same
temperature. What is the enthalpy of sublimation of methanol
at this temperature?
Solution
Since DHsub = DHvap + DHfus,
DHsub = (38 kJ mol-1) + (3 kJ mol -1) = 41 kJ mol-1
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7.12 Heating Curves
- Heating curve is the graph showing the variation in
the temperature of a sample as it is heated at a
constant rate at constant pressure and therefore at a
constant rate of increase in enthalpy.
-The steeper the slope of a heating curve, the lower
the heat capacity.
-The horizontal sections correspond to phase changes:
melting and boiling.
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Heating curve of water (Fig. 7.26)
- In
water, the slope for liquid < those for solid or vapor,
the high heat capacity of the liquid is due largely to the
extensive hydrogen bonding network.
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Example 7.41
The following data were collected for a new compound used in
cosmetics: DHfus = 10.0 kJ·mol-1, DHvap = 20.0 kJ·mol-1; heat
capacities: 30 J·mol-1 for the solid; 60 J·mol-1 for the liquid; 30 J·mol-1
for the gas. Which heating curve below best matches the data for
this compound?
Solution: (c)
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Chapter 7. THERMODYNAMICS:
THE FIRST LAW
THE ENTHALPY OF CHEMICAL CHANGE
7.13 Reaction Enthalpies
7.14 The Relation Between DH and DU
7.15 Standard Reaction Enthalpies
7.16 Combining Reaction Enthalpies: Hess’s Law
7.17 The Heat Output of Reactions
7.18 Standard Reaction Enthalpies
7.19 The Born-Haber Cycle
7.20 Bond Enthalpies
7.21 The Variation of Reaction Enthalpy with
Temperature
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THE ENTHALPY OF CHEMICAL CHANGE
(Sections 7.13-7.21)
7.13 Reaction Enthalpies
-Thermochemical equation: consisting of a chemical
equation together with a statement of the reaction
enthalpy, the corresponding enthalpy change.
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Self-Test 7.11A
When 0.231 g of phosphorus reacts with chlorine to form
phosphorus trichloride, PCl 3, in a constant pressure
calorimeter of heat capacity 216 J (oC)-1, the temperature
of the calorimeter rises by 11.06 o C. What is the thermo-chemical equation for the reaction?
Solution
The molar mass of P is 30.97 g mol -1, hence 0.231 g
corresponds to 7.46 x 10-3 mol.
Heat transferred to calorimeter, qcal = CcalDT
= (216 J (oC)-1) x (11.06 oC) = 2.39 kJ
If the equation for the reaction is
2 (mol) x qcal
2P(s) + 3Cl2(g)
2PCl 3(l), then DH = _
no. mol
_ (2 mol) x 2.39 kJ = _ 641 kJ
=
7.46 x 10-3 mol
_
Therefore 2P(s) + 3Cl2(g)
2PCl 3(l), DH = 641 kJ
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Example 7.43
7.43 Carbon disulfide can be prepared from coke (an impure form of
carbon) and elemental sulfur:
4 C(s) + S8(s) → 4 CS2(l)
DHo = +358.8 kJ
(a) How much heat is absorbed in the reaction of 1.25 mol S8?
(b) Calculate the heat absorbed in the reaction of 197 g of carbon
with an excess of sulfur.
(c) If the heat absorbed in the reaction was 415 kJ, how much CS2
was produced?
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7.14 The Relation Between DH and DU
- For reactions in liquids and solids only,
- If a gas is formed in the reaction,
DH = Hfinal – Hinitial = DU + (nfinal – ninitial)RT
= DU + DngasRT
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Example 7.51
Oxygen difluoride is a colorless, very poisonous gas that reacts
rapidly with water vapor to produce O2, HF, and heat:
OF2(g) + H2O(g) → O2(g) + 2HF(g) DH = -318
kJ
What is the change in internal energy for the reaction of 1.00 mol OF2?
Solution
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Self-Test 7.12A
The thermochemical equation for the combustion of
cyclohexane, C6H12, is
C6H12(l) + 9O2(g)
6CO2(g) + 6H2O(l), DH = -3920 kJ
at 298 K. What is the change in internal energy for the
combustion of 1.00 mol C6H12((l) at 298 K?
Solution
_
Dngas = nfinal _ ninitial = 3
Since
(
_
DH = DU + DngasRT
3920 kJ) = DU
DU =
_
_
3 x (8.315 J K-1 mol -1) x (298 K)
3910 kJ or _ 3.91 x 103 kJ
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7.15 Standard Reaction Enthalpies
- Reaction enthalpies depend on the physical states of
the reactants and products
88 kJ = enthalpy of vaporization
of water (44 kJ·mol-1) × 2
It is therefore useful if enthalpies
(DHo) can be referred to a
standard state.
