Download Chapter Six

Chapter ?
1
Thermochemistry
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Chapter Six
2
Energy
• Energy is the capacity to do work (to
displace or move matter).
• Energy literally means “work within”;
however, an object does not contain work.
• Potential energy is energy of position or
composition.
• Kinetic energy is the energy of motion.
Ek = ½ mv2
Energy has the units of joules (J or kg . m2/s2)
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Chapter Six
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Potential Energy and
Kinetic Energy
At what point in each
bounce is the potential
energy of the ball at a
maximum?
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Chapter Six
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Thermochemistry: Basic Terms
• Thermochemistry is the study of energy
changes that occur during chemical
reactions.
• System: the part of the universe being
studied.
• Surroundings: the rest of the universe.
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Chapter Six
5
Types of Systems
• Open: energy and
matter can be
exchanged with the
surroundings.
• Closed: energy can
be exchanged with
the surroundings,
matter cannot.
• Isolated: neither
energy nor matter
can be exchanged
with the
surroundings.
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A closed system;
energy (not matter)
can be exchanged.
After the lid of the jar
is unscrewed, which
kind of system is it?
Chapter Six
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Internal Energy (E)
• Internal energy (E) is the total energy
contained within a system
• Part of E is kinetic energy (from molecular
motion)
– Translational motion, rotational motion,
vibrational motion.
– Collectively, these are sometimes called thermal
energy
• Part of E is potential energy
– Intermolecular and intramolecular forces of
attraction, locations of atoms and of bonds.
– Collectively these are sometimes called
chemical energy
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Chapter Six
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Heat (q)
• Technically speaking,
heat is not “energy.”
• Heat is energy transfer
between a system and its
surroundings, caused by
a temperature difference.
• Thermal equilibrium
occurs when the system
and surroundings reach
the same temperature
and heat transfer stops.
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More energetic
molecules …
… transfer energy to
less energetic molecules.
How do the root-mean-square
speeds of the Ar atoms and the N2
molecules compare at the point of
thermal equilibrium?
Chapter Six
Exchange of Heat between System
and Surroundings
8
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
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Chapter Six
Exchange of Heat between System
and Surroundings
9
• 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.
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Work (w)
• Like heat, work is an energy transfer between a system and
its surroundings.
• Unlike heat, work is caused by a force moving through a
distance (heat is caused by a temperature difference).
• A negative quantity of work signifies that the system loses
energy.
• A positive quantity of work signifies that the system gains
energy.
• There is no such thing as “negative energy” nor “positive
energy”; the sign of work (or heat) signifies the direction of
energy flow.
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Chapter Six
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Pressure-Volume Work
For now we will consider only
pressure-volume work.
work (w) = –PDV
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How would the magnitude of DV
compare to the original gas
volume if the two weights (initial
and final) were identical?
Chapter Six
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What is work?
•
•
•
•
•
•
Work is a force acting over a distance.
w= F x Dd
P = F/ area
d = V/area
w= (P x area) x D (V/area)= PDV
Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
• units of liter - atm L-atm
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Chapter Six
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Work needs a sign
• If the volume of a gas increases, the system
has done work on the surroundings.
• work is negative
• w = - PDV
• Expanding work is negative.
• Contracting, surroundings do work on the
system w is positive.
• 1 L atm = 101.3 J
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Chapter Six
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Examples
• What amount of work is done when 15 L of
gas is expanded to 25 L at 2.4 atm pressure?
• If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
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Chapter Six
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Same rules for heat and work
• Heat given off is negative.
• Heat absorbed is positive.
• Work done by system on surroundings is
negative.
• Work done on system by surroundings is
positive.
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State Functions
• The state of a system: its exact condition at a fixed instant.
• State is determined by the kinds and amounts of matter
present, the structure of this matter at the molecular level,
and the prevailing pressure and temperature.
• A state function is a property that has a unique value that
depends only the present state of a system, and does not
depend on how the state was reached (does not depend on
the history of the system).
• Law of Conservation of Energy – in a physical or
chemical change, energy can be exchanged between a
system and its surroundings, but no energy can be created
or destroyed.
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Chapter Six
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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 DE is the same.
– But q and w are different
in the two cases.
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Chapter Six
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First Law of Thermodynamics
• “Energy cannot be created or destroyed.”
• Inference: the internal energy change of a system
is simply the difference between its final and
initial states:
DE = Efinal – Einitial
• Additional inference: if energy change occurs only
as heat (q) and/or work (w), then:
DE = q + w
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Chapter Six
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First Law: Sign Convention
• Energy entering a system carries a positive
sign:
– heat absorbed by the system, or
– work done on the system
• Energy leaving a system carries a negative
sign
– heat given off by the system
– work done by the system
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Chapter Six
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Direction
•
Every energy measurement has three
parts.
1. A unit ( Joules of calories).
2. A number how many.
3. and a sign to tell direction.
• negative - exothermic
• positive- endothermic
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Chapter Six
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Surroundings
System
Energy
DE <0
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Chapter Six
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Surroundings
System
Energy
DE >0
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Chapter Six
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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, DE = q + w.
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Chapter Six
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Example 6.1
A gas does 135 J of work while expanding, and at the
same time it absorbs 156 J of heat. What is the change
in internal energy?
Example 6.2: A Conceptual Example
The internal energy of a fixed quantity of an ideal gas
depends only on its temperature. If a sample of an ideal
gas is allowed to expand against a constant pressure at a
constant temperature,
(a) what is DU for the gas? (b) Does the gas do work? (c)
Is any heat exchanged with the surroundings?
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Chapter Six
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DE, q, w, and Their Signs
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Chapter
Six
Inc.
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Heats of Reaction (qrxn)
• qrxn is the quantity of heat exchanged between a
reaction system and its surroundings.
• An exothermic reaction gives off heat
– In an isolated system, the temperature increases.
– The system goes from higher to lower energy; qrxn is
negative.
• An endothermic reaction absorbs heat
– In an isolated system, the temperature decreases.
– The system goes from lower to higher energy; qrxn is
positive.
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Chapter Six
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Conceptualizing an Exothermic Reaction
Surroundings are at 25 °C
25 °C
Typical situation:
some heat is released
to the surroundings,
some heat is absorbed
by the solution.
Hypothetical situation: all heat
is instantly released to the
surroundings. Heat = qrxn
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32.2 °C
35.4 °C
In an isolated system, all heat is
absorbed by the solution.
Maximum temperature rise.
Chapter Six
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Internal Energy Change at Constant Volume
• For a system where the
reaction is carried out at
constant volume, DV = 0
and DE = qV.
• All the thermal energy
produced by conversion
from chemical energy is
released as heat; no P-V
work is done.
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Chapter Six
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lnternal Energy Change at Constant Pressure
• For a system where the
reaction is carried out at
constant pressure,
DE = qP – PDV or
DE + PDV = qP
• Most of the thermal energy
is released as heat.
• Some work is done to
expand the system against
the surroundings (push
back the atmosphere).
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30
Calorimetry
• We measure heat flow using calorimetry.
• A calorimeter is a device used to make this
measurement.
• A “coffee cup” calorimeter may be used for
measuring heat involving solutions.
A “bomb” calorimeter is used to
find heat of combustion; the
“bomb” contains oxygen and a
sample of the material to be burned.
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Chapter Six
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Calorimetry
n Measuring heat.
n Use a calorimeter.
n Two kinds
n Constant pressure calorimeter (called a
coffee cup calorimeter)
n heat capacity for a material, C is calculated
n C= heat absorbed/ DT = DH/ DT
n specific heat capacity = C/mass
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Calorimetry, Heat Capacity, Specific Heat
• Heat evolved in a reaction is absorbed by the calorimeter
and its contents.
• In a calorimeter we measure the temperature change of
water or a solution to determine the heat absorbed or
evolved by a reaction.
• The heat capacity (C) of a system is the quantity of heat
required to change the temperature of the system by 1 °C.
C = q/DT (units are J/°C)
• Molar heat capacity is the heat capacity of one mole of a
substance.
• The specific heat (s) is the heat capacity of one gram of a
pure substance (or homogeneous mixture).
s = C/m = q/(mDT)
q = s m DT
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Calorimetry
n molar heat capacity = C/moles
n heat = specific heat x m x DT
n heat = molar heat x moles x DT
n Make the units work and you’ve done the
problem right.
n A coffee cup calorimeter measures DH.
n An insulated cup, full of water.
n The specific heat of water is 1 cal/gºC
n Heat of reaction= DH = sh x mass x DT
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Chapter Six
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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.
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Chapter Six
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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.
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Chapter Six
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Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
s=
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heat transferred
mass  temperature change
q
m  DT
© 2009, Prentice-Hall,
Chapter
Six
Inc.
37
More on Specific Heat
•
•
•
•
•
•
q = mass x specific heat x DT
If DT is positive (temperature increases), q is
positive and heat is gained by the system.
If DT is negative (temperature decreases), q is
negative and heat is lost by the system.
The calorie, while not an SI unit, is still used to
some extent.
Water has a specific heat of 1 cal/(g oC).
4.184 J = 1 cal
One food calorie (Cal or kcal) is actually equal to
1000 cal.
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Chapter Six
38
Many metals have
low specific heats.
The specific heat of
water is higher than
that of almost any
other substance.
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Chapter Six
39
Heat Capacity: A Thought Experiment
• Place an empty iron pot weighing 5 lb on the
burner of a stove.
• Place an iron pot weighing 1 lb and containing 4
lb water on a second identical burner (same total
mass).
• Turn on both burners. Wait five minutes.
• Which pot handle can you grab with your bare
hand?
• Iron has a lower specific heat than does water. It
takes less heat to “warm up” iron than it does
water.
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Chapter Six
40
Example
Calculate the heat capacity of an aluminum block
that must absorb 629 J of heat from its
surroundings in order for its temperature to rise
from 22 °C to 145 °C.
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Chapter Six
41
Example 6.7
How much heat, in joules and in kilojoules, does
it take to raise the temperature of 225 g of water
from 25.0 to 100.0 °C?
Example 6.8
What will be the final temperature if a 5.00-g
silver ring at 37.0 °C gives off 25.0 J of heat to its
surroundings? Use the specific heat of silver
listed in Table 6.1.
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Chapter Six
42
Examples
n The specific heat of graphite is 0.71 J/gºC.
Calculate the energy needed to raise the
temperature of 75 kg of graphite from 294
K to 348 K.
n A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and the
copper is 21.8ºC. What is the specific heat
of copper?
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Chapter Six
43
Calorimetry
n Constant volume calorimeter is called a
bomb calorimeter.
n Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
n The heat capacity of the calorimeter is
known and tested.
n Since DV = 0, PDV = 0, DE = q
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Chapter Six
44
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.
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Chapter Six
45
Constant Pressure Calorimetry
Because the specific heat
for water is well known
(4.184 J/g-K), we can
measure DH for the
reaction with this equation:
q = m  s  DT
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Chapter Six
46
Bomb Calorimeter
n thermometer
n stirrer
n full of water
n ignition wire
n Steel bomb
n sample
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Chapter Six
47
Properties
n intensive properties not related to the
amount of substance.
n density, specific heat, temperature.
n Extensive property - does depend on the
amount of stuff.
n Heat capacity, mass, heat from a reaction.
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Chapter Six
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Enthalpy
n Symbol is H
n Change in enthalpy is DH
n delta H
n If heat is released the heat content of the
products is lower
n DH is negative (exothermic)
n If heat is absorbed the heat content of the
products is higher
n DH is positive (endothermic)
48
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Chapter Six
49
Enthalpy
n Since DE = q + w and w = -PDV, we can
substitute these into the enthalpy
expression:
DH = DE + PDV
DH = (q+w) − w
DH = q
n So, at constant pressure, the change in
enthalpy is the heat gained or lost.
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© 2009, Prentice-Hall,
Chapter
Six
Inc.
50
Enthalpy and Enthalpy Change
Enthalpy is the sum of the
internal energy and the
pressure-volume product
of a system:
H = E + PV
For a process carried out
at constant pressure,
qP = DE + PDV
so
qP = DH
The evolved H2 pushes
back the atmosphere;
work is done at
constant pressure.
Mg + 2 HCl  MgCl2 + H2
Most reactions occur at constant
pressure, so for most reactions, the heat
evolved equals the enthalpy change.
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Chapter Six
51
Properties of Enthalpy
• Enthalpy is an extensive property.
– It depends on how much of the
substance is present.
• Since E, P, and V are all state
functions, enthalpy H must be a
state function also.
• Enthalpy changes have unique
values. DH = qP
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Two logs on a fire give
off twice as much heat
as does one log.
Enthalpy change depends
only on the initial and
final states. In a chemical
reaction we call the initial
state the ____ and the
final state the ____.
Chapter Six
52
Enthalpy Diagrams
• Values of DH are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.
A decrease in enthalpy
during the reaction; DH
is negative.
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An increase in enthalpy
during the reaction; DH
is positive.
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53
Enthalpy of Reaction
The change in enthalpy,
DH, is the enthalpy of
the products minus the
enthalpy of the
reactants:
DH = Hproducts − Hreactants
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Chapter Six
54
Enthalpy of Reaction
This quantity, DH, is called the enthalpy of
reaction, or the heat of reaction.
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Chapter Six
55
The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. DH for a reaction in the forward direction
is equal in size, but opposite in sign, to DH
for the reverse reaction.
3. DH for a reaction depends on the state of
the products and the state of the reactants.
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Chapter
Six
Inc.
56
Reversing a Reaction
• DH changes sign when a process is reversed.
• Therefore, a cyclic process has the value DH = 0.
Same magnitude; different signs.
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Chapter Six
57
Example 6.3
Given the equation
(a) H2(g) + I2(s)  2 HI(g)
DH = +52.96 kJ
calculate DH for the reaction
(b) HI(g)  ½ H2(g) + ½ I2(s).
Example 6.4
The complete combustion of liquid octane, C8H18, to
produce gaseous carbon dioxide and liquid water at 25
°C and at a constant pressure gives off 47.9 kJ of heat
per gram of octane. Write a chemical equation to
represent this information.
Prentice Hall © 2005
Chapter Six
ΔH in Stoichiometric Calculations
58
• For problem-solving, heat evolved (exothermic reaction) can be
thought of as a product. Heat absorbed (endothermic reaction)
can be thought of as a reactant.
• We can generate conversion factors involving DH.
• For example, the reaction:
H2(g) + Cl2(g)  2 HCl(g)
DH = –184.6 kJ
can be used to write:
–184.6 kJ
————
1 mol H2
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–184.6 kJ
————
1 mol Cl2
–184.6 kJ
————
2 mol HCl
Chapter Six
59
Example 6.5
What is the enthalpy change associated with the
formation of 5.67 mol HCl(g) in this reaction?
H2(g) + Cl2(g)  2 HCl(g)
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DH = –184.6 kJ
Chapter Six
60
Measuring Enthalpy Changes for
Chemical Reactions
For a reaction carried out in a calorimeter, the heat evolved by
a reaction is absorbed by the calorimeter and its contents.
qrxn = – qcalorimeter
qcalorimeter = mass x specific heat x DT
By measuring the temperature change that occurs in a
calorimeter, and using the specific heat and mass of the
contents, the heat evolved (or absorbed) by a reaction can be
determined and the enthalpy change calculated.
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Chapter Six
61
Example 6.11
A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to
50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a
calorimeter. After mixing, the solution temperature rises to
21.21 °C. Calculate the heat of this reaction.
Example 6.12
Express the result of Example 6.11 for molar amounts of
the reactants and products. That is, determine the value
of DH that should be written in the equation for the
neutralization reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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DH = ?
Chapter Six
Bomb Calorimetry:
Reactions at Constant
Volume
62
• Some reactions, such as combustion,
cannot be carried out in a coffee-cup
calorimeter.
• In a bomb calorimeter, a sample of
known mass is placed in a heavywalled “bomb,” which is then
pressurized with oxygen.
• Since the reaction is carried out at
constant volume,
–qrxn = qcalorimeter = DE
… but in many cases the value of DE
is a good approximation of DH.
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Chapter Six
63
Hess’s Law of Constant
Heat Summation
• Some reactions cannot be carried out “as written.”
• Consider the reaction:
C(graphite) + ½ O2(g)  CO(g).
• If we burned 1 mol C in ½ mol O2, both CO and CO2
would probably form. Some C might be left over.
However …
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64
Hess’s Law of Constant
Heat Summation
• … enthalpy change is a state function.
• The enthalpy change of a reaction is the same
whether the reaction is carried out in one step or
through a number of steps.
• Hess’s Law: If an equation can be expressed as the
sum of two or more other equations, the enthalpy
change for the desired equation is the sum of the
enthalpy changes of the other equations.
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Chapter Six
65
Hess’s Law
n Enthalpy is a state function.
n It is independent of the path.
n We can add equations to to come up with
the desired final product, and add the DH
n Two rules
n If the reaction is reversed the sign of DH is
changed
n If the reaction is multiplied, so is DH
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66
Hess’s Law
Hess’s law states that
“[i]f a reaction is carried
out in a series of steps,
DH for the overall
reaction will be equal to
the sum of the enthalpy
changes for the
individual steps.”
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Chapter Six
67
Hess’s Law
Because DH is a state
function, the total
enthalpy change depends
only on the initial state of
the reactants and the final
state of the products.
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Chapter Six
68
Example 6.14
Calculate the enthalpy change for reaction (a) given the data
in equations (b), (c), and (d).
(a) 2 C(graphite) + 2 H2(g)  C2H4(g)
DH = ?
(b) C(graphite) + O2(g)  CO2(g)
DH = –393.5 kJ
(c) C2H4(g) + 3 O2  2 CO2(g) + 2 H2O(l)
DH = –1410.9 kJ
(d) H2(g) + ½ O2  H2O(l)
DH = –285.8 kJ
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Chapter Six
69
Standard Enthalpies of Formation
• It would be convenient to be able to use the simple relationship
ΔH = Hproducts – Hreactants
to determine enthalpy changes.
• Although we don’t know absolute values of enthalpy, we don’t
need them; we can use a relative scale.
• We define the standard state of a substance as the state of the pure
substance at 1 atm pressure and the temperature of interest (usually
25 °C).
• The standard enthalpy change (ΔH°) for a reaction is the enthalpy
change in which reactants and products are in their standard states.
• The standard enthalpy of formation (ΔHf°) for a reaction is the
enthalpy change that occurs when 1 mol of a substance is formed
from its component elements in their standard states.
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70
Standard Enthalpies of Formation
n Hess’s Law is much more useful if you
know lots of reactions.
n Made a table of standard heats of formation.
The amount of heat needed to for 1 mole of
a compound from its elements in their
standard states.
n Standard states are 1 atm, 1M and 25ºC
n For an element it is 0
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Chapter Six
71
Standard Enthalpies of Formation
Standard enthalpies of formation, DHf°, are
measured under standard conditions (25 °C and
1.00 atm pressure).
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Chapter Six
72
Standard Enthalpy of Formation
When we say “The standard enthalpy of formation of
CH3OH(l) is –238.7 kJ”, we are saying that the reaction:
C(graphite) + 2 H2(g) + ½ O2(g)  CH3OH(l)
has a value of ΔH of –238.7 kJ.
We can treat ΔHf° values as though they were absolute
enthalpies, to determine enthalpy changes for reactions.
Question: What is ΔHf° for an element in its standard
state [such as O2(g)]? Hint: since the reactants are
the same as the products …
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Chapter Six
73
Calculations Based on
Standard Enthalpies of Formation
•
•
•
DH°rxn = Snp x DHf°(products) – Snr x DHf°(reactants)
The symbol S signifies the summation of several terms.
The symbol n signifies the stoichiometric coefficient used
in front of a chemical symbol or formula.
In other words …
1. Add all of the values for DHf° of the products.
2. Add all of the values for DHf° of the reactants.
3. Subtract #2 from #1
(This is usually much easier than using Hess’s Law!)
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Chapter Six
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Example 6.15
Synthesis gas is a mixture of carbon monoxide and hydrogen that is
used to synthesize a variety of organic compounds. One reaction for
producing synthesis gas is
3 CH4(g) + 2 H2O(l) + CO2(g)  4 CO(g) + 8 H2(g) ΔH° = ?
Use standard enthalpies of formation from Table 6.2 to calculate the
standard enthalpy change for this reaction.
Example 6.16
The combustion of isopropyl alcohol, common rubbing alcohol, is
represented by the equation
2 (CH3)2CHOH(l) + 9 O2(g)  6 CO2(g) + 8 H2O(l) ΔH° = –4011 kJ
Use this equation and data from Table 6.2 to establish the standard
enthalpy of formation for isopropyl alcohol.
Example 6.17: A Conceptual Example
Without performing a calculation, determine which of these two
substances should yield the greater quantity of heat per mole upon
complete combustion: ethane, C2H6(g), or ethanol, CH3CH2OH(l).
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Chapter Six
75
Ionic Reactions
in Solution
• We can apply
thermochemical concepts
to reactions in ionic
solution by arbitrarily
assigning an enthalpy of
formation of zero to
H+(aq).
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Chapter Six
76
Example 6.18
H+(aq) + OH–(aq)  H2O(l)
ΔH° = –55.8 kJ
Use the net ionic equation just given, together with ΔHf°
= 0 for H+(aq), to obtain ΔHf° for OH–(aq).
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Chapter Six
77
Looking Ahead
• A reaction that occurs (by itself) when the reactants are
brought together under the appropriate conditions is
said to be spontaneous.
• A discussion of entropy is needed to fully understand
the concept of spontaneity, and will be discussed in
Chapter 17.
• A spontaneous reaction isn’t necessarily fast (rusting;
diamond  graphite; etc. are slow).
• The difference between the tendency of a reaction to
occur and the rate at which a reaction occurs will be
discussed in Chapter 13.
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Chapter Six