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Standard state: refers to a pure form at exactly 1 bar
or for a solute in a liquid solution: concentration
of 1 mol·L-1.
 Standard reaction enthalpy, DHo is the reaction
enthalpy when reactants in their standard states
change into products in their standard states.
-Most thermochemical data is reported for 25 oC
(298.15 K) but the temperature is not part of
standard states.
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7.16 Combining Reaction Enthalpies: Hess’s
Law
- Enthalpy is a state function; DH is independent of the path.
 Hess’s law: the overall reaction enthalpy is the sum of
the reaction enthalpies of the steps into which the reaction
can be divided.
DHo = -393.5 kJ
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EXAMPLE 7.9
Consider the synthesis of propane, C3H8,
a gas use as camping fuel:
It is difficult to measure the enthalpy change of this reaction directly.
However, standard enthalpies of combustion reactions are easy to
measure. Calculate the standard enthalpy of this reaction from the
following experimental data:
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Self-Test 7.13B
Methanol is a clean-burning liquid fuel proposed as a
replacement for gasoline. Suppose it could be produced
by the controlled reaction of oxygen with methane. Find
the standard reaction enthalpy for the formation of 1 mol
CH3OH(l) from methane and oxygen, given the following
information.
(1) CH4(g) + H2O(g)
(2) 2H2(g) + CO(g)
(3) 2H2(g) + O 2(g)
CO(g) + 3H2(g) DHo = +206.10 kJ
CH3OH(l) DHo = _ 128.33 kJ
_
2H2O(g) DHo =
483.64 kJ
Solution
The required equation is 2CH4(g) + O2(g)
2CH3OH(l)
Multiply equation (1) and (2) by 2, and add both to equation (3):
this gives the required equation.
_
DHo = 2DHo(1) + 2DHo(2) + DHo (3) = 328.10 kJ
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Example 7.67
Calculate the reaction enthalpy for the synthesis of hydrogen
chloride gas, H2(g) + Cl2(g) → 2 HCl(g), from the following data:
NH3(g) + HCl(g) → NH4Cl(s)
N2(g) + 3 H2(g) → 2 NH3(g)
DHo = -176.0 kJ
DHo = -92.22 kJ
N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s)
DHo = -628.86 kJ
Solution
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7.17 The Heat Output of Reactions
Combustion reactions are important throughout
chemistry - they are always exothermic.
 the standard enthalpy of combustion, DHco
is defined as the change in enthalpy per mole of
a substance that is burned in a combustion reaction
under standard conditions.
There is also the specific enthalpy of combustion:
the enthalpy of combustion per gram
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Standard Enthalpies of Combustion at 25 oC
(Table 7.4)*
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Self-Test 7.14A
The thermochemical equation for the combustion of
propane is:
C H (g) + 5O (g)
3CO (g) + 4H O(l) DHo = _ 2220. kJ
3 8
2
2
2
What mass of propane must be burned to supply 350. kJ
as heat? Would it be easier to pack propane rather than
butane?
Solution
The above equation gives the heat supplied by burning
1 mol of propane, hence 350. kJ of heat can be obtained
from
350
= 0.1577 mol of propane
2220
Since the molar mass of propane is 44.07 g mol-1, the
mass of propane needed to supply 350. kJ of heat is
(44.07 g mol-1) x (0.1577 mol) = 6.95 g
Yes, propane would be slightly easier to pack, because
the liquid is slightly less dense than butane.
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7.18 Standard Enthalpies of Formation
 Standard enthalpy of formation, DHfo is defined as
the standard reaction enthalpy per mole of formula
units for the formation of a substance from its
elements in their most stable form.
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Standard Enthalpies of Formation at 25 oC (kJ mol-1)
(Table 7.5)
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 Standard reaction enthalpy can be calculated from
standard enthalpies of formation of reactants and
products.
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Self-Test 7.15A
Calculate the standard enthalpy of combustion of
glucose from the standard enthalpies of formation
in Table 7.5 and Appendix 2A.
Solution
The combustion equation is
C6H12O 6(s) + 6O2(g)
 nDH
f
 nDH
_
(products) = {(6 mol) x ( 393.51 kJ mol-1)}
_
+ {(6 mol) x ( 285.83 kJ mol-1)}
= _ 4076.04 kJ
o
f
(reactants) = _ 1268 kJ + 0 kJ = _ 1268 kJ
o
DHo (combustion) =
=
_
6CO2(g) + 6H2O(l)

nDH fo(products)
_
nDHfo(reactants)

2808 kJ mol-1
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EXAMPLE
7.12
Use the information of standard enthalpies of formation and the
enthalpy of combustion of propane gas to calculate the enthalpy
of formation of propane, a gas commonly used for camping stoves
and outdoor barbecues.
= -2323.85 kJ
= -2220 kJ
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Self-Test 7.16A
Calculate the standard enthalpy of formation of
ethyne, the fuel used in oxyacetylene welding
torches, from the information in Tables 7.4 and 7.5.
Solution
The thermochemical equation for the combustion
of ethyne is:
_
C2H2(g) + 5/2O2(g)
2CO 2(g) + H2O(l) DHo = 1300 kJ mol -1


nDHfo (reactants)
DHor =
nDHfo (products) _
_
_
1300 kJ mol -1 = [(2 mol) x ( _ 393.51 kJ mol-1) + ( 285.83 kJ mol -1)
_ [(DH o (ethyne)]
f
DHfo(ethyne) = + 227 kJ mol-1
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273s
Example 7.73
Calculate the standard enthalpy of formation of PCl5(s) from the
standard enthalpy of formation of PCl3(l) (DHof(PCl3,l) = -319.7 kJ·mol-1)
and PCl3(l) + Cl2(g) → PCl5(s), DHo = -124 kJ.
Solution
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7.19 The Born-Haber Cycle
 Lattice enthalpy of the solid, DHL
- Lattice
enthalpies at 25oC (kJ·mol-1)(Table 7.)
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 Born-Haber cycle: a closed path of steps,
one of which is the formation of a solid
lattice from the gaseous ions
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EXAMPLE
7.13
Lattice enthalpy of KCl
DHf (K, atoms)
DHf (Cl, atoms)
I(K)
-Eea(Cl)
-DHf(KCl)
K+Cl-(s)
K+(g) + Cl-(g)
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Self-Test 7.17A
Calculate the lattice enthalpy of calcium chloride, CaCl2,
by using data in Appendices 2A and 2D.
Solution
Ca2+(g) + 2e-(g) + 2Cl(g)
_ 698
The Born-Haber cycle is:
(not to scale)
D HL = 795.8 + 178.2
+ 244 + 1735 _ 698
= 2255 kJ
+1735
H
kJ mol-1
Ca2+(g) + 2Cl-(g)
Ca(g) + 2Cl(g)
+178.2 + 244
Ca(s) + Cl2(g)
- DHf o = +795.8
Ca2+(Cl -)2(s)
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- DHL
7.20 Bond Enthalpies
 Bond enthalpy, DHB is the difference between the
standard molar enthalpies of a molecule, X-Y, and
its fragments X and Y in the gas phase.
E.g.
- Bond breaking is always
endothermic and bond
formation is always
exothermic.
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 Mean bond enthalpy: small variations of a
specific bond enthalpy in polyatomic molecules
give a guide to the average bond strength.
- The enthalpy change from a liquid (or solid) sample,
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EXAMPLE
7.14
Estimate the enthalpy of the reaction between gaseous iodoethane
and water vapor:
- Breaking the bonds (reactants)
DHBo(C-I) + DHBo(O-H) =
- Forming the bonds (products)
DHBo(C-O) + DHBo(H-I) =
- The overall enthalpy change:
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Self-Test 17.18B
Use bond enthalpies to estimate the standard enthalpy of
reaction in which 1.00 mol of gaseous CH4 reacts with
gaseous F2 to form gaseous CH2F2 and HF.
Solution
The relevant equation is
CH4(g) + 2F2(gas)
CH2F2(g) + 2HF(g)
In this reaction, 2 mol of F-F bonds and 2 mol of C-H bonds
are broken (reactants), whereas 2 mol of C-F bonds and 2 mol
of H-F bonds are formed.
DHo =
mean DH (reactants) _
mean DH (products)

B

B
= [(2 x +412 kJ mol -1) + (2 x +158 kJ mol-1)]
_ [(2 x +484 kJ mol-1) + (2 x +565 kJ mol -1)
=
_
958 kJ mol -1
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277s
Example 7.81
7.81 Use the bond enthalpies in the table to estimate the reaction enthalpy
for 3 C2H2(g) → C6H6(g)
Solution
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7.21 The Variation of Reaction Enthalpy with
Temperature
- The enthalpies of both reactants and products
increase with temperature.
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 Kirchhoff’s law
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EXAMPLE
7.15
The standard enthalpy of reaction of N2(g) + 3H2(g) → 2NH3(g) is
-92.22 kJ·mol-1 at 298 K. The industrial synthesis takes place at 450 oC.
What is the standard reaction enthalpy at the latter temperature?
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Self-Test 17.19B
The standard enthalpy of formation of ammonium nitrate
_
is 365.56 kJ mol -1 at 298 K. Estimate its value at 250. oC.
Solution
The equation of formation is
N2(g) + 3/2O 2(g) + 2H2(g)
NH4NO 3(s)
250. oC = 523 K
_
Use Kirchhoff's law: DHo (T2) = DHo (T1) + (T2 T1)DCp
Values of Cp (J K-1 mol-1): NH4NO3(s) 84.1 (product);
N2(g) 29.12; 3/2O 2(g) 44.04; 2H2(g) 57.64 (reactants)
DHo (T2) = (_ 365.56 kJ mol -1) + (225 K)( _ 46.7 J K-1 mol -1)
_
= 376 kJ mol-1
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Example 7.87
(a) Calculate the enthalpy of vaporization of benzene (C6H6) at
298.2 K. The standard enthalpy of formation of gaseous benzene
is +82.93 kJ·mol-1, and that of liquid benzene is +49.0 kJ·mol-1.
Solution
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(b) Given that, for liquid benzene, CP,m = 136.1 J·mol-1·K-1 and that,
for gaseous benzene, CP,m = 81.67 J·mol-1·K-1, calculate the
enthalpy of vaporization of benzene at its boiling point (353.2 K).
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(c) Compare the value obtained in part (b) with the real value of
30.8 kJ·mol-1. What is the source of difference between these
numbers?
